information for students in math 329 2005 01

McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 329 2005 01

THEORY OF INTEREST

Information for Students(Winter Term, 2004/2005)

Pages 1 - 10 of these notes may be considered theCourse Outline for this course.

W. G. Brown

April 12, 2005

Information for Students in MATH 329 2005 01

Contents

1 General Information 11.1 Instructor and Times . . . . . 11.2 Course Description . . . . . . 1

1.2.1 Calendar Description . 11.2.2 Syllabus (in terms of sec-

tions of the text-book) 11.2.3 “Verbal” arguments . 4

1.3 Evaluation of Your Progress . 41.3.1 Term Mark . . . . . . 41.3.2 Assignments. . . . . . 41.3.3 Class Test . . . . . . . 41.3.4 Final Examination . . 51.3.5 Supplemental Assessments 51.3.6 Machine Scoring . . . 51.3.7 Plagiarism . . . . . . . 5

1.4 Published Materials . . . . . 61.4.1 Required Text-Book . 61.4.2 Website . . . . . . . . 61.4.3 Reference Books . . . 6

1.5 Other information . . . . . . 71.5.1 Prerequisites . . . . . 71.5.2 Calculators . . . . . . 71.5.3 Self-Supervision . . . . 71.5.4 Escape Routes . . . . 71.5.5 Showing your work; good

mathematical form; sim-plifying answers . . . . 7

2 Timetable 9

3 First Problem Assignment 11

4 Second Problem Assignment 13

5 Solutions, First Problem Assign-ment 16

6 Third Problem Assignment 22

7 Solutions, Second Problem As-signment 25

8 Class Tests 348.1 Class Test, Version 1 . . . . . 348.2 Class Test, Version 2 . . . . . 408.3 Class Test, Version 3 . . . . . 468.4 Class Test, Version 4 . . . . . 52

9 Solutions, Third Problem Assign-ment 58

10 Fourth Problem Assignment 67

11 Solutions to Problems on the ClassTest 7011.1 Problems on rates of interest and

discount . . . . . . . . . . . . 7011.2 Problems on the values of annu-

ities and perpetuities with con-stant payments . . . . . . . . 73

11.3 Problems on increasing and de-creasing annuities and perpetu-ities . . . . . . . . . . . . . . 76

11.4 Problems on combinations of an-nuities and perpetuities . . . 78

11.5 Problems on drop and balloonpayments . . . . . . . . . . . 79

12 Fifth Problem Assignment 82

13 Solutions, Fourth Problem Assign-ment 84

14 Solutions, Fifth Problem Assign-ment 92

15 References 901

A Supplementary Lecture Notes 2001A.1 Supplementary Notes for the Lec-

tures of January 4th and Janu-ary 5th, 2005 . . . . . . . . . 2001A.1.1 These notes . . . . . . 2001A.1.2 §1.1 INTRODUCTION 2001


A.1.3 §1.2 THE ACCUMULA-TION AND AMOUNTFUNCTIONS . . . . . 2002

A.2 Supplementary Notes for the Lec-ture of January 7th, 2005 . . 2007A.2.1 §1.2 THE ACCUMULA-

TION AND AMOUNTFUNCTIONS (conclusion) 2007

A.2.2 §1.3 THE EFFECTIVERATE OF INTEREST 2007

A.3 Supplementary Notes for the Lec-ture of January 10th, 2005 . . 2009A.3.1 §1.3 THE EFFECTIVE

RATE OF INTEREST (con-clusion) . . . . . . . . 2009

A.3.2 §1.4 SIMPLE INTEREST 2010A.3.3 §1.5 COMPOUND INTER-

EST . . . . . . . . . . 2012A.4 Supplementary Notes for the Lec-

ture of January 12th, 2005 . . 2018A.4.1 §1.6 PRESENT VALUE 2018

A.5 Supplementary Notes for the Lec-ture of January 14th, 2005 . . 2021A.5.1 §1.7 THE EFFECTIVE

RATE OF DISCOUNT 2021A.5.2 §1.8 NOMINAL RATES

OF INTEREST AND DIS-COUNT (barely begun 2025

A.6 Supplementary Notes for the Lec-ture of January 17th, 2005 . . 2026A.6.1 §1.8 NOMINAL RATES

OF INTEREST AND DIS-COUNT (continued) . 2026

A.6.2 §1.9 FORCES OF INTER-EST AND DISCOUNT(barely begun) . . . . 2030

A.7 Supplementary Notes for the Lec-ture of January 19th, 2005 . . 2031A.7.1 §1.9 FORCES OF INTER-

EST AND DISCOUNT 2031A.7.2 §1.10 VARYING INTER-

EST . . . . . . . . . . 2032

A.7.3 §1.11 SUMMARY OF RE-SULTS . . . . . . . . 2032

A.7.4 §2.1 INTRODUCTION 2033A.7.5 §2.2 OBTAINING NUMER-

ICAL RESULTS . . . 2033A.8 Supplementary Notes for the Lec-

ture of January 21st, 2005 . . 2036A.8.1 §2.2 OBTAINING NUMER-

ICAL RESULTS (contin-ued) . . . . . . . . . . 2036

A.8.2 §2.3 DETERMINING TIMEPERIODS . . . . . . . 2037

A.8.3 §2.4 THE BASIC PROB-LEM . . . . . . . . . . 2039

A.8.4 §2.5 EQUATIONS OF VALUE 2039A.9 Supplementary Notes for the Lec-

ture of January 24th, 2005 . . 2041A.9.1 §2.5 EQUATIONS OF VALUE

(continued) . . . . . . 2041A.9.2 §2.6 UNKNOWN TIME 2042

A.10 Supplementary Notes for the Lec-tures of January 26th, 2005 . 2045A.10.1 §2.7 UNKNOWN RATE

OF INTEREST . . . . 2045A.10.2 §2.8 PRACTICAL EX-

AMPLES . . . . . . . 2046A.10.3 §2.9 MISCELLANEOUS

PROBLEMS . . . . . 2049A.10.4 §3.1 INTRODUCTION 2050

A.11 Supplementary Notes for the Lec-ture of January 28rd, 2005 . . 2052A.11.1 §3.2 ANNUITY-IMMEDIATE 2052

A.12 Supplementary Notes for the Lec-ture of January 31st, 2005 . . 2056A.12.1 §3.3 ANNUITY-DUE 2056

A.13 Supplementary Notes for the Lec-ture of February 2nd, 2005 . . 2058A.13.1 §3.3 ANNUITY-DUE (con-

tinued) . . . . . . . . 2058A.13.2 §3.4 ANNUITY VALUES

ON ANY DATE . . . 2059


A.14 Supplementary Notes for the Lec-ture of February 4th, 2005 . . 2063A.14.1 §3.4 ANNUITY VALUES

ON ANY DATE (contin-ued) . . . . . . . . . . 2063

A.14.2 §3.5 PERPETUITIES 2066A.15 Supplementary Notes for the Lec-

ture of February 7th, 2005 . . 2067A.15.1 §3.5 PERPETUITIES (con-

tinued) . . . . . . . . 2067A.15.2 §3.6 NONSTANDARD TERMS

AND INTEREST RATES 2068A.15.3 §3.7 UNKNOWN TIME 2068

A.16 Supplementary Notes for the Lec-ture of February 9th, 2005 . . 2070A.16.1 §3.7 UNKNOWN TIME

(continued) . . . . . . 2070A.16.2 §3.8 UNKNOWN RATE

OF INTEREST . . . . 2073A.16.3 §3.9 VARYING INTER-

EST . . . . . . . . . . 2081A.16.4 §3.10 ANNUITIES NOT

INVOLVING COMPOUNDINTEREST . . . . . . 2081

A.17 Supplementary Notes for the Lec-ture of February 11th, 2005 . 2082A.17.1 §4.1 INTRODUCTION 2082A.17.2 §4.2 ANNUITIES PAYABLE

AT A DIFFERENT FRE-QUENCY THAN INTER-EST IS CONVERTIBLE 2082

A.18 Supplementary Notes for the Lec-ture of February 14th, 2005 . 2086A.18.1 §4.3 FURTHER ANALY-

SIS OF ANNUITIES PAYABLELESS FREQUENTLY THANINTEREST IS CONVERT-IBLE . . . . . . . . . 2086

A.18.2 §4.4 FURTHER ANALY-SIS OF ANNUITIES PAYABLEMORE FREQUENTLYTHAN INTEREST IS CON-VERTIBLE . . . . . . 2090

A.19 Supplementary Notes for the Lec-ture of February 16th, 2005 . 2093A.19.1 §4.5 CONTINUOUS AN-

NUITIES . . . . . . . 2093A.19.2 §4.6 BASIC VARYING

ANNUITIES . . . . . 2093A.20 Supplementary Notes for the Lec-

ture of February 18th, 2005 . 2094A.20.1 §4.6 BASIC VARYING

ANNUITIES (continued) 2094A.20.2 §4.7 MORE GENERAL

VARYING ANNUITIES 2099A.21 Supplementary Notes for the Lec-

ture of February 28th, 2005 . 2100A.21.1 §4.6 BASIC VARYING

ANNUITIES (conclusion) 2100A.21.2 §4.8 CONTINUOUS VARY-

ING ANNUITIES . . 2101A.21.3 §4.9 SUMMARY OF RE-

SULTS . . . . . . . . 2101A.21.4 §5.1 INTRODUCTION 2101A.21.5 §5.2 DISCOUNTED CASH

FLOW ANALYSIS . . 2101A.21.6 §5.3 UNIQUENESS OF

THE YIELD RATE . 2101A.22 Supplementary Notes for the Lec-

ture of March 2nd, 2005 . . . 2103A.22.1 §5.4 REINVESTMENT

RATES . . . . . . . . 2103A.22.2 §5.5 INTEREST MEA-

SUREMENT OF A FUND 2106A.22.3 §5.6 TIME-WEIGHTED

RATES OF INTEREST 2106A.22.4 §5.7 PORTFOLIO METH-

ODS AND INVESTMENTYEAR METHODS . . 2106


A.22.5 §5.8 CAPITAL BUDGET-ING . . . . . . . . . . 2106

A.22.6 §5.9 MORE GENERALBORROWING/LENDINGMODELS . . . . . . . 2106

A.23 Supplementary Notes for the Lec-ture of March 4th, 2005 . . . 2107A.23.1 §6.1 INTRODUCTION 2107A.23.2 §6.2 FINDING THE OUT-

STANDING LOAN BAL-ANCE . . . . . . . . . 2107

A.24 Supplementary Notes for the Lec-ture of March 7th, 2005 . . . 2112A.24.1 §6.3 AMORTIZATION SCHED-

ULES . . . . . . . . . 2112A.25 Supplementary Notes for the Lec-

ture of March 11th, 2005 . . . 2115A.25.1 §6.3 AMORTIZATION SCHED-

ULES (continued) . . 2115A.25.2 §6.4 SINKING FUNDS 2119

A.26 Supplementary Notes for the Lec-ture of March 14th, 2005 . . . 2123A.26.1 §6.4 SINKING FUNDS

(concluded) . . . . . . 2123A.26.2 §6.5 DIFFERING PAY-

MENT PERIODS ANDINTEREST CONVERSIONPERIODS . . . . . . . 2127

A.26.3 §6.6 VARYING SERIESOF PAYMENTS . . . 2127

A.26.4 §6.7 AMORTIZATION WITHCONTINUOUS PAYMENTS 2127

A.26.5 §6.8 STEP-RATE AMOUNTSOF PRINCIPAL . . . 2127

A.27 Supplementary Notes for the Lec-ture of March 16th, 2005 . . . 2128A.27.1 §7.1 INTRODUCTION 2128A.27.2 §7.2 TYPES OF SECU-

RITIES . . . . . . . . 2128A.28 Supplementary Notes for the Lec-

ture of March 21st, 2005 . . . 2130

A.28.1 §7.2 TYPES OF SECU-RITIES (conclusion) . 2130

A.28.2 §7.3 PRICE OF A BOND 2131A.29 Supplementary Notes for the Lec-

ture of March 21st, 2005 . . . 2138A.29.1 §7.2 TYPES OF SECU-

RITIES (conclusion) . 2138A.29.2 §7.3 PRICE OF A BOND 2140

A.30 Supplementary Notes for the Lec-ture of March 23rd, 2005 . . . 2144

A.31 Supplementary Notes for the Lec-ture of March 30th, 2005 . . . 2146A.31.1 §7.4 PREMIUM AND DIS-

COUNT . . . . . . . . 2146A.31.2 §7.7 CALLABLE BONDS 2149

A.32 Supplementary Notes for the Lec-ture of April 1st, 2005 . . . . 2153A.32.1 §7.6 DETERMINATION

OF YIELD RATES . 2157A.32.2 §7.5 VALUATION BE-

TWEEN COUPON PAY-MENT DATES . . . . 2157

A.32.3 §7.8 SERIAL BONDS 2159A.32.4 §7.9 SOME GENERAL-

IZATIONS . . . . . . 2159A.32.5 §7.10 OTHER SECURI-

TIES . . . . . . . . . . 2159A.32.6 §7.11 VALUATION OF

SECURITIES . . . . . 2159

B Problem Assignments, Tests, andExaminations from Previous Years 3001B.1 2002/2003 . . . . . . . . . . . 3001

B.1.1 First 2002/2003 ProblemAssignment, with Solu-tions . . . . . . . . . . 3001

B.1.2 Second 2002/2003 Prob-lem Assignment, with So-lutions . . . . . . . . . 3006

B.1.3 Third 2002/2003 Prob-lem Assignment, with So-lutions . . . . . . . . . 3011


B.1.4 Fourth 2002/2003 Prob-lem Assignment, with So-lutions . . . . . . . . . 3018

B.1.5 Fifth 2002/2003 ProblemAssignment, with Solu-tions . . . . . . . . . . 3028

B.1.6 2002/2003 Class Tests, withSolutions . . . . . . . 3036

B.1.7 Final Examination, 2002/2003 3042B.2 2003/2004 . . . . . . . . . . . 3045

B.2.1 First 2003/2004 ProblemAssignment, with Solu-tions . . . . . . . . . . 3045

B.2.2 Second 2003/2004 Prob-lem Assignment, with So-lutions . . . . . . . . . 3048

B.2.3 Third 2003/2004 Prob-lem Assignment, with So-lutions . . . . . . . . . 3058

B.2.4 Fourth 2003/2004 Prob-lem Assignment, with So-lutions . . . . . . . . . 3068

B.2.5 Fifth 2003/2004 ProblemAssignment, with Solu-tions . . . . . . . . . . 3073

B.2.6 2003/2004 Class Test, Ver-sion 1 . . . . . . . . . 3079




B.2.10 Solutions to Problems onthe 2003/2004 Class Tests 3085

B.2.11 Final Examination, 2003/2004 3094B.2.12 Supplemental/Deferred Ex-

amination, 2003/2004 3097

Information for Students in MATH 329 2005 01 1

1 General Information

Distribution Date: Friday, January 7th, 2005subject to correction

(All information is subject to change, either by announcements at lectures,on WebCT, or in print.)

An updated version may be placed, from time to time, on the Math/Statwebsite (cf. §1.4.2 below), and will also be accessible via a link from WebCT.)

The Course Outline for MATH 329 2005 01 can be considered to be pages1 through 10 of these notes.

1.1 Instructor and Times

INSTRUCTOR: Prof. W. G. BrownOFFICE: BURN 1224OFFICE HRS. W 13:20→14:15 h.;(subject to F 10→11 h.;change) and by appointment

TELEPHONE: 398–3836E-MAIL: [email protected]: BURN 1B36

(room change effective 7 Jan 05)CLASS HOURS: MWF 14:35–15:25 h.

Table 1: Instructor and Times

1.2 Course Description

1.2.1 Calendar Description

THEORY OF INTEREST. (3 credits) (Prerequisite: MATH 141.) Simple and com-pound interest, annuities certain, amortization schedules, bonds, depreciation.

1.2.2 Syllabus (in terms of sections of the text-book)

The central part of the course consists of many of the topics in the first nine chapters ofthe textbook [1]1; section numbers, where shown, refer to that book. In the list below

1[n] refers to item n in the bibliography, page 901.

UPDATED TO April 12, 2005


we show the chapters and appendices of the textbook. Following each is a descriptionas of the date of this revision, of the sections to be excluded. This list will be updatedduring the semester, as becomes apparent that certain sections are not appropriate tothe level of the course or the lecture time available.

Chapter 1. The Measurement of Interest §§1.1-1.8. Portions of §1.9, will beomitted. For the present §1.10 will also be omitted.

Chapter 2. Solution of Problems in Interest In §2.6 You may omit the discussion[1, pp. 45-46] of the “method of equated time”.

Chapter 3. Elementary Annuities You may omit [1, §3.6 Nonstandard terms andinterest rates], [1, §3.10 Annuities not involving compound interest, pp. 82-88] and cor-responding exercises. We will also omit [1, §3.8 Unknown rate of interest] and [1, §3.9Varying interest] for the present (possibly to return).

Chapter 4. More General Annuities In the following section we shall consider theproblems strictly on an ad hoc basis: students are not expected to derive nor to applythe identities obtained: [1, §4.2 Annuities payable at a different frequency than interestis convertible; §4.3 Further analysis of annuities payable less frequently than interest isconvertible; §4.4 Further analysis of annuities payable more frequently than interest isconvertible]. Omit [1, §4.5 Continuous annuities] for the present. In [1, §4.6 Nonstandardterms and interest rates] we shall consider the derivation of formulæ for (Ia)n, (Is)n,(Da)n, (Ds)n, and their due and perpetual variants, also the question of annuities ingeometric progression. Omit [1, §4.7 More general varying annuities, §4.8 Continuousvarying annuities, §4.9 Summary of results] for the present, together with their exercises.

Chapter 5. Yield Rates Omit [1, §5.2 Discounted cash flow analysis], except for thedefinition [1, p. 131] of yield rate. Omit [1, §5.3 Uniqueness of the yield rate] except youshould read and understand the example [1, p. 133] of a problem where the yield rateis not unique. Omit [1, §5.5 — §5.9] and accompanying exercises; but we will study [1,§5.4 Reinvestment rates] in preparation for Chapter 6.

Chapter 6. Amortization Schedules and Sinking Funds In [1, §6.4] omit pages178-179, where the function an i&j is introduced. Omit [1, §6.5 — §6.8] and accompanyingexercises.


Chapter 7. Bond and Other Securities Omit [1, §7.6 Determination of yield rates],[1, §7.8 Serial bonds], [1, §7.9 Some generalizations], [1, §7.10 Other securities], [1, §7.11Valuation of securities].

Chapter 8. Practical Applications Omit this chapter.

Chapter 9. More Advanced Financial Analysis Omit this chapter.

Chapter 10. A Stochastic Approach to Interest Omit this chapter.

Appendix I. Table of compound interest functions While most calculations willbe done using calculators, these tables may prove useful.

Appendix II. Table numbering the days of the year

Appendix III. Basic mathematical review Topics that are beyond the requiredprerequisites will be explained if, as, and when they are used.

Appendix IV. Statistical background Omit this section: no background in prob-ability is prerequisite to Math 329.

Appendix V. Iteration methods

Appendix VI. Further analysis of varying annuities Omit this Appendix, whichis concerned with the formula for “Summation by Parts”, analogous to integration byparts for functions of a continuous variable.

Appendix VII. Illustrative mortgage loan amortization schedule

Appendix VIII. Full immunization Omit this Appendix, which is related to [1,§9.9], which is not in the syllabus.

Appendix IX. Derivation of the variance of an annuity Omit this Appendix.

Appendix X. Derivation of the Black-Scholes formula Omit this Appendix.



1.2.3 “Verbal” arguments

An essential feature of investment and insurance mathematics is the need to be able tounderstand and to formulate “verbal” arguments; that is, explanations of the truth ofan identity presented verbally i.e., a proof in words, rather then an algebraic proof. Ina verbal argument we seek more than mathematically correctness: we wish to see anexplanation that could be presented to a layman who is not competent in the mathe-matical bases of this subject, but is still possessed of reason, and needs to be assuredthat he is not being exploited. This facet of the course will be seen, at first, to be quitedifficult. When the skill has been mastered it can be used to verify the correctness ofstatements proved mathematically. Verbal arguments require some care with the under-lying language; students who have difficulty with expression in English are reminded thatall students have the right to submit any written materials in either English or French.2

1.3 Evaluation of Your Progress

1.3.1 Term Mark

The Term Mark will be computed one-third from the assignment grades, and two-thirdsfrom the class test. The Term Mark will count for 30 of the 100 marks in the finalgrade, but only if it exceeds 30% of the final examination percentage; otherwise the finalexamination will be used exclusively in the computation of the final grade.

1.3.2 Assignments.

A total of about 6 assignments will be worth 10 of the 30 marks assigned to Term Work.

1.3.3 Class Test

A class test, will be held on Wednesday, March 09th, 2005, at the regular class time,counting for 20 of the 30 marks in the Term Mark. This date may be changed, afterdiscussion with the class at any scheduled lecture date. Students who don’t come toclass should ensure that they are aware of any changes in the date of the test. Therewill be no “make-up” test for persons who miss the test.

2For a lexicon of actuarial terms in English/French, see The Canadian Institute of Actuaries English-French lexicon [8], at

http://www.actuaries.ca/publications/lexicon/



1.3.4 Final Examination

Written examinations form an important part of the tradition of actuarial mathematics.The final examination in MATH 329 2004 01 will count for either 70% or 100% of thenumerical grade from which the submitted final letter grade will be computed. Wherea student’s Final Examination percentage is superior to her Term Mark percentage, theFinal Examination grade will replace the Term Mark grade in the calculations.

A 3-hour-long final examination will be scheduled during the regular examinationperiod for the winter term (April 14th, 2005 through April 29th, 2005). You are advisednot to make any travel arrangements that would prevent you from being present oncampus at any time during this period. Students who have religious or other constraintsthat could affect their ability to write examinations at particular times should watch forthe Preliminary Examination Timetable, as their rights to apply for special considerationat their faculty may have expired by the time the final examination timetable is published.

1.3.5 Supplemental Assessments

Supplemental Examination. For eligible students who obtain a Final Grade of For D in the course there will be a supplemental examination. (For information aboutSupplemental Examinations, see the McGill Calendar, [3].)

There is No Additional Work Option. “Will students with marks of D, F, or Jhave the option of doing additional work to upgrade their mark?” No. (“AdditionalWork” refers to an option available in certain Arts and Science courses, but not availablein this course.)

1.3.6 Machine Scoring

“Will the final examination be machine scored?” While there could be Multiple Choicequestions on quizzes, and/or the Final Examination, such questions will not be machinescored.

1.3.7 Plagiarism

While students are not discouraged from discussing assignment problems with their col-leagues, the work that you submit — whether through homework, the class test, or ontutorial quizzes or the final examination should be your own. The Handbook on StudentRights and Responsibilities states in ¶15(a)3 that

3http://ww2.mcgill.ca/students-handbook/chapter3secA.html


“No student shall, with intent to deceive, represent the work of another personas his or her own in any academic writing, essay, thesis, research report,project or assignment submitted in a course or program of study or representas his or her own an entire essay or work of another, whether the material sorepresented constitutes a part or the entirety of the work submitted.”

You are also referred to the following URL:

http://www.mcgill.ca/integrity/studentguide/

1.4 Published Materials

1.4.1 Required Text-Book

The textbook for the course this semester is [1] Stephen G. Kellison, The Theory ofInterest, Second Edition. Irwin/McGraw-Hill, Boston, etc. (1991), ISBN 0-256-09150-1.

1.4.2 Website

These notes, and other materials distributed to students in this course, will be accessibleat the following URL:

http://www.math.mcgill.ca/brown/math329b.html

The notes will be in “pdf” (.pdf) form, and can be read using the Adobe Acrobat reader,which many users have on their computers. This free software may be downloaded fromthe following URL:

http://www.adobe.com/prodindex/acrobat/readstep.html 4

Where revisions are made to distributed printed materials — for example these informa-tion sheets — it is expected that the last version will be posted on the Web.

The notes will also be available via a link from the WebCT URL:

http://webct.mcgill.ca

but not all features of WebCT will be implemented.

1.4.3 Reference Books

The textbook used for 2001-2003 may be used as a reference: [5] Michael M. Parmen-tier, Theory of Interest and Life Contingencies, with Pension Applications: A Problem-Solving Approach, 3rd edition. ACTEX Publications, Winstead, Conn. (1999), ISBN0-56698-333-9.

4At the time of this writing the current version is 5.1.


1.5 Other information

1.5.1 Prerequisites

It is your responsibility as a student to verify that you have the necessary calculusprerequisites. It would be foolish to attempt to take the course without them.

1.5.2 Calculators

The use of non-programmable, non-graphing calculators only will be permitted in home-work, tests, or the final examination in this course. Students may be required to convinceexaminers and invigilators that all memories have been cleared. The use of calculatorsthat are either graphing or programmable will not be permitted during test or examina-tions, in order to “level the playing field”.

1.5.3 Self-Supervision

This is not a high-school course, and McGill is not a high school. The monitoring ofyour progress before the final examination is largely your own responsibility. While theinstructor is available to help you, he cannot do so unless and until you identify theneed for help. While the significance of the homework assignments and class test in thecomputation of your grade is minimal, these are important learning experiences, andcan assist you in gauging your progress in the course. This is not a course that canbe crammed for: you must work steadily through the term if you wish to develop thefacilities needed for a strong performance on the final examination.

Working Problems on Your Own. You are advised to work large numbers of prob-lems from your textbook. The skills you acquire in solving textbook problems could havemuch more influence on your final grade than either the homework or the class test.

1.5.4 Escape Routes

At any time, even after the last date for dropping the course, students who are experi-encing medical or personal difficulties should not hesitate to consult their advisors or theStudent Affairs office of their faculty. Don’t allow yourself to be overwhelmed by suchproblems; the University has resource persons who may be able to help you.

1.5.5 Showing your work; good mathematical form; simplifying answers

When, in a quiz or examination problem, you are explicitly instructed to show all yourwork, failure to do so could result in a substantial loss of marks — possibly even allof the marks; this is the default . The guiding principle should be that you want to be


able to communicate your precise reasoning to others and to yourself. You are alwaysexpected to “simplify” any algebraic or numerical expressions that arise in your solutionsor calculations. Verbal proofs are expected to be “convincing”: it will not be sufficientto simply describe mathematical expressions verbally.


2 Timetable

Distribution Date: (original version) Tuesday, January 4th, 2005this revision as of April 12, 2005

(Subject to correction and change.)Section numbers refer to the text-book.5

MONDAY WEDNESDAY FRIDAY

JANUARY04 (TUESDAY!) §§1.1–

1.205 §§1.1–1.2 07 §§1.2–1.3

10 §§1.3–§§1.5 12 §1.6 14 §1.7, §1.8Course changes must be completed on MINERVA by Jan. 16, 2005

17 §§1.8, 1.9 (part) 19 §1.9, §§2.1, 2.2 21 §§2.2, 2.3, 2.4Deadline for withdrawal with fee refund = Jan. 23, 2005

24 §§2.5, 2.6 26 §§2.7, 2.8, §3.1 28 §§3.2, 3.3Verification Period: January 31–February 04, 2005

31 §3.3 1©FEBRUARY

02 §3.4 04 §§3.4, 3.5Deadline for withdrawal (with W) from course via MINERVA = Feb. 13, 2005

07 §3.7 09 §3.7 11 §§4.1, 4.214 §§4.3, 4.4 2© 16 §4.6 18 §4.6

Study Break: February 21–25, 2005No lectures, no regular office hours

21 NO LECTURE 23 NO LECTURE 25 NO LECTURE28 §5.3,§5.4

5

Notation: n© = Assignment #n due todayR© = Read OnlyX = reserved for eXpansion or review


MONDAY WEDNESDAY FRIDAYMARCH

02 §5.4 04 §§6.1, 6.207 §6.3 3© 09 CLASS TEST 11 §§6.3, 6.414 §6.4 16 §7.1 18 §7.221 §7.3 23 §7.3, §7.4 25 NO LECTURE28 NO LECTURE 30

APRIL01

04 Discussion of oldexam

06 Discussion of oldexam

08 NO LECTURE

11 X 13 X


3 First Problem Assignment

Mounted on the Web on Sunday, January 16th, 2005Distributed in hard copy on Wednesday, January 25th, 2005

Full solutions are to be submitted by Monday, January 31st, 2005

These problems are to be solved with full solutions, modelled either on the solutions toproblems in the textbook, or in the notes on the Web for this or previous years. Theessence is that the reader should be able to reconstruct every step of the proof from whatyou have written: getting the right answer is never enough. You are not being gradedfor elegance, but simply for the proof being logical, without serious gaps.

1. (a) Show that the function 225− (t− 10)2 cannot be used as an amount functionfor t ≥ 10.

(b) For the interval 0 ≤ t ≤ 10, determine the accumulation function a(t) thatcorresponds to A(t) = 225− (t− 10)2.

(c) Determine, for the above functions, I3, i6, d7.

(d) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent effective annual rate of compound interest,compounded every year, which would yield the same accumulation after 10years.

(e) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of simple interest which wouldyield the same accumulation after 10 years.

(f) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of simple discount.

(g) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of compound discount.

(h) Verify that the rates i and d of compound interest and discount have theproperty that (1 + i)(1− d) = 1.

2. Determine the accumulated value of 100 at the end of 2 years if

(a) the nominal annual rate of interest is 8% convertible quarterly

(b) the nominal annual rate of discount is 8% convertible once every 6 years

(c) interest is compounded instantaneously at the rate of 8%

(d) an annual effective compound discount rate of 8% is applied

(e) an annual simple discount rate of 8% is applied


(f) interest is earned during the first year at a nominal annual rate of 8% con-vertible quarterly, and during the second year at a nominal annual rate ofdiscount of 8% convertible quarterly.

3. Today is New Year’s Day. In return for payments of 1500 at the end of January,February, and March, and of 3000 at the end of May, July, and September, aninvestor agrees to pay now the total value of the 6 payments, and to either makeor receive an additional payment at the end of December. Find the amount ofthat additional payment if it is known that the nominal annual interest rate is 6%,compounded monthly. (First set up an equation of value.)

4. Analogously to the “rule of 72”, you are asked to develop a rule of n to approximatehow long it takes for money to increase to 11

2times its initial value. (That is, to

determine a 2-digit integer N = 10n1 + n0 (where n0, n1 are decimal digits), forwhich (0.01)N

iis a good approximation to the number of years required.) Your

approximation should be best around 8%.

5. (a) Find the smallest nominal rate of interest convertible monthly at which theaccumulated value of 15,000 at the end of 3 years is at least 24,000.

(b) Find the smallest6 nominal rate of discount convertible semi-annually at whichthe accumulated value of 15,000 at the end of 3 years is at least 24,000.

6. Let x be a positive real number, 0 ≤ x < 1. Prove that

1

1− x≥ 1 + x

for all such x. Conclude that the accumulated value of 100 after 10 years at aninterest rate of x% is always less than or equal to the accumulated value of 100after 10 years at a discount rate of x%, with equality holding only when x = 0.

6Added 24 Jan 05: One student observed that the original wording of this problem, with largest , wasclearly not what was intended.



4 Second Problem Assignment

Mounted on the Web on Sunday, January 30th, 2005Distributed in hard copy on Wednesday, February 2nd, 2005

Full solutions are to be submitted by Monday, February 14th, 2005.(subject to correction)


1. (a) At a certain rate of compound interest an investment of 1000 will grow to1500 at the end of 12 years. Determine its value at the end of 5 years.

(b) At a certain rate of compound interest an investment of 1000 will grow to1500 at the end of 12 years. Determine precisely when its value is exactly1200.

(c) A debt of 7000 is due at the end of 5 years. If 2000 is paid at the end of1 year, what single payment should be made at the end of the 2nd year toliquidate the debt, assuming interest at the rate of 6.5% per year, compoundedquarterly.

(d) George agrees to buy his brother’s car for 7000. He makes a down payment of4000, and agrees to pay two equal payments, one at the end of 6 months, andthe other at the end of a year. If interest is being charged at 5% per annumeffective, how large should each of the equal payments be?

(e) A bill for 1500 is purchased for 1000 15 months before it is due. Determinethe nominal rate of discount, compounded monthly, which the purchaser ispaying.

(f) Bills for 1500 are regularly purchased for 1000 15 months before they are due.The purchaser knows that, among 10 such bills, he will be unable to collectanything on one of them, will have to pay his lawyers 500 each to effectcollection on 2 others, and will collect the others without any impediment.Lumping all of these together, what is the effective annual interest rate earnedby the purchaser on his investment, if it is assumed that the lawyers’ accountsare due at the same time as the bills?

2. A government offers savings bonds in multiples of $1000, which mature in 10 yearsat $2000, but pay no interest until they are redeemed.


(a) Assuming interest compounded semi-annually, what nominal annual rate ofinterest does the bond holder earn?

(b) Suppose that the bonds earn interest at the nominal rate of i − 0.01 forthe first 5 years, compounded semi-annually; and that they earn interest atthe nominal rate of i + 0.01 compounded semi-annually for the last 5 years.Determine i.

(c) The government has contracted with a bank to market the bonds, at a costof $40 per $1000 bond. What interest rate, compounded semi-annually, is thegovernment paying for the net proceeds it receives for each $1000 bond?

3. The cash price of a new automobile is 18,000 plus 15.025% tax. The purchaseris prepared to finance the car and taxes at 18% convertible semi-monthly, and tomake payments of 230 at the end of every half-month for 3.5 years, with the firstpayment to be made one half-month after delivery. The dealer requires a downpayment upon delivery, both to make up the gap in the financing, and from whichto pay his immediate costs (commission to the salesperson, preparation costs, salestaxes).

(a) Determine the value of this down payment.

(b) Determine the value of the down payment if the purchaser decides, instead ofsemi-monthly payments, to make a payment of 460 at the end of every month,first payment a month after delivery, last payment to be made 3.5 years afterdelivery. The interest rate and compounding period do not change.

(c) Determine the value of the down payment if the original conditions are changedso that the first payment of 230 is made 1.5 months after delivery, but thesame number of payments of 230 are made as originally planned. The interestrate and compounding period do not change.

4. An employee aged exactly 40 decides to accumulate a fund for retirement at age 65by depositing 200 at the beginning of each month for 25 years. When she reachesage 65, she plans to withdraw a fixed amount at the beginning of each year for 15years. Assuming that all payments are made, determine the amount of the annualpayments that she will be able to withdraw:

(a) if the interest rate is always taken to be (a nominal annual rate of) 6% perannum, compounded monthly.

(b) if the interest rate is taken to be 6% per annum compounded monthly duringthe time when the fund is being built up, and 8% per annum effective whenshe reaches age 65.


(c) if the interest rate is taken to be 4% per annum compounded monthly for thecoming 5 years, then 6% per annum compounded monthly for the next 20years while the fund is being built up, then 8% per annum effective when thefund is paying out annual payments.

5. In his will, a benefactor of McGill contributes a large sum of money to establisha fund, whose proceeds are to be used to provide annual bursaries of 5000 to6 actuarial students. The principal of the fund is to remain constant, and thebursaries are funded by the interest earned, forever.

(a) If the effective annual interest rate is assumed to be 6% forever, determine thelump sum that the benefactor’s estate needs to contribute to McGill a yearbefore the first payment of bursaries.

(b) If the effective annual interest rate is assumed to be 6% forever, determinethe lump sum that the benefactor’s estate needs to contribute to McGill ifthe first bursaries are to be issued immediately.

(c) Suppose that each student receives not a bursary of 5000, but an annualannuity-due of 1250 for 4 years. Under these changed conditions, determinethe lump sum payment needed now if the first bursaries are to be awardedimmediately or 1 year hence respectively. (The number of awards will remainthe same — each year 6 students are awarded a 4-year sequence of bursarypayments of 1250, the first payment to be made immediately.)

6. Let m and n be positive integers. Consider an annuity-immediate which pays 1 atthe end of every period for mn periods. Explain in words why each of the followingformulæ represents the value of this annuity m + 1 years before the first paymentis made. (If you have doubts about the truth of this claim, you may wish to verifyalgebraically that the claim is correct before you attempt to explain it in words.)

(a) vm · amn

(b) am(n+1)

− am

(c) vm · am(n+1)

− vm(n+2) · sm


5 Solutions, First Problem Assignment

Mounted on the Web on Thursday, February 10th, 2005(Caveat lector! 7 There could be some undetected misprints or errors.)Full solutions were to be submitted by Monday, January 31st, 2005

Students were advised that “These problems are to be solved with full solutions, modelledeither on the solutions to problems in the textbook, or in the notes on the Web for thisor previous years. The essence is that the reader should be able to reconstruct everystep of the proof from what you have written: getting the right answer is never enough.You are not being graded for elegance, but simply for the proof being logical, withoutserious gaps.”

1. (a) Show that the function 225− (t− 10)2 cannot be used as an amount functionfor t ≥ 10.

(b) For the interval 0 ≤ t ≤ 10, determine the accumulation function a(t) thatcorresponds to A(t) = 225− (t− 10)2.

(c) Determine, for the above functions, I3, i6, d7.

(d) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent effective annual rate of compound interest,compounded every year, which would yield the same accumulation after 10years.

(e) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of simple interest which wouldyield the same accumulation after 10 years.

(f) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of simple discount.

(g) Suppose that A(0) is invested at time t = 0 and accumulates to A(10) at timet = 10. What is the equivalent annual rate of compound discount.

(h) Verify that the rates i and d of compound interest and discount have theproperty that (1 + i)(1− d) = 1.

Solution:

(a) By condition [1, 2, p. 2], an accumulation function must be increasing (or atleast non-decreasing); the same property must hold for an amount function,

7Let the reader beware!


since it is a positive multiple of its corresponding accumulation function. ButA(t) = 225− (t− 10)2 has the property that

d

dtA(t) = 2(10− t)

which is increasing only when 20− t ≥ 0, i.e., when t ≤ 10.

(b) a(t) =A(t)

A(0)=

125 + 20t− t2

125=

(5 + t)(25− t)

125.

(c)

I3 = A(3)− A(2) = 15

i6 = a(6)− a(5) =A(6)− A(5)

A(5)=

9

200

d7 =I7

A(7)=

A(7)− A(6)

A(7)=

7

216

(d) Let i be the equivalent annual rate of compound interest. Since A(10) = 225,A(0) = 125,

125(1 + i)10 = 225 ⇒ 1 + i =

(225

125

) 110

⇒ i =10√

1.8− 1 = 6.0540482%.

(e) Let i be the equivalent annual rate of simple interest. Then

125(1 + 10i) = 225 ⇒ 1 + 10i =225

125⇒ i = 8%.

(f) Let d be the equivalent annual rate of simple discount. Then

125 = (1− 10d)225 ⇒ d =2

45= 4.4444%.

(g) Let d be the equivalent annual rate of simple discount. Then

125 = 225(1− d)10 ⇒ d = 1− 10

√5

9= 5.70845551%.

(h) (1 + 0.060540482)(1− 0.0570845551) = 1 .


2. Determine the accumulated value of 100 at the end of 2 years if

(a) the nominal annual rate of interest is 8% convertible quarterly

(b) the nominal annual rate of discount is 8% convertible once every 6 years

(c) interest is compounded instantaneously at the rate of 8%

(d) an annual effective compound discount rate of 8% is applied

(e) an annual simple discount rate of 8% is applied

(f) interest is earned during the first year at a nominal annual rate of 8% con-vertible quarterly, and during the second year at a nominal annual rate ofdiscount of 8% convertible quarterly.

Solution:

(a) A nominal annual interest rate of 8% convertible quarterly is equivalent to a3-month rate of 8

4% = 2%. The accumulated value of 100 after 2

14

= 8 quarter

years is100(1.02)8 = 117.17 .

(b) We are given that

d( 16) = 8% .

The accumulated value of 100 after 26

= 13

6-year periods

100 (1− (6× 0.08))−26 = 124.36 .

(c) Interest compounded instantaneously at the rate of 8% yields an accumulationfactor of

limm→∞

(1 +

0.08

m

)m

= e0.08 .

After 2 years 100 accumulates to

100e2×0.08 = 117.35 .

(d) At an annual effective compound discount rate of 8%, 100 accumulates after2 years to

100(1− 0.08)−2 = 118.15 .

(e) At an annual simple discount rate of 8%, 100 accumulates after 2 years to

100

1− 2(0.08)= 119.05 .


(f) The accumulated value of 100 after the first year will be 100(1.02)4. Duringthe second year this amount will accumulate to a final amount of

100(1.02)4

(1− 0.08

4

)−4

= 100

(1.02

0.98

)4

= 117.36 .

3. Today is New Year’s Day. In return for payments of 1500 at the end of January,February, and March, and of 3000 at the end of May, July, and September, aninvestor agrees to pay now the total value of the 6 payments, and to either makeor receive an additional payment at the end of December. Find the amount ofthat additional payment if it is known that the nominal annual interest rate is 6%,compounded monthly. (First set up an equation of value.)

Solution: Let us denote the final payment by the investor by X. The effectivemonthly interest rate is 6

12% = 1

2%. We will take as the comparison date the end

of December (although any date chosen would yield the same information). Thevalue on 31 December of the payments paid to the investor is

1500((1.005)11 + (1.005)10 + (1.005)9

)

+3000((1.005)7 + (1.005)5 + (1.005)3

)= 13, 957.74 .

The value on 31 December (just after she has paid the last payment) of the pay-ments paid out by the investor is

(3(1500) + 3(3000))(1.005)12 + X = 14332.65 + X .

Equating the two values yields X = −374.91. Thus the investor is entitled toreceive a final payment of 374.91 at the end of December. (The fact that the finalpayment would be received by the investor, rather than paid by her could havebeen reasoned without calculation; we didn’t need the mathematical calculation totell us that from the sign of the answer.)

4. Analogously to the “rule of 72”, you are asked to develop a rule of n to approximatehow long it takes for money to increase to 11

2times its initial value. (That is, to

determine a 2-digit integer N = 10n1 + n0 (where n0, n1 are decimal digits), forwhich (0.01) · N

iis a good approximation to the number of years required.) Your

approximation should be best around 8%.

Solution: We have to approximate a solution to the equation

(1 + i)n = 1.5,

which is equivalent to

n =ln 1.5

i· i

ln(1 + i).

We find that


i (ln 1.5) · i

ln(1 + i)7.0% 0.4198.0% 0.4219.0% 0.423

so we take N = 42.

5. (a) Find the smallest nominal rate of interest convertible monthly at which theaccumulated value of 15,000 at the end of 3 years is at least 24,000.

(b) Find the smallest nominal rate of discount convertible semi-annually at whichthe accumulated value of 15,000 at the end of 3 years is at least 24,000.

Solution:

(a) Let i denote the nominal interest rate sought. Then

15000

(1 +

i

12

)36

≥ 24000

⇔(

1 +i

12

)36

≥ 1.6000

⇔ 1 +i

12≥ (1.6)

136 = 1.013141254

so i =≥ 15.7695048%. The smallest interest rate is 15.8%.

(b) This could be considered a trick question; but the fact is that the word largestwas not intended. I “solve” it both as written, and with the intended wording.Let d denote the nominal discount rate.

As written, “largest” Then

15000

(1− d

2

)−6

≥ 24000

⇔(

1− d

2

)6

≤ 5

8

⇔ 1− d

2≤ 6

√5

8= .9246555971

implying that d ≥ 15.068880%. Thus any discount rate d ≥ 15.1% hasthe desired property. There is no largest rate, since, as the rate d(2)

approaches 100% from below, the accumulated value approaches ∞. We


haven’t attached a meaning to discount rates above 100%. (As for adiscount rate of 100% — that does make sense, but would not permit theaccumulation of funds: if we discount a sum X of money back from time1 to time 0 at an effective discount rate of 100%, we obtain a presentvalue of 0.)

Corrected to “smallest”

15000

(1− d

2

)−6

≤ 24000

⇔(

1− d

2

)6

≤ 5

8

⇔ 1− d

2≤ 6

√5

8= .9246555971

implying that d ≥ 15.068880%. Thus the smallest discount rate is 15.1%.

6. Let x be a positive real number, 0 ≤ x < 1. Prove that

1

1− x≥ 1 + x

for all such x. Conclude that the accumulated value of 100 after 10 years at aninterest rate of x% is always less than or equal to the accumulated value of 100after 10 years at a discount rate of x%, with equality holding only when x = 0.

Solution: Since x2 is a square, it cannot be negative, and can be 0 only when x = 0;hence 1− x2 ≤ 1, with equality holding precisely when x = 0. But this inequalityis equivalent to

1 ≥ (1− x)(1 + x)

or to1

1− x≥ 1 + x

since the inequality is preserved when we divide both sides by the positive number1− x; again, equality holds when x = 0.

The accumulated value of 100 at the interest rate of x% is 100(1 + x

100

)10. We

apply the preceding argument 10 times, after replacing x by x100

: this cannot exceed

100(1− x

100

)−10, which is the accumulated value of 100 at the discount rate of x%.

Equality holds when x100

= 0, i.e., when x = 0.


6 Third Problem Assignment

Mounted on the Web on Sunday, February 13th, 2005Distributed in hard copy on Wednesday, February 16th, 2005

Full solutions were to be submitted by March 7th, 2005.


1. McGill plans to create a scholarship fund that will eternally pay 50 students amonthly stipend of $200 at the beginning of months September through April,plus an amount of $300 on the following May 1st.

(a) If interest is assumed to be at a nominal annual rate of 6% per annum, com-pounded monthly, determine the amount that is needed in this fund on Sep-tember 1st just before the fund begins making payments.

(b) Determine the amount that will be in the fund just after the December schol-arship payments in the first year, and on September 1st of the following year,just before the September payments.

(c) Suppose that at the beginning of September, 8 years after the fund is estab-lished, it is decided to increase the capital in the fund because the interestrate has changed to 4% per annum compounded monthly. Determine howmuch additional capital needs to be added to the fund.

(d) Suppose that 2 years after the interest rate is changed to 4%, it changes again,this time to 8%. This time it is decided to leave the capital unchanged, butto increase the payments to students by a lump sum of $M to each student,payable on December 1st, together with the regular December payment underthe scholarship. Determine the amount of that lump sum payment.

2. A loan of 10,000 is to be repaid by regular, half-yearly payments of 1,000, the firstto be made at the end of the 3rd year. The loan will be paid off by a final paymentwhich must be at least 1,000.

(a) If the interest rate is to be 8% per annum, compounded semi-annually, deter-mine the amount of the final payment, and when it is made.

(b) Suppose that the final payment is now not more than 1,000. and the inter-est rate remains 8% per annum compounded semi-annually. Determine theamount of the final payment, and when it is made.


(c) Suppose that, after signing his original commitment, the borrower decidesthat he would like to make the first payment 6 months from the date ofborrowing, and that all the payments under this loan should be exactly equal.If the interest rate is 6% per annum, compounded semi-annually, determinethat payment level that would be closest to 1000 per half-year, and determineexactly when the loan will be paid off.

3. (An acknowledgement of source of this problem will be contained in the solutions,when published.) Katherine, 25 years old, deposits 10,000 at the beginning ofevery 4-year period into an RSSP account. The account pays compound interestannually, at the effective annual rate i. The accumulated amount in the account atthe end of 40 years is X, which is 6 times the accumulated amount in the accountat the end of 20 years. Determine X.

4. For each of the following sequences of payments as of the time stated,

• express the value using standard symbols, as simple as possible;

• give a formula for the value (in terms of i, v, d, etc.);

• evaluate using a calculator or computer — not with tables.

(a) a perpetuity-immediate paying 1 per year at effective annual interest rate4.25%, evaluated 1 year before the first payment;

(b) a perpetuity paying 1 per year at effective annual interest rate 4.25%, evalu-ated 2 years before the first payment;

(c) a perpetuity-due paying 1 per half-year at nominal interest rate 5%, com-pounded semi-annually, evaluated just before the first payment;

(d) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the3rd year, etc., at an interest rate of 7% per year effective, evaluated just afterthe last payment;

(e) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the3rd year, etc., at an interest rate of 7% per year effective, evaluated at thetime of the 2nd payment;

(f) a 10-year decreasing annuity paying 1 less each quarter-year, evaluated justafter the last payment, where the nominal interest rate is a 8% annual, com-pounded quarterly;

5. X and Y have sold their home for 400,000, and wish to purchase an annuity-immediate so that they can spread the proceeds over the next 10 years. The firstpayment will be one month from the date of purchase of the annuity.


(a) Determine the level monthly payments they will receive if the effective annualinterest rate is 6%.

(b) X and Y live frugally, and don’t think they need as much income now asthey will as time passes. Accordingly they plan to receive monthly paymentswhich will gradually increase. If the first payment is 3,000, and the paymentsincrease each month by the same dollar amount K, determine K. The effectiveannual interest rate remains 6%. What is the final monthly payment?

(c) Suppose that the payments remain constant in any year. The first year’smonthly payments are all 3000, the next year’s 3000+L, the next years’ 3000+2L, 3000 + 3L, . . . , 3000 + 9L. Determine L if the nominal interest rate is6%, compounded monthly.


7 Solutions, Second Problem Assignment

Mounted on the Web on March 3rd, 2005Full solutions were to be submitted by February 14th, 2005.

(Subject to Correction)

These problems were to be solved with full solutions, modelled either on the solutionsto problems in the textbook, or in the notes on the Web for this or previous years. Theessence is that the reader should be able to reconstruct every step of the proof from whatyou have written: getting the right answer is never enough. You are not being gradedfor elegance, but simply for the proof being logical, without serious gaps.

1. (a) At a certain rate of compound interest an investment of 1000 will grow to1500 at the end of 12 years. Determine its value at the end of 5 years.

(b) At a certain rate of compound interest an investment of 1000 will grow to1500 at the end of 12 years. Determine precisely when its value is exactly1200.

(c) A debt of 7000 is due at the end of 5 years. If 2000 is paid at the end of1 year, what single payment should be made at the end of the 2nd year toliquidate the debt, assuming interest at the rate of 6.5% per year, compoundedquarterly.

(d) George agrees to buy his brother’s car for 7000. He makes a down payment of4000, and agrees to pay two equal payments, one at the end of 6 months, andthe other at the end of a year. If interest is being charged at 5% per annumeffective, how large should each of the equal payments be?

(e) A bill for 1500 is purchased for 1000 15 months before it is due. Determinethe nominal rate of discount, compounded monthly, which the purchaser ispaying.

(f) Bills for 1500 are regularly purchased for 1000 15 months before they are due.The purchaser knows that, among 10 such bills, he will be unable to collectanything on one of them, will have to pay his lawyers 500 each to effectcollection on 2 others, and will collect the others without any impediment.Lumping all of these together, what is the effective annual interest rate earnedby the purchaser on his investment, if it is assumed that the lawyers’ accountsare due at the same time as the bills?

Solution:

(a) Let I be the effective annual rate of compound interest. Then

1000(1 + i)12 = 1500 (1)


implies that

(1 + i)5 =

(1500

1000

) 512

= 1.184053587

so the investment is worth 1184.05 after 5 years.

(b) Let t be the time in years when the investment is worth exactly 1200. Then

1000(1 + i)t = 1200 ,

where i is the effective annual rate of compound interest. By equation (1),

ln(1 + i) =ln 1.5

12.

Hence

t =ln 1.2

ln(1 + i)=

12× ln 1.2

ln 1.5= 5.395923442 :

the investment has the desired value after 5.40 years.

(c) Let x denote the payment that must be made at the end of the 2nd year. Theequation of value at that time is

x + 2000

(1 +

0.065

4

)4

= 7000

(1 +

0.065

4

)−12

implying that

x = −2000

(1 +

0.065

4

)4

+ 7000

(1 +

0.065

4

)−12

= 3635.67

should be paid at the end of the second year to liquidate the debt.

(d) Let x denote the amount of each of the payments that should be made at theends of 6 months and 1 year. Then the equation of value at time 0 is

x((1.05)−

12 + (1.05)−1

)= 7000− 4000 = 3000 ,

implying that

x =3000

(1.05)−12 + (1.05)−1

= 1555.79 .

(e) Let d be the effective monthly rate of discount. Then

1000 = (1− d)151500 ,


implying that

d = 1−(

1000

1500

) 115

= 2.66689392% .

This is the effective monthly rate. The nominal annual rate of discount,compounded monthly, is, therefore 12d = 32%.

(f) Of 10 bills which mature in 15 months, the total return that will enure to thepurchaser at that time is

7(1500) + 2(1500− 500) + 0(1500) = 12500 .

If we denote the effective annual interest rate by i, then the discounted valueof 12500 at time t = 0 will be 12500(1 + i)−

1512 . The equation of value at time

t = 0 is, therefore,

12500(1 + i)−1512 = 10(1000) = 10000 ,

implying that (1 + i)54 = 1.25, so i = 1.250.8 − 1 = 19.5440625%.

2. A government offers savings bonds in multiples of $1000, which mature in 10 yearsat $2000, but pay no interest until they are redeemed.

(a) Assuming interest compounded semi-annually, what nominal annual rate ofinterest does the bond holder earn?

(b) Suppose that the bonds earn interest at the nominal rate of i − 0.01 forthe first 5 years, compounded semi-annually; and that they earn interest atthe nominal rate of i + 0.01 compounded semi-annually for the last 5 years.Determine i.

(c) The government has contracted with a bank to market the bonds, at a costof $40 per $1000 bond. What interest rate, compounded semi-annually, is thegovernment paying for the net proceeds it receives for each $1000 bond?

Solution:

(a) Let i be the nominal interest rate, compounded semi-annually, earned by thepurchaser. The equation of value at time 10 is then

1000

(1 +

i

2

)20

= 2000

implying that

i = 2(2

120 − 1

)= 7.0529848%.


(b) The equation of value is now

1000

(1 +

i− 1

2

)10 (1 +

i + 1

2

)10

= 2000

implying that(

1 +i− 0.01

2

)(1 +

i + 0.01

2

)= 2

110

⇒(

1 +i

2

)2

− (0.01)2

4= 2

110

⇒ 1 +i

2=

(2

110 +

(0.01)2

4

) 12

⇒ i = 2

(2

110 +

(0.01)2

4

) 12

− 2

= 7.0553996%

(c) Let i be the nominal interest rate, compounded semi-annually. From thegovernment’s point of view the equation of value at maturity is

960

(1 +

i

2

)20

= 2000

implying that

i = 2

((2000

960

) 120

− 1

)= 7.4760322%

3. The cash price of a new automobile is 18,000 plus 15.025% tax. The purchaseris prepared to finance the car and taxes at 18% convertible semi-monthly, and tomake payments of 230 at the end of every half-month for 3.5 years, with the firstpayment to be made one half-month after delivery. The dealer requires a downpayment upon delivery, both to make up the gap in the financing, and from whichto pay his immediate costs (commission to the salesperson, preparation costs, salestaxes).

(a) Determine the value of this down payment.

(b) Determine the value of the down payment if the purchaser decides, instead ofsemi-monthly payments, to make a payment of 460 at the end of every month,first payment a month after delivery, last payment to be made 3.5 years afterdelivery. The interest rate and compounding period do not change.


(c) Determine the value of the down payment if the original conditions are changedso that the first payment of 230 is made 1.5 months after delivery, but thesame number of payments of 230 are made as originally planned. The interestrate and compounding period do not change.

Solution:

(a) The present value of the future payments is

230 · a(3.5)(24) 0.18

24= 230

(1− (1.0075)−84

0.0075

)

= 14295.41185

The excess of the purchase price and taxes over the present value of the futurepayments is, therefore,

(1.0015025)(18000)− 14295.41185 = 6409.08815 .

Thus the down payment will be 6,409.09.

(b) We can still interpret the payments as constituting an annuity-immediate.But the interval of payments has paid, while the compounding interval hasnot. So we need to determine the interest rate that corresponds to one month.This is

(1.0075)2 − 1 = 0.01505625 .

The present value of the future payments is

460a(3.5)(12) 0.01505625

= 460 · 1− (1.01505625)−42

0.01505625= 14242.00433

The excess of the purchase price and taxes over the present value of the futurepayments is, therefore,

(1.0015025)(18000)− 14242.00433 = 6462.49567 .

Thus the down payment will be 6,462.50.

(c) Since the repayment schedule is delayed by 2 payments, which are added atthe end, the present value of the future payments is

230a(3.5)(24)+2 0.18

24− 230a2 0.18

24= 230

(1− (1.0075)−86

0.0075− 1− (1.0075)−2

0.0075

)

= 230(63.20976257− 1.977722907)

= 14083.36912


The excess of the purchase price and taxes over the present value of the futurepayments is now

(1.0015025)(18000)− 14083.36912 = 6621.13088 .

Thus the down payment will now be 6,621.13.

4. An employee aged exactly 40 decides to accumulate a fund for retirement at age 65by depositing 200 at the beginning of each month for 25 years. When she reachesage 65, she plans to withdraw a fixed amount at the beginning of each year for 15years. Assuming that all payments are made, determine the amount of the annualpayments that she will be able to withdraw:

(a) if the interest rate is always taken to be (a nominal annual rate of) 6% perannum, compounded monthly.

(b) if the interest rate is taken to be 6% per annum compounded monthly duringthe time when the fund is being built up, and 8% per annum effective whenshe reaches age 65.

(c) if the interest rate is taken to be 4% per annum compounded monthly for thecoming 5 years, then 6% per annum compounded monthly for the next 20years while the fund is being built up, then 8% per annum effective when thefund is paying out annual payments.

Solution:

(a) The effective interest rate per month is 6%12

= 12%. The number of contributions

will be 25× 12 = 300. The value of the fund at maturity, when the employeereaches age 65, will be

200s25×12 12% = 200

((1.005)300 − 1

(0.005)(

11.005

))

= 139291.7864

The withdrawals from the fund will be once a year. If we wish to interpretthem as payments under an annuity-due, we need to determine the rate ofinterest that corresponds to the period between the payments, i.e. to a 1-yearperiod; that is, we need to determine i, knowing i(12). This gives us an annualrate of (1.005)12 − 1 = 6.1677812%. If the annual withdrawal from the fund,at the beginning of each year, be denoted by X, then the equation of valuejust before the first withdrawal is

X · a15 6.1677812% = 139291.7864


which we can solve to determine

X =139291.7864

1−(1.061677812)−15

0.061677812( 11.061677812)

= 15393.80469

so the withdrawals will be 15393.80, at the beginning of each of 15 years.

(b) Proceeding as in the previous case, we replace the interest rate of 6.1677812%by 8%. Now we solve

X · a15 8% = 139291.7864

to determine

X =139291.7864

1−(1.08)−15

0.08( 11.08)

= 15067.95928

so the withdrawals will be 15,067.96, at the beginning of each of 15 years.

(c) We need to recompute the value at the time of maturity of the fund of thecontributions. The last 20 years of payments accumulate at maturity to

200s20×12 12% = 200

((1.005)240 − 1

(0.005)(

11.005

))

= 92870.21994

The first five years of payments accumulate to

200 · s60 412

%

at the end of 5 years; at maturity these payments have accumulated to

((1.005)20×12

)200s60 4

12% =

((1.005)20×12

)200

((1.003333333)60 − 1

(0.003333333)(

11.00333333333

))

= 44, 038.93886

The sum of the two components of the fund is

44, 038.93886 + 92, 870.21994 = 136, 909.1588

so the annual withdrawal will be

136909.1588

a15 8%

=136909.1588(0.08)(1.08)−1

1− (1.08)−15

= 14810.21733

or 14,810.22.


5. In his will, a benefactor of McGill contributes a large sum of money to establisha fund, whose proceeds are to be used to provide annual bursaries of 5000 to6 actuarial students. The principal of the fund is to remain constant, and thebursaries are funded by the interest earned, forever.

(a) If the effective annual interest rate is assumed to be 6% forever, determine thelump sum that the benefactor’s estate needs to contribute to McGill a yearbefore the first payment of bursaries.

(b) If the effective annual interest rate is assumed to be 6% forever, determinethe lump sum that the benefactor’s estate needs to contribute to McGill ifthe first bursaries are to be issued immediately.

(c) Suppose that each student receives not a bursary of 5000, but an annualannuity-due of 1250 for 4 years. Under these changed conditions, determinethe lump sum payment needed now if the first bursaries are to be awardedimmediately or 1 year hence respectively. (The number of awards will remainthe same — each year 6 students are awarded a 4-year sequence of bursarypayments of 1250, the first payment to be made immediately.)

Solution:

(a) The lump sum payment must equal the present value of the perpetuity-immediate of annual payments of 30,000, i.e.

6(5000)a∞ 6% =30000

0.06= 500, 000.

(b) The lump sum payment must equal the present value of the perpetuity-dueof annual payments of 30,000, i.e.

6(5000)a∞ 6% =30000

0.06× 11.06

= 530, 000.

Alternatively, this may be viewed as the cost of funding the bursaries of thepreceding question, to which must be added the cost of the bursaries thatmust be paid out immediately in the amount of 6× 5, 000.

(c) We must replace each payment of 5000 by

1250 · a5 = (1250)(1.06) · 1− 1.06−40.06 = 4591.264936.

i. The lump sum payment when the bursaries are awarded beginning oneyear hence is

4591.264936

5000× 500, 000 = 459126.4936

or 459,126.49.


ii. The lump sum payment when the bursaries are awarded beginning im-mediately is

4591.264936

5000× 530, 000 = 486674.0832

or 486,674.08.

6. Let m and n be positive integers. Consider an annuity-immediate which pays 1 atthe end of every period for mn periods. Explain in words why each of the followingformulæ represents the value of this annuity m + 1 years before the first paymentis made. (If you have doubts about the truth of this claim, you may wish to verifyalgebraically that the claim is correct before you attempt to explain it in words.)

(a) vm · amn

(b) am(n+1)

− am

(c) vm · am(n+1)

− vm(n+2) · sm

Solution:

(a) One period before the first payment, the value of an annuity of 1 per periodpayable for mn periods is amn. At a time m periods earlier the value must bediscounted by a factor of 1

1+iper year, or vm for m years.

(b) am(n+1)

is the value of an annuity-immediate of 1 payable for mn+n periods.

At a time m periods ago, this represents the value then of the annuity underpresent consideration, augmented by an annuity of m payments of 1, the lastto be made now. If we subtract the value of these m payments as of m periodsago, we are left with the value of the mn payments under consideration.

(c) Consider an annuity that consists of all the payments under present consider-ation, extended by an additional m payments. Such a scheme is worth a

m(n+1)

now, and vm · am(n+1)

m periods ago. The m payments added at the end are

worth sm at the time of the last of them, which is mn + m periods fromnow. We may discount that back to a time m periods ago by multiplying byvmn+m+m = vm(n+2).


8 Class Tests

8.1 Class Test, Version 1

McGILL UNIVERSITY, FACULTY OF SCIENCECLASS TEST in MATH 329, THEORY OF INTEREST

EXAMINER: Professor W. G. Brown DATE: Wednesday, 9 March, 2005.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:35→15:20

FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions

• The time available for writing this test is about 45 minutes.

• This test booklet consists of this cover, Pages 35 through 38 containing questions togetherworth 66 marks; and Page 39, which is blank.

• Show all your work. All solutions are to be written in the space provided on the page wherethe question is printed. When that space is exhausted, you may write on the facing page, onthe blank page, or on the back cover of the booklet, but you must indicate any continuationclearly on the page where the question is printed! (Please inform the instructor if you findthat your booklet is defective.)

• All your writing — even rough work — must be handed in.

• Calculators. While you are permitted to use a calculator to perform arithmetic and/or ex-ponential calculations, you must not use the calculator to calculate such actuarial functionsas ani, sni, etc. without first stating a formula for the value of the function in terms ofexponentials and/or polynomials involving n and the interest rate. You must not use yourcalculator in any programmed calculations. If your calculator has memories, you are expectedto have cleared them before the test.

• In your solutions to problems on this test you are expected to show all your work. You areexpected to simplify algebraic and numerical answers as much as you can.

• Your neighbours may be writing a version of this test which is different from yours.

PLEASE DO NOT WRITE INSIDE THIS BOX

1(a) 1(b) 1(c) 2(a) 2(b) 2(c) 3(a) 3(b) 4 Total

/4 /4 /4 /6 /6 /6 /6 /12 /18 /66


1. Showing your work in detail, determine each of the following; the rates you de-termine should be accurate to 4 decimal places, or as a percentage accurate to 2decimal places:

(a) [4 MARKS] the nominal annual interest rate, compounded monthly, corre-sponding to an effective annual interest rate of i = 6%

(b) [4 MARKS] the effective annual interest rate corresponding to a nominal dis-count rate, compounded quarterly, of d = 3.6%

(c) [4 MARKS] the effective monthly interest rate corresponding to a force ofinterest of δ = 0.06.


2. For each of the following sequences of payments, determine, as of the given time,and for the given interest or discount rate, the value, showing all of your work.Before determining the numeric value you are expected to express the value usingstandard symbols.

(a) [6 MARKS] the value now of 20 payments of 1 at the end of every year startingone year from now, at an interest rate of 4%

(b) [6 MARKS] the value one year ago of 10 payments of 1 at the end of everyhalf-year, the first to be paid 5 years from now, at a nominal interest rate of8% compounded semi-annually

(c) [6 MARKS] 300 payments of 1 at the end of every month, as of the dateof the 100th payment, which has just been made; the interest rate is 12%compounded monthly


3. (a) [6 MARKS] An annuity at interest rate i consists of payments of 10 now, 12at the end of one year, 14 at the end of two years, increasing by a constantamount until the last payment in the amount of 40, is to be evaluated as of 1year ago. Express its value in terms of symbols (Ia)n, (Is)n, an, sn, i, v, butdo not evaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 15-year annuity of 1000 per year, paying interest at the rate of 8% per annumeffective, is to be used to purchase a perpetuity, first payment to be made2 years after the last payment under the annuity. Showing all your work,determine the size of the payments under the perpetuity, assuming that theinterest rate from the time of the last payment under the 15-year annuity is5%.


4. [18 MARKS] A loan of 6000 is to be repaid by annual payments of 350 to commenceat the end of the 1st year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be larger thanthe regular payments. Assume i = 4%.


continuation page for problem number

You must refer to this continuation page on the page where the problem is printed!





FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions









1 2(a) 2(b) 2(c) 3(a) 3(b) 3(c) 4(a) 4(b) Total

/18 /4 /4 /4 /6 /6 /6 /6 /12 /66


1. [18 MARKS] A loan of 7000 is to be repaid by annual payments of 450 to commenceimmediately, and to continue at the beginning of each year for as long as necessary.Find the time and amount of the final payment if the final payment is to be nolarger than the regular payments. Assume i = 6%.



(a) [4 MARKS] the effective annual discount rate corresponding to a nominalinterest rate, compounded quarterly, of i = 2.4%

(b) [4 MARKS] the nominal annual interest rate, compounded quarterly, equiva-lent to an effective semi-annual discount rate of d = 4%

(c) [4 MARKS] the effective semi-annual interest rate corresponding to a force ofinterest of δ = 0.04.



(a) [6 MARKS] the value of 25 payments of 1 at the end of every year, the lastone having just been made, at an interest rate of 6%

(b) [6 MARKS] the value 4 months from now of 30 payments of 1 at the end ofevery 4 months, the first to be paid 2 years from now, at a nominal interestrate of 9% compounded 3 times a year

(c) [6 MARKS] 120 payments of 1 at the end of every 2 months, as of the dateof the 90th payment, which has just been made; the interest rate is 6% com-pounded every 2 months.


4. (a) [6 MARKS] An annuity at interest rate i consists of payments of 100 now, 95at the end of 1 year, 90 at the end of 2 years, decreases by a constant amountuntil the last payment in the amount of 25, is to be evaluated as of the timeof the payment of 95. Express its value then in terms of symbols (Ia)n, (Is)n,an, sn, i, v, but do not evaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 12-year annuity of 1000 per year, paying interest at the rate of 6% per annumeffective, is to be used to purchase a 10-year annuity at an interest rate of 7%,first payment to be made 4 years after the last payment under the annuity.Showing all your work, determine the size of the payments under the 10-yearannuity. Assume that the 7% rate is in effect from the time of the last paymentunder the 12-year annuity.





FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions









1(a) 1(b) 2 3(a) 3(b) 3(c) 4(a) 4(b) 4(c) Total

/6 /12 /18 /4 /4 /4 /6 /6 /6 /66


1. (a) [6 MARKS] An annuity at interest rate i consists of payments of 10 now, 13at the end of one year, 16 at the end of twq years, increasing by a constantamount until the last payment in the amount of 40, and is to be evaluated asof two years ago. Express its value then in terms of symbols (Ia)n, (Is)n, an,sn, i, v, but do not evaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 15-year annuity of 1000 per year, paying interest at the rate of 9% per annumeffective, is to be used to purchase a perpetuity at an interest rate of 8%,first payment to be made at the same time as the last payment under theannuity. Showing all your work, determine the size of the payments under theperpetuity.


2. [18 MARKS] A loan of 8000 is to be repaid by annual payments of 500 to commenceat the end of the 1st year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be smaller thanthe regular payments. Assume i = 5%.



(a) [4 MARKS] the effective annual interest rate, corresponding to an nominalannual interest rate of i = 6% compounded every 4 months.

(b) [4 MARKS] the effective monthly interest rate corresponding to a force ofinterest of δ = 0.12.

(c) [4 MARKS] the effective annual interest rate corresponding to d(4) = 3.2%



(a) [6 MARKS] the value one half-year ago of 12 payments of 1 at the end ofevery half-year, the first to be paid 4 years from now, at a nominal interestrate of 6% compounded semi-annually

(b) [6 MARKS] 180 payments of 1 at the end of every month, as of the dateof the 100th payment, which has just been made; the interest rate is 18%compounded monthly

(c) [6 MARKS] the value now of 16 payments of 1 at the end of every year starting1 year from now, at an interest rate of 4%





FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions









1(a) 1(b) 1(c) 2(a) 2(b) 3 4(a) 4(b) 4(c) Total

/6 /6 /6 /6 /12 /18 /4 /4 /4 /66



(a) [6 MARKS] 90 payments of 1 at the end of every 2 months, as of the date of the18th payment, which has just been made; the interest rate is 9% compoundedevery 2 months.

(b) [6 MARKS] the value of 20 payments of 1 at the end of every year, the lastone having just been made, at an interest rate of 5%

(c) [6 MARKS] the value 4 months from now of 45 payments of 1 at the end ofevery 4 months, the first to be paid 3 years from now, at a nominal interestrate of 6% compounded 3 times a year


2. (a) [6 MARKS] An annuity at interest rate i consists of payments of 118 now,112 at the end of 1 year, 106 at the end of 2 years, decreasing until the lastpayment in the amount of 34, the totality to be evaluated as of the time ofthe payment of 112. Express its value then in terms of symbols (Ia)n, (Is)n,an, sn, i, v, but do not evaluate.

(b) [12 MARKS] The accumulated value just after the last payment under a 14-year annuity of 1000 per year, paying interest at the rate of 10% per annumeffective, is to be used to purchase a 12-year annuity-immediate at an interestrate of 7%, first payment to be made 1 year after the last payment under the14-year annuity. Showing all your work, determine the size of the paymentsunder the 12-year annuity. Assume that the 7% rate applies from the time ofthe last payment under the 10% annuity.


3. [18 MARKS] A loan of 9000 is to be repaid by annual payments of 800 to commenceimmediately, and to continue at the beginning of each year for as long as necessary.Find the time and amount of the final payment if the final payment is to be nosmaller than the regular payments. Assume i = 8%.



(a) [4 MARKS] the nominal annual interest rate, compounded quarterly, equiva-lent to an effective semi-annual discount rate of d = 4%

(b) [4 MARKS] the effective semi-annual interest rate corresponding to a force ofinterest of δ = 0.04.

(c) [4 MARKS] the effective annual discount rate corresponding to a nominalinterest rate, compounded quarterly, of i = 2.5%


9 Solutions, Third Problem Assignment

Mounted on the Web on March 7th, 2005Full solutions were to be submitted by March 7th, 2005.(Caveat lector! There could be misprints or other errors!)


1. McGill plans to create a scholarship fund that will eternally pay 50 students amonthly stipend of $200 at the beginning of months September through April,plus an amount of $300 on the following May 1st.

(a) If interest is assumed to be at a nominal annual rate of 6% per annum, com-pounded monthly, determine the amount that is needed in this fund on Sep-tember 1st just before the fund begins making payments.

(b) Determine the amount that will be in the fund just after the December schol-arship payments in the first year, and on September 1st of the following year,just before the September payments.

(c) Suppose that at the beginning of September, 8 years after the fund is estab-lished, it is decided to increase the capital in the fund because the interestrate has changed to 4% per annum compounded monthly. Determine howmuch additional capital needs to be added to the fund.

(d) Suppose that 2 years after the interest rate is changed to 4%, it changes again,this time to 8%. This time it is decided to leave the capital unchanged, butto increase the payments to students by a lump sum of $M to each student,payable on December 1st, together with the regular December payment underthe scholarship. Determine the amount of that lump sum payment.

Solution:

(a) Since monthly payments are not completely regular, we cannot interpret thisfund as a monthly perpetuity with regular payments. We could still developits properties from first principles, but it is easier to interpret it as an annualperpetuity-due. The payments made to each student in an academic year areworth

200a8 0.5% + 300(1.005)−8 = 200(1.005)

(1− (1.005)−8

0.005

)+ 300(1.005)−8

= 1860.680368 .


Alternatively, we can think of an annuity-due of 9 payments of 200 togetherwith an additional payment of 100 alongside the last payment:

200a9 0.5% + 100(1.005)−8 = 200(1.005)

(1− (1.005)−9

0.005

)+ 100(1.005)−8

= 1860.680368 .

The effective annual interest rate is

(1.005)12 − 1 = 6.1677812% .

It follows that the amount in the fund at the beginning of September, justbefore the September payments, must be

50(1860.680368)a∞ (1.005)12−1 = 50(1860.680368)(1.005)12

(1.005)12 − 1= 1, 601, 421.16 .

(b) We could evaluate the fund at the beginning of each month, determining thegrowth in principal, and then subtracting the payments that would need tobe made on that day. Instead, we will shift all payments to the dates underexamination. On December 1st, just after the December payments, the fundwill be worth

1, 601, 421.16(1.005)3 − 50(200)s4 0.005

= 1, 601, 421.16(1.005)3 − 50(200)

((1.005)4 − 1

0.005

)

= 1, 585, 261.78

On the following September 1st, just before the first payments of the newacademic year, the value of the fund must be

1, 585, 261.78(1.005)9 − 50(200)s4 0.005(1.005)5 − 50(300)(1.005)4

= 1, 585, 261.78(1.005)9 − 50(200)

((1.005)4 − 1

0.005

)(1.005)5 − 50(300)(1.005)4

= 1, 601, 421.16

This is no surprise — we constructed this fund so that the amount just beforethe September 1st payments would be constant.

(c) The amount in the fund on September 1st, just before payments are made,has been constant. Because of the revised interest rate, the annual costs per


participant will be, as of the time of the September payment,

200a8 13% + 300(1.0033333333)−8

= 200(1.0033333333)

(1− (1.0033333333)−8

0.00333333333

)+ 300(1.0033333333)−8

= 1873.636976 .

The effective annual interest rate is now

(1.0033333333)12 − 1 = 4.0741539% .

The amount of the fund will, therefore, need to be

50× 1873.636976× 1.040741539

0.040741539= 2, 393, 100.36 .

The additional capital required is, therefore,

2, 393, 100.36− 1, 601, 421.16 = 791, 679.20 .

(d) Since this is a perpetuity, the amount in the account is periodic. Unless theinterest rate or other conditions change, the amount is always the same on 1September, and it does not matter how many years have passed. We repeatthe calculations of the preceding part, this time for a nominal rate of 8%.Because of the revised interest rate, the annual costs per participant will be,as of the time of the September payment,

200a8 23% + 300(1.0066666667)−8

= 200(1.0066666667)

(1− (1.0066666667)−8

0.0066666667

)+ 300(1.0066666667)−8

= 1847.870679 .

The effective annual interest rate is now

(1.0066666667)12 − 1 = 8.2999511% .

The amount of the fund to support the previous obligations will, therefore,need to be

50× 1847.870679× 1.082999511

0.082999511= 1, 205, 575.19 .

The capital no longer required for the previously defined purposes is, therefore,

2, 393, 100.36− 1, 205, 575.19 = 1, 187, 525.17


on September 1st, prior to the payments due on that date. On December 1stthis excess is worth

1, 187, 525.17(1.0066666667)3 = 1, 211, 434.36

in total, or 24,228.68728 for each of the 50 participants at any time. This canpurchase a perpetuity-due of

M =24228.68728

a∞ 8.2999511%

=24228.68728× 0.082999511

1.082999511= 1856.851435 .

(The problem was somewhat ambiguous: some students thought the intentionwas that there would be a single lump sum payment in just one December. Ifthat had been the intention, the value would have been 24,228.69 determinedabove.)

2. A loan of 10,000 is to be repaid by regular, half-yearly payments of 1,000, the firstto be made at the end of the 3rd year. The loan will be paid off by a final paymentwhich must be at least 1,000.

(a) If the interest rate is to be 8% per annum, compounded semi-annually, deter-mine the amount of the final payment, and when it is made.

(b) Suppose that the final payment is now not more than 1,000. and the inter-est rate remains 8% per annum compounded semi-annually. Determine theamount of the final payment, and when it is made.

(c) Suppose that, after signing his original commitment, the borrower decidesthat he would like to make the first payment 6 months from the date ofborrowing, and that all the payments under this loan should be exactly equal.If the interest rate is 6% per annum, compounded semi-annually, determinethat payment level that would be closest to 1000 per half-year, and determineexactly when the loan will be paid off.

Solution:

(a) Let n be the number of the last payment (counting in half-years from the dateof the loan). Then n will be the largest integer such that

10000(1.04)n ≥ 1000sn−5 4%

since the first 5 opportunities for payments are missed. The inequality isequivalent to

10000(1.04)5

1000≥ sn−5

(1.04)n−5= an−5


and to(1.04)−(n−5) ≥ 1− 0.4(1.04)5

which implies that

n− 5 ≤ − ln (1− 0.04(1.04)5)

− ln(1.04)= 17.00170887

so the last payment is made at n = 22. The excess of the amount of thepayment over 1000 will be

10000(1.04)22 − 1000s17 = 1.67552

rounding to 1.68.

(b) This time the final payment will be #23; the amount of the payment will bethe accumulation of the residue remaining unpaid after the payment at timen = 22, i.e., (1.04)1.67552 = 1.74254 rounding to 1.74.

(c) If the last payment is n half-years after the date of borrowing, the paymentswill be in the amount

10000

an

.

The object is to minimize ∣∣∣∣10000

an

− 1000

∣∣∣∣or to minimize ∣∣∣∣

10

an

− 1

∣∣∣∣Since an is an increasing function of n, its reciprocal is decreasing. We needto find the value of n such that

an 3% ≤ 10 < an+1 3%

i.e., the largest n such that an 3% ≤ 10, i.e., such that vn ≥ 0.7, i.e., such that

n ≤ − ln(0.7)

ln(1.03)= 12.06662371. So the only candidates are n = 12 and n = 13,

and the corresponding semiannual payments are

10000

a12

= 1004.620855

and10000

a13

= 940.2954396

Taking the payment closest to 1000, we find that the last payment will be atthe end of 6 years, and the amount of the regular payments will be 1004.62.


3. (An acknowledgement of the source of this problem will be contained in the solu-tions, when published.) Katherine, 25 years old, deposits 10,000 at the beginningof every 4-year period into an RSSP account. The account pays compound interestannually, at the effective annual rate i. The accumulated amount in the account atthe end of 40 years is X, which is 6 times the accumulated amount in the accountat the end of 20 years. Determine X.

Solution: This problem was modelled on Problem 8 of Course 2, May, 2003, of theSociety of Actuaries.

There are several ways of attacking a problem like this. One method is to determinefrom i the interest rate that would apply to the period of payments, i.e. 4 years;another is to determine the annual payment in advance that would correspond toa payment of 10,000 in advance every 4 years.

The problem on which this is modelled was posed as a multiple-choice problem,with 5 possible numerical answers. None of the answers was expressed in terms ofi; so, in fact, there was no need for the question to even mention i. You may judgefor yourself whether that inclusion made the problem harder or easier.

Denote by Y the annual payment in advance that is equivalent to a payment every4-years in advance of 10,000. Then

Y · a4 i = 10000 ,

so Y =10000

a4 i

. The values of the account at the ends of 20 years and 40 years are,

respectively

X = Y · s40 i = Y · (1 + i) · (1 + i)40 − 1

i

and Y · s20 i = Y · (1 + i) · (1 + i)20 − 1

i

From the hypothesis that these amounts are in the ratio of 6 : 1, we conclude that

6 =Y · (1 + i) · (1+i)40−1

i

Y · (1 + i) · (1+i)20−1i

= (1 + i)20 + 1

which implies that (1 + i)20 = 5. We were not asked to compute the interest rate,and don’t actually require it. (If you did require it, you could solve the preceding

equation to show that 1 + i = 5120 = 1.083798, so i = 8.38%.) Then

X = 10000 · (1 + i) · (1+i)40−1i

(1 + i) · 1−v4

i


= 10000 · (1 + i)40 − 1

1− v4=

10000(25− 1)

1− 5−15

=240000

1− 0.72478=

240000

0.275220336= 872, 028.58

4. For each of the following sequences of payments as of the time stated,

• express the value using standard symbols, as simple as possible;

• give a formula for the value (in terms of i, v, d, etc.);

• evaluate using a calculator or computer — not with tables.

(a) a perpetuity-immediate paying 1 per year at effective annual interest rate4.25%, evaluated 1 year before the first payment;

(b) a perpetuity paying 1 per year at effective annual interest rate 4.25%, evalu-ated 2 years before the first payment;

(c) a perpetuity-due paying 1 per half-year at nominal interest rate 5%, com-pounded semi-annually, evaluated just before the first payment;

(d) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the3rd year, etc., at an interest rate of 7% per year effective, evaluated just afterthe last payment;

(e) a 20-year increasing annuity paying 1 the first year, 2 the second year, 3 the3rd year, etc., at an interest rate of 7% per year effective, evaluated at thetime of the 2nd payment;

(f) a 10-year decreasing annuity paying 1 less each quarter-year, evaluated justafter the last payment, where the nominal interest rate is a 8% annual, com-pounded quarterly;

Solution:

(a) a∞ 4.25% =1

i=

1

0.00425= 23.52941176 .

(b) v · a∞ 4.25% =v

i=

1

1.0425(0.0425)= 22.57017915 .

(c) a∞ 0.025 =1

iv=

1

0.025(1.025)−1= 41.

(d) (Is)20 0.07 =s21 0.07 − 21

0.07=

(1.07)21−10.07

− 21

0.07= 340.9310971 .

(e) v18 · (Is)20 0.07 = (1.07)−18 · (340.9310971) = 100.8692096 .


(f) (Ds)40 0.02 =n(1 + i)n − sn

i=

40(1.02)40 − (1.02)40−10.02

0.02= 1395.980168.

5. X and Y have sold their home for 400,000, and wish to purchase an annuity-immediate so that they can spread the proceeds over the next 10 years. The firstpayment will be one month from the date of purchase of the annuity.

(a) Determine the level monthly payments they will receive if the effective annualinterest rate is 6%.

(b) X and Y live frugally, and don’t think they need as much income now asthey will as time passes. Accordingly they plan to receive monthly paymentswhich will gradually increase. If the first payment is 3,000, and the paymentsincrease each month by the same dollar amount K, determine K. The effectiveannual interest rate remains 6%. What is the final monthly payment?

(c) Suppose that the payments remain constant in any year. The first year’smonthly payments are all 3000, the next year’s 3000+L, the next years’ 3000+2L, 3000 + 3L, . . . , 3000 + 9L. Determine L if the nominal annual interestrate is 6%, compounded monthly.

Solution:

(a) The effective monthly interest rate is (1.06)112 − 1 = 0.4867551%, which we

shall denote by j below. The payments will be

400000

a120 (1.06)

112−1

=400000((1.06)

112 − 1)

1− (1.06)−12012

=400000((1.06)

112 − 1)

1− (1.06)−10= 4408.961439

(b) We solve for K the equation of value

400000 = K · (Ia)120 j + (3000−K) · a120 j

⇔ 400000 = K

(a120 j − 120(1 + j)−120

j

)+ (3000−K) · a120j

= K

(a120 j − 120(1 + j)−120

j− a120j

)+ 3000 · a120j

Hence

K =j(400000− 3000 · a120 j

)

a120 j − 120(1 + j)−120 − j · a120j


=j(400000− 3000 · a120 j

)

a120 j − 120(1 + j)−120

= 26.23459805

so the final payment will be

3, 000.00 + 119(26.23459805) = 6, 121.92 .

(c) Consider payments of L at the end of every month for a year. Discountedback to the beginning of the year, these payments are worth

L · a12 0.005 = L

(1− 1.005−12

0.005

)= L(11.61893206) .

The value of all of the payments as of the date of purchase of the annuity is,therefore,

3000 · a1200.005 + L · a12 0.005 · (Ia)9 (1.005)12−1 = 400000 .

Now

(Ia)9 (1.005)12−1 =a9 (1.005)12−1 − 9(1.005)−9(12)

(1.005)12 − 1

=(1.005)12 · a9 (1.005)12−1 − 9(1.005)−9(12)

(1.005)12 − 1

=a9 (1.005)12−1 − 9(1.005)−10(12)

1− (1.005)−12

=

1−(1.005)−9(12)

(1.005)12−1− 9(1.005)−10(12)

1− (1.005)−12

= 31.08041107 .

Solving for L, we obtain

L =400000− 3000 · a120 1

2%

11.61893206(31.08041107)

= 0.002769153521(400000− 3000 · a120 0.005

)

= 0.002769153521

(400000− 3000 · 1− (1.005)−120

0.005

)

= 359.3797471.

The monthly payments in the first year will be 3,000; those in the last yearwill be

3, 000 + 9(359.3797471) = 6, 234.42 .


10 Fourth Problem Assignment

Mounted on the Web on March 7th, 2005Distributed in hard copy on Wednesday, March 9th, 2005Full solutions are to be submitted by March 21st, 2005.


1. A loan of 20,000 is to be repaid by 15 annual payments beginning one year afterthe loan was received, payments which increase by a factor of 1.03, each over thepreceding. If the effective annual interest rate is 7%, determine

(a) the amount of the first payment

(b) the amount of the loan which is outstanding immediately after the 10th pay-ment, determined using the Retrospective Method

(c) the amount of the loan which is outstanding immediately after the 10th pay-ment, determined using the Prospective Method

2. Mary always dreamed of owning a Maserati, and now had the chance. Her employerwas buying a new Ferrari, and was willing to part with his used Maserati for a mere20,000. Mary agreed to repay the loan by equal monthly payments over 3 years at4% interest, compounded monthly, first payment one month after the sale. Twoand one-half years after the sale Mary’s employer declared bankruptcy, and Mary’sfuture payments became one of the assets to be administered by the trustee.

(a) Determine Mary’s regular monthly payment.

(b) The trustee in bankruptcy needs to know the present value of Mary’s futurepayments. Determine this value just after the payment made on the date ofbankruptcy if

i. the interest rate remains a nominal 4% compounded monthly;

ii. the interest rate is now 6% compounded monthly.

(c) Complete an amortization table beginning on the date of bankruptcy, whenthe rate is 4% compounded monthly, under the headings


Payment Payment Interest Principal OutstandingNumber Amount paid repaid loan balance

3031. . .

(d) Suppose that Mary’s future payments, discounted at 6%, have been assignedto one of the employer’s creditors. You have computed the present value ofthese payments. Now set up an amortization table to show how his outstand-ing debt in this amount will be amortized at 6%.

3. On January 1st, 2005, John receives a loan of 50,000 from his brother, and agreesto repay it by semi-annual payments: 2,000 every July 1st, and 2,500 every January1st; the payments begin on July 1st, 2005, and continue until one last, possiblyirregular payment which will be either less than 2,000 if it is on July 1st, or under2,500 if it is on January 1st. The effective interest rate is 6% per annum.

(a) Determine the outstanding principal just after the payment on January 1st,2015.

(b) Complete an amortization schedule beginning with January 1st, 2020, underthe following headings

Date Payment Interest Principal OutstandingAmount paid repaid loan balance

January 1, 2020etc.

4. A 30-year loan is amortized by level payments at the end of every month, at anominal interest rate of 16.8%, compounded monthly. The amount of interest paidin the 120th payment is 486.72. Determine

(a) the amount of interest in the 300th payment;

(b) the original amount of the loan;

(c) the amount due at the end of 30 years if the lender defaults on8 the last 6payments.

5. A loan of 100,000 is to be maintained by a semi-annual interest payment, with theprincipal being repaid in a single payment at the end of 25 years. The borrower’stotal cost is 5,000 at the end of each 6-month period, being made up of the interestand a contribution to a sinking fund maintained by a 3rd party earning 4% effective:the proceeds will be paid to the lender upon maturity. Find

8=fails to pay


(a) the effective semi-annual interest rate received by the lender of the loan.

(b) the effective interest rate paid out by the borrower. (For this purpose it willsuffice to determine an equation for the interest rate, and to use the tablesin your book to locate the rate between two tabulated values; you are notexpected to interpolate.)

6. Pierre borrows 18,000 for 8 years, and agrees to make level annual payments. Thelender receives 6% on the investment for each of the first 5 years, and 8% for thelast 3 years. Throughout the 8 years, the balance of each payment is invested in asinking fund earning 7%.

(a) Determine the amount of the lender’s total annual payment.

(b) Complete the following table.

Period Total Interest Interest Earned Contribution to Balance inPayment to Lender in Sinking Fund Sinking Fund Sinking Fund

0etc.


11 Solutions to Problems on the Class Test

Distribution Date: Mounted on the Web on Saturday, March 12th, 2005Distributed in hard copy on Wednesday, March 16th, 2005

(Subject to correction)

There were 4 versions of this test. The final grades of all were scaled upwards slightly,by slightly different factors, in an attempt to equalize the difficulty of the versions.

11.1 Problems on rates of interest and discount

1. Problem 1 on Version 1 Showing your work in detail, determine each of thefollowing; the rates you determine should be accurate to 4 decimal places, or as apercentage accurate to 2 decimal places:

(a) [4 MARKS] the nominal annual interest rate, compounded monthly, corre-sponding to an effective annual interest rate of i = 6%

(b) [4 MARKS] the effective annual interest rate corresponding to a nominal dis-count rate, compounded quarterly, of d = 3.6%

(c) [4 MARKS] the effective monthly interest rate corresponding to a force ofinterest of δ = 0.06.

Solution:

(a) The effective interest rate per month is (1.06)112 − 1 = 0.4867551%; the nom-

inal annual interest rate, compounded monthly, i(12), is 12 times this rate,5.8410612%.

(b) 1−d =(1− d(4)

4

)4

= (1−0.009)4 = 0.9644830906. 1+i = 11−d

= (0.9644830906)−1 =

1.036824813, so the equivalent effective annual interest rate is 3.6824813%.

(c) Since δ = 0.05 = ln(1 + i), 1 + i = e0.06.

i(12) = 12(e

0.0612 − 1

)= 12× 0.5012521% = 6.0150252%

The effective monthly interest rate isi(12)

12= 0.5012521.



(a) [4 MARKS] the effective annual discount rate corresponding to a nominalinterest rate, compounded quarterly, of i = 2.4%

(b) [4 MARKS] the nominal annual interest rate, compounded quarterly, equiva-lent to an effective semi-annual discount rate of d = 4%

(c) [4 MARKS] the effective semi-annual interest rate corresponding to a force ofinterest of δ = 0.04.

Solution:

(a) The effective interest rate per quarter is 14× 2.4% = 0.006. If i and d be the

effective annual interest and discount rates, then

1− d =1

1 + i=

1

(1.006)4,

so

d = 1− 1

(1.006)4= 2.36442751%

(b) The annual discount factor corresponding to an effective semi-annual discountrate of 4% is (1 − 0.04)2; the annual accumulation factor will, therefore, be(0.96)−2. The accumulation factor corresponding to a quarter of a year will be

(0.96)−24 , and so the equivalent effective quarterly interest rate is (0.96)

−24 −1;

the nominal annual interest rate, compounded quarterly, is 4 times this, i.e.

4((0.96)

−24 − 1

)= 8.2482904%.

(c) If i be the equivalent effective annual interest rate, then ln(1 + i) = 0.04, so1 + i = e0.04. The effective semi-annual interest rate corresponding is

√1 + i− 1 = e0.02 − 1 = 2.0201340% .


(a) [4 MARKS] the effective annual interest rate, corresponding to an nominalannual interest rate of i = 6% compounded every 4 months.

(b) [4 MARKS] the effective monthly interest rate corresponding to a force ofinterest of δ = 0.12.

(c) [4 MARKS] the effective annual interest rate corresponding to d(4) = 3.2%


Solution:

(a) A nominal interest rate of 6% compounded every 4 months is equivalent to aneffective interest rate of 2% for a 4-month period. The accumulation factor for1 year will be (1.02)3, and the equivalent annual interest rate will, therefore,be (1.02)3 − 1 = 6.1208%.

(b) If i be the equivalent annual interest rate, ln(1 + i) = 0.12, so 1 + i = e0.12.

The accumulation factor for 1 month will be (1 + i)112 = e0.01, so the effective

interest rate for 1 month will be e0.01 − 1 = 1.0050167%.

(c) Let i and d denote the effective annual interest and discount rates. Then

1 + i =1

1− d=

1(1− d(4)

4

)4 =1

(1− 0.008)4= 1.032650385

so the effective annual interest rate is 3.2650385%.


(a) [4 MARKS] the nominal annual interest rate, compounded quarterly, equiva-lent to an effective semi-annual discount rate of d = 4%

(b) [4 MARKS] the effective semi-annual interest rate corresponding to a force ofinterest of δ = 0.04.

(c) [4 MARKS] the effective annual discount rate corresponding to a nominalinterest rate, compounded quarterly, of i = 2.5%

Solution:

(a) The discount factor for a year is (1− 0.04)2 = (0.96)2. If i is the effectiveannual interest rate, then 1 + i = (0.96)−2, and the accumulation factor for

one quarter of a year is (1+ i)14 = (0.96)−

24 . The effective interest rate for one

quarter of a year is, therefore, (0.96)−24 − 1, and the nominal annual interest

rate, compounded quarterly, is

4((0.96)−

24 − 1

)= 4× 2.0620726% = 8.2482904%.

(b) If i be the equivalent effective annual interest rate, then ln(1 + i) = 0.04, so

1 + i = e0.04, and (1 + i)12 = e0.02 is the accumulation factor for half a year.

The effective semi-annual interest rate is, therefore

e0.02 − 1 = 2.0201340%.


(c) The accumulation factor for a year is (1+ 0.0254

)4 = 1.025235353, so the effectiveannual interest rate is 2.5235353%. If d is the effective annual discount rate,then

1 + d =

(1 +

0.025

4

)−4

= 0.9753857951

so d = 2.46142049%.

11.2 Problems on the values of annuities and perpetuities withconstant payments

1. Problem 2 on Version 1 For each of the following sequences of payments, deter-mine, as of the given time, and for the given interest or discount rate, the value,showing all of your work. Before determining the numeric value you are expectedto express the value using standard symbols.

(a) [6 MARKS] the value now of 20 payments of 1 at the end of every year starting1 year from now, at an interest rate of 4%

(b) [6 MARKS] the value 1 year ago of 10 payments of 1 at the end of everyhalf-year, the first to be paid 5 years from now, at a nominal interest rate of8% compounded semi-annually

(c) [6 MARKS] 300 payments of 1 at the end of every month, as of the dateof the 100th payment, which has just been made; the interest rate is 12%compounded monthly

Solution:

(a) This question was ambiguous. One reading was that the payments wouldstart one year from now, in which case the value is

a20 4% =1− (1.04)−20

0.04= 13.59032635 .

Another reading was that the first payment would be at the end of the yearthat starts one year from now, in which case the preceding value should be

multiplied by v =1

1.04, yielding

v · a20 4% =1− (1.04)−20

(0.04)(1.04)= 13.06762149 .


(b) The effective half-yearly interest rate is 12·8% = 4%. The value of the payments

4.5 years from now is a10 4%. The value 1 year ago — i.e. 5.5 years prior tothe value just given, is

v11 · a10 4% = (1.04)−11 · 1− (1.04)−10

0.04= 5.268683237 .

(c) The effective monthly interest rate is 112·12% = 1%. The value of the payments

just after the last of them is s300 1%. Precisely 200 months earlier the value is

v−200 · s300 1% =(1.01)100 − (1.01)−200

0.01= 256.8127449 .


(a) [6 MARKS] the value of 25 payments of 1 at the end of every year, the lastone having just been made, at an interest rate of 6%

(b) [6 MARKS] the value 4 months from now of 30 payments of 1 at the end ofevery 4 months, the first to be paid 2 years from now, at a nominal interestrate of 9% compounded 3 times a year

(c) [6 MARKS] 120 payments of 1 at the end of every 2 months, as of the dateof the 90th payment, which has just been made; the interest rate is 6% com-pounded every 2 months.

Solution:

(a) s25 6% =(1.06)25 − 1

0.06= 54.86451200 .

(b) The effective interest rate per 4 months is 13(9%) = 3%. The first payment is

due 24 months from now; if we wish to interpret the payments as an annuity-immediate, then the value 20 months from now will be a30 3%. The value 4months from now, i.e., 16 months, or 4

3of a year before “the clock starts” will

be

v4a30 3% = (1.03)−4 · 1− (1.03)−30

0.03

=(1.03)−4 − (1.03)−34

0.03= 17.41473827 .


(c) The effective interest rate for a 2-month period is 212· 6% = 1%. At the time

of the 120th payment the value of all the payments is s120 1%. 30 paymentsprior to that date, the value is

v30s120 1% = (1.01)−30 · (1.01)120 − 1

0.01= 170.6709757 .


(a) [6 MARKS] the value one half-year ago of 12 payments of 1 at the end ofevery half-year, the first to be paid 4 years from now, at a nominal interestrate of 6% compounded semi-annually

(b) [6 MARKS] 180 payments of 1 at the end of every month, as of the dateof the 100th payment, which has just been made; the interest rate is 18%compounded monthly

(c) [6 MARKS] the value now of 16 payments of 1 at the end of every year starting1 year from now, at an interest rate of 4%

Solution:

(a) The effective semi-annual interest rate is 12(6%) = 3%. Three and one-half

years from now the payments will be worth a12 3%. One half-year ago, theywere worth

v8 · a12 3% =v8 − v20

i=

(1.03)−8 − (1.03)−20

0.03= 7.857782670 .

(b) The effective monthly interest rate is 112

(18%) = 1.5%. As of the last paymentthe payments are worth s180 1.5%; 80 payments earlier the value is

v80 · s180 1.5% =(1.015)100 − (1.015)−80

0.015= 275.2103668 .

(c) This problem is ambiguous. Does the word “starting” refer to the paymentsor to the years? If it refers to the payments, the value is

a16 4% =1− (1.04)−16

0.04= 11.65229561 .


If the word refers to the years, then one has an annuity-immediate which hasbeen deferred one year, and the value is

v · a16 4% =1− (1.04)−16

(0.04)(1.04)= 11.20413039 .


(a) [6 MARKS] 90 payments of 1 at the end of every 2 months, as of the date of the18th payment, which has just been made; the interest rate is 9% compoundedevery 2 months.

(b) [6 MARKS] the value of 20 payments of 1 at the end of every year, the lastone having just been made, at an interest rate of 5%

(c) [6 MARKS] the value 4 months from now of 45 payments of 1 at the end ofevery 4 months, the first to be paid 3 years from now, at a nominal interestrate of 6% compounded 3 times a year

Solution:

(a) The effective interest rate for a 2-month period is 212

(9%) = 1.5%. Two monthsbefore the first payment the value of all the payments is a90 1.5%. Immediatelyafter the 18th payment the value of all the payments is

(1.015)18 · a90 1.5% =(1.015)18 − (1.015)−72

0.015= 64.33404251 .

(b) s20 5% = (1.05)20−10.05

= 33.06595410 .

(c) The effective interest rate per 4-month period is 412

(6%) = 2%. Two and two-thirds years from now the value of the payments is a45 2%. Hence one-thirdyear from now the value will be

(1.02)−7 · a45 2% =(1.02)−7 − (1.02)−52

0.02= 25.67295884 .

11.3 Problems on increasing and decreasing annuities and per-petuities

1. Problem 3(a) on Version 1 [6 MARKS] An annuity at interest rate i consists ofpayments of 10 now, 12 at the end of 1 year, 14 at the end of 2 years, increasing by


a constant amount until the last payment in the amount of 40, is to be evaluatedas of 1 year ago. Express its value in terms of symbols (Ia)n, (Is)n, an, sn, i, v,but do not evaluate.

Solution: The total increase of 40− 10 = 30 will be spread over 40−102

= 15 years:the last payment will be 15 years from now. The value 1 year ago was

8 · a16 + 2(Ia)16

2. Problem 4(a) on Version 2 [6 MARKS] An annuity at interest rate i consists ofpayments of 100 now, 95 at the end of 1 year, 90 at the end of 2 years, decreasingby a constant amount until the last payment in the amount of 25, is to be evaluatedas of the time of the payment of 95. Express its value then in terms of symbols(Ia)n, (Is)n, an, sn, i, v, but do not evaluate.

Solution: The total decrease of the annuity is 75, at 5 per year; so the last paymentwill be 75

5= 15 years from now. The value is

100(1 + i) + 95(1 + a14

)− 5(Ia)14

3. Problem 1(a) on Version 3 [6 MARKS] An annuity at interest rate i consistsof payments of 10 now, 13 at the end of 1 year, 16 at the end of 2 years, increasingby a constant amount until the last payment in the amount of 40, and is to beevaluated as of 2 years ago. Express its value in terms of symbols (Ia)n, (Is)n, an,sn, i, v, but do not evaluate.

Solution: The increase of 40 − 10 = 30 will be spread over 303

= 10 years. Thevalue of the annuity today is 10a11 + 3(Ia)10. As of 2 years ago the value was

10v · a11 + 3v2 · (Ia)10

4. Problem 2(a) on Version 4 [6 MARKS] An annuity at interest rate i consists ofpayments of 118 now, 112 at the end of 1 year, 106 at the end of 2 years, decreasinguntil the last payment in the amount of 34, the totality to be evaluated as of thetime of the payment of 112. Express its value in terms of symbols (Ia)n, (Is)n, an,sn, i, v, but do not evaluate.

Solution: The total decrease of 118 − 34 = 84 is to be spread over 846

= 14 years.The value of the payments at time 1 is

118(1 + i) + 112 + 112a14 − 6(Ia)13


11.4 Problems on combinations of annuities and perpetuities

1. Problem 3(b) on Version 1 [12 MARKS] The accumulated value just after thelast payment under a 15-year annuity of 1000 per year, paying interest at the rateof 8% per annum effective, is to be used to purchase a perpetuity, first paymentto be made 2 years after the last payment under the annuity. Showing all yourwork, determine the size of the payments under the perpetuity, assuming that theinterest rate from the time of the last payment under the 15-year annuity is 5%.

Solution: Let the payment under the perpetuity be X. Then the equation of valueat time 15 is

1000 · s15 8% =X

0.05− X

1.05=

X

(0.05)(1.05)

since the first payment of the perpetuity-immediate “whose clock begins to tick attime 15” is not made. Solving the equation yields X = 1425.49.

2. Problem 4(b) on Version 2 [12 MARKS] The accumulated value just afterthe last payment under a 12-year annuity of 1000 per year, paying interest at therate of 6% per annum effective, is to be used to purchase a 10-year annuity at aninterest rate of 7%, first payment to be made 4 years after the last payment underthe annuity. Showing all your work, determine the size of the payments under the10-year annuity. Assume that the 7% rate begins from the time of the last paymentunder the 12-year annuity.

Solution: Let X denote the payment under the 10-year annuity. The value of thepayments under the 12-year annuity at the time of the last of them is 1000 ·s12 6% =(1.07)−3a10 7%. The value of the payments under the deferred 10-year annuity, asof that same date, is X

(a13 7% − a3 7%

). Equating these amounts and solving, we

obtain

X =1000 · s12 6%

a13 7% − a3 7%

=100(1.07)3s12 6%

a10 7%

= 2942.43

3. Problem 1(b) on Version 3 [12 MARKS] The accumulated value just after thelast payment under a 15-year annuity of 1000 per year, paying interest at the rateof 9% per annum effective, is to be used to purchase a perpetuity at an interestrate of 8%, first payment to be made at the same time as the last payment underthe annuity. Showing all your work, determine the size of the payments under theperpetuity.

Solution: Denote the amount of the payments under the perpetuity by X. Theequation of value at time 15 is

1000 · s15 9% = X · a∞ 8% = X · 1.08

0.08


implying that X = 0.08×1000×((1.09)15−1)1.08×0.09

= 2174.88

4. Problem 2(b) on Version 4 [12 MARKS] The accumulated value just after thelast payment under a 14-year annuity of 1000 per year, paying interest at the rate of10% per annum effective, is to be used to purchase a 12-year annuity-immediate atan interest rate of 7%, first payment to be made 1 year after the last payment underthe 14-year annuity. Showing all your work, determine the size of the paymentsunder the 12-year annuity. Assume that the 7% rate applies from the time of thelast payment under the 10% annuity.

Solution: If we denote the amount of the payments under the 12-year annuity byX, the equation of value at time 14 is

1000 · s14 10% = X · a12 7%

implying that X = 3522.11.

11.5 Problems on drop and balloon payments

1. Problem 4 on Version 1 [18 MARKS] A loan of 6000 is to be repaid by annualpayments of 350 to commence at the end of the 1st year, and to continue thereafterfor as long as necessary. Find the time and amount of the final payment if the finalpayment is to be larger than the regular payments. Assume i = 4%.

Solution: The balloon payment will take place at the time — call it n — which isthe largest solution of the inequality

350 · an 4% < 6000

i.e., of the inequalities

1− (1.04)−n

0.04<

6000

350

(1.04)−n >110

350

n <ln 350− ln 110

ln 1.04= 29.51126321 .

It follows that the last payment is at the end of the 29th year. The amount of thepayment is

6000(1.04)29 − 350(1.04)

((1.04)28 − 1

0.04

)= 523.70851


2. Problem 1 on Version 2 [18 MARKS] A loan of 7000 is to be repaid by annualpayments of 450 to commence immediately, and to continue at the beginning ofeach year for as long as necessary. Find the time and amount of the final paymentif the final payment is to be no larger than the regular payments. Assume i = 6%.

Solution: The drop payment will take place at the time — call it n — which is thesmallest solution of the inequality

450 · an+1 6% ≥ 7000

i.e., of the inequalities

(1.06) · 1− (1.06)−(n+1)

0.06)≥ 7000

450

(1.06)−(n+1) ≤ 1− 7000

450· 6

106=

19

159

n + 1 ≥ − ln 19159

ln 1.06= 36.45967106

implying that n = 36. The amount of the final drop payment will be

7000(1.06)36 − 450s36 6% = 7000(1.06)36 − (450) · 1.06

0.06

((1.06)36 − 1

)

= 210.11059

so the final payment is in the amount of 210.11.

3. Problem 2 on Version 3 [18 MARKS] A loan of 8000 is to be repaid by annualpayments of 500 to commence at the end of the 1st year, and to continue thereafterfor as long as necessary. Find the time and amount of the final payment if the finalpayment is to be smaller than the regular payments. Assume i = 5%.

Solution: Let the final payment be made at time n, which will be the smallestinteger such that

500 · an 5% > 8000

⇔ 1− (1.05)−n > 16(0.05) = 0.8

⇔ n > − ln 0.2

ln 1.05= 32.98693373 ,

implying that n = 33. If the present value of the deficiency of the last paymentfrom 500 be denoted by X, then

X = 500 · an 5% − 8000

= 500

(1− (1.05)−33

0.05− 16

)= 1.27460500


so the final payment is

500.00− (1.27460500)(1.05)33 = 493.6229109

or 493.62.

4. Problem 3 on Version 4 [18 MARKS] A loan of 9000 is to be repaid by annualpayments of 800 to commence immediately, and to continue at the beginning ofeach year for as long as necessary. Find the time and amount of the final paymentif the final payment is to be no smaller than the regular payments. Assume i = 8%.

Solution: Let the final payment be at time n, i.e., be the (n + 1)st payment. Thenn will be the largest integer such that

800 · an+1 8% ≤ 9000

⇔ an+1 8% ≤90

8

⇔ 1− 1.08−(n+1) ≤ 90

8· 0.08

1.08

⇔ n + 1 ≤ ln 6

ln 1.08= 23.28138292

from which it follows that n = 22. The excess of the final payment over 800 willbe

9000(1.08)22 − 800 · s23 = 9000(1.08)22 − 800 · (1.08)23 − 1

0.08= 214.22726 ,

The final payment will, therefore, be 800 + 214.22726 = 1014.23.


12 Fifth Problem Assignment

Mounted on the Web on March 20th, 2005Revised on March 21st, 2005

Hard copy distributed on March 23rd, 2005Full solutions are to be submitted by April 4th, 2005.

1. (a) A 25,000 7% bond matures on June 30th, 2020. Interest is payable semian-nually on June 30th and December 31st. Determine the price to be paid forthe bond on June 30th, 2005, in order to earn the investor a yield of 6%.(Remember the convention for bonds — interest is normally interpreted as anominal (annual) rate, compounded semi-annually, even when not explicitlystated.)

(b) Repeat problem 1a if the purchase date is December 31st, 2012.

(c) A 10,000 bond has coupon rate 8% payable semiannually, and is redeemableafter a certain number of years at 11,250. The bond is purchased to yield 7%convertible semiannually. If the present value of the redemption value of thebond is 4,927 at the given yield rate, determine the purchase price.

(d) A 1,000 bond matures after a m years at par, and has a coupon rate of 10%convertible semiannually. It is purchased at a price to yield 7% convertiblesemi-annually. If the term of the bond is 2m years, the price of the bond willincrease by 107. Determine the price of the m-year bond to the nearest dollar.

2. An investor is considering the purchase of two 1,000 face value bonds which areredeemable at the end of the same number of years, and are both to be bought toyield 7.5% convertible semiannually. One bond costs 907, and pays coupons at therate of 6% per year, convertible semiannually. The second bond pays coupons at7% per half-year.

(a) Determine the price he should pay for the second bond so that it is an equiv-alent investment to the first.

(b) Determine the price the investor should pay for the second bond if the firstbond costs 960.45, and pays a premium of 100 upon redemption, while thesecond pays a premium of 150 upon redemption.

3. A loan of 96,000 is being repaid by a monthly payment of interest at a nominalrate of 6.0% compounded monthly, together with a monthly payment into a sinkingfund which earns interest at the rate of 4.0% per annum, also compounded monthly.The total monthly payment is constant, at 1,000, with the exception of the verylast payment, which could be less than 1,000, in order to bring the fund up to its



target level of 96,000; at that time the entire sinking fund will be paid over to thelender and the loan will have been repaid. Determine when the final deposit intothe sinking fund will be made, and its amount.

4. A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at theend of the 12th through the 17th years at par.

(a) Find the price to yield 6% convertible semiannually.

(b) Find the price to yield 8% convertible semiannually.

(c) Find the price to yield 8% convertible semiannually, if the bond pays a pre-mium of 250 if it is called.

(d) Find the price to yield 6% convertible semiannually, if the bond pays a pre-mium of 250 if it is called at the end of years ##12, 13, 14, and a premiumof 150 if it is called at the end of years 15 or 16. (It may still be called atthe end of year 17 without premium, and otherwise will mature at the end ofyear 18.)

5. (a) Construct a bond amortization schedule for a 4 year bond of face amount5,000, redeemable at 5,250 with semiannual coupons, if the coupon rate is 6%and the yield rate is 7% — both converted semiannually. Use the format

Time Coupon Interest Principal BookValue Adjustment Value

0...

(b) Construct a bond amortization schedule for a 4 year bond of face amount5,000, redeemable at 5,250 with semiannual coupons, if the coupon rate is 7%and the yield rate is 6% — both converted semiannually.

6. An 8-year bond of face value 10,000 with semiannual coupons, redeemable at par,is purchased at a premium to yield 6% convertible semiannually.

(a) If the book value (just after the payment of the coupon) six months before theredemption date is 10,024.27, find the total amount of premium or discountin the original purchase price.

(b) Determine the nominal annual coupon rate of the bond, compounded semi-annually.

(c) Give the amortization table for the last 2 years.


13 Solutions, Fourth Problem Assignment

Mounted on the Web on March 31st, 2005Full solutions were to be submitted by March 21st, 2005.


1. A loan of 20,000 is to be repaid by 15 annual payments beginning one year afterthe loan was received, payments which increase by a factor of 1.03, each over thepreceding. If the effective annual interest rate is 7%, determine

(a) the amount of the first payment

(b) the amount of the loan which is outstanding immediately after the 10th pay-ment, determined using the Retrospective Method

(c) the amount of the loan which is outstanding immediately after the 10th pay-ment, determined using the Prospective Method

Solution:

(a) Denote the amount of the first payment by X. The value at time 0 of allfuture payments is

(1.07)−1(1.03)0X + . . . + (1.07)−r(1.03)r−1X + . . . + (1.07)−15(1.03)14X

=X

1.07

(1 +

(1.03

1.07

)+

(1.03

1.07

)2

+ . . . +

(1.03

1.07

)14)

=X

1.07· 1− (

1.031.07

)15

1− (1.031.07

)1

=X

0.04

(1−

(1.03

1.07

)15)

= 10.8830229X .

Equating to 20,000, we find that X = 1837.72472 is the first payment.

(b) By the Retrospective Method, the amount outstanding after the 10th paymentis

10, 000(1.07)10 −X((1.03)9 + (1.03)8(1.07) + . . . + (1.03)0(1.07)9

)


= (1.07)10(20, 000)

(1− 1− (

103107

)10

1− (103107

)15

)

= 10, 709.669 .

(c) By the Prospective Method, the amount outstanding after the 10th paymentis

(1.03)10 X

1.07+ (1.03)11 X

(1.07)2+ . . . + (1.03)14 X

(1.07)5

=(1.03)10

1.07·X ·

(1 +

103

107+

(103

107

)2

+ . . . +

(103

107

)4)

= 10, 709.669 .

2. Mary always dreamed of owning a Maserati, and now had the chance. Her employerwas buying a new Ferrari, and was willing to part with his used Maserati for a mere20,000. Mary agreed to repay the loan by equal monthly payments over 3 years at4% interest, compounded monthly, first payment one month after the sale. Twoand one-half years after the sale Mary’s employer declared bankruptcy, and Mary’sfuture payments became one of the assets to be administered by the trustee.

(a) Determine Mary’s regular monthly payment.

(b) The trustee in bankruptcy needs to know the present value of Mary’s futurepayments. Determine this value just after the payment made on the date ofbankruptcy if

i. the interest rate remains a nominal 4% compounded monthly;

ii. the interest rate is now 6% compounded monthly.

(c) Complete an amortization table beginning on the date of bankruptcy, whenthe rate is 4% compounded monthly, under the headings


3031

(d) Suppose that Mary’s future payments, discounted at 6%, have been assignedto one of the employer’s creditors. You have computed the present value ofthese payments. Now set up an amortization table to show how his outstand-ing debt in this amount will be amortized at 6%.

Solution:


(a) The effective interest rate per month is 4%12

= 13%. Let X be the regular

monthly payment. Then

X· a36 13% = 20000

⇒ X =20000

1300

1−(

1 +1

300

)−36

=20000

300(1− (

300301

)36) = 590.4797061

so the monthly payment is 590.48.

(b) i. By the prospective method, the unpaid balance is

X · a6 13% = 20, 000 ·

a6 13%

a36 13%

= 20, 000 · 1− (300301

)6

1− (300301

)36 = 3501.909296

so the unpaid balance is 3,501.91.

ii. By the prospective method, the unpaid balance is

X · a6 13% = 20, 000 ·

a6 12%

a36 13%

= 20, 000 · 1− (200201

)6

1− (300301

)36 ·2

3= 3481.695351

so the unpaid balance is 3,481.70.

(c) The unpaid balance before the 30th payment was

3, 501.91 + 590.48 = 4, 092.39 .

One month earlier, this was worth4, 092.39

1.00333333= 4, 078.79, so interest earned

in the 30th period was the difference,

4, 092.39− 4, 078.79 = 13.60 .

Thus the principal reduction in the 30th payment was

590.48− 13.60 = 576.88 .



30 590.48 13.60 576.88 3,501.9131 590.48 11.67 578.81 2,923.1032 590.48 9.74 580.74 2,342.3633 590.48 7.81 582.67 1,759.6934 590.48 5.87 584.61 1,175.0835 590.48 3.92 586.56 585.5236 590.48 1.96 588.52 0.00

(d) The new interest rate takes effect after the 30th payment.


30 590.48 13.60 576.88 3,481.7031 590.48 17.41 573.07 2,908.6332 590.48 14.54 575.94 2,332.6933 590.48 11.66 578.81 1,753.8734 590.48 8.77 581.71 1,172.1635 590.48 5.86 584.62 587.5436 590.48 2.94 587.54 0.00

3. On January 1st, 2005, John receives a loan of 50,000 from his brother, and agreesto repay it by semi-annual payments: 2,000 every July 1st, and 2,500 every January1st; the payments begin on July 1st, 2005, and continue until one last, possiblyirregular payment which will be either less than 2,000 if it is on July 1st, or under2,500 if it is on January 1st. The effective interest rate is 6% per annum.

(a) Determine the outstanding principal just after the payment on January 1st,2015.

(b) Complete an amortization schedule beginning with January 1st, 2020, underthe following headings


January 1, 2020

Solution:

(a) Just after a January 1st payment of 2,500, that payment and the payment of2,000 from the preceding July 1st are together worth

2, 500 + 2, 000√

1.06 = 4, 559.26028 .


By the retrospective method, the outstanding principal just after the paymentof January 1st, 2015, is

50, 000(1.06)10 −(2, 500 + 2, 000

√1.06

)· s10 6%

= 50, 000(1.06)10 − (4559.126028) · (1.06)10 − 1

0.06= 29, 449.47952

so the outstanding balance is 29,449.48.

(b) We need to begin the table with January 1st, 2020. It is convenient to repeatthe preceding computation: the outstanding balance as of January 1st, 2019,is

50, 000(1.06)14 − (4, 559.126028) · (1.06)14 − 1

0.06= 17, 234.86372 .

or 17,234.86. The effective interest rate per half year is 1.0612 − 1 = 2.9563%.


January 1, 2019 2,500.00 17,234.86July 1, 2019 2,000.00 509.51 1,490.49 15,744.37

January 1, 2020 2,500.00 465.45 2,034.55 13,709.82July 1, 2020 2,000.00 405.30 1,594.70 12115.12

January 1, 2021 2,500.00 358.16 2,141.84 9,973.28July 1, 2021 2,000.00 294.84 1,705.16 8,268.12

January 1, 2022 2,500.00 244.43 2,255.57 6,012.55July 1, 2022 2,000.00 177.75 1,822.26 4,190.29

January 1, 2023 2,500.00 123.88 2,376.12 1,814.17July 1, 2023 1,867.80 53.63 1,814.17 0.00

4. A 30-year loan is amortized by level payments at the end of every month, at anominal interest rate of 16.8%, compounded monthly. The amount of interest paidin the 120th payment is 486.72. Determine

(a) the amount of interest in the 300th payment;

(b) the original amount of the loan;

(c) the amount due at the end of 30 years if the lender defaults on9 the last 6payments.

Solution:

9=fails to pay


(a) Suppose that the original amount of the loan was X. The level payments

are thenX

a360 1.4%

, which I will denote by Y . By the prospective method, the

amount owing just after the rth payment is Y · a360−r 1.4% , so the interestcomponent of the r + 1st payment is

(0.014)Y · a360−r 1.4% = Y(1− (1.014)r−360

).

The data tell us that

Y(1− (1.014)119−360

)= 486.72

so Y = 504.4061464: the monthly payments are 504.41. It follows that theinterest component of the 300th payment is

Y(1− (1.014)299−360

)= 285.8812253

or 288.40.

(b) The original amount of the loan was

504.4061464 · a360 1.4% =504.41

0.014

(1− (1.014)−360

)= 35, 787.47073

or 35,787.47.

(c) At the due date of the last payment, the value of the last 6 payments is504.4061464 · s6 1.4% = 3, 134.360302 or 3,134.36.

5. A loan of 100,000 is to be maintained by a semi-annual interest payment, with theprincipal being repaid in a single payment at the end of 25 years. The borrower’stotal cost is 5,000 at the end of each 6-month period, being made up of the interestand a contribution to a sinking fund maintained by a 3rd party earning 4% effective:the proceeds will be paid to the lender upon maturity. Find

(a) the effective semi-annual interest rate received by the lender of the loan.

(b) the effective interest rate paid out by the borrower. (For this purpose it willsuffice to determine an equation for the interest rate, and to use the tablesin your book to locate the rate between two tabulated values; you are notexpected to interpolate.)

Solution:


(a) The sinking fund earns interest at an effective annual rate of 4%; the equivalentsemi-annual rate is √

1.04− 1 = 1.9803903%.

The semi-annual payments into the sinking fund will each be in the amountof

100000

s50√

1.04−1

=100000(

√1.04− 1)

(1.04)25 − 1= 1188.826454

or 1,188.83. This leaves a residue of 5,000-1,188.83=3,811.17. The interestrate per half year is, therefore

3, 811.17

100, 000= 3.811% .

(b) Let i be the effective semi-annual interest rate being paid by the borrower.Then

5, 000 · a50 i = 100, 000

so a50 i = 20. We see from the tables that

a50 4.5% = 19.7620

a50 4% = 21.4822

(i is approximately 4.427%.)

NOTE TO THE GRADER: STUDENTS WERE NOT EXPECTED TO IN-TERPOLATE OR ITERATE TO OBTAIN A GOOD APPROXIMATION.

6. Pierre borrows 18,000 for 8 years, and agrees to make level annual payments. Thelender receives 6% on the investment for each of the first 5 years, and 8% for thelast 3 years. Throughout the 8 years, the balance of each payment is invested in asinking fund earning 7%.

(a) Determine the amount of the lender’s total annual payment.

(b) Complete the following table.


0

Solution:


(a) Let Pierre’s annual payment be X. The interest component in this paymentwill be

.06× 18, 000 = 1, 080 during the first 5 years

.08× 18, 000 = 1, 440 during the last 3 years

Hence the annual contributions in arrears to the sinking fund will be

X-1,080 during the first 5 yearsX-1,440 during the last 3 years

This information enables us to infer an equation that determines X:

(X − 1, 080)s8 7% − 360 · s3 7% = 18, 000

from which it follows that

X = 1, 080 +18, 000 + 360 · s3 7%

s8 7%

= 1, 080 +1, 260 + 360 ((1.07)3 − 1)

(1.07)8 − 1= 2, 947.2254

(b) Growth of the Sinking Fund:


0 0.001 2,947.23 1,080.00 0.00 1,867.23 1,867.232 2,947.23 1,080.00 130.71 1,867.23 3,865.173 2,947.23 1,080.00 270.56 1,867.23 6,002.964 2,947.23 1,080.00 420.21 1,867.23 8,290.405 2,947.23 1,080.00 580.33 1,867.23 10,737.966 2,947.23 1,440.00 751.66 1,507.23 12,996.857 2,947.23 1,440.00 909.78 1,507.23 15,413.868 2,947.23 1,440.00 1,078.96 1,507.23 18,000.06

Is it surprising that the balance is 18,000? Of course not — the repaymentscheme was designed so that Pierre would accumulate in the fund just theamount needed to repay the loan after 8 years.


14 Solutions, Fifth Problem Assignment

Mounted on the Web on April 12th, 2005Full solutions were to be submitted by April 4th, 2005.

1. (a) A 25,000 7% bond matures on June 30th, 2020. Interest is payable semian-nually on June 30th and December 31st. Determine the price to be paid forthe bond on June 30th, 2005, in order to earn the investor a yield of 6%.(Remember the convention for bonds — interest is normally interpreted as anominal (annual) rate, compounded semi-annually, even when not explicitlystated.)

(b) Repeat problem 1a if the purchase date is December 31st, 2012.

(c) A 10,000 bond has coupon rate 8% payable semiannually, and is redeemableafter a certain number of years at 11,250. The bond is purchased to yield 7%convertible semiannually. If the present value of the redemption value of thebond is 4,927 at the given yield rate, determine the purchase price.

(d) A 1,000 bond matures after a m years at par, and has a coupon rate of 10%convertible semiannually. It is purchased at a price to yield 7% convertiblesemi-annually. If the term of the bond is 2m years, the price of the bond willincrease by 107. Determine the price of the m-year bond to the nearest dollar.

Solution:

(a) F = C = 25, 000, r = 12× 7% = 3.5%, i = 3%, n = 2× 15 = 30. By the Basic

Formula

P = (25, 000)(0.035) · a15 3% + (25, 000)(1.03)−30

= 25, 000

((0.035)

(1− (1.03)−30

0.03

)+ (1.03)−30

)

= 27, 450.06

(b) F = C = 25, 000, r = 3.5%, i = 3%, n = 15.

P = (25, 000)(0.035) · a15 3% + (25, 000)(1.03)−15

= 25, 000

((0.035)

(1− (1.03)−15

0.03

)+ (1.03)−15

)

= 26, 492.24


(c) F = 10, 000, C = 11, 250, r = 4%, i = 3.5%, K = 4927. This last value of Kimplies that K = Cvn = 11, 250(1.025)−n, so (1.035)n = 11,250

4,927, where n is the

number of half-years remaining before maturity. By the Basic Formula,

P = Fr · an 0.035 + K

= 400

(1− (1.035)−n

0.035

)+ 4, 927

= 400

(1− 4,927

11,250

0.035

)+ 4, 927 = 11, 350.37

(d) For the m-year bond, F = C = 1, 000, n = 2m, r = 5%, i = 3.5%. For the2m-year bond, n = 4m. Denote the price of the m-year bond by P . Then,using the Premium/Discount formula, we have

P = 1, 000 + (50− 35) · a2m 3.5% (2)

P + 107 = 1, 000 + (50− 35) · a4m 3.5% (3)

Subtracting yields107 = 15

(a4m 3.5% − a2m 3.5%

)

which implies that

0.2496666 =1.035−2m − (1.035)−4m

0.035

yielding, approximately,

(1.035)−2m = 0.5182574184 or 0.4817425816

so, by (2),

P = 1, 000 + 15

(1− 0.5182574184

0.035

)= 1206.46

or

P = 1, 000 + 15

(1− 0.4817425816

0.035

)= 1, 222.11;

to the nearest dollar there are two answers, 1,206 and 1,222.

2. An investor is considering the purchase of two 1,000 face value bonds which areredeemable at the end of the same number of years, and are both to be bought toyield 7.5% convertible semiannually. One bond costs 907, and pays coupons at therate of 6% per year, convertible semiannually. The second bond pays coupons at7% per half-year.


(a) Determine the price he should pay for the second bond so that it is an equiv-alent investment to the first.

(b) Determine the price the investor should pay for the second bond if the firstbond costs 960.45, and pays a premium of 100 upon redemption, while thesecond pays a premium of 150 upon redemption.

Solution: The earlier references in the problem to interest rates always stated thatthe rate was “convertible semiannually”. But, in this case, those words were notmentioned. It appears that the rate of 7% stated is intended to be an effectivesemi-annual rate, equivalently that the nominal annual rate is 14%. However, thatwould be a much different rate than the others given in the problem. Because ofthis ambiguity, I will solve the problem with both interpretations.

Assuming 3.5% effective per half year: (a) For both bonds we know that F =C = 1000, i = 0.0375. By the Premium/Discount Formula applied to thefirst bond we obtain

907 = 1000 + (30− 37.5)an0.0375 ,

implying that an 0.0375 = 12.4. Substitution of this value into the sameformula applied to the second bond yields a price for that bond of

1000 + (Fr − Ci)an 0.0375 = 1000 + (35− 37.5)an 0.0375

= 1000− 31 = 961.

(b) In this case the Premium/Discount Formula applied to the first bondgives

960.45 = 1100 + (30− 41.25)an 0.0375 ,

which implies that an 0.0375 = 12.404444. Substitution of this value intothe same formula applied to the second bond yields a price for that bondof

1150 + (Fr − Ci)an 0.0375 = 1150 + (35− 43.125)an 0.035

= 1150− 8.125(12.404444) = 1049.21

Assuming 7% effective per half year: (a) For both bonds we know that F =C = 1000, i = 0.0375. By the Premium/Discount Formula applied to thefirst bond we obtain

907 = 1000 + (30− 37.5)an0.0375 ,


implying that an 0.0375 = 12.4. Substitution of this value into the sameformula applied to the second bond yields a price for that bond of

1000 + (Fr − Ci)an 0.0375 = 1000 + (70− 37.5)an 0.0375

= 1000 + 403 = 1403.

(b) In this case the Premium/Discount Formula applied to the first bondgives

960.45 = 1100 + (30− 41.25)an 0.0375 ,

which implies that an 0.0375 = 12.404444. Substitution of this value intothe same formula applied to the second bond yields a price for that bondof

1150 + (Fr − Ci)an 0.0375 = 1150 + (70− 43.125)an 0.035

= 1150 + 26.875(12.404444) = 1483.37

3. A loan of 96,000 is being repaid by a monthly payment of interest at a nominalrate of 6.0% compounded monthly, together with a monthly payment into a sinkingfund which earns interest at the rate of 4.0% per annum, also compounded monthly.The total monthly payment is constant, at 1,000, with the exception of the verylast payment, which could be less than 1,000, in order to bring the fund up to itstarget level of 96,000; at that time the entire sinking fund will be paid over to thelender and the loan will have been repaid. Determine when the final deposit intothe sinking fund will be made, and its amount.

Solution: The monthly interest payment to the lender will be

1

12(6%)× 96, 000 = 480,

so the monthly contribution to the sinking fund will be 1, 000−480 = 520. Regularcontributions of 520 will be made for as long as

520 · sn 13% ≤ 96, 000

⇔ 520

((1.00333333)n − 1

0.00333333

)≤ 96, 000

⇔ (1.0033333)n ≤ 1.615385

⇔ n ≤ 144.1116636

i.e., until n = 144. Immediately after that payment the balance in the sinking fundwill be 520 · s144 1

3% = 95, 906.43598. But, by the time for the next payment, this

will have grown to1.0033333× 95, 906.44 = 96, 226.12


so, the amount needed to bring the balance up to the target is −226.12: the lenderwill receive a refund! (Note that this refund is from the sinking fund only — hestill has to pay interest of 480 to the lender.)

4. A 5000 par value 18-year bond has 7% semiannual coupons, and is callable at theend of the 12th through the 17th years at par.

(a) Find the price to yield 6% convertible semiannually.

(b) Find the price to yield 8% convertible semiannually.

(c) Find the price to yield 8% convertible semiannually, if the bond pays a pre-mium of 250 if it is called.

(d) Find the price to yield 6% convertible semiannually, if the bond pays a pre-mium of 250 if it is called at the end of years ##12, 13, 14, and a premiumof 150 if it is called at the end of years 15 or 16. (It may still be called atthe end of year 17 without premium, and otherwise will mature at the end ofyear 18.)

Solution:

(a) F = C = 5, 000, n = 36, r = 3.5%, i = 3%. Let m be the coupon number atwhose date the bond is called or matures. Then, by the Premium/DiscountFormula

P = 5000 + (175− 150)am 3% , (n = 24, 26, 28, 30, 32, 34, 36).

Without knowing which will be the date of call — if any — we take theworst possible date in order to minimize the price; since am 3% is an increasingfunction of m, and is multiplied by a positive number, 25, we minimize bymaking m as small as possible, i.e., 2× 12 = 24:

P = 5000 + (175− 150)a24 3% = 5000.00 + 25× 16.9355 = 5, 423.39 .

(b) When the semi-annual yield rate is 4%, the multiplier is negative, 175−200 =−25, and we must choose the largest value of m, i.e., m = 36, for a price of

P = 5000 + (175− 200)a36 4% = 5000.00− 25× 18.9083 = 4, 527.29 .

(c) If the bond should be called at the time of any of coupons ##24, . . . , 34,C = 5, 250 and

P = 5250 + (175− 210)am 4%


which is minimized when m = 34: P = 4, 605.61. We must compare this withthe price if not called,

5000 + (175− 200)a36 4% = 4527.293 ,

so the price will be the latter, 4,527.29.

(d) We will have to determine the minimum of

5250 + (175− 157.50)am 3% if m = 24, 26, 28 (4)

5150 + (175− 154.50)am 3% if m = 30, 32 (5)

5000 + (175− 150)am 4% if m = 34, 36 (6)

i.e., of

5, 250 + (175− 157.51)a24 3% = 5, 250 + 17.50× 16.9355 = 5546.37

5, 150 + (175− 154.50)a30 3% = 5, 150 + 20.50× 19.6004 = 5551.81

5, 000 + (175− 150)a34 3% = 5, 000 + 25× 21.1318 = 5528.30

so the price should be 5528.30.

5. (a) Construct a bond amortization schedule for a 4 year bond of face amount5000, redeemable at 5250 with semiannual coupons, if the coupon rate is 6%and the yield rate is 7% — both converted semiannually. Use the format


0...

(b) Construct a bond amortization schedule for a 4 year bond of face amount5000, redeemable at 5250 with semiannual coupons, if the coupon rate is 7%and the yield rate is 6% — both converted semiannually.

Solution:

(a) By the Basic Formula the purchase price of the bond will be

5, 250(1.035)−8 + 150a83.5% = 3, 986.910670 + 150(6.8740) = 5, 018.01



0 5,018.011 150.00 175.63 -25.63 5,043.64 .2 150.00 176.53 -26.53 5,070.173 150.00 177.46 -27.46 5,097.634 150.00 178.42 -28.42 5,126.055 150.00 179.41 -29.41 5,155.466 150.00 180.44 -30.44 5,185.907 150.00 181.51 -31.51 5,217.418 150.00 182.61 -32.61 5,250.02

(b) By the Basic Formula the purchase price of the bond will now be

5, 250(1.03)−8 + 175a83% = 4, 144.398480 + 175(7.019692190) = 5, 372.84 .


0 5,372.841 175.00 161.19 13.81 5,359.032 175.00 160.77 14.23 5,344.803 175.00 160.34 14.66 5,330.144 175.00 159.90 15.10 5,315.045 175.00 159.45 15.55 5,299.496 175.00 158.98 16.02 5,283.477 175.00 158.50 16.50 5,266.978 175.00 158.01 16.99 5,249.98

6. An 8-year bond of face value 10,000 with semiannual coupons, redeemable at par,is purchased at a premium to yield 6% convertible semiannually.

(a) If the book value (just after the payment of the coupon) six months before theredemption date is 10024.27, find the total amount of premium or discount inthe original purchase price.


(c) Give the amortization table for the last 2 years.

Solution:


(a) The book value just after the pænultimate10 coupon is

10, 024.27 = 10, 000v + Fr · a10.03

= 10, 000v + Fr · v =10000 + Fr

1.03so

Fr = 1.03× 10, 024.27− 10000 = 325 .

Knowing the amount of each coupon we can now evaluate the purchase priceof the bond to have been

10, 000(1.03)−16 + 325a160.03 = 6, 231.67 + 4, 082.36

= 10, 314.03 .

The bond was purchased at a premium of 10, 314.03− 10, 000 = 314.03.

(b) The rate per period was 32510,000

= 3.25%; hence the nominal rate compounded

semi-annually, is 2× 3.25% = 6.5%.

(c) For convenience we could compile this table backwards, beginning with Time=16.We were given that B15 = 10, 024.27. Hence the Principal Adjustment con-tained in the 16th coupon is

10, 024.27− 10, 000 = 24.27.

The book value at Time=15 will be

(10, 000)(1.03)−2 + 325((1.03)−1 + (1.03)−2

)= 10, 047.84 .

The book value at Time=14 will be

(10, 000)(1.03)−3 + 325(1.03−1 + (1.03)−2 + (1.03)−3) = 10, 070.72 .

etc. Alternatively, we could observe by the Basic Formula that the book valuejust after the payment of the 12th coupon is

10, 000(1.03)−4 + 325a4 0.03 = 10, 092.93

and so we may construct the table:


12 325.00 . . . . . . 10,092.9313 325.00 302.79 22.21 10,070.7214 325.00 302.12 22.88 10,047.8415 325.00 301.44 23.56 10,024.2816 325.00 300.73 24.27 10,000.01

10preceding the last, i.e., 2nd last


• References to these sources are often given in the notes for completeness. Studentsare not expected to look up sources, but may wish to do so out of curiosity.

• The entries in this list may not be in alphabetical order. As the notes are con-structed, new entries will be added at the end, so as not to upset the earliernumbering of references.

15 References

[1] S. G. Kellison, The Theory of Interest, Second Edition. Irwin/McGraw Hill, Inc.,Boston, etc. (1991). ISBN 0-256-04051-1.

[2] S. G. Kellison, The Theory of Interest . Richard D. Irwin, Inc., Homewood, Ill. (1970).ISBN ???-083841.

[3] McGill Undergraduate Programs Calendar 2004/2005. Also accessible athttp://www.coursecalendar.mcgill.ca/Frameset.html.

[4] R. Muksian, Mathematics of Interest Rates, Insurance, Social Security, and Pen-sions . Pearson Education, Inc., Upper Saddle River, NJ. (2003). ISBN 0-13-009425-0.

[5] M. M. Parmenter, Theory of Interest and Life Contingencies, with Pension Appli-cations. A Problem-Solving Approach, 3rd Edition. ACTEX Publications, WinstedCT, (1999). ISBN 1-56698-333-9.

[6] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition.Brooks/Cole (1999). ISBN 0-534-35563-3.

[7] H. S. Hall and S. R. Knight, Higher Algebra, Fourth Edition, MacMillan & Co.(London, 1891).

[8] The Canadian Institute of Actuaries English-French lexicon,http://www.actuaries.ca/publications/lexicon/


A Supplementary Lecture Notes

A.1 Supplementary Notes for the Lectures of January 4th andJanuary 5th, 2005

Distribution Date: Tuesday, January 4th, 2005, revised Friday, January 7th, 2005(subject to further revision)

A.1.1 These notes

These notes will contain some of the material that I propose to discuss in the lectures, andalso some material that will not make it to the lectures. I will be following the textbookvery closely, sometimes explaining statements that I find require some elaboration. Inthe first chapter of the book my notes will be detailed; I do not know yet whether I willbe able to continue this detailed a set of notes for subsequent chapters. Much of theclass time will be spent in discussing problems from the textbook, for which I will beincluding sketches of solutions wherever possible.

Textbook Chapter 1. The measurement of interest.

A.1.2 §1.1 INTRODUCTION

While some of the concepts in this course may be applied in non-financial contexts, thelanguage of the course involves growth of amounts of money under the passage of time.Most of our discussions will concern one or two specific ways in which amounts of moneyare, in practice, assumed to grow; but, initially, we will define very general concepts.

Definition A.1 1. Principal is the initial amount of money borrowed, lent, or in-vested.

2. Interest is the compensation paid by a borrower of capital to the lender for the useof the capital.

3. Interest is said to be earned by the lender, or to accrue to the lender.

4. A distinction may be made between when and how interest is earned , and when itis paid or credited to the lender. When the word interest is used, without furtherelaboration, payment is normally at the end of each time period. When interestis paid at the beginning of a time period it is usually called discount ; however, thecompensation itself may still be called interest


A.1.3 §1.2 THE ACCUMULATION AND AMOUNT FUNCTIONS

Definition A.2 The growth of a given, specific amount of principal — as a function oftime, t — is often denoted by the amount function, denoted by A(t).

We will always assume that the rate of growth does not depend on the amount of moneythat is invested.11 This scalability assumption permits us to “normalize” many of ourdiscussions, by speaking about the growth of a fund of 1, and then just scaling theresulting figures by multiplying by the correct initial or terminal value of the fund. Wewill take this as our first

Axiom A.1 Invested amounts are normally scalable.

Under this postulate it is often convenient to study amounts that are “normalized” tohave value 1 at some convenient time. We will sometimes design our notation aroundthis postulate, by defining pairs of functions with similar names, one for the actual sizeof the fund, the other for the “normalized” account. The first example of this is thenormalized amount function, which we call the “accumulation” function.

Definition A.3 Corresponding to a specific amount function, which has value A(0) attime t = 0, we define a normalized amount function or accumulation function a byA(t) = A(0) · a(t) or

a(t) =A(t)

A(0). (7)

A simple consequence of this equation is

Theorem A.1 a(0) = 1.

(Most mathematicians don’t like to use the word “theorem” for results as trivial as this;we would be more likely to call it a “proposition” or use some other term that reservesthe word “theorem” for more serious results.)

In practice certain types of amount functions would be absurd: for example, weusually wouldn’t normally want to permit funds to get smaller with increasing time,although there are situations where such shrinkage could be given some sense. But, forthe present, we will follow the textbook in assuming

11This is an assumption that is not completely realistic, since, in the real financial world, a personwho has, say $1,000,000 to invest can often obtain a higher interest rate on his investment than theperson who has only $100. But this model is used throughout the textbook, and we will follow it in thecourse.


Axiom A.2 The function a is non-decreasing; i.e.,

t1 < t2 ⇒ a(t1) ≤ a(t2) .

(The textbook uses the word increasing, but many mathematicians use that word in aslightly restrictive way. Here I want to permit the amount to stay unchanged for a time;I am assuming only that it cannot get smaller as time increases.)

The third axiom of the textbook [1, p. 2], concerning continuity of the function a(t) isnot clear. It appears to say, tautologically, that a(t) is continuous if a(t) is continuous.I will ignore it until I know what the author means.

Definition A.4 1. The nth period of time is defined to be the period of time betweent = n − 1 and t = n. More precisely, the period normally will consist of the timeinterval n− 1 < t ≤ n.

2. Where n is a non-negative integer, the interest earned by the amount A(t) in thenth period of time is denoted by In, and defined by

In = A(n)− A(n− 1) . (8)

We may relate this notation to a “standard” convention in the “finite difference calculus”,which would define the increment in A(n) by

4A(n) = A(n + 1)− A(n) .

Thus In = 4A(n− 1).12

Example A.2 The textbook [1, p. 3] gives graphical examples of 4 kinds of amountfunctions.

1. The first has a straight line graph; the graph is sloping upward because the deriv-ative of the function is positive, in order that the function should be increasing ata constant rate. In this case the fund is earning a fixed amount which is propor-tional to the time that has elapsed, even if the time is not an integer number oftime-units; this is the situation we will later call simple interest .

2. The second example is an exponential curve, where A(t) is a function of the form

A(t) = k · e`t , (9)

` being a real number. If we set t = 0 in (9), we see that k = A(0). Since thefunction must be non-decreasing, we know by the calculus that ` ≥ 0. This is thecase that we will be calling compound interest .

12Note that the name of the function is 2 characters long: 4A. In situations like this, where thereader might be at risk of not knowing where the name of the function begins or ends, one might useparentheses, and write (4A)(n− 1).


3. The third case has a graph which is a horizontal line. This is simply the specialcase of the preceding where ` = 0, and A(t) is constant; the amount neither growsnor shrinks.

4. The fourth case is a “step” function, whose graph is horizontal line segments of unitlength, each a fixed distance above the preceding one. This function is “piecewisecontinuous”, having points of discontinuity at the integer times. This is a realisticmodel, for example, where an amount earns interest only if the amount remains inthe account for a full time period13, and where the interest is credited at the endof the period.

To be more precise, the function is “continuous from the right” everywhere, but itis “discontinuous from the left” at integer points; this is because the limit from theleft is not equal to the function value there — i.e., because the value you wouldinfer from the size of the amount preceding an integer time is not equal to theactual size of the account at integer time.

[1, Exercise 1, p. 29] “Consider the amount function A(t) = t2 + 2t + 3.”

1. “Find the corresponding accumulation function a(t).”

2. (modified) Show that a(0) = 1 and that a is a non-decreasing function of time.

3. “Find In.”

Solution:

1.

a(t) =A(t)

A(0)=

t2 + 2t + 3

02 + 2(0) + 3=

1

3t2 +

2

3t + 1 .

2. (a) a(0) = 13· 02 + 2

3· 0 + 1 = 1

(b) If the author intends that we consider this a function of a real variablethat may take any values on the non-negative real axis, then one way toprove the function is non-decreasing is to show that the derivative is notnegative. (This can be used only for a function that is differentiable, butthe present function is a polynomial, and polynomials have that property.)

a′(t) =1

3· 2 · t +

2

3· 1 =

2

3(t + 1) .

We see that a(t) is a non-decreasing function provided t ≥ −1.

13In the language of the textbook [1, p. 6] interest is accrued only for completed periods, with nocredit for fractional periods.


This could also have been approached from “first principles”. Supposethat t1 < t2. Then

a(t2)− a(t1) =1

3

(t22 − t21

)+

2

3(t2 − t1)

= (t2 − t1)

(t23

+t13

+2

3

)

and both factors are positive for non-negative t2 and t1 and t1 < t2.

3. In = A(n)− A(n− 1) = (n2 + 2n + 3)− ((n− 1)2 + 2(n− 1) + 3) = 2n + 1.

[1, Exercise 2, p. 30] 1. “Prove that A(n)− A(0) = I1 + I2 + . . . + In.”

2. “Verify verbally the result obtained” above.

Solution:

1. Summing the equations A(m)− A(m− 1) = Im for 1 ≤ m ≤ n, we obtain

A(n)− A(0) =n∑

m=1

Im .

2. Verbally, the (amount of) interest earned over the concatenation of n periodsis the sum of the interest earned in each of the periods separately.

[1, Exercise 3, p. 30] “Find the amount of interest earned between time t and timen, where t < n, if

1. Ir = r

2. Ir = 2r.”

Solution: In the preceding problem we showed that the amount earned betweentime 0 and time n was the sum of the values of I. The interest earned betweentime t and time n will be the total interest from time 0 to time n diminished bythe interest earned from time 0 to time t, so it must be

n∑m=1

Im −t∑

m=1

Im =n∑

m=t+1

Im

When Im = m, this is14

n∑m=1

m−t∑

m=1

m =n(n + 1)

2− t(t + 1)

2=

(n− t)(n + t + 1)

2.

14using the familiar formula for the sum of the first natural numbers


When Im = 2m, the sum is

n∑m=t+1

2m = 2t+1

(2n−t − 1

2− 1

)= 2n+1 − 2t+1 .


A.2 Supplementary Notes for the Lecture of January 7th, 2005

Distribution Date: Friday, January 7th, 2005, subject to further revision

A.2.1 §1.2 THE ACCUMULATION AND AMOUNT FUNCTIONS (con-clusion)

Summation of Geometric Progressions and Series (These results were appliedin the solution of [1, Exercise 3, p. 30], shown in the notes for the preceding lecture.)

The reader is reminded of the easily derived formula [1, Appendix III, A.2, p. 394]

n−1∑

k=0

ark = a

(rn − 1

r − 1

)= a

(1− rn

1− r

)(r 6= 1) . (10)

The textbook overlooks the essential restriction that r 6= 1 for the use of this formula;when r = 1 the sum is

n−1∑

k=0

a1k = a · n (r 6= 1) . (11)

We shall be using these formulæ repeatedly. We shall also be interested in the behaviorof (10) in the limit as n →∞:

∞∑

k=0

ark = a

(1

1− r

)(|r| < 1) . (12)

(When |r| ≥ 1, the limit does not exist.)

[1, Exercise 4, p. 30] “It is known that a(t) is of the form ct2 + b. If $100 investedat time 0 accumulates to $172 at time 3, find the accumulated value at time 10 of$100 invested at time 5.”

Solution: Let’s begin by normalizing the function: a(0) = 1 ⇔ c·02+b = 1, so b = 1.The given growth condition implies that 100(c32 + 1) = 172, so c = 72

900= 0.08.

Consequently 100 invested at time 5 grows to 100((0.08) · 52 + 1) =300.

A.2.2 §1.3 THE EFFECTIVE RATE OF INTEREST

When we speak of a rate of interest, we mean the amount of interest per unit investment .As this rate is normalized, we follow the convention we began with the pair A and a, byusing a lower case letter to represent it.


Definition A.5 The effective rate of interest during the nth period, in, is the ratio ofthe amount of interest earned during, and paid at the end of the period to the amountof principal invested at the beginning of the period. When the effective rate is constantover all unit time intervals, we may suppress the subscript, and simply write i.

We may use the term effective for periods of lengths other than one unit; but themost common use will be for terms of one year, where we will speak of the effectiveannual rate.

Note that, when we use the unqualified word interest alone, the listener may not knowwhether we are referring to the amount of interest, the rate of interest, or the generalconcept of payment for the use of capital. There may also be doubts about the accrualperiod.

Theorem A.3 In symbols,

in =In

A(n− 1)=

A(n)− A(n− 1)

A(n− 1)=

a(n)− a(n− 1)

a(n− 1)(13)

for integer n ≥ 1. And, when the rate is constant, i, recalling that a(0) = 1, we have

a(n) = (1 + i)a(n− 1) = (1 + i)2a(n− 2) = . . . = (1 + i)na(0) = (1 + i)n . (14)

In all of this discussion the interest rate is associated with one time interval; this will becontrasted later with rates — called “nominal” — that are stated for one time interval,but need to be applied to another.


A.3 Supplementary Notes for the Lecture of January 10th,2005

Distribution Date: Monday, January 10th, 2005, subject to further revision

A.3.1 §1.3 THE EFFECTIVE RATE OF INTEREST (conclusion)

[1, Exercise 5, p. 30] “Assume that A(t) = 100 + 5t.” Find i5 and i10.

Solution:

i5 =A(5)− A(4)

A(4)=

5 · 1100 + 5 · 4 =

1

24.

i10 =A(10)− A(9)

A(9)=

5 · 1100 + 5 · 9 =

1

29.

[1, Exercise 6, p. 30] “Assume A(t) = 100(1.1)t.” Find i5 and i10.

Solution:

i5 =A(5)− A(4)

A(4)=

100(1.1)4(0.1)

100(1.1)4= 0.1 .

i10 =A(10)− A(9)

A(9)=

100(1.1)9(0.1)

100(1.1)9= 0.1 .

[1, Exercise 7, p. 30] “Show that A(n) = (1 + in) A(n− 1).”

Solution: (1 + in)A(n − 1) =(1 + A(n)−A(n−1)

A(n−1)

)A(n − 1) = A(n)

A(n−1)· A(n − 1) =

A(n). ¤A verbal solution is also feasible.

[1, Exercise 8, p. 30] “If A(4) = 1000 and in = 0.01n, find A(7).”

Solution: I(n) = in · A(n− 1) = (0.01)nA(n− 1). It follows that

A(n) = In + A(n− 1) = (0.01(n) + 1)A(n− 1)

A(7) = (0.01(7) + 1)A(6)

= (0.01(7) + 1)(0.01(6) + 1)A(5)

= (0.01(7) + 1)(0.01(6) + 1)(0.01(5) + 1)A(4) = (1.07)(1.06)(1.05)1000

= 1190.91 .


A.3.2 §1.4 SIMPLE INTEREST

We saw in the preceding discussion that a(0) = 1 and a(1) = 1 + i1. Under simpleinterest we assume the linear function we saw in the first graphical example above: theamount of interest earned in a time interval of length t and starting at a fixed time, liket = 0, is assumed to be proportional to t; we obtain the formula

a(t) = 1 + i1t

over the 1st time interval, or any longer interval where the so-called simple interest rateremains equal to i1. Suppose that the rate of interest is i, and that this rate is to remainin effect over an interval of n or more successive time periods without adjusting the baseprincipal from which it is computed. We can determine the rate of interest during thenth period under this assumption; it is

in =a(n)− a(n− 1)

a(n)=

[1 + i1n]− [1 + i1(n− 1)]

1 + i1(n− 1)=

i11 + i1(n− 1)

(15)

where we see that, even though the amount of interest earned in any period of time isa constant multiple of the length of the period, the rate of interest per period is notconstant — it is decreasing from each period to the next.15 Thus a constant rate ofsimple interest implies that the effective rates of interest for the successive time intervalsform a decreasing sequence. What is constant for simple interest is the absolute amountof interest earned in each time interval; under compound interest which we shall meetin the next section, it is the relative amount of interest that is constant — that is, theratio of interest earned to principal.

Note that, when we speak of simple interest, we must specify not only the rate, butalso give enough information so that the reader will know when the principal amount isadjusted to incorporate the accrued interest.

A “rigorous” mathematical derivation of simple interest I modify here a deriva-tion in the textbook. The object is to show that the formula assumed above for simpleinterest is a consequence of a “reasonable”, general assumption.

Axiom A.3 (Defining postulate of simple interest) Under simple interest the rateof interest in all time periods of length t + s is the same; and it is equal to the sum ofthe interest rates for periods of lengths t and s. Symbolically,

a(t + s)− a(0) = [a(t)− a(0)] + [a(s)− a(0)] , (16)

or, more compactly,a(t + s)− a(0) = a(t) + a(s)− 2a(0) , (17)

for all non-negative real numbers t and s.

15But the derivative of the amount function will be constant.


Note that the axiom assumes the rule is to hold for periods of any non-negative length,not just of integer length. The textbook then attempts16 to show that this hypothesisimplies that

d

dta(t) =

d

dta(0) .

Thus the time derivative of a(t) is shown to be constant. We know from elementarycalculus17 that a(t) must have the form

a(t) = a′(0) · t + C

where C is a constant; and we can determine that constant by assigning to t the particularvalue 0, so that

C = a(0) = 1

anda(t) = 1 + a′(0) · t .

If, in this last equation, we assign t = 1, we find that

a′(0) = a(1)− a(0)

which is what was defined to be i1, so

a(t) = 1 + i1 · t for t ≥ 0 . (18)

The textbook writes i in place of i1 in the preceding equation, but this is confusing,since we saw in equation (15) above that the rate of interest is not constant under simpleinterest.

[1, Exercise 9, p. 30] 1. “At what rate of simple interest will $500 accumulate to$615 in 21

2years?

2. “In how many years will $500 accumulate to $630 at 7.8% simple interest?”

Solution:

1. Under simple interest at a rate i, 615 = 500a(2.5) = 500(1 + i2.5); solving for

i, we obtain i =615500

−1

2.5= 0.092 = 9.2%.

16To a mathematician the proof appears slightly defective, although it can be repaired: the definitionof derivative requires a 2-sided limit, and the textbook only considers a “right” derivative. The sameidea used in the proof can be used to supply the needed result for the limit from the left. You are notexpected to be concerned with this degree of subtlety in this course.

17The Mean Value Theorem


2. Under a rate of 7.8% simple interest for a period of t years, 630 = 500a(t) =

500(1 + 0.078t); solving for t, we obtain t =630500

−1

0.078= 10

3years.

[1, Exercise 10, p. 30] “If ik is the rate of simple interest for period k, where k =1, 2, . . . , n, show that a(n)− a(0) = i1 + i2 + . . . + in.”

Solution: Note that there is an implicit assumption that the earned interest is notincorporated into the principle during the period 0 ≤ t ≤ n. What is denoted byit in this problem is not the effective rate of interest for the tth period, shown inequation (1.6). Rather, it is the total amount earned during the tth period by theamount that began as 1 at time t = 0.

Under simple interest an initial investment of a(0) = 1 grows by the prescribedamounts in each of the following years: a(n) = a(0)+i1+i2+. . .+in, so a(n)−a(0) =i1 + i2 + . . . + in .

[1, Exercise 11, p. 30] “At a certain rate of simple interest $1000 will accumulate to$1100 after a certain period of time. Find the accumulated value of $500 at a rateof simple interest three fourths as great over twice as long a period of time.”

Solution: It is known that, at the rate of interest i, and for a prescribed length oftime t, 1000(1 + it) = 1110, so it = 0.11. Then 500

(1 + 3i

4· 2t) = 500

(1 + 3

2it

)=

500(1 + 0.165) = 582.50.

[1, Exercise 12, p. 30] “Simple interest of i = 4% is being credited to a fund. Inwhich period is this equivalent to an effective rate of 21

2%?”

Solution: We have to solve for n the equation

0.025 = in =0.04

1 + (0.04)(n− 1)

and find that n = 16.

A.3.3 §1.5 COMPOUND INTEREST

Under compound interest, the earned interest is incorporated into the principal after eachaccrual period, following which the interest in the next period is based on the updatedprincipal. Thus the earned interest is reinvested together with the original principal. Thisis what we observed in equation (14) above, where the interest rate remained constantover successive time periods.

The growth of the principal can be analyzed in various ways. Thus, for example, thevalue a(2) = (1 + i)2a(0) = (1 + i)2 may be interpreted

• as (1 + i)a(1) = (1 + i) · (1 + i) as we did in the discussion preceding (14)


• as 1+2i+i2, where the principal of 1 is augmented by 2i, representing the “simple”interest earned by the principal over two successive time periods, augmented bythe interest at rate i earned by the interest i from the first period and credited atthe end of that period.

Under compound interest at an effective interest rate of i we assume that an amount of 1grows to (1+i)t during a time interval of length t, where t is not necessarily a non-negativeinteger . In most of the sequel it is compound interest that will be assumed to be acting.Where we assume simple interest, it will often be for fractions of a time period. In someways this assumption is obsolete now — a relic of the days when exact computing wasdifficult because of the unavailability of reliable and inexpensive calculators. However,some of the practices of using simple interest have become entrenched in the financialsystem, and so we are obliged to continue using them.

A “rigorous” mathematical derivation of compound interest (This is parallelto the derivation given earlier for simple interest; again I modify here a derivation in thetextbook.)

Axiom A.4 ((Defining postulate of compound interest) Under compound interest

the ratioa(t + s)

a(t)is constant for all s, as a function of t; and the product of these ratios

for times t1 and t2 is equal to the ratio for time t1 + t2. Symbolically,

a(t1 + t2)

a(0)=

a(t1)

a(0)· a(t2)

a(0)(19)

or, more compactly,a(t1 + t2) = a(t1) · a(t2) , (20)

for all non-negative real numbers t1 and t2.

The textbook then infers thata′(t)a(t)

= a′(0),

again a constant for all t. This equation may be solved in a standard way. The fractionon the left is the derivative of ln a(t), so

ln a(t) = a′(0) · t + C for all t. (21)

Since a(0) = 1, setting t = 0 yields C = 0, so

a(t) = ea′(t) . (22)


Setting t = 1, and recalling that

a(1)− a(0) = i1 ,

yieldsa′(0) = ln(1 + i1) (23)

soa(t) = (1 + i1)

t . (24)

Here we can show that in is constant for all n, so we can legitimately suppress thesubscripts and write simply i:

a(t) = (1 + i)t . (25)

We thus see the graphical distinction between simple and compound interest: the graphof an amount function under simple interest is a straight line — a “linear” function;the graph of an amount function under compound interest is an exponential function.Some of the variations we will consider will be “piecewise linear”, where the graph ismade up of line segments attached end-to-end, where the points of attachment lie on anexponential curve.

[1, Exercise 13, p. 30] (This solution is not examination material, but you shouldremember the result.)

“Assuming that 0 < i < 1, show that:

1. (1 + i)t < 1 + it if 0 < t < 1;

2. (1 + i)t = 1 + it if t = 1;

3. (1 + i)t > 1 + it if t > 1.

“This exercise verifies the relative magnitudes of accumulated values at simple andcompound interest over various periods of time.”

Solution: This problem supplies the proof of a statement made in the text: thatcompound interest provides a greater return than simple interest for periods greaterthan 1; it also proves the complementary statement, that compound interest pro-vides a lower return than simple interest for periods shorter than 1. Observe, forthe function f(t) = (1 + i)t − (1 + it),

f ′(t) = (1 + i)t ln(1 + i)− i (26)

f ′′(t) = (1 + i)t(ln(1 + i))2 (27)

f(0) = f(1) = 0 (28)


From (27) we see that, at any critical point t,

f ′′(t) > 0 : (29)

thus any critical point is a local minimum for f . Hence, on the closed interval0 ≤ t ≤ 1, f attains its global maximum only at one or both of the end points, atboth of which, by (28), it is 0. This proves that

(1 + i)t < 1 + it if 0 < t < 1

which is part (a). Equation (28) proves part (b).

Finally, consider the case t > 1. Define g(x) = (1+x) ln(1+x)−x, observing that

g′(x) = ln(1 + x) (30)

g(0) = 0 , (31)

f ′(t) > f ′(1) = g(i) . (32)

Suppose that there existed a point a > 0 where g(a) ≤ 0. Then, by the MeanValue Theorem, there would exist a point c such that 0 < c < a and

0 < ln(1 + c) = g′(c) =g(a)− g(0)

a≤ 0

a= 0 ,

which is a contradiction. It follows that g(x) > 0 for x > 0, and f ′(t) > 0 for allt > 1. Suppose that it happened that f(b) ≤ 0 for b > 1, and apply the MeanValue Theorem to f(t) on the interval 1 ≤ t ≤ b. There would exist a point k,1 < k < b, such that

f ′(k) =f(b)− f(1)

b− 1< 0 ,

contradicting (32). We conclude that f(t) > 0 for t > 1, which implies (c).

[1, Exercise 14, p. 30] “It is known that $600 invested for two years will earn $264in interest. Find the accumulated value of $2000 invested at the same rate ofcompound interest for three years.”

Solution: What is known is that, at the given interest rate i, 600(1+i)2 = 600+264,so (1+ i)2 = 1.44 and (1+ i)1 = 1.2 (i.e. i = 20%). At the same rate, 2000(1+ i)3 =2000(1.2)3 = 2000× 1.728 = 3456.

[1, Exercise 15, p. 31] “Show that the ratio of the accumulated value of 1 invested atrate i for n periods, to the accumulated value of 1 invested at rate j for n periods,i > j, is equal to the accumulated value of 1 invested for n periods at rate r. Findan expression for r as a function of i and j.”


Solution:1(1 + i)n

1(1 + j)n=

(1 +

i− j

1 + j

)n

.

Thus r =i− j

1 + j.

[1, Exercise 16, p. 31] “At a certain rate of compound interest, 1 will increase to 2 ina years, 2 will increase to 3 in b years, and 3 will increase to 14 in c years. If 6 willincrease to 10 in n years, express n as a function of a, b, and c.”

Solution: If the common rate is i, the hypotheses are that

1(1 + i)a = 2

2(1 + i)b = 3

3(1 + i)c = 15

6(1 + i)n = 6 .

Taking logarithms and solving, we obtain

a =ln 2

ln(1 + i)(33)

b =ln 3− ln 2

ln(1 + i)(34)

c =ln 5

ln(1 + i)(35)

n =ln 5− ln 3

ln(1 + i)

= c− a− b (36)

Note that this answer is “overdetermined”, in the sense that a, b and c are notindependent: by equations (33), (34), (35), a : b : c :: ln 2 : ln 3

2: ln 5, that is, a, b,

and c are related by proportionality equations of the form

a

ln 2=

b

ln 32

=c

ln 5.

Put another way, we didn’t need all 3 of these equations to solve the problem —any 2 of them would have been sufficient!

[1, Exercise 17 p. 31] “An amount of money is invested for one year at a rate ofinterest of 3% per quarter. Let D(i) be the difference between the amount of


interest earned on a compound interest basis, and on a simple interest basis forquarter k, where k = 1, 2, 3, 4. Find the ratio of D(4) to D(3).”

Solution: Let the amount of money invested be x. The interest rate is 3% perquarter . We have

D(1) = x((1.03)0((1.03)− 1)− 0.03

)

D(2) = x((1.03)1((1.03)− 1)− 0.03

)

D(3) = x((1.03)2((1.03)− 1)− 0.03

)

D(4) = x((1.03)3((1.03)− 1)− 0.03

)


D(4)

D(3)=

(1.03)3((1.03)− 1)− 0.03

(1.03)2((1.03)− 1)− 0.03=

1.033 − 1

1.032 − 1= 1.522610837 .



Distribution Date: Thursday, January 13th, 2005, subject to revision

A.4.1 §1.6 PRESENT VALUE

Definition A.6 1. Under an interest rate of i the sum 1+i is called the accumulationfactor which, when applied to an amount, yields the value of the amount at theend of 1 time period.

2. When we view an amount from the past, asking what should be today’s value ofan amount receivable some time in the future, we call the value today the presentor current value or the value discounted (back) to the present.

3. To obtain the amount which, at the beginning of a time period, would yield, at theinterest rate i, an amount of 1 at the end of the period, we may divide by 1 + i or

multiply by the discount factor1

1 + i, which is often denoted by v.

The textbook denotes the reciprocal1

a(t)by a−1(t). This is an unfortunate notation,

since it appears to involve an inverse function, and that is not the author’s intention.

You are advised not to use this notation: write1

a(t)or (a(t))−1.

Definition A.7 Analogous to the accumulation function a(t), we can call (a(t))−1 thediscount function. It represents the amount that needs to be invested today to yield anamount of 1 at the end of t time periods.

In the following problems there will be a gradual transition from the general formulationof the preceding definitions to the specific case of compound interest.

[1, Exercise 18 p. 31] “Find an expression for the discount factor during the nth pe-riod from the date of investment, i.e., (1 + in)−1, in terms of the amount function.”

Solution: 11+in

= A(n−1)A(n)

.

[1, Exercise 19 p. 31] “The sum of the present value of 1 paid at the end of n periodsand 1 paid at the end of 2n periods is 1. Find (1 + i)2n.”

Solution:

(1 + i)−n + (1 + i)−2n = 1 ⇔ ((1 + i)−n

)2+ (1 + i)−n − 1 = 0


⇔(

(1 + i)−n +1

2

)2

=5

4

(completing the square)

⇔ (1 + i)−n =−1±√5

2

But the negative sign would be associated with a negative value of (1 + i)n, whichis impossible. Hence

(1 + i)−n =−1 +

√5

2

(1 + i)−2n = 1− (1 + i)−n =3−√5

2

(1 + i)2n =2

3−√5=

2

3−√5· 3 +

√5

3 +√

5

=2(3 +

√5)

9− 5=

3 +√

5

2

[1, Exercise 20, p. 31] “Show that the current value of a payment of 1 made n periodsago and a payment of 1 to be made n periods in the future is greater than 2, ifi > 0.”

Solution: The payment from the past is currently worth (1 + i)n, while the futurepayment is currently (=presently) worth vn = (1 + i)−n. The excess of this sumover 2 is

(1 + i)n + (1 + i)−n − 2 =((1 + i)

n2 − (1 + i)−

n2

)2, (37)

which, being a square, cannot be negative. Hence

(1 + i)n + (1 + i)−n ≥ 2 ,

which equality holding precisely when (1+i)n2 = (1+i)−

n2 , i.e., when (1+i)n−1 = 0,

i.e. if either i = 0 or n = 0.

[1, Exercise 21 p. 31] “It is known that an investment of $500 will increase to $4000at the end of 30 years. Find the sum of the present values of three payments of$10,000 each which will occur at the end of 20, 40, and 60 years.”

Solution: At the rate i, 500(1 + i)30 = 4000, so (1 + i)30 = 8, and (1 + i)10 = 2.Hence the present value of payments of 10,000 to occur at the ends of 20, 40, and


60 years is

10000(v20 + v40 + v60

)= 10000

(1

4+

1

16+

1

64

)

=210000

64= 3281.25.



Distribution Date: Friday, January 14th, 2005, subject to revision

A.5.1 §1.7 THE EFFECTIVE RATE OF DISCOUNT

We will try to develop, in parallel to the interpretation of amount and accumulationfunctions from the point of view of the addition of interest, paid at the ends of timeperiods, a discount interpretation under which the amount and accumulation functionsare discounted by subtracting an amount of discount. This will lead to several symbolsanalogous to those in the interest model. This parallel development will be incomplete,and most of the time we will be working only with the interest model. For example,we could interpret In given in equation (8) as an amount of discount . Then we couldparallel equation (13) with

In

A(n)=

A(n)− A(n− 1)

A(n)=

a(n)− a(n− 1)

a(n)(38)

Definition A.8 The common amount of the members of equation (38)is denoted by dn,and called the effective rate of discount .

Note the differences between the two models:

• Under the interest model, the payment for the use of the money is made at theend of the period, based on the balance at the beginning of the period.

• Under the discount model, the “payment” is deducted at the beginning of theperiod from the final amount which will be present at the end of the period.

We will use the word equivalent to describe schemes of interest and/or discount thatproduce the same ultimate accumulated value from an initially invested amount. In thepresent application, which is only the first where we will meet this concept, we can, forexample, equate

1

1 + i= 1− d

from which we obtain various other relationships, e.g.,

d =i

1 + i(39)

i =d

1− d(40)

d = 1− v (41)

i− d = id (42)


You should try to explain each of these identities “verbally”, i.e., in words , in a way thata person lacking your technical knowledge can find the equation plausible.

If a discount rate d is applied t times to a final amount of 1, the current value will be

1

a(t)= vt = (1− d)t.

The textbook makes several observations about the use of the discount model, and aboutthe term discount . For these and other reasons, we shall not be devoting equal time tothe discount model in this course. But you should be aware that most of what we docould be redeveloped from the discount point of view.

Definition A.9 The accumulation function for simple discount at a discount rate d ≥ 0

is given by a(t) =1

1− dt, for t <

1

d.

Note that, unlike the situation for simple interest , we must restrict the length of timeover which we propose to apply simple discount.

Definition A.10 The accumulation function for compound discount at a discount rate

d is given by a(t) =1

(1− d)t.

While we need not restrict t in the compound discount case, we must restrict the effectivediscount rate d, requiring that 0 ≤ d < 1.

[1, Exercise 22 p. 31] “The amount of interest earned on A for one year is $336, whilethe equivalent amount of discount is $300. Find A.”

Solution: Denote the equivalent rates of interest and discount by i and d respec-tively. The hypotheses are that

iA = 336

dA = 300

subject to the relationship between i and d, which may be expressed in variousways. We wish to obtain A, not i or d. So let us solve the preceding equations fori and d in terms of A,

i =336

Ad =

330

A

and substitute into one of the equations relating i and d, e.g., into i− d = id:

336

A− 300

A=

336× 300

A2⇔ 36A = 336(300)

⇔ A = 2800


[1, Exercise 23 p. 31] 1. “Find d5 if the rate of simple interest is 10%.”

2. “Find d5 if the rate of simple discount is 10%.”

Solution:

1. Under simple interest at 10%, In = 0.1×A(0) for all non-negative integers n.So

d5 =I5

A(5)=

(0.1)A(0)

(1 + 5(0.1))A(0)=

0.1

1.5=

1

15.

2. But, if the rate of simple discount is 10%, then A(n) =A(0)

1− 0.01n.

d5 =A(5)− A(4)

A(5)=

1

0.05− 1

0.061

0.5

=1

6.

[1, Exercise 24 p. 31] 1. “Assuming compound discount, show that dn is constantfor all n.’

2. Assuming simple discount, show that dn is increasing for increasing n if 0 <

n− 1 <1

d.”

Solution:

1. Under compound discount at a rate d of discount, A(n− 1) = (1− d)A(n), so

dn =A(n)− A(n− 1)

A(n)= d for all n. (This result is the discount analogue of

the equation at [1, p. 8, l. 5].)

2. Under simple discount at a rate d of discount, A(0) = (1− dn) ·A(n). Hence

dn =A(n)− A(n− 1)

A(n)=

1

1− dn− 1

1− d(n− 1)1

1− dn

=1

1

d− (n− 1)

.

So long as 1d

> n−1; thus dn is an increasing function of d in this interval. (Ifthe inequality were to fail the principal would have been depleted as discount.This result is the discount analogue of equation [1, (1.6), p. 5].)

[1, Exercise 25 p. 31] (As I stated in [1, Exercise 13, p. 30], this solution also is notexamination material, but you should remember the result.) “Assuming that 0 <d < 1, show that:


1. (1− d)t < 1− dt if 0 < t < 1;

2. (1− d)t = 1− dt if t = 1;

3. (1− d)t > 1− dt if t > 1.

“This exercise verifies the relative magnitudes of present values at simple andcompound discount over various periods of time.”

Solution: (We adapt the solution given above to [1, Exercise 16, p. 31] on 2016 ofthese notes.) Observe, for the function f(t) = (1− d)t − (1− dt),

f ′(t) = (1− d)t ln(1− d) + d (43)

f ′′(t) = (1− d)t(ln(1− d))2 (44)

f(0) = f(1) = 0 (45)

From (44) we see that, at any critical point t,

f ′′(t) > 0 : (46)

thus any critical point is a local minimum for f . Hence, on the closed interval0 ≤ t ≤ 1, f attains its global maximum at one or both of the end points, at bothof which, by (45) it is 0. This proves that

(1− d)t < 1− dt if 0 < t < 1

which is part (a). Equation (45) proves part (b).

Finally, consider the case t > 1. Define g(x) = (1−x) ln(1−x)+x, observing that

g′(x) = − ln(1− x) (47)

g(0) = 0 , (48)

f ′(t) > f ′(1) = g(d) . (49)

Suppose that there existed a point a > 0 where g(a) ≤ 0. Then, by the MeanValue Theorem, there would exist a point c such that 0 < c < a and

0 < − ln(1− c) = g′(c) =g(a)− g(0)

a≤ 0

a= 0 ,

which is a contradiction. It follows that g(x) > 0 for x > 0, and f ′(t) > 0 for allt > 1. Suppose that it happened that f(b) ≤ 0 for b > 1, and apply the MeanValue Theorem to f(t) on the interval 1 ≤ t ≤ b. There would exist a point k,1 < k < b, such that

f ′(k) =f(b)− f(1)

b− 1< 0 ,

contradicting (32). We conclude that f(t) > 0 for t > 1, which implies (c).


[1, Exercise 26 p. 32] “If i and d are equivalent rates of simple interest and simplediscount over t periods, show that i− d = idt.”

Solution: The discounted value (at simple interest) at time 0 of 1 at time t is 1−dt.At simple interest this accumulates to (1−dt)(1+ it) = 1. Expanding and dividingby t yields

i− d = idt . ¤

[1, Exercise 27 p. 32] “Show that

d3

(1− d)2=

(i− d)2

1− v.”

Solution:

1. Algebraic proof: In an algebraic proof of an identity we can, in the mostuninspired proof, prove both members of the alleged equation are equal tothe same quantity. Usually many possible proofs exist, but some may bemore interesting than others, because they may be easier to explain in words,i.e., verbally. Such proofs are more satisfying if we can start with the quantityon one side of the equal sign, and, by applying familiar results, transform itinto the quantity on the other side. For example

d3

(1− d)2= d

(1

1− d

)2

= di2 [1, (1.14), p. 14]

=(id)2

d=

(i− d)2

1− v[1, (1.17), p. 15] .

2. Verbal proof: HOPEFULLY TO BE SUPPLIED

A.5.2 §1.8 NOMINAL RATES OF INTEREST AND DISCOUNT (barelybegun



Distribution Date: Monday, January 17th, 2005, subject to revision

A.6.1 §1.8 NOMINAL RATES OF INTEREST AND DISCOUNT (contin-ued)

Definition A.11 [1, p. 14] Two rates of interest or discount are equivalent if an amountof principal invested for the same length of time under either of the rates accumulatesto the same value.

We have already applied this concept in the preceding section, when we related equiv-alent rates of interest and discount over the same time interval. In this section we willrelate equivalent rates of interest and/or discount over different time intervals, where oneinterval is a multiple of the other.

When we speak of a nominal rate of interest, we will normally describe a paymentscheme that is more or less frequent than the usual payment interval; the interest ratethat we will describe will represent the total of all payments made over that usualinterval. Summing the payments does not take into account the interest or discount toshift all payments to one specific time. We will determine relationships between nominalrates and effective rates. The definitions are particularly unintuitive, so they must bememorized.

Definition A.12 Let m be either a positive integer or the reciprocal of a positive integer.

1. A nominal rate of interest i(m) payable m times per period represents m times theamount of compound interest that will have to be paid at the end of each 1

mth of

a period which is equivalent to an effective interest rate of i per period.

2. A nominal rate of discount d(m) charged m times per period represents m timesthe amount of compound discount that will have to be charged at the beginning ofeach 1

mth of a period which is equivalent to d, the amount charged at the beginning

of the period under an effective discount rate of d per period.

Theorem A.4 1. A nominal rate of interest of i(m) payable and compounded at theend of every 1

mth of a year is equivalent to18 an effective rate of interest of i(m)

m

payable every mth of a year. Thus a principal of 1 accumulates at the end of a full

year to(1 + i(m)

m

)m

. We have, therefore, the relationship

(1 +

i(m)

m

)m

= 1 + i , (50)

18Indeed, it has been defined to be


or, equivalently,

i(m) = m((1 + i)

1m − 1

). (51)

2. A nominal rate of discount of d(m) payable and compounded at the beginning ofevery 1

mth of a year is equivalent to an effective rate of discount of d(m)

mpayable

in advance, every mth of a year. Thus an accumulation of 1 at the end of a full

year has value(1− d(m)

m

)m

at the beginning of the year. We have, therefore, the

relationship (1− d(m)

m

)m

= 1− d , (52)

or, equivalently,

d(m) = m(1− (1− d)

1m

)= m

(1− v

1m

). (53)

We can combine equations (50) and (52) into

(1 +

i(m)

m

)m

= 1 + i =

(1− d(p)

p

)−p

; (54)

when m = p, this yields the equation

i(m)

m− d(m)

m=

i(m)

m· d(m)

m(55)

of which (42) is the special case m = 1.

[1, Exercise 28 p. 32] 1. “Express d(4) as a function of i(3).”

2. “Express i(6) as a function of d(2).”

Solution:

1. Since(

1 +i(3)

3

)3

= 1 + i

and

(1− d(4)

4

)4

= 1− d ,

(1 + i)(1− d) = 1 ⇒ 1− d(4)

4=

(1 +

i(3)

3

)− 34

⇒ d(4)

4= 4

(1− i(3)

3

)− 34

.


2.

(1 +

i(6)

6

)6

= 1 + i = (1− d)−1 =

(1− d(2)

2

)−2

⇒

i(6) = 6

((1− d(2)

2

)− 13

− 1

).

[1, Exercise 29 p. 32] “On occasion, interest is convertible less frequently than once a

year. Define i(1m) and d( 1

m) to be the nominal annual rates of interest and discountconvertible once every m years. Find a formula analogous to formula (1.22a) forthis situation.”

Solution: (1 + i(

1m)m

) 1m

=(1− d( 1

p)p) 1

p

[1, Exercise 30 p. 32] “Find the accumulated value of $100 at the end of two years:

1. “if the nominal annual rate of interest is 6%, convertible quarterly; and

2. “if the nominal annual rate of discount is 6%, convertible once every fouryears.”

1. 100(1 + 0.06

4

)2×4= 100(1.015)8

2. 100 (1− 4× 0.06)−24 = 100(1− 0.24)−

12

[1, Exercise 31 p. 32] “Derive the formula

i(m)

m− d(m)

m=

i(m)

m· d(m)

m.” (56)

Solution: For 1m

th of a year, the effective interest and discount rates are, respec-

tively i(m)

mand d(m)

m. We apply the formula

i− d = id [1, (1.17), p. 15]

for 1m

th of a year.

Another derivation of this equation would be from (18) with p = m.

[1, Exercise 32 p. 32] 1. “Show that i(m) = d(m) · (1 + i)1m .

2. “Verbally interpret the result obtained above.”

Solution:



1.

d(m) · (1 + i)1m = m

(1− v

1m

)· (1 + i)

1m

= m((1 + i)

1m − v

1m · (1 + i)

1m

)

= m((1 + i)

1m − (v(1 + i))

1m

)

= m((1 + i)

1m − 1

1m

)

= m((1 + i)

1m − 1

)= i

1m

2. i(m) =interest payable on loan of 1 at the end of 1m

period. If prepaid, d(m)

accumulates for 1m

period at (1 + i)1m .

[1, Exercise 33 p. 32] “Given that i(m) = 0.1844144, and d(m) = 0.1802608, find m.”

Solution: (1 +

i(m)

m

)(1− d(m)

m

)= 1

⇒ m =i(m)d(m)

−d(m) + i(m)=

0.1844144× 0.1802608

0.0041536= 8.0033434...

[1, Exercise 34 p. 32] “It is known that 1 +i(n)

n=

1 + i(4)

4

1 + i(5)

5

. Find n.”

Solution:

1 +i(n)

n=

1 + i(4)

4

1 + i(5)

5

⇒ (1 + i)1n = (1 + i)

14− 1

5 = (1 + i)120

⇒ n = 20 .

[1, Exercise 35 p. 32] “If r = i(4)

d(4) , express v in terms of r.”

Solution:

i(4) = 4((1 + i)

14 − 1

)

d(4) = 4(1− (1− d)

14

)

r =i(4)

d(4)=

1

(1− d)14

= v−14

Hence v = r−4.


A.6.2 §1.9 FORCES OF INTEREST AND DISCOUNT (barely begun)



Distribution Date: Wednesday, January 19th, 2005, subject to revision

A.7.1 §1.9 FORCES OF INTEREST AND DISCOUNT

This section of the textbook begins with the definition of the force of interest, δt at timet, defined as

δt =A′(t)A(t)

=a′(t)a(t)

.

A definition in this generality is appropriate if we are considering very general interestschemes. However, in the context of compound interest we can simply define

Definition A.13 δ = limm→∞

i(m) = ln(1 + i)

and we can then prove that

Theorem A.5 δ = limm→∞

d(m) = ln(1 + i)

[1, Exercise 36 p. 32] Derive the formula limm→∞

d(m) = δ.

Solution:

limm→∞

i(m) = limm→∞

m((1 + i)

1m − 1

)

= limm→∞

(1 + i)1m − 1

1m

= limx→0+

(1 + i)x − 1

x

=d

dx((1 + i)x)

∣∣∣∣x=0

= (1 + i)x ln(1 + i)|x=0 = ln(1 + i)

Analogously,

limm→∞

d(m) = limm→∞

m(1− (1− d)

1m

)

= limm→∞

1− (1− d)1m

1m

= limx→0+

1− (1− d)x

x


= − d

dx((1− d)x)

∣∣∣∣x=0

= − ln(1− d) = ln(1 + i)

[1, Exercise 37 p. 32] Use relationship19 (1.23) to give a third proof of the result thatδ′ = δ.

Solution: By the given equation,

d(m) =i(m)

1 + i(m)

m

→ δ

1 + 0= δ

as m →∞.

A.7.2 §1.10 VARYING INTEREST

Omit this section for the present.

A.7.3 §1.11 SUMMARY OF RESULTS

Table A.7.3, page 2033 is reproduced from the textbook [1, Table 1.1, p. 29].

19these notes, equation (56)


Rate of interest The accumulated value The present value of 1

or discount of 1 at time t = a(t) at time t =1

a(t)Compound interest

i (1 + i)t vt = (1 + i)−t

i(m)

(1 +

i(m)

m

)mt (1 +

i(m)

m

)−mt

d (1− d)−t (1− d)t

d(m)

(1− d(m)

m

)−mt (1− d(m)

m

)mt

δ eδt e−δt

Simple interest

i 1 + it (1 + it)−1

Simple discount

d (1− dt)−1 1− dt

Table 2: Summary of Relationships in Chapter 1

Textbook Chapter 2. Solution of problems in interest


Chapter 2 “discusses general principles to be followed in the solution of problems ininterest. The purpose of this chapter is to develop a systematic approach by which thebasic principles from Chapter 1 can be applied to more complex financial transactions.”

“Successive chapters have two main purposes:

1. “to familiarize the reader with more complex types of financial transactions, in-cluding definitions of terms, which occur in practice;

2. “to provide a systematic analysis of these financial transactions, which will oftenlead to a more efficient handling of the problem than resorting to basic principles.”

A.7.5 §2.2 OBTAINING NUMERICAL RESULTS

Numerical results will be obtained by

• direct calculation


• use of compound interest tables, as in [1, Appendix I]; in this case we may inter-polate between entries in the tables; more information will be provided — you arenot expected to have any prior knowledge about interpolation;

• direct calculation by hand, typically using series expansions. Students in MATH329 to not often have background from Calculus 3, where the computation usingseries is normally introduced. Thus any applications of this type will have to bejustified to students whose background cannot be assumed to include more thanCalculus 1 and Calculus 2. Examples of series of this type are

(1 + i)k = 1 + ki +k(k − 1)

2!i2 +

k(k − 1)(k − 2)

3!i3 + . . . (57)

ex = 1 + x +x2

2!+

x3

3!+ . . . (58)

ln(1 + x) = x− x2

2+

x3

3− . . . (59)

which are “MacLaurin” series (a special case of “Taylor” series), and have restric-tions on the numbers for which they are valid. (The first is always valid if |i| < 1,and also for any i if k is a non-negative integer; the second is valid for all x; andthe third is valid for −1 < x ≤ 2.) The error that occurs when one “truncates” aseries — i.e., when one stops adding at a particular term — can be estimated.

“Using series expansions for calculation purposes is cumbersome and should beunnecessary except in unusual circumstances.”

[1, Exercise 1, p. 53] “10000 is invested for 4 months at 12.6%, where interest is com-puted using a quadratic to approximate an exact calculation. Find the accumulatedvalue.”

Solution: The exact value using compound interest is

10000(1.126)412 = 10403.50 .

Alternatively, when we approximate by a quadratic, we truncate the MacLaurinseries after the 2nd degree term (cf. (57))

(1 + i)13 = 1 +

1

3· i +

1

3·(−2

3

)· 1

2!· i2 +

1

3·(−2

3

)·(−5

3

)· 1

3!· i3 + . . .

≈ 1 +1

3· i +

1

3·(−2

3

)· 1

2!· i2

= 1.040236


hence the accumulated value is approximately 10402.36.

Note that a linear approximation, which would be equivalent to simple interest,would give 10420: as the period of time is less than one unit, a linear approximationis higher than the value given by compound interest — equivalently, the line joiningthe points (x, y) = (0, (1.043)0) and (x, y) = (1, (1.043)1), passes over the graph ofy = (1.043)x.


A.8 Supplementary Notes for the Lecture of January 21st, 2005

Distribution Date: Friday, January 21st, 2005, subject to further revision

A.8.1 §2.2 OBTAINING NUMERICAL RESULTS (continued)

[1, Exercise 2, p. 53] “Find the present value of 5000, to be paid at the end of 25months, at a rate of discount of 8% convertible quarterly:

1. assuming compound discount throughout;

2. assuming simple discount during the final fractional period.”

Solution: The nominal discount rate compounded quarterly corresponds to aneffective discount rate of d(4)

4= 2% per quarter.

1. Under compound discount the present value of 5000 is (1−0.02)253 = 4225.27.

(Twenty-five months is 253

quarter-years.)

2. We discount through 8 full quarter-years by multiplying by (1 − 0.02)8: thisbrings the payment to one month from the present. For that month we areinstructed to discount by simple discount; as the time is one-third of a periodof 3 months, the discount factor is 1− 0.02

3= 0.9933333. We obtain a present

value of5000(0.98)8(0.9933333) = 4225.46.

[1, Exercise 3, §2.2, p. 53] “Find an expression for the fraction of a period at whichthe excess of present values computed at simple discount over compound discountis a maximum.”

Solution: Define f(x) = (1 − xd) − (1 − d)x, where 0 ≤ x ≤ 1. Then f ′(x) =−d − (1 − d)x · ln(1 − d), and f ′′(x) = −(1 − d)x · (ln(1 − d))2 < 0: the graph off is concave downward, and so the point where f ′(x) = 0, i.e. x = ln δ−ln d

δ, is the

global maximum point on the given interval. This result is analogous to that of [1,Exercise 44 §1.9, p. 33].

[1, Exercise 4, §2.2, p. 53] (not discussed in the lectures) “Method A assumes simpleinterest over final fractional periods, while Method B assumes simple discount overfinal fractional periods. The annual effective rate of interest is 20%. Find the ratioof the present value of a payment to be made in 1.5 years computed under MethodA to that computed under Method B.”

Solution: Under Method A, the present value of a principal of 1 after 1.5 years at20% effective interest is 1

(1.2)(1.1)= 1

1.32. Under Method B, the corresponding rate


of discount is 1− 11.2

= 16; the present value will be 1

(1− 12· 16)(1.2)

= 1112· 1

1.2. The ratio

of values under Method A to Method B is, therefore, 120121

.

A.8.2 §2.3 DETERMINING TIME PERIODS

This section is part of the course, but was not discussed in the lectures. Students areasked to read the text, and to remember the definitions. There is nothing particularlylogical about the definitions — they just have to be memorized. You should be able tosolve problems of the following types, for which (unchecked) solutions are included.

A number of schemes are in common use for calculating interest for fractions of aperiod:

• Under exact simple interest or “actual/actual” one counts the exact number ofdays, and assumes the year has 365 days.

• Under ordinary simple interest or “30/360” one assumes that each calendar monthhas 30 days, and the year has 360 days.

• Under Banker’s Rule or “actual/360” one uses the exact number of days but treatsa year as having 360 days. The treatment of February 29th in leap years (like thepresent) is not completely standardized. These calculation bases can be used foreither simple or compound interest.

Example A.6 Here is a statement I found last year at the bottom of a bank statementthat I received today from a large Canadian bank: “(name of bank) calculates interestdaily using a 365-day year, including leap years. The interest rate charged in a leap yearwill be equal to the Annual Interest Rate in effect on each day in that year multiplied by366 and divided by 365. Although this results in slightly more interest being charged,the effective annual rate is the same when rounded to the nearest 1/8th of 1%.”

[1, Exercise 5, p. 53] “If an investment was made on the date the United States en-tered World War II, i.e., December 7, 1941, and was terminated at the end of thewar on August 8, 1945, for how many days was the money invested:

1. “on the actual/actual basis?

2. on the 30/360 basis?”

Solution:

1. “From December 7, 1941 to December 7, 1944, 3 years passed, the last ofwhich was a leap year. The number of days is 3(365) + 1. From December 7,1944 to August 7, 1945, the numbers of days in the months December through



July were 31 + 31 + 28 + 31 + 30 + 31 + 30 + 31. Add one day, from 7th to8th August. The total is 1340.

2. From December 7, 1941 to December 7, 1944, 3 years passed, the last of whichwas a leap year. The number of days is 3(360). From December 7, 1944 toAugust 7, 1945, the numbers of days in the months December through Julywere 8(30). Add one day, from 7th to 8th August. The total is 1321.

[1, Exercise 6, p. 54] “A sum of 10,000 is invested for the months of July and Augustat 6% simple interest. Find the amount of interest earned:

1. “Assuming exact simple interest.

2. “Assuming ordinary simple interest.

3. “Assuming the Banker’s Rule.”

Solution:

1. The number of days is 31+31 = 62; exact simple interest is 621365·10000·(0.06) =

101.92.

2. Ordinary simple interest is 60360· 10000 · (0.06) = 100.

3. Interest under the Banker’s Rule is 62360· 10000 · (0.06) = 103.33.

[1, Exercise 7, p. 54] 1. “Show that the Banker’s Rule is always more favourableto the lender than is exact simple interest.

2. “Show that the Banker’s Rule is usually more favourable to the lender thanis ordinary simple interest.

3. “Find a counterexample to (the preceding) for which the opposite relationshipholds.”

Solution:

1. Under the Banker’s Rule the denominator of the fraction is decreased, so thefraction is increased.

2. Suppose that k months have passed. The Banker’s Rule counts the days as30k. How does this compare with the actual number of days? To answer thisquestion correctly, we need to compare all terms of lengths under a year. Butthe question is not precisely stated, and it’s not clear how we would have toweight the results. As the computations are extensive, we will not bother.

3. A loan between February 1 and March 1 provides a counterexample.


A.8.3 §2.4 THE BASIC PROBLEM

The author states that “...an interest problem involves four basic quantities:

1. the principal originally invested

2. the length of the investment period

3. the rate of interest

4. the accumulated value of the principal at the end of the investment period”,

and observes that 3 of these variables can be used to determine the 4th.Read the author’s comments on language. He observes that many problems have two

point of view — the borrower’s point of view, and the lender’s; you should be comfortablewith both, and with the terminology of both. He also observes that some of the acceptedterminology of the industry is ambiguous, or not intuitive, or even counter-intuitive. If astatement “doesn’t make sense” it could be that you are interpreting a definition literally,instead of using its definition, which may be misleading.

Read paragraph [1, d), p. 42], where the author discusses the reality that languagein the subject is not always completely unambiguous. As you work problems in thetextbook, and see solutions which will be given, you will gradually become familiar withthe usual conventions.

A.8.4 §2.5 EQUATIONS OF VALUE

We will often express the relationship between different sums in a problem in an “equationof value”, which is simply a statement obtained by bringing all amounts to a particulartime (by accumulating or discounting according to whatever accumulation functions areappropriate)”. Sometimes we will have an inequality rather than an equation. The timewhere the various amounts are compared is the comparison date. Under compound inter-est equations of value for the same payments, made a different comparison dates, shouldbe equivalent : you should be able to extract one from the other simply by multiplyingor dividing by the appropriate accumulation factors.

Time diagrams A time diagram is a one-dimensional diagram where the only variableis time, shown on a single coordinate axis. We may show above or below the coordinateof a point on the time-axis values of money intended to be associated with differentfunds. As in any mathematical exercises, the diagram is not a formal part of a solution,but may be very helpful in visualizing the solution. Some authors use variants of timediagrams where there may be several parallel horizontal axes, representing several funds;alternatively, you could use the usual type of graph that you have seen in calculus,


where the values of the fund will be shown as points in the plane, with the horizontalaxis representing time, and the second coordinate giving the function value. (The systemI am using for these notes accommodates figures with great difficulty; I will usually notattempt to show time diagrams for that reason.)

[1, Exercise 8, p. 55] “In return for payments of 2000 at the end of 4 years and 5000at the end of ten years, an investor agrees to pay 3000 immediately and to make anadditional payment at the end of three years. Find the amount of the additionalpayment if i(4) = 0.06.”

Solution: The equation of value at time t = 0 is

3000 + x

(1 +

0.06

4

)−4×3

= 2000

(1 +

0.06

4

)−4×4

+ 5000

(1 +

0.06

4

)−4×10

implying that

x =2000(1.015)−16 + 5000(1.015)−40 − 3000

(1.015)12

= 2000(1.015)−4 + 5000(1.015)−28 − 3000(1.015)12

= 1593.01 .



Distribution Date: Monday, January 24th, 2005(subject to revision)

Assignment 1 is due next Monday. No problems will be added to the list thatwas posted on the Web. A hard copy of the assignment should be distributed onWednesday, January 26th.

A.9.1 §2.5 EQUATIONS OF VALUE (continued)

[1, Exercise 9, p. 54] “At a certain interest rate the present values of the followingtwo payment patterns are equal:

(i) 200 at the end of 5 years plus 500 at the end of 10 years;

(ii) 400.94 at the end of 5 years.

At the same interest rate 100 invested now plus 120 invested at the end of 5 yearswill accumulate to P at the end of 10 years. Calculate P .

Solution: Payment schemes (i) and (ii), both with comparison date t = 0, give riseto the following equations of value:

200(1 + i)−5 + 500(1 + i)−10 = 400.94(1 + i)−5 (60)

100 + 120(1 + i)−5 = P (1 + i)−10 (61)

Equation (60) implies that

(1 + i)−5 = 0.40188 . (62)

Substituting in (61) yields

P =100

(0.40188)2+

120

0.40188= 917.76.

[1, Exercise 10, p. 54] “An investor makes three deposits into a fund, at the end of1, 3, and 5 years. The amount of the deposit at time t is 100(1.025)t. Find thesize of the fund at the end of 7 years, if the nominal rate of discount convertiblequarterly is 4

41.”

Solution: The nominal rate of discount of d(4) = 441

convertible quarterly produces

an annual accumulation factor of(1− 1

41

)−4= (1.025)4. At the end of 7 years the

3 payments accumulate to

100(1.025)1+(4×6) + 100(1.025)3+(4×6) + 100(1.025)5+(4×2) = 483.11 .


[1, Exercise 11, p. 54] “Whereas the choice of a comparison date has no effect onthe answer obtained with compound interest, the same cannot be said of simpleinterest. Find the amount to be paid at the end of 10 years which is equivalentto two payments of 100 each, the first to be paid immediately, and the second tobe paid at the end of 5 years. Assume 5% simple interest is earned from the dateeach payment is made, and us a comparison date of

1. The end of 10 years.

2. The end of 15 years.”

Solution:

1. With a comparison date at t = 10, the payment at the end of 10 years will be100(1 + 10× 0.05) + 100(1 + 5× 0.05) = 275.

2. With a comparison date at t = 15, the amount P that must be paid at theend of 10 years satisfies the equation of value

100(1 + 15× 0.05) + 100(1 + 10× 0.05) = P · (1 + 5× 0.05) ,

which implies that P = 260.

A.9.2 §2.6 UNKNOWN TIME

You may omit the discussion [1, pp. 45-46] of the “method of equated time”.Read the discussion of the Rule of 72 . The textbook shows that, at a given rate i of

compound interest, money will double in

n =ln 2

ln(1 + i)=

ln 2

i· i

ln(1 + i)

measurement periods. He then approximates the last ratio when i = 8%, so

time for money to double ≈ ln 2

i· 0.8

ln(1.08)≈ 0.72

i

and observes that the approximation is “surprisingly accurate over a wide range of in-terest rates.” To justify this rule we could investigate the behavior of i

ln(1+i), whose

MacLaurin expansion begins 1 + i2− i2

12+ . . ., but that investigation is beyond this

course.

[1, Exercise 12, p. 55] Solve the following equation for t:(

n∑r=1

sr

)vt =

n∑r=1

srvtr


Solution: Taking logarithms of both sides yields

ln

(n∑

r=1

sr

)+ t ln v = ln

(n∑

r=1

srvtr

).

Solving this equation for t yields

t =

ln

(n∑

r=1

srvtr

)− ln

(n∑

r=1

sr

)

ln v

=

ln

(n∑

r=1

srvtr

)− ln

(n∑

r=1

sr

)

−δ

[1, Exercise 13, p. 55] “Find how long 1000 should be left to accumulate at 6% effec-tive in order that it will amount to twice the accumulated value of another 1000deposited at the same time at 4% effective.”

Solution: Equation (??) implies that vt =

nPk=1

(skvtk)nP

k=1sk

, so

t =

ln

(n∑

k=1

(skvtk)

)− ln

(n∑

k=1

sk

)

ln v=

ln

(n∑

k=1

(skvtk)

)− ln

(n∑

k=1

sk

)

−δ.

If we let t denote the number of years, the equation of value at t = n is

1000(1.06)t = 2000(1.04)t

so

n =2

ln 106− ln 104= 36.39 .

[1, Exercise 14, p. 55] “The present value of two payments of 100 each to be made atthe end of n years and 2n years is 100. If i = 0.08, find n.”

Solution: Solving the equation of value, 100v2n + 100vn = 100, we obtain vn =−1±√5

2, in which only the + sign is acceptable, since vn > 0. Taking logarithms

gives

n =ln −1+

√5

2

− ln(1.08)= 6.2527 years.


[1, Exercise 16, p. 55] “You are asked to develop a rule of n to approximate how longit takes money to triple. Find n.”

Solution: On page 2042 of these notes we have considered the “Rule of 72.” Let usconsider the rationale of that rule:

(1 + i)n = 2

⇒ n =ln 2

ln(1 + i)

=ln 2

i· i

ln(1 + i)

=ln 2

i· i

i− i2

2+ i3

3− i4

4+ . . .

=ln 2

i· 1

1− i2

+ i2

3− i3

4+ . . .

=ln 2

i· 1

1− (i2− i2

3+ i3

4− . . .

)

=ln 2

i·(

1 +

(i

2− i2

3+

i3

4− . . .

)+

(i

2− i2

3+

i3

4− . . .

)2

+ . . .

)

=ln 2

i·(

1 +i

2− i2

12+ . . .

)

≈(ln 2)

(1 + i

2− i2

12+ . . .

)

i

The only change if we wish to obtain a rule for the time for money to triple is thatwe obtain

n ≈(ln 3)

(1 + i

2− i2

12+ . . .

)

i.

If we approximate i in the numerator by 8%, we obtain a numerator of approxi-mately 1.14; the rule could be called the “Rule of 114”.


A.10 Supplementary Notes for the Lectures of January 26th,2005

Distribution Date: Wednesday, January 26th, 2005(subject to revision)

(Some parts of these notes were not discussed in the lecture, which soonmoved from Chapter 2 to an introduction to Chapter 3.)

A.10.1 §2.7 UNKNOWN RATE OF INTEREST

The textbook describes 4 methods for determining an unknown rate of interest:

• Sometimes — particularly if only one payment is involved — it may be possible tosolve the equation of value by using the fact that the logarithm and exponentialfunctions are mutually inverse.

• Sometimes the equation may be solved by algebraic techniques, for example whenthe equation is equivalent to a polynomial equation that factorizes.

• Sometimes the equation can be solved by interpolation in interest tables.

• When all else fails, it may be necessary to solve by successive approximation oriteration. There are various methods, and some converge much faster than others.The justification of the better methods is beyond this course, but is the subjectmatter of courses like MATH 317 (Numerical Analysis).

[1, Exercise 19, p. 55] “Find the nominal rate of interest convertible semiannually atwhich the accumulated value of 1000 at the end of 15 years is 3000.”

Solution: The equation of value at time t = 15 is 1000(1 + i

2

)2(15)= 3000, implying

that i = 2(3

130 − 1

)= 7.4598394% .

[1, Exercise 20, p. 55] “Find an expression for the exact effective rate of interest atwhich payments of 300 at the present, 200 at the end of one year, and 100 at theend of two years will accumulate to 700 at the end of two years.”

Solution: For the unknown interest rate i, which we assume to be non-negative,the equation of value at time 2 is

300(1 + i)2 + 200(1 + i)1 + 100 = 700 ,

yielding a quadratic equation in 1 + i:

3(1 + i)2 + 2(1 + i)− 6 = 0 ,

whose only positive solution is 1 + i = −1+√

193

= 1.119632981, so i = 11.9632981%.


[1, Exercise 21, p. 55] “It is known that an investment of 1000 will accumulate to1825 at the end of 10 years. If it is assumed that the investment earns simpleinterest at rate i during the 1st year, 2i during the second year, . . . , 10i during the10th year, find i.”

Solution: A misreading of the information given was interpreted as stating that

(1 + i)(1 + 2i)...(1 + 10i) = 1.825 .

This is a polynomial equation of degree 10. We shall apply successive approxima-tions. We begin by ignoring powers greater than the 1st, and approximating

1 + i

10∑

k=1

k ≈ 1.825 ,

yielding a first approximation i0 = 0.82555

= 0.015. Let’s define f(i) = (1 + i)(1 +2i)...(1+10i)−1.825. Then f(i0) = 0.37. We compute f(0.016) = 0.48, f(0.014) =0.26, f(0.013) = 0.16, f(0.012) = 0.06, f(0.011) = −0.03. This suggests try-ing f(0.0113) = −0.006, f(0.0114) = 0.003, f(0.01136) = −0.00044, f(0.01137) =0.00048, f(0.011365) = 0.000024, f(0.011364) = −0.000068, f(0.0113647) = −0.000000589,f(0.01136473) = −7.97× 10−7, etc.

But this was not the intended interpretation. In fact, we have read the problemas though interest was compounded at varying rates, which was not intended. Theintended interpretation was that the equation of value is

1 + i + 2i + 3i + ... + 10i = 1.825 ,

which implies that i = 0.15 exactly .

[1, Exercise 23, p. 55] “The sum of the accumulated value of 1 at the end of threeyears at a certain effective rate of interest, and the present value of 1 to be paidat the end of three years at an effective rate of discount numerically equal to i is2.0096. Find the rate.”

Solution: The equation of value at time t = 3 of a payable at time 0 and 1 payableat time 6 is (1 + i)3 + (1 − i)3 = 2.0096, which implies that i = 0.04 (the onlypositive solution).

A.10.2 §2.8 PRACTICAL EXAMPLES

[1, Exercise 25, p. 56] “A bill for 100 is purchased for 96 three months before it isdue. Find


1. “The nominal rate of discount convertible quarterly earned by the purchaser.

2. “The annual effective rate of interest earned by the purchaser.”

Solution:

1. 100(1− d(4)

4

)= 96 ⇒ d(4) = 16%.

2. 96(1 + i)14 = 100 ⇒ i =

(10096

)4 − 1 = 17.7375701%.

[1, Exercise 26 p. 56] “A two-year certificate of deposit pays an annual effective rateof 9%. The purchaser is offered two options for prepayment penalties in the eventof early withdrawal:

A - a reduction in the rate of interest to 7%

B - loss of three months’ interest.

In order to assist the purchaser in deciding which option to select, compute theratio of the proceeds under Option A to those under Option B if the certificate ofdeposit is surrendered:

1. at the end of 6 months

2. at the end of 18 months”

Solution:

1. Under Option A the effective annual interest rate is now 7%, so a principal of 1returns (1.07)

12 . Under Option B a principal of 1 returns (1.09)

12− 1

4 . The ratio

of returns under Option A to those under option B is (1.07)12

(1.09)12−

14

= 1.012360669.

2. Under Option A the effective annual interest rate is now 7%, so a principal of 1returns (1.07)

32 . Under Option B a principal of 1 returns (1.09)

32− 1

4 . The ratio

of returns under Option A to those under option B is (1.07)32

(1.09)32−

14

= .9937852442.

[1, Exercise 27, p. 56] “A savings and loan association pays 7% effective on depositsat the end of each year. At the end of every 3 years a 2% bonus is paid on thebalance at that time. Find the effective rate of interest earned by an investor ifthe money is left on deposit

1. “Two years.

2. “Three years.

3. “Four years.”


Solution: Let i denote the effective rate of interest in each case.

1. There being no bonus, i = 7%.

2. Here (1 + i)3 = (1.07)3(1.02), so i = (1.07)(1.02)13 − 1 = 7.7086300%.

3. This time i = (1.07)(1.02)14 − 1 = 7.5410377%.

[1, Exercise 28 p. 56] “A bank offers the following certificates of deposit:

Nominal annual interest rateTerm in years (convertible semiannually)

1 5%2 6%3 7%4 8%

The bank does not permit early withdrawal. The certificates mature at the end ofthe term. During the next six years the bank will continue to offer these certificatesof deposit. An investor deposits 1000 in the bank. Calculate the maximum amountthat can be withdrawn at the end of six years.”

Solution: It is instructive to begin first with an interpretation that resulted froma misreading of the problem. In this misreading, I overlooked the fact that therates were convertible semi-annually. the terms are correct, but, where I usedfactors 1.05, 1.06, 1.07, 1.08, I should have used (1.025)2, (1.03)2, (1.035)2, and(1.04)2 respectively; but fractional years were still not permitted. “We considerthe partitions of 6 into sums of integers: 6 = 4+2 = 4+1+1 = 3+3 = 3+2+1 =3 + 1 + 1 + 1 = 2 + 2 + 2 = 2 + 2 + 1 + 1 = 2 + 1 + 1 + 1 + 1 = 1 + 1 + 1 +1 + 1 + 1. Because the rates increase for periods of increased length, there is noadvantage to partitioning a summand further; so we need not consider any partitioncontaining 1 + 1, since 2 will always be better, etc. This leaves the partitions6 = 4 + 2 = 3 + 3 = 3 + 2 + 1 to consider. These three partitions correspond toaccumulations of 1000(1.08)4(1.06)2, 1000 ((1.07)3)

2, and 1000(1.07)3(1.06)2(1.05)1,

which are respectively equal to 1528.645395, 1500.730352, and 1445.281231, so thebest partition is into 4 + 2 years, and the best return is 1528.645395.”

Now I repeat the computations, this time correctly. I must compare 1000(1.04)8(1.03)4,1000 ((1.035)6)

2, and 1000(1.035)6(1.03)4(1.025)2, which are respectively equal to

1540.336523, 1511.068657, 1453.579295; thus the best partition is again into 4 + 2years, and the best return is 1540.336523.


A.10.3 §2.9 MISCELLANEOUS PROBLEMS

[1, Exercise 29, p. 56] “A manufacturer sells a product to a retailer who has the op-tion of paying 30% below the retail price immediately, or 25% below the retail pricein six months. Find the annual effective rate of interest at which the retailer wouldbe indifferent between the two options.

Solution: We solve the equation of value, 70% = 75% · (1 + i)12 . i =

(7570

)2 − 1 =14.7959183%.

[1, Exercise 30, p. 56] “If an investment will be doubled in 8 years at a force of interestδ, in how many years will an investment be tripled at a nominal rate of interestnumerically equal to δ and convertible once every three years?”

Solution: The given information implies that(eδ

)= 2, so δ = ln 2

8. We need

to determine the number of years n with the property that (1 + 3δ)n3 = 3, so

n = 3 ln 3ln(1+3δ)

= 14.26421450 years.

[1, Exercise 31, p. 56] “Fund A accumulates at 6% effective, and Fund B accumulatesat 8% effective. At the end of 20 years the total of the two funds is 2000. At theend of 10 years the amount in Fund A is half that in Fund B. What is the total ofthe two funds at the end of 5 years.?”

Solution: Denote the principals in funds A and B by A and B respectively. Thenwe know that

A(1.06)20 + B(1.08)20 = 2000

A(1.06)10 =1

2·B(1.08)10

and wish to determine A(1.06)5 + B(1.08)5. Solving the equations yields

A =2000(1.06)−10

(1.06)10 + 2(1.08)10= 182.8195812

B =2000(1.08)−10

12(1.06)10 + (1.08)10

= 303.3009728 .

Hence

A(1.06)5 + B(1.06)5 =2000(1.06)−5

(1.06)10 + 2(1.08)10+

2000(1.08)−5

12(1.06)10 + (1.08)10

= 182.8195812(1.06)5 + 303.3009728(1.08)5 = 690.3024748.


[1, Exercise 32, p. 57] “An investor deposits 10,000 in a bank. During the first yearthe bank credits an annual effective rate of interest i. During the second year thebank credits an annual effective interest i − 0.05%. At the end of two years theaccount balance is 12,093.75. What would the account balance have been at theend of three years if the annual effective rate of interest were i + 0.09 for each ofthe three years?”

Solution: The equation of value is

10000(1 + i)(1 + i− 0.05) = 12093.75 ,

which we interpret as a quadratic equation in 1 + i:

(1 + i)2 − 0.05(1 + i)− 1.209375 = 0

whose only positive solution is

1 + i =0.05 +

√(0.05)2 + 4(1.209375)

2= 1.125000000

from which we conclude that i = 12.5%, and that the account balance after 3 yearswould be 10000(1.125 + 0.09)3 = 10000(1.215)3 = 17936.13.

[1, Exercise 33, p. 57] “A signs a one-year note for 1000 and receives 920 from thebank. At the end of six months, A makes a payment of 288. Assuming simplediscount, to what amount does this reduce the face amount of the note?”

Solution: The rate of simple discount is 801000

= 8%. At the due date, 1 year fromnow, the 288 accumulates to 288

1−0.04= 300, which is the reduction in the face value

of the note, reducing the value to 700.

Textbook Chapter 3. Basic Annuities


Definition A.14 1. An annuity is a series of payments, usually made at equal inter-vals of time.

2. The total period during which payments will be made is called the term.

3. The interval between payments is called the payment period ; the name “annuity”suggests a default period of 1 year, but, in some specific applications, the defaultperiod may have some other length; there can, in fact, be two distinct periodsinvolved — one associated with the interest calculations, and one associated withthe payments under the annuity.


4. An annuity-certain consists of payments that are all certain to be made. Otherwisean annuity is a contingent annuity. An annuity whose payments are contingent onwhether a given individual is alive is a life annuity ; the study of such annuitiesis concerned with “life contingencies”, and is outside of this course; following thetextbook, we will usually suppress the suffix -certain.


A.11 Supplementary Notes for the Lecture of January 28rd,2005

Distribution Date: Friday, January 28rd, 2005, subject to further revision

A.11.1 §3.2 ANNUITY-IMMEDIATE

In an annuity-immediate the payments are made at the end of each period. The presentvalue of an annuity-certain for n periods is denoted by an , or, if the interest rate needsto be stated explicitly an i . The author uses a diagram to denote annuities, with 2arrows, one sometimes labelled t1 and the other sometimes labelled t2. The first showsthe beginning of the first period, at the end of which a payment is due under the annuity-certain. The second arrow, labelled t2, indicates the last payment date — more precisely,just after the payment has been made. I shall usually not include diagrams in these notes,as they are very time-consuming to arrange using the text-preparation software that Iam using.

The accumulated value of the payments as of time t2 is denoted by sn , or, if theinterest rate needs to be stated explicitly sn i . We can prove the following basic formulæ:

an = v + v2 + . . . + vn−1 + vn

= v · 1− vn

1− vwhen i 6= 0

=1− vn

i(63)

sn = 1 + (1 + i) + (1 + i)2 + . . . + (1 + i)n−1

=(1 + i)n − 1

(1 + i)− 1when i 6= 0

=(1 + i)n − 1

i(64)

Note that we have assumed — as is usually the case — that i 6= 0.20 We can also provethe following identities, both algebraically and verbally:

1 = ian + vn (65)

sn = an · (1 + i)n (66)1

sn

=1

an

+ i (67)

Remember to sketch a time diagram as you read each of the following problems.(Not all of these problems was discussed in the lecture.)

20When i = 0 an = n = sn.


[1, Exercise 1, p. 88] “A family wishes to accumulate 50,000 in a college educationfund (by) the end of 20 years. If they deposit 1000 into the fund at the end of eachof the first 10 years, and 1000 + X at the end of each of the second 10 years, findX to the nearest unit if the fund earns 7% effective.”

Solution: The equation of value at time t2 = 20 is

1000s20 + X · s10 = 50000 ,

which we may solve to yield

X =50000

s10 0.07

− 1000s20 0.07

s10 0.07

=50000

s10 0.07

− 1000(1.07)20 − 1

(1.07)10 − 1

= 50000s−1

10 0.07− 1000((1.07)10 + 1)

= 50000(0.072378)− 1000(2.96715) from the tables

= 651.75.

A more precise computation would yield 651.724; the difference is due to roundingerrors and/or the limited precision of the tables.

[1, Exercise 2, p. 88] “The cash price of a new automobile is 10,000. The purchaseris willing to finance the car at 18% convertible monthly and to make payments of250 at the end of each month for 4 years. Find the down payment which will benecessary.”

Solution: Let X be the down payment. Then the equation of value at time t = 0is X + 250a48 1.5% = 10000, from which we determine the down payment, X =1489.361590. (Here again, the answer obtained using the tables in the text-bookis slightly incorrect, at 1489.35.)

[1, Exercise 3, p. 88] “An annuity provides a payment of n at the end of each yearfor n years. The annual effective interest rate is 1

n. What is the present value of

the annuity?”

Solution: n · an 1n

= n · 1−(1+ 1n)−n

1n

= n2(1− (

nn+1

)n).

[1, Exercise 4, p. 88] “If an = x and a2n = y, express d as a function of x and y.”

Solution:an = x ⇒ 1− ix = vn ,


a2n = y ⇒ 1− iy = (vn)2 .

Eliminating vn between the equations yields i = 2x−yx2 . Hence

d = iv =i

1 + i=

2x− y

x2 + 2x− y.

[1, Exercise 5, p. 88] 1. Show that

am+n = am + vman = vnam + an

2. Show thatsm+n = sm + (1 + i)msn = (1 + i)nsm + sn

3. Interpret the results in (a) and (b) verbally.

Solution:

1. (supply an algebraic proof)

2. (supply an algebraic proof)

3. (a) The present value of the first m payments of an (m + n)-year annuity-immediate of 1 is am . The remaining n payments have value an at timet = m; discounted to the present, these are worth vman at time t = 0.Since m + n = n + m, the same argument shows that am+n = vnam + an .

(b) We can obtain similar results for the s functions by analogous arguments;or by multiplying both sides of the previous equations by (1 + i)m+n.

[1, Exercise 7, p. 88] “You are given the following annuity values. Find i:

a7 i = 5.153 a11 i = 7.036 a18 i = 9.180.”

Solution: Let’s first prove a useful identity:

am+n = am + an − iaman (68)

This identity follows from

am+n =1− vm · vn

i=

1− (1− iam)(1− ian)

i.

For a verbal explanation, consider an annuity-immediate of 1 per year for m + nyears, the first payment at the end of year #1. The present value of the first mpayments is am. The value of the remaining payments at time t = m is vman, which


may be interpreted as the value of an n-year annuity-immediate whose paymentsare each vm. The proof may be completed by replacing vm by 1 − iam, for whicha verbal proof is given in [1, p. 60]

The identity implies that

i =am + an − am+n

am · an

.

With m = 7 and n = 11, we obtain

i =5.153 + 7.036− 9.180

(5.153)(7.036)= .08299199691

or approximately 8.3%.

Note that we didn’t need three equations to find i — we could have determined itfrom any one of the equations. For example, since a7 i is a continuous, decreasingfunction of i, and since it is 7 when i = 0 and it approaches 0 as i becomes large,the function must take on all intermediate values, so a7 i−5.153 takes on the value0 for just one positive number, which may be found by Successive Bisection [1, p.399].

[1, Exercise 8, p. 88] “Show that

1

1− v10=

1

s10

(s10 +

1

i

)”

Solution:

1

1− vn=

(1 + i)n

(1 + i)n − 1

=((1 + i)n − 1) + 1

(1 + i)n − 1

=i

(1 + i)n − 1· ((1 + i)n − 1) + 1

i

=1i

(1+i)n−1

·(

(1 + i)n − 1

i+

1

i

)

=1

sn

(sn +

1

i

)

¤


A.12 Supplementary Notes for the Lecture of January 31st,2005

Distribution Date: Monday, January 31st, 2005, subject to further revision

A.12.1 §3.3 ANNUITY-DUE

In an annuity-due the payments are made at the beginning of each period. The presentvalue of an annuity-certain for n periods is denoted by an , or, if the interest rate needsto be stated explicitly an i . Again the author uses a diagram to denote annuities, with2 arrows, one labelled t1 and the other labelled t2. The first shows the beginning of thefirst period, just before a payment is due under the annuity-due. The second arrow,labelled t2, indicates the end of the last period at the beginning of which a payment wasmade. The accumulated value of the payments as of time t2 is denoted by sn , or, if theinterest rate needs to be stated explicitly sn i . We can prove the following basic formulæ:

an = 1 + v + v2 + . . . + vn−2 + vn−1

=1− vn

1− vwhen i 6= 0

=1− vn

d(69)

sn = (1 + i) + (1 + i)2 + . . . + (1 + i)n

=(1 + i)n − 1

dwhen i 6= 0 . (70)

Note that we have assumed — as is usually the case — that i 6= 0. We can also provethe following identities, both algebraically and verbally:

an = (1 + i)an (71)

sn = (1 + i)sn (72)

an = 1 + an−1 (73)

sn = −1 + sn+1 (74)

sn = (1 + i)nan (75)

There will not be time to discuss all of the following problems in the lecture:

[1, Exercise 9, p. 89] “Find the present value of payments of 200 every six monthsstarting immediately and continuing through four years from the present, and 100every six months thereafter through ten years from the present, if i(2) = 0.06.”

Solution: The reader is likely to make assumptions in a casual reading that werenot intended by the author. Reading carefully, one might see that he intends that,


though the payments start immediately, they continue to the end of 4 years fromnow — that is, that there be 9 payments of 200, not 8. Similarly, it is the paymentsthat end after 10 years, not the years for which they are prepaid — so there arepayments over 21 half-years in all. With this interpretation, at the effective rateof 1

2i(2) = 3% per half-year,

Present value = 100a21 3% + 100a9 3%

= 100(1.03)(100a21 3% + 100a9 3%

)

= 2389.716705

[1, Exercise 10, p. 89] “A worker aged 40 wishes to accumulate a fund for retirementby depositing 1000 at the beginning of each year for 25 years. Starting at age 65the worker plans to make 15 annual withdrawals at the beginning of each year.Assuming that all payments are certain to be made, find the amount of each with-drawal starting at age 65 to the nearest dollar, if the effective rate of interest is 8%during the first 25 years, but only 7% thereafter.”

Solution: Let the constant amount of the withdrawals beginning at age 65 be X.The equation of value at age 65 is

1000 · s25 8% = X · a15 7%

⇒ X =1000 · s25 8%

a15 7%

= 1000 · 1.08

1.07· 0.07

0.08· (1.08)25 − 1

1− (1.07)−15

= 8101.654558

so, to the nearest dollar, the annual withdrawals will be 8102.

[1, Exercise 11, p. 89] “Find a8 if the effective rate of discount is 10%.”

Solution: Since (1− d)(1 + i) = 1, v = 1− d = 0.9 when d = 0.1.

a8 =1− (0.9)8

0.1= 5.695327900.


A.13 Supplementary Notes for the Lecture of February 2nd,2005

Distribution Date: Wednesday, February 2nd, 2005, subject to further revision

A.13.1 §3.3 ANNUITY-DUE (continued)

[1, Exercise 12, p. 89] Prove that

1

an

=1

sn

+ d .

Solution: I suspect that the textbook was looking only for an algebraic proof. Thatis straightforward, and I leave that to the student.

Here is a verbal proof: We can consider two different payment schemes that areworth 1 today. One of these is an annuity-due of n payments, each in the amountof 1

an. The second scheme is more complicated. Consider an annuity-immediate of

n payments, each in the amount of 1sn

. These payments will be worth 1 at the end

of n years — not today. But we can account for the n years’ delay in the repaymentif we pay for the use of that 1 by an annual payment i of interest at the end of theyear; or, alternatively, by the annual prepayment of d of discount at the beginningof each year. It is the latter that we propose to combine with the n payments of1

sn. Thus the annual payments are to be the sum

1

sn

+ d

and this annual payment must, therefore, be equal to the annual level payment 1an

.

[1, Exercise 13, p. 89]] (not worked in the lecture) “If ap = x and sq = y, show that

ap+q =vx + y

1 + iy.”

Solution:

ap = x ⇒ 1− vp

d= x ⇒ vp = 1− dx

sq = y ⇒ (1 + i)q − 1

i= y ⇒ vq =

1

1 + iy

⇒ vp+q =1− dx

1 + iy


⇒ 1− vp+q =iy + dx

1 + iy=

iy + ivx

1 + iy

⇒ ap+q =y + vx

1 + iy¤

A.13.2 §3.4 ANNUITY VALUES ON ANY DATE

Three cases are considered: the date is always an integral number of periods from eachpayment date.

1. present values more than one period before the first payment date

2. accumulated values more than one period after the last payment date;

3. current values at a date strictly between the first and last payment dates.

Present values more than one period before the first payment date An annuityis said to be deferred by m time units if the first payment is m time units later thanthe type of annuity in question would normally be paid, and the subsequent paymentsoccur at the expected intervals. The notation of the earlier edition of the textbook was“standard”: the value of an annuity that is deferred through k periods is denoted by theearlier symbol, prefixed by k|. Thus k |an is the present value of an annuity-immediatewhose first payment has been deferred through k time periods; analogous symbols canbe used for an. It can be seen that

k |an = vkan

= an+k − ak ,

k |an = vkan

= an+k − ak .

Accumulated values more than 1 period after the last payment date Herewe can express the current values by multiplying by the appropriate power of 1 + i; or,alternatively, by taking the difference of two annuity values.

Summary You are urged to follow the textbook’s suggestion, “The reader should nottry to work problems by memorizing formulas...” Remember the reasoning that was usedto derive the formulæ, and apply that reasoning “from first principles” in each case.


Some exercises from the textbook The textbook asks you to prove several formulæ.There are two levels at which such exercises should be approached:

• an algebraic proof

• a verbal justification

Usually an algebraic proof should not be difficult; I will not normally include proofs inthese notes, but you can see me if you have difficulty working through a proof. Theissue is not to find an elegant proof — just to show that the two sides of the equationare equal. As for a verbal proof, that will be much harder, and I will spend increasingamounts of time at the lectures discussing problems of this type.

[1, Exercise 17, p. 89] “Payments of 100 per quarter are made from June 7, yearZ through December 7, year Z + 11, inclusive. If the nominal rate of interest,convertible quarterly, is 6%:

1. find the present value on September 7, year Z − 1;

2. find the current value on March 7, year Z + 8;

3. find the accumulated value on June 7, year Z + 12”.

Solution:

1. As of September 7, year Z−1, no payments have yet been made. The presentvalue is21

(1.015)−2100a44+3 1.5% = 3256.879998 .

2. As of March 7, year Z + 8, the value as of March 7, year Z will accumulateby a factor (1.015)(8×4), for an accumulated value of 5403.152103.

3. As of June 7, year Z + 12, the originally computed value will accumulate bya factor (1.015)(13×4)−1, for an accumulated value of 6959.369761.

The next 3 problems were not discussed in the lecture:

[1, Exercise 18, p. 89] To prove that

15∑t=10

(st − st

)= s16 − s10 − 6 .

21Why is the exponent −2 and not −3 even though the the evaluation is being made 9 months beforethe first payment? Because I am viewing the payments as an annuity-immediate: the clock startsticking one period before the first payment. And why have 3 periods been added, even though Juneand December are only 2 quarter-years apart? Again because the first payment is associated with the3-month period ending with the payment, so the actual length of the annuity period is 3 months longer.



Solution: Here is a verbal proof: st − st can be interpreted as the excess over 1 ofthe value after t + 1 years of a payment of 1 at time t = 0, which is (1 + i)t+1 − 1;for the present purposes, let us interpret this as the excess over 1 of the present

value of a payment t + 1 years ago. The sum15∑

t=10

is, therefore, the excess over 6 of

the value accumulated to the present of an annuity of 6 annual payments of 1, thefirst 16 years ago, the last 11 years ago; hence the sum is equal to

s16 − s10 − 6 .

[1, Exercise 19, p. 89] “Annuities X and Y provide the following payments:

End of Year Annuity X Annuity Y1–10 1 K11–20 2 021–30 1 K

“Annuities X and Y have equal present values at an annual effective interest ratei such that v10 = 1

2. Determine K.”

Solution: As of today, the present values of annuities X and Y are respectively1 ·a30 +1

(a20 − a10

)and K ·a30−K

(a20 − a10

). Setting these amounts equal and

solving, we obtain

K =a30 + a20 − a10

a30 − a20 + a10

=1− v30 − v20 + v10

1− v30 + v20 − v10

=1− 1

8− 1

4+ 1

2

1− 18

+ 14− 1

2

=9

5= 1.8.

[1, Exercise 20, p. 90] “At an annual effective interest rate i it is known that

1. The present value of 2 at the end of each year for 2n years, plus an additional1 at the end of each of the first n years, is 36.

2. The present value of an n-year deferred annuity-immediate paying 2 per yearfor n years is 6.

“Find i.”

Solution: It is convenient, because of a situation that will develop later in theproof, to distinguish two cases.


Case i 6= 0: From hypotheses 1 we have an equation of value

36 = 2 · a2n + 1 · an ; (76)

from (ii) we have the equation of value

6 = vn · 2 · an = 2(a2n − an

). (77)

Solving these equations, we obtain a2n = 13, an = 10, implying that

1− v2n

1− vn=

1.3

1

⇒ (vn)2 − 1.3vn + 0.3 = 0

⇒ vn =1.3± 0.7

2= 0.3 or 1 .

The value vn = 1 corresponds to i = 0, which we shall treat in the next case.Since vn = 0.3, we can substitute in the second equation of value:

6 = (0.3) · 2 · 1− 0.3

i

and solve for i, obtaining i = 7%.

Case i = 0: Here Equations (76) and (77) become

36 = 2(2n) + n

6 = 2n = 2(2n− n)

which are inconsistent. Thus this case is impossible.


A.14 Supplementary Notes for the Lecture of February 4th,2005

Distribution Date: Friday, February 4th, 2005, subject to further revision

A.14.1 §3.4 ANNUITY VALUES ON ANY DATE (continued)

[1, Exercise 21, p. 90] “It is known that

a7

a11

=a3 + sx

ay + sz

. (78)

Find x, y, and z.”

Solution: Intuitively we usually expect that the determination of 3 variables re-quires 3 constraints. While this is not always the case, the fact that only oneequation has been presented here should ring an alarm bell. The variable i hasnot been mentioned, and we might also wish to know whether the solution we areasked to find should be dependent on i.22 It appears from the textbook that thevalues given above are the only solution the textbook was seeking — but thereexist others!.

At time t = 0 the present values of annuities-due of 1 per year for respectively 7

and 11 years have value in the ratioa7

a11

. Both of these values should increase by

a factor of 1 + i per year under compound interest. Hence, 4 years later, the ratio

will not have changed. But, at that time, it can be interpreted asa3 + s4

a7 + s4

. Thus

one solution to the problem is

(x, y, z) = (4, 7, 4), (79)

and this solution is valid for all i. I would not expect students to be able to generatethe rest of this solution!

Having found one solution, which we can see from the solutions [1, p. 418] tobe the solution the author is seeking, we might be expected to stop. But, to amathematician, an instruction like “Find x, y, and z” means, implicitly,

“Find all possible sets of values for x, y, and z.”

22This can explain the apparent paucity of equations: to assume that the solution holds for all i isequivalent to assuming an equation for every value of i — infinitely many equations, for the 3 unknownswe are trying to determine. In such a situation we should not be surprised if there is no solution at all.


So let’s investigate whether we have all solutions. This raises new issues. Thesolution we found above in (79) is valid for all i. We have two subquestions:

• Could there be other solutions for all i?

• Could there be solutions that hold for specific values of i, but not for all i?

Could there be other solutions for all i? When i = 0, equation

a7

a11

=a3 + sx

ay + sz

(80)

becomes7

11=

3 + x

y + z,

implying that11x− 7y − 7z = 33 . (81)

As i →∞, v → 0, and the left side of equation (80) approaches

limi→∞

1− (1 + i)−7

1− (1 + i)−11= 1 ;

the limit of the right side will be 1 only if

x = z . (82)

Then (81) becomes4x− 7y = −33 . (83)

Our earlier solution (79) satisfies this last “diophantine” equation; some othersolutions are

(x, y, z) = (11, 11, 11), (18, 15, 18), (−3, 3,−3) .

If we define

w =y − 3

4=

x + 3

7,

the conditions we have determined on x, y, z have the general solution

(x, y, z) = (7w − 3, 4w + 3, 7w − 3)

and the case (4, 7, 4) corresponds to w = 1. What about the other solutions? Allwe have shown is that, if there are any solutions valid for all i, then they have to be


of the preceding form. But, when we substitute these general values into equation(78), we may reduce the equation to the following condition on w:

1− v7

1− v11=

1− v7w

1− v11w.

All values of w other than w = 1 produce a polynomial equation which constrainsv, so the equality will not hold for all interest rates. Thus solution (79) is the onlysolution valid for all i.

You are not expected to be able to reproduce this detailed argument.

[1, Exercise 22, p. 90] “Simplify a15 (1 + v15 + v30) to one symbol.”

Solution: Here is an algebraic solution:

a15

(1 + v15 + v30

)=

1− v15

i· (1 + v15 + v30

)

=1− v45

i= a45

A verbal proof could be based on the equation

a15

(1 + v15 + v30

)= a15 + v15 · a15 + v30 · a15

= a15 + 15

∣∣a15 + 30

∣∣a15

In the last member the first summand is the present value of an annuity of 15annual payments of 1, the first one year from now and the last 15 years from now;the second summand is the present value of a sequence of 15 payments which beginone year after the last represented by the first summand and the last being paid30 years from now; and the last summand is the present value of a final subseriesof 15 payments of 1, the first to be paid one year after the last of the 30 paymentsmentioned above, and the last to be paid 45 years from now. We know that thepresent value of such a series of 45 payments of 1 is a45.

[1, Exercise 23, p. 90] (not discussed in the lecture) “Find the present value to thenearest dollar on January 1 of an annuity which pays 2000 every six months forfive years. The first payment is due on the next April 1, and the rate of interest is9% convertible semiannually.”

Solution: On April 1st the annuity is worth 2000a10 4.5%

. Discounting back toJanuary 1st, i.e. through half of a half-year period, we obtain a value of

(1.045)−12 2000a10 4.5% = (1.045)

12 2000a10 4.5%

= (1.045)12 · 2000 · 1− (1.045)−10

0.045= 16177.59053

or 16178 to the nearest dollar.


A.14.2 §3.5 PERPETUITIES

The symbol a∞ is used for the present value of a perpetual annuity — an infinite sequenceof payments, the first one period from now; where the interest rate is not clear from thecontext, we may write a∞ i. When the interest rate i is 0 the present value would beinfinite. Otherwise we sum an infinite geometric series, obtaining

v + v2 + v3 + . . . =v

1− v=

v

iv=

1

i.

It can be seen that this is the limit of the usual formula for an as n → ∞. An infiniteannuity of this type is called a perpetuity-immediate, or usually just a perpetuity .

Analogously we may define a perpetuity-due to be an infinite sequence of equal pay-ments, where the first is made immediately. We use the symbol a∞ , and can prove thatits present value is 1

d. While perpetuities do exist in the real world, for example in the

bond market, they are also useful in providing verbal explanations of identities. Forexample, we can explain the formula an = 1−vn

iby expressing the quotient on the right

as a difference, 1i− vn 1

iand interpreting the n-payment annuity as the difference of 2

perpetuities, one beginning a year from now, and the other n + 1 years from now; i.e.,

an = a∞ − (n|a∞)



Distribution Date: Monday, February 7th, 2005, subject to further revision

A.15.1 §3.5 PERPETUITIES (continued)

[1, Exercise 24, p. 90] “A sum P is used to buy a deferred perpetuity-due of 1 payableannually. The annual effective rate of interest is i > 0. Find an expression for thedeferred period.”

Solution: Denote the deferral period by n. Then P = vn 1d⇒ n = 1 + ln P+ln i

ln v=

1− ln P+ln iln δ

= 1− ln Piln δ

.

[1, Exercise 25, p. 90] “Deposits of 1000 are placed into a fund at the beginning ofeach year for the next 20 years. After 30 years, annual payments commence, andcontinue forever, with the first payment at the end of the 30th year. Find anexpression for the amount of each payment.”

Solution: The deposits accumulate to a fund worth 1000s20 a year after the lastpayment, and 1000(1 + i)10s20 10 years after the last payment. A perpetuity-duebought with this amount will have annual payments of

1000s20(1 + i)10d = 1000(1 + i)10 · (1 + i)21 − (1 + i)

i· iv

= 1000((1 + i)30 − (1 + i)10

).

Alternatively, we could set up an equation of value at any other time, for exampleat time t = 0. The value of the deposits will be 1000a20. Suppose that the levelamount of the withdrawals is X. These withdrawals may be interpreted as eitheran perpetuity-due for which the clock starts at the time of the first payment, 30years from now; or as a perpetuity-immediate for which the clock starts 29 yearsfrom now in order that the first payment occur at the end of that year. This leadsto equations, either

1000a20 = Xv30a∞

or1000a20 = Xv29a∞

which may be solved for the same value of X as determined earlier.

[1, Exercise 26, p. 90] (not discussed in detail in the lecture) “A benefactor leaves aninheritance to 4 charities, A, B, C, and D. The total inheritance is a series of levelpayments at the end of each year forever. During the first n years, A, B, and C


share each payment equally. All payments after n years revert to D. If the presentvalues of the shares of A, B, C, and D are all equal, find (1 + i)n.”

Solution: Imposing the condition that the sum of the first n payments is equal to 3times the present value of payments ##n + 1, n + 2, . . ., we obtain an = 3vna∞ ⇒vn = 1

4⇒ (1 + i)n = 4.

[1, Exercise 27, p. 90] “A level perpetuity-immediate is to be shared by A, B, C,and D. A receives the first n payments, B the second n payments, C the third npayments, and D the payments thereafter. It is known that the ratio of the presentvalue of C’s share to A’s share is 0.49. Find the ratio of the present value of B’sshare to D’s share.”

Solution: The present values of the shares of A, B, C, D are, respectively, an, vnan,v2nan, and v3na∞. The fact that C’s share, divided by A’s share is to equal 0.49implies that vn = 0.7. The ratio of B’s share to D’s is then seen to be

vnan

v3na∞=

0.7× 0.3

(0.7)3

=30

49.

A.15.2 §3.6 NONSTANDARD TERMS AND INTEREST RATES

Omit this section.

A.15.3 §3.7 UNKNOWN TIME

In this section the textbook considers situations where n, the number of annuity pay-ments, is not known. Rather than solving such problems for some non-integer n —which is not practical because there would not likely be universal agreement about theinterpretation — such problems are usually resolved by having a final payment in anotherwise regular sequence of payments cover the amount necessary to make up the dif-ference. Where the final payment needed to meet a goal is larger than a regular payment,it is called a balloon payment; where it is smaller, it is called a drop payment. Solvingproblems in this section typically involves two phases:

• Determination of the number of regular payments.

• Determination of the value of the final payment, subject to given constraints (e.g.that it be a balloon or drop payment).

The author uses tables in the first phase, but we can solve such problems without ta-bles, by first solving an inequality. Read [1, Example 3.7, pp. 74-75], where the authorconsiders a problem where a planned final payment turns out to be negative.


[1, Exercise 32, p. 91] “A loan of 1000 is to be repaid by annual payments of 100 tocommence at the end of the 5th year, and to continue thereafter for as long asnecessary. Find the time and amount of the final payment if the final payment isto be larger than the regular payments. Assume i = 4.5%.”

Solution: Let the time of the last — balloon — payment be n, and let the amountof the last payment be X. Then n is the largest integer solution to the inequality

1000 ≥ 100(1.045)−4an−4 = 100(1.045)−4 · 1− (1.045)−(n−4)

0.045

⇔ (1.045)−(n−4) ≥ 1− 1000× 0.045× (1.045)4

100⇔ −(n− 4) ln 1.045 ≥ ln

(1− 10× 0.045× (1.045)4

)

⇔ −(n− 4) ≥ ln (1− 0.45(1.045)4)

ln 1.045

⇔ n ≤ 4− ln (1− 0.45(1.045)4)

ln 1.045= 21.47594530.

Thus we conclude that the balloon payment is made at time t = 21. The equationof value at time t = 21 is

1000(1.045)21 = 100s17 + (X − 100)

implying that

X = 100 + 1000(1.045)21 − 100

0.045

((1.045)17 − 1

)= 146.070467 .




Distribution Date: Wednesday, February 9th, 2005, subject to further revision

A.16.1 §3.7 UNKNOWN TIME (continued)

[1, Exercise 33, p. 91] “A fund of 2000 is to be accumulated by n annual paymentsof 50, followed by n annual payments of 100, plus a smaller final payment made 1year after the last regular payment. if the effective rate of interest is 4.5%, find nand the amount of the final irregular payment.”

Solution: We shall interpret the payments to be made under two annuities-due:the first, for 2n years, consists of an annual deposit of 50 in advance; the second,for n years, deferred n years after the first, also consists of an annual deposit of 50in advance. It is at the end of year 2n that the final, drop payment is to be made,and it is to be under 100. (Note that this is the type of problem where the droppayment could turn out to be negative. We seek the smallest n for which

50s2n + 50sn > 2000− 100

⇔ 50(1.045) · (1.045)2n + (1.045)n − 2

0.045> 1900

⇔ (1.045)2n + (1.045)n − 2 >1900

50· 0.045

1.045

⇔(

(1.045)n +1

2

)2

>1900

50· 0.045

1.045+ 2.25 = 3.886363636

Since the exponential is positive, the preceding inequality is equivalent to (1.045)n >1.471386222, and, in turn, to

n >ln 1.471386222

ln 1.045= 8.774018446 .

Thus the drop payment will be when t = 2 × 9, i.e., 18 years after the first pay-ment under the annuity with payments of 50. Just before the drop payment theaccumulated value of all previous payments is

50(s9 + s18

)= 50(1.045) · (1.045)18 + (1.045)9 − 2

0.045= 1967.588591

so the drop payment at time t = 18 is 2000− 1967.588591 = 32.411409.

Note that there is an error in the answers in the textbook: while n = 9 is correct,the payment of 32.41 is not “at time n = 9”.


(If tables like those in the textbook were available, one could determine the valueof n by inspecting the value of s2n + sn. We observe from the 4.5% tables thefollowing values:

n s2n + sn

8 32.09939 37.6572

10 43.6596

We seek the smallest n such that

50s2n + 50sn > 2000− 100

i.e., such thats2n + sn > 38 ,

equivalently,

s2n + sn >38

1.045= 36.37 ,

and so can conclude that n = 9.)

[1, Exercise 34, p. 91] “One annuity pays 4 at the end of each year for 36 years.Another annuity pays 5 at the end of each year for 18 years. The present valuesof both annuities are equal at effective rate of interest i. If an amount of moneyinvested at the same rate i will double in n years, find n.”

Solution: If i 6= 0, the equation of value for the two annuities may be solved asfollows:

4a36 = 5a18 ⇔ 4(1− v36

)= 5

(1− v18

)

⇔ 4(v18

)2 − 5(v18

)+ 1 = 0

⇔ v18 = 1 or v18 =1

4

Of these equations, the first is inadmissible, since it corresponds to the excludedcase i = 0; the second equation is the only valid conclusion, and it implies that(1 + i)18 = 4 = 22, so money doubles in 18

2= 9 years.

As for the case i = 0, the equation of value can be transformed as follows:

4a36 = 5a18 ⇔ 4× 36 = 5× 18 (84)

which is a contradiction; hence i 6= 0, and the preceding conclusion is the onlyvalid one.


[1, Exercise 35, p. 91] “A fund earning 8% effective is being accumulated with pay-ments of 500 at the beginning of each year for 20 years. Find the maximum numberof withdrawals of 1000 which can be made at the ends of years under the conditionthat once withdrawals start they must continue through the end of the 20-yearperiod.”

Solution: Suppose that the first withdrawal occurs at time t = n. We must deter-mine the smallest n satisfying the following sequence of equivalent inequalities:

500s20 ≥ 1000s21−n

⇔ (1.08)((1.08)20 − 1

) ≥ 2((1.08)21−n − 1

)

⇔ (1.08)21 − 1.08 ≥ 2(1.08)21

(1.08)n− 2

⇔ (1.08)n ≥ 2

1 + 0.92(1.08)21

⇔ n ≥ 6.825449633 ,

hence n = 7. The maximum number of withdrawals of 1000 is, therefore, 21− 7 =14.

[1, Exercise 36, p. 91] “A borrower has the following two options for repaying a loan:

(i) Sixty monthly payments of 100 at the end of each month.

(ii) A single payment of 6000 at the end of K months.

Interest is at the nominal annual rate of 12% convertible monthly. The two optionshave the same present value. Find K.”

Solution: The equation of value at the present time t = 0 leads to the followingsequence of equivalent equations:

100a60 1% = 6000(1.01)−K

⇔ 1− (1.01)−60

0.01=

60

(1.01)K

⇔ (1.01)K =0.01× 60

1− (1.01)−60

⇔ K =ln 0.01×60

1−(1.01)−60

ln 1.01= 29.01227333 .

To the precision of the problem, K = 29.


A.16.2 §3.8 UNKNOWN RATE OF INTEREST

The details of this section can be omitted for examination purposes.

Approximating compound interest for fractions of a measurement periodThe textbook observes in [1, §2.2] that a method that is often used when a fractionof a measurement period is involved is to approximate compound interest by simpleinterest. This amounts to using series (57) above, but stopping after the 1st degreeterm:

(1 + i)k ≈ 1 + ki .

It can also be interpreted as linear interpolation between values of (1 + i)x for integerx. Suppose that we are interested in the accumulation factor for compound interestbetween t = n and t = n + k, where 0 ≤ k ≤ 1. If we think of a line joining the points(n, (1 + i)n) and (n + 1, (1 + i)n+1), and take as a value approximating (1 + i)n+k, theordinate of the point where this line meets the line x = n + k, we have similar triangles,leading to the equation

k

1=

approximation− (1 + i)n

(1 + i)n+1 − (1 + i)n

from which we determine that the approximation is

k(1 + i)n+1 + (1− k)(1 + i)n = (1 + i)n (k(1 + i) + (1− k)) = (1 + i)n(1 + ki).

The textbook remarks that linear approximation between successive integer values of vx

is equivalent to simple discount . We can obtain a better approximation by truncatingthe MacLaurin expansion after a higher term than the first degree term. In the followingexample the accumulation factor is approximated by a quadratic — not a linear —function.

Unknown rate of interest Where it is the interest rate that is not known, there canbe several different approaches:

• Algebraic methods:

– Where possible, one tries to find a solution using algebraic methods. Sincethe various formulæ we work with are usually polynomial in i or ratios ofpolynomials in i, it may be possible to find the particular solution(s) we seekby algebraic means.

– Algebraic means may still be available where solving for i fails, if we can findanother convenient intermediary variable.


• It may be possible to solve by interpolation on tables, provided the functions weare interested in are tabulated.

• The favoured method is by successive approximation, to obtain a solution to anydesired accuracy.

Linear Interpolation Suppose that we know the values of f at distinct points x = x1

and at x = x2, and that f(x1) 6= f(x2). If |x1 − x2| is small, it may be reasonable toassume that the graph of f is approximately linear between x1 and x2; or, more generally,near x1 and x2. That is equivalent to assuming that

f(x) = f(x1) +f(x2)− f(x1)

x2 − x1

· (x− x1)

=f(x2)− f(x1)

x2 − x1

· x +x2f(x1)− x1f(x2)

x2 − x1

.

If now we wish to determine an approximate value of x where f(x) has a specific valuey0, we may use this equation for that purpose provided f(x2)− f(x1) is not too small:

x = x1 + (x2 − x1) · y0 − f(x1)

f(x2)− f(x1).

In particular, if we wish to find a zero of f near the given points, and if f(x1) 6= f(x2),a good approximation could be

x ≈ x1 + (x2 − x1) · 0− f(x1)

f(x2)− f(x1)

=x1f(x2)− x2f(x1)

f(x2)− f(x1).

When we apply these formulæ for points between two points x1 and x2, we speak of linearinterpolation; otherwise linear extrapolation. Interpolation may be used to improve theefficiency of the Method of Successive Bisection [1, p. 349, Appendix V], or to provide afirst approximation prior to the application of other approximation methods.

Newton-Raphson iteration method In [1, Appendix V, pp. 399-400] the authorconsiders several Iteration methods . In all of these methods one is interested in solvingan equation

f(x) = 0 , (85)

i.e. in finding the zeroes of f ; usually there are several solutions, and one is interestedin a zero in a certain interval. For example, if x is to represent an interest rate, or


an accumulation factor, we would not normally be interested in a solution that wasnegative. One of the iteration methods discussed is the method of Successive bisection[1, §B, pp. 399-400] which you have already seen in an assignment problem. Another isthe Newton-Raphson method [1, §D, p. 400].

The general idea we are applying to (85) is to find a fixed point for a mappingx → h(x). It can be shown that if, in an interval where f ′ is continuous and less than1 in magnitude, we start with an approximation — call it x0 — and replace it by asuccession of approximations [1, p. 76]:

x1 = h(x0)

x2 = h(x1)

x3 = h(x2)

· · ·xn = h(xn−1)

xn+1 = h(xn)

· · ·this iteration algorithm will converge in the interval within which x0 is chosen; that is,the sequence x0, x1, . . . will approach a limit x which satisfies equation

x = h(x) . (86)

In applying the Newton-Raphson method to solve (85) for a function f , we take

h(x) = x− f(x)

f ′(x).

The condition |h′(x)| < 1 that must be satisfied is equivalent to∣∣∣∣f(x) · f ′′(x)

(f ′(x))2

∣∣∣∣ < 1 . (87)

Any solution to (85) will be a fixed point x for the transformation h. Here the successiveapproximations are:

x0

x1 = x0 − f(x0)

f ′(x0)

x2 = x1 − f(x1)

f ′(x1). . .

xn+1 = xn − f(xn)

f ′(xn). . .


and the sequence converges to a point x with the property that

x = x− f(x)

f ′(x),

so f(x) = 0. This method can be shown to produce “an extremely fast rate of conver-gence, called ‘second-order convergence’”23. While the theory of approximations of thistype is beyond this course, we shall use the method when it is useful.

Newton-Raphson iteration to solve an i = k for i The given equation must berecast as an equivalent equation in the form f(i) = 0. Our usual formula for an i gives

1− (1− i)n

i= k

so we might express this as an equivalent constraint on a function

f(i) =1− (1− i)n

i− k ,

asking that

f(i) =1− (1− i)n

i− k = 0 .

It can be shown that condition (87) is satisfied for this function. The iteration is givenby [1, (3.28), p. 77]

ir+1 = ir

[1 +

1− (1 + ir)−n − kir

1− (1 + ir)−(n+1) (1 + (n + 1)ir)

](88)

Newton-Raphson iteration to solve sn i = k for i Analogously to the preceding,it may be shown that the iteration is given by [1, (3.28), p. 78]

ir+1 = ir

[1 +

(1 + ir)n − 1− kir

(1 + ir)n−1 (1− ir(n− 1))− 1

](89)

Some remaining exercises from [1, Chapter 3] (not all were discussed in thelecture):

[1, Exercise 38, p. 92] “If a2 = 1.75, find an exact expression for i.”

23cf., http://mathworld.wolfram.com/NewtonsMethod.html


Solution:

a2 = 1.75 ⇔ (1 + i)2 − 1

i= 1.75(1 + i)2

⇔ 1.75i2 + 2.5i− 0.25 = 0 (90)

⇔ 7i2 + 10i− 1 = 0

⇔ i =−10 +

√100 + 28

14=−5− 4

√2

7or−5 + 4

√2

7. (91)

Since i > 0, the second solution is inadmissible, and i = 0.09383632129.

[1, Exercise 41, p. 92] “A fund of 17,000 is to be accumulated at the end of 5 yearswith payments at the end of each half-year. The first 5 payments are 1,000 each,while the second 5 payments are 2,000 each. Find the nominal rate of interestconvertible semiannually earned on the fund.”

Solution: The equation of value at time 5, just after the last payment, is

1000(s10 + s5

)= 17000 (92)

⇔ ((1 + i)5

)2+ (1 + i)5 − 2 = 17i. (93)

There is a unique solution to this problem, since the sum on the left side of equation(92) is an increasing function of positive i.

Definef(x) = (1 + i)10 + (1 + i)5 − 2− 17i .

We can approximate with successive bisection, beginning with f(0.03) = −0.006809547and f(0.04) = 0.016897187.

f(0.035) = 0.016897187

f(0.0325) = −0.002194300

f(0.03375) = 0.000436424

f(0.033125) = −0.000906060

f(0.0334375) = −0.000241612

f(0.03359375) = 0.000095705

f(0.033515625) = −0.000073378

f(0.033554687) = 0.000011056

f(0.033535156) = −0.000031188

f(0.033544922) = −0.000010071

f(0.033549805) = 0.000000491


f(0.033547364) = −0.000004790

f(0.033548585) = −0.000002148

f(0.033549195) = −0.0000008286

f(0.033549500) = −0.000000169

f(0.033549653) = 0.000000162

f(0.033549577) = −0.000000001

f(0.033549615) = 0.000000080

f(0.033549587) = 0.000000020

f(0.033549582) = 0.000000009

f(0.033549580) = 0.000000005

f(0.033549579) = 0.000000002

f(0.033549578) = 0.000000000

This is the effective interest rate for a half-year. The nominal annual rate, com-pounded semi-annually will be 2× 0.033549578 = 6.7099156%.

[1, Exercise 42, p. 92] “A beneficiary receives a 10,000 life insurance benefit. If thebeneficiary uses the proceeds to buy a 10-year annuity-immediate, the annual pay-out will be 1,538. If a 20-year annuity-immediate is purchased, the annual payoutwill be 1,072. Both calculations are based on an annual effective interest rate of i.Find i.”

Solution: The equations of value at time 0 are

10000 = 1538 · a10 = 1072 · a20

The last equation alone implies that 1 + v10 = 1.434701492, which implies thati = 8.6878222%. I don’t see why the amount of the insurance benefit was given(although it is consistent with the other other information.)

[1, Exercise 50, p. 93] 1. “Show that

sn = n +n(n− 1)

2!· i +

n(n− 1)(n− 2)

3!· i2 + . . . ”

2. “Show that1

sn

=1

n

[1− n− 1

2· i +

n2 − 1)

12· i2 + . . .

]”

Solution:



1. The sum on the right — a finite sum — is obtained from the expansion ofthe 10th power of the binomial 1 + i by

• deleting the term for i0, which is 1;

• dividing by i.

2. Unlike the preceding expansion, this one is infinite. It can be obtained fromthe preceding result as follows:

1sn

=i

(1 + i)n − 1

=1n

[1 + i

(n− 1

2+

(n− 1)(n− 2)6

· i +(n− 1)(n− 2)(n− 3)

24· i2 + . . .

)]−1

=1n

[1−

(n− 1

2+

(n− 1)(n− 2)6

· i +(n− 1)(n− 2)(n− 3)

24· i2 + . . .

)i

+(

n− 12

+(n− 1)(n− 2)

6· i +

(n− 1)(n− 2)(n− 3)24

· i2 + . . .

)2

i2 + . . .

]

=1n

[1− n− 1

2· i +

n2 − 112

· i2 +(n− 1)(n2 − n− 1)

24· i3 + . . .

]

[1, Exercise 51, p. 93] “A loan of 1,000 is to be repaid with annual payments at theend of each year for the next 20 years. For the next 5 years the payments are kper year; the second 5 years, 2k per year; the third 5 years, 3k per year; and thefourth 5 years, 4k per year. Find an expression for k.”

Solution: The equation of value at time 0 is

kv20(s20 + s15 + s10 + s5 = 1000

),

implying that

k =1000

s20 + s15 + s10 + s5

=1000i · a5

s20

[1, Exercise 52, p. 93] “The present value of an annuity-immediate which pays 200every 6 months during the next 10 years and 100 every 6 months during the follow-ing 10 years is 4,000. The present value of a 10-year deferred annuity-immediatewhich pays 250 every 6 months for 10 years is 2,500. Find the present value of anannuity-immediate which pays 200 every 6 months during the next 10 years and300 every 6 months during the following 10 years. (Hint: Payments made during


the first 10 years are discounted at a different rate than payments made during thesecond 10 years.)”

Solution: The “Hint” is objectionable, as it reveals a datum that should have beenstated in the problem.

Denote the effective semi-annual interest rates during the first and second 10 yearsby i and j respectively. The equations of value at time 0 are

200 · a20 i + 100 · a20 j(1 + i)20 = 4000

250 · a20 j(1 + i)20 = 2500

Solving these equations, we infer that

a20 i = 15

a20 j(1 + i)20 = 10

200 · a20 i + 300 · a20 j(1 + i)20 = 6000

where the last sum is the present value of the desired annuity-immediate.

[1, Exercise 53, p. 94] “A depositor puts 10,000 into a bank account that pays anannual effective interest rate of 4% for 10 years. If a withdrawal is made duringthe first 51

2years, a penalty of 5% of the withdrawal amount is made. The depositor

withdraws K at the end of each of years 4, 5, 6, and 7. The balance in the accountat the end of year 10 is 10,000. Find K.”

Solution: We shall assume that the account is debited by each penalty after thewithdrawal that triggers it. An equation of value at time 10 is

10000(1.04)10 −K(1.05)((1.04)6 + (1.04)5

)−K((1.04)4 + (1.04)3

)= 10000 .

Hence

K =(1.04)10 − 1

(1.04)3(2.04) ((1.05)(1.04)2 + 1)= 979.9317732 .

[1, Exercise 54, p. 94] “Simplify40∑

n=15

sn .”

Solution:

1

i

(40∑

n=15

(1 + i)n −40∑

n=15

1

)=

1

i

((1 + i)41 − (1 + i)15

i− 26

)

=1

i

(s41 − s15 − 26

)


[1, Exercise 55, p. 94] “Show that sn · an > n2 if i > 0 and n > 1.”

Solution:

sn · an =

((1 + i)n − 1

i

)2

· 1

(1 + i)n−1= (sn)2 · (1 + i)n−1 .

But sn is a strictly increasing function of i; its minimum value is when i = 0, whereit is equal to n. And 1 + i > 1 ⇒ (1 + i)n−1 > 1.

A.16.3 §3.9 VARYING INTEREST

Omit this section for now.

A.16.4 §3.10 ANNUITIES NOT INVOLVING COMPOUND INTEREST

Omit this section.




Textbook Chapter 4. More General Annuities.


The textbook considers

• annuities whose payments do not have the same frequency as the interest conversionperiods;

• annuities whose payments are not constant.

For the first of these topics, contained in the following three sections of the textbook, weshall not study all the material carefully, but shall consider ad hoc solutions to problems.

A.17.2 §4.2 ANNUITIES PAYABLE AT A DIFFERENT FREQUENCYTHAN INTEREST IS CONVERTIBLE

A typographical error. The following example, given in the textbook [1, Example4.1, p. 96] contains a serious error. “Example 4.1 Find the accumulated value at theend of 45 years of an investment fund in which 100 is deposited at the beginning of eachquarter for the first 2 years, and 200 is deposited at the beginning of each quarter forthe second 2 years, if the fund earns 12% convertible quarterly.”Solution: If the interest rate were as stated the problem would be of a type studied inChapter 2. A reading of the solution shows that the author intended the last word to bemonthly .

Some Exercises

[1, Exercise 1, p. 122] “Find the accumulated value 18 years after the first paymentis made of an annuity on which there are 8 payments of 2000 each made at 2-yearintervals. The nominal rate of interest convertible semiannually is 7%. Answer tothe nearest dollar.”

Solution:


1. First Solution: Working with 2-year periods. The nominal biennial

interest rate i(12) is given by

(1 +

i(12)

12

) 12

= 1 + i =

(1 +

i(2)

2

)2

=

(1 +

0.07

2

)2

⇒ i(12) = 2

((1.035)4 − 1

)

and the effective biennial rate — call it j — is

j =i(

12)

12

= (1.035)4 − 1 .

Using this interest rate and a scale with time unit of a 2 year interval, thepayments — 2 intervals after the last — are worth

2000(s10 − s2

)=

2000

j

((1 + j)10 − (1 + j)8)

= 2000

((1.035)40 − (1.035)32

(1.035)4 − 1

)

= 35824.25354

which is 35824 to the nearest unit. (Why, then, does the textbook give thevalue as 35825? Because the author is using his tables. If we work with the3.5% tables [1, p. 383], we obtain

2000(s10 − s2

)= 2000

(s40 0.035 − s8 0.035

s4 0.035

)(94)

= 2000

(84.5503− 9.0517

4.2149

)(95)

= 35824.62217 . (96)

The tables also contain values of the inverse of s4 0.035:

2000

(s40 0.035 − s8 0.035

s4 0.035

)= 2000s−1

4 0.035

(s40 0.035 − s8 0.035

)

= 2000(0.237251)(84.5503− 9.0517)

= 35824.23670 ,

which gives an answer which is closer to the correct one.)


2. Second Solution: Working with half-year periods. The effective inter-est rate for half-year periods is 1

2(7%) = 3.5%. A payment of 2000 at the end

of a 2-year period could be replaced by 4 equal payments at the ends of thefirst, second, third, and 4th half year (which is the time of the payment of2000). If we denote the amount of these equal payments by X, we have theequation of value at the end of the 2-year period,

X · s4 3.5% = 2000

so X =2000

s4 3.5%

. There will be 4×8 = 32 such payments of X, the last of them

8 half-years before the date at which we require the accumulated value. Wecan, instead think of a series of 32 + 8 = 40 payments, and then subtract theaccumulated value of the last 8 payments that we have added. We obtain asthe accumulated value

X(s40 3.5% − s8 3.5%

)

=2000

s4 3.5%

· (s40 3.5% − s8 3.5%

)

which agrees with equation (94) above.

[1, Exercise 2, p. 122] (not discussed in the lecture) “Find the present value of a 10-year annuity which pays 400 at the beginning of each quarter for the first 5 years,increasing to 600 per quarter thereafter. The annual effective rate of interest is12%. Answer to the nearest dollar.”

Solution: With 1 + i = (1.12)14 ,

Present Value = 600a40 − 200a20

= (600− 200) + 600

(1− v39

i

)− 200

(1− v19

i

)

= 11466.12687

[1, Exercise 3, p. 122] (not discussed in the lecture) “A sum of 100 is placed into afund at the beginning of every other year for 8 years. If the fund balance at theend of 8 years is 520, find the rate of simple interest earned by the fund.”

Solution: The wording of this problem is not as precise as it could be — while itis clear that the payments into the fund are 2 years apart, it is not clear whetherthey are at the beginnings of years ##0, 2, 4, 6 or the beginnings of years ##1, 3,5, 7. From the author’s answer we see that he intended the former interpretation.


Let i be the annual rate of simple interest. Then the equation of value at timet = 8 is

100[(1 + 8i) + (1 + 6i) + (1 + 4i) + (1 + 2i)] = 520 ⇒ i = 6%

(The other interpretation mentioned would have given

100[(1 + 7i) + (1 + 5i) + (1 + 3i) + (1 + 1i)] = 520 ⇒ i = 7.5% .)




A.18.1 §4.3 FURTHER ANALYSIS OF ANNUITIES PAYABLE LESS FRE-QUENTLY THAN INTEREST IS CONVERTIBLE

Some Exercises

[1, Exercise 5, p. 122] “Give an expression in terms of functions assuming a rate ofinterest per month for the present value, 3 years before the first payment is made,of an annuity on which there are payments of 200 every 4 months for 12 years:

1. expressed as an annuity-immediate;

2. expressed as an annuity-due.”

Solution: In order to use these functions, we need to replace the regular paymentevery 4 months by an equivalent monthly payment. Let i represent the effec-tive monthly interest rate, and A and B respectively be the monthly paymentsat the end of the month/in advance equivalent to a payment under an annuity-immediate/annuity-due of 200 every 4 months. Then

As4 i = 200

Ba4 i = 200

Both replacement annuities will pay 12× 12 = 144 monthly payments.

1. The annuity is deferred 3 years, i.e. 3 × 12 = 36 months, before the firstpayment is made — that means 32 months before the beginning of annuity-immediate at 4-month intervals, and also 32 months before the beginning ofthe replacement annuity-immediate with monthly payments of A.

Value = A · (32

∣∣a144i

)

= 200 · v32a144

s4

= 200 · a176 − a32

s4


2. This annuity is deferred 36 months.

Value = B · (36

∣∣a144i

)

= 200 · v36a144

a4

= 200 · a180 − a36

a4

= 200 · a180 − a36

a4

[1, Exercise 6, p. 122] “Show that the present value at time 0 of 1 payable at times7, 11, 15, 19, 23, and 27 is

a28 − a4

s3 + a1

.

Solution: s3 + a1 is the present value — just after the third payment — of a 4-payment annuity of 1 for which the first payment was 1 unit ago. Hence a single

payment of 1 at time 3 is equivalent to payments of1

s3 + a1

at times 1, 2, 3, 4.

Hence an n-payment annuity that pays 1 every 4 years, starting in 2 years with firstpayment in 3 years, may be replaced by a 4n-payment annuity-immediate startingnow, with first payment in 4 years, whose present value is

a4n

s3 + a1

.

In the problem the first payment of the given annuity is excluded; this correspondsto the first 4 payments of the replacement annuity. Thus the present value of theannuity described is

a28 − a4

s3 + a1

.

¤

[1, Exercise 7, p. 122] “A perpetuity of 750 payable at the end of every year, and aperpetuity of 750 payable at the end of every 20 years are to be replaced by anannuity of R payable at the end of every year for 30 years. If i(2) = 0.04, show that

R = 37500 ·(

1

s2

+v40

a40

)· s2

a60

where all functions are evaluated at 2% interest.”


Solution: The given nominal annual interest rate of 4%, compounded semi-annually,is equivalent to an effective semi-annual rate of 2%.

1. Each payment of 750 at the end of a year is equivalent to 2 payments of 750s2 2%

,

one at the end of 6 months, the other at the end of the year. Thus the present

value of the first perpetuity is750

s2 2%

· a∞ 2%.

2. The perpetuity of 750 payable at the end of 20-year intervals, is, analogously,

worth at present750

s40 2%

· a∞ 2%.

3. Hence the amount available to purchase the annuity is

750 · a∞ 2% ·(

1

s2 2%

+1

s40 2%

)= 750 · 1

0.02

(1

s2 2%

+v40

v40s40 2%

)

= 37500 ·(

1

s2 2%

+v40

a40 2%

)(97)

4. A payment of R at the end of a year is equivalent to a payment of Rs2

at the

end of every 6 months. The present value of the 30-year annuity is, therefore,

R · a60

s2

(98)

since we are replacing it by a 60-half-year annuity-immediate.

5. The equation of value equates amounts (97) and (98), implying that

R =

37500 ·(

1

s2 2%

+v40

a40 2%

)

a60

s2

= 37500 ·(

1

s2 2%

+v40

a40 2%

)· s2

a60

¤

[1, Exercise 8, p. 122] “Find an expression for the present value of an annuity-due of600 per annum payable semiannually for 10 years, if d(12) = 0.09.”

Solution: The effective discount rate per month is 0.0912

= 34%. The effective discount

rate per 6 months is, therefore, d′ = d(2)

2= 1− (1− 0.0075)6. The textbook wishes


us to interpret the statement “...600 per annum payable semiannually” to mean“...300 paid per half-year”. With this interpretation the present value of the 20-half-year annuity-due is, therefore,

300 · a20 = 600 · 1− (1− d′)20

d′

= 300 · 1− (1− 0.0075)120

1− (1− 0.0075)6

[1, Exercise 9, p. 122] “The present value of a perpetuity paying 1 at the end of every

3 years is125

91. Find i.”

Solution: A payment of 1 at the end of the year for 3 years is equivalent to apayment of s−1

3at the end of every year. The equation of value is

125

91= s−1

3· 1

i

⇒ 125

91=

1

(1 + i)3 − 1

⇒ (1 + i)3 =

(6

5

)3

⇒ i = 20%.

[1, Exercise 10, p. 123] “Find an expression for the present value of an annuity onwhich payments are 100 per quarter for 5 years, just before the first payment ismade, if δ = 0.08.”

Solution: The effective interest rate per quarter is i = e0.084 − 1, so v = e−0.02.

Accordingly, the present value of the annuity-due is

100a20 i = 100 · (1 + i) · (1− v20)

i

= 100 · 1− e−0.4

e−0.02 · (−1 + e0.02)

[1, Exercise 11, p. 123] “A perpetuity paying 1 at the beginning of each year has apresent value of 20. If this perpetuity is exchanged for another perpetuity paying Rat the beginning of every 2 years, find R so that the values of the two perpetuitiesare equal.”

Solution: An equation of value now for the perpetuity is 1d

= 20, implying thati = 1

19. At an effective annual discount rate of d, a payment of 1 now is equivalent to


a 2-year annuity-due paying 11+v

= 1+i2+i

each year, in advance. The perpetuity-duepaying R at the beginning of each 2-year period is equivalent to a perpetuity-due paying R · 1+i

2+iat the beginning of every year. Equating this to 20 yields

R = 3920

= 1.95.

[1, Exercise 12, p. 123] “Find an expression for the present value of an annuity onwhich payments are 1 at the beginning of each 4-month period for 12 years, as-suming a rate of interest per 3-month period.”

Solution: If the effective interest rate per 3-month period is i, then the equivalenteffective rate per 4-month period will be j = (1 + i)

43 − 1. The present value of the

annuity-due at this rate for 12× 3 4-month periods will be

a36 j =(1 + j)− (1 + j)−35

j

=(1 + i)

43 − (1 + i)−

1403

(1 + i)43 − 1

=1− v48

1− v43

A.18.2 §4.4 FURTHER ANALYSIS OF ANNUITIES PAYABLE MOREFREQUENTLY THAN INTEREST IS CONVERTIBLE

In §4.4 we have a new notation. A right-superscribed (m) indicates that the annuity ispayable m times in a time unit for compounding of interest. More than that, we willassume that the unit previously associated with one payment is divided into m parts;this usage is analogous to the use of the upper right parentheses in i(m). Thus a

(m)n i

represents the present value of an annuity of 1m

payable at m times regularly spacedthrough an interest period. There will be immediate and due versions of the symbol,analogous symbols for perpetuities, and a corresponding symbol for s.

Some Exercises

[1, Exercise 14, p. 123]

a(m)n =

1

m

(1 + v

1m + v

2m + . . . + vn− 1

m

)

=1

m· 1− vn

d(m)

[1, Exercise 20, p. 123] “A sum of 10000 is used to buy a deferred perpetuity-duepaying 500 every 6 months forever. Find an expression for the deferred periodexpressed as a function of d.”


Solution: The effective discount rate per 6 months is 1− v12 . If the perpetuity-due

has been deferred for n full years, the equation of value is

10000 = 500 · vn · 1

1−√v


n =ln(20(1−√v))

ln v

=ln(20(1−√1− d))

ln(1− d)

[1, Exercise 21, p. 124] “If 3 · a(2)n = 2 · a(2)

2n= 45 · s(2)

1, find i.”

Solution:

3 · a(2)n = 2 · a(2)

2n= 45 · s(2)

1

⇔ 3

2(1− vn) =

2

2

(1− v2n

)=

45

2i

The first equation implies that 1 + vn = 32⇒ (1 + i)n = 2. Substitution in the

equations yields i = 130

.

[1, Exercise 22, p. 124] “Find an expression for the present value of an annuity whichpays 1 at the beginning of each 3-month period for 12 years, assuming a rate ofinterest per 4-month period.”

Solution: If i be the effective interest rate per 4-month period, the effective rateper 3-month period will be j = (1 + i)

34 − 1. Accordingly the value of the desired

annuity is

a48 j = (1 + j) · 1− (1 + j)−48

j

= (1 + i)34 · 1− (1 + i)−36

(1 + i)34 − 1

=1− (1 + i)−36

1− (1 + i)−34

An alternative approach to this problem would be to define A to be the amountof payment that would have to be made under an annuity-due every four monthsfor 12 years to give the same present value as the annuity-due described in the


problem. We can look at a single year in order to determine A. The present valueof payments of A now, 4 months from now, and 8 months from now, is

A(1 + (1 + i)−1 + (1 + i)−2

)= A · (1 + i) · 1− (1 + i)−3

i

and the value of the 4-payment annuity-due at 3 month intervals is

1(1 + (1 + j) + (1 + j−2 + (1 + j)−3 = (1 + j) · 1− (1 + j)−4

j.

Equating the two, and recalling that (1 + i)3 = (1 + j)4, gives A =i(1 + j)

j(1 + i). The

value of the annuity will then be

a48 j = A · a36 i

=i(1 + j)

j(1 + i)· (1 + i) · 1− (1 + i)−36

i

=1− (1 + i)−36

1− (1 + i)−34

as found previously.



Distribution Date: Wednesday, February 16th, 2005, subject to further revision

A.19.1 §4.5 CONTINUOUS ANNUITIES

Omit this section for the present.

A.19.2 §4.6 BASIC VARYING ANNUITIES

Payments varying in arithmetic progression. A general formula will be developedfor an annuity-immediate with a term of n periods, with payments beginning at P andincreasing by Q per period thereafter, at interest rate i, showing that its present valueA is

A = P · an + Q · an − nvn

i

and the accumulated value just after the last payment is

S = P · sn + Q · sn − n

i.

When P = Q = 1, a specific symbol is used:

(Ia)n =an − nvn

i(99)

(Is)n =sn − n

i=

sn+1 − (n + 1)

i(100)

Similarly, when P = n and Q = −1, we have a decreasing annuity for which the presentand accumulated values are given by

(Da)n =n− an

i,

(Ds)n =n(1 + i)n − sn

i.

Other symbols of interest are (Ia)n , (Is)n , (Da)n , (Ds)n , (Ia)∞ , (Ia)∞ .




A.20.1 §4.6 BASIC VARYING ANNUITIES (continued)

You may omit the discussion of Fn, Gn, Hn on [1, p. 113], and any exercises based onthese functions. However, you should be able to derive formulæ

(Ia)∞ = limn→∞

an − nvn

i=

1 + i

i2

(Ia)∞ = (1 + i)(Ia)∞ =(1 + i)2

i2

using the calculus.

Verbal Interpretations A number of the identities we can develop in connection withincreasing and decreasing annuities admit interesting verbal interpretations.

1.an = i(Ia)n + nvn (101)

This equation can be obtained from (99): that provides an algebraic proof. Oneverbal interpretation is as follows. The annual payments of 1 under an n-paymentannuity-due of present value an may be each invested for repayment n years afterthe first of the payments. The interest earned by these investments will be multiplesof i: 1i at the end of year 1, 2i at the end of year 2, . . . , ni at the end of year n;the principal of n repaid after n years is worth nvn at time 0.

2. Equation (99) can be interpreted directly as describing a decomposition of theincreasing annuity. Think of a perpetuity paying an−nvn at the end of every year.That annual payment may be interpreted as the value of an n-payment annuity-dueof regular payments of 1, the first payment at that time, followed by a charge of nthe year after the last payment. These payments, when summed, yield 0 for yearsn + 1, . . ., and yield an increasing annuity with payments 1, 2, . . . , n for the firstn payments.

Payments varying in geometric progression. We shall work some problems toillustrate that problems of this type are not hard to solve, since the effect of the geometricprogression is equivalent to altering the interest rate.


Some Exercises

[1, Exercise 29, p. 124] In [1, Example 4.13, p. 116] it is shown that the present valueof an annuity-immediate such that payments start at 1, increase by annual amountsof 1 to a payment of n, and then decrease by annual amounts of 1 to a final paymentof 1, is an · an . The present exercise is to justify this value verbally.

Solution: Consider a sequence of n annuities-due, each of them consisting of npayments of 1. The first of these annuities is to make its first payment 1 yearfrom now, the second 2 years from now, ..., the nth n years from now. The totalpayments made will increase from 1 to n, then decrease to a payment of 1, 2n− 1years from now. The value of the payments under each of these annuities-due is an

just before the first payment. These values, when discounted to the present, havevalue an · an .

[1, Exercise 31, p. 124] “Show algebraically, and by means of a time diagram, thefollowing relationship between (Ia)n and (Da)n :

(Da)n = (n + 1) · an − (Ia)n .”

Solution: In this method of presenting notes it is difficult to present a time diagram;instead, I give a verbal explanation. To simplify, I shall move the subtracted termfrom the right side to an added term on the left side of the identity. Then we aresumming two variable annuities: as the amount of one decreases by 1 unit, theamount of the other takes up the slack and increases by one unit, so the sum ofthe two remains constant for n payments. And that sum begins with value n + 1,so that is the amount that remains constant, giving an annuity-immediate of thatconstant payment for n payments.

Algebraically, we have

(n + 1) · an − (Ia)n = (n + 1) · 1− vn

i− an − nvn

i

=(n + 1)− vn − an

i

=(n + 1)− vn − 1− an−1

i

=(n + 1− 1)− (

an−1 + vn)

i

=n− an

i= (Da)n


[1, Exercise 32, p. 124] “The following payments are made under an annuity: 10 atthe end of the 5th year, 9 at the end of the 6th year, decreasing by 1 each yearuntil nothing is paid. Show that the present value is

10− a14 + a4 (1− 10i)

i.”

Solution: The payments decrease until a payment of 1 at the end of the 14th year.We can think of a decreasing annuity starting with a payment of 14 at the end ofthe 1st year, and then make corrections. We can subtract a 4-payment decreasingannuity-immediate beginning with a payment of 4 at the end of the year, and a4-payment annuity-immediate with a constant payment of 10. Thus we have

(Da)14 − (Da)4 − 10a4 =

(14− a14

)− (4− a4

)− 10i · a4

i

=10− a14 + a4 (1− 10i)

i

[1, Exercise 33, p. 124] “Find the present value of a perpetuity under which a pay-ment of 1 is made at the end of the 1st year, 2 at the end of the 2nd year, increasinguntil a payment of n is made at the end of the nth year, and thereafter paymentsare level at n per year forever.”

Solution: We can begin with a perpetuity-immediate of n per year, and subtractfrom it a decreasing annuity-immediate which begins with a payment of n− 1 anddecreases by 1 unit per year.

n · a∞ − (Da)n−1 =n− (

(n− 1)− an−1

)

i

=1 + an−1

i

=an

i

=(1 + i)an

i

=an

vi=

an

d

[1, Exercise 34, p. 124] “A perpetuity-immediate has annual payments of 1, 3, 5, 7,. . . . If the present value(s) of the 6th and 7th payments are equal, find the presentvalue of the perpetuity.”


Solution: We equate the values of the 6th and 7th payments:

(1 + (6− 1)2)v6 = (1 + (7− 1)2)v7

⇔ 1 + i =13

11

⇔ i =2

11

The perpetuity can be viewed as the sum of an increasing perpetuity with pay-ments increasing by 2 each year, i.e., 2(Ia)∞ diminished by a constant perpetuity-immediate a∞ :

2 · (Ia)∞ − a∞ = 2lim

n→∞

(an − n

(1+i)n

)

i− 1

i

=2a∞ − 1

iby l’Hopital’s Rule

= 66

[1, Exercise 35, p. 124] “If X is the present value of a perpetuity of 1 per year withthe first payment at the end of the 2nd year and 20X is the present value of aseries of annual payments 1, 2, 3, . . . with the first payment at the end of the 3rdyear, find d.”

Solution: The constraints are:

X =1

i− v =

v

i20X = (Ia)∞ − 2a∞ + v

=v

i2,

implying that i = 120

, d = 121

.

[1, Exercise 36, p. 125] “An annuity-immediate has semiannual payments of 800, 750,700, . . . , 350, at i(2) = 0.16. If a10 0.08 = A, find the present value of the annuity interms of A.”

Solution: We will be working with an effective semi-annual interest rate of j =i(2)

2= 8%. The truncated decreasing annuity-immediate that we wish to evaluate

has present value

50(Da)10 − 50v10(Da)6 = 50 ·(16− a16

)− v10(6− a6

)

i


=50

i· ((16− 6v10

)− (a16 − v10a6

))

=50

i· (10 + 6ia10 − a10

)

=50

i· (10 + (6i− 1)a10

)

= 625(10− 0.52A) = 6250− 325A .

Alternatively we could express the payments as a constant annuity of payment size300, superimposed on a decreasing annuity of payments 500, 450, 400, . . . , 50. Thishas value

300 · a10 8% + 50(Da)10 8% = 300A + 50 · 10− A

0.08= 300A + 6250− 625A = 6250− 325A.

[1, Exercise 58, p. 127] “There are two perpetuities. The first has level payments ofp at the end of each year. The second is increasing such that the payments are q,2q, 3q, . . . . Find the rate of interest that will make the difference in present valuebetween these perpetuities

a) “Zero;

b) “A maximum.”

Solution: I assume that the second perpetuity is a perpetuity-immediate, like

the first. The present value of the first perpetuity-immediate isp

i. The second

perpetuity-immediate is worth

q(Ia)∞ = q · a∞i

=q

di.

a) For the present values to be equal,

p

i=

q

id

⇒ p

i=

q(1 + i)

i2

⇒ i =q

p− q

b) The difference is

p

i− q

id=

(p− q)2

4q− q

(1

i− p− q

2q

)2


which is maximized when1

i− p− q

2q= 0

i.e., when i =2q

p− q.

A.20.2 §4.7 MORE GENERAL VARYING ANNUITIES

We will not formally study this section and its interesting generalizations, e.g.,(I(m)a

)(m)

n

is an increasing annuity with payments of 1m

per interest conversion period at the end ofthe first mth of an interest conversion period, 2

mper interest conversion period at the end

of the second mth of an interest conversion period, etc. Thus the first payment is 1m2 ,

the second is 2m2 , etc. If you meet any problems of these types, they should be solvable

“by first principles”.




A.21.1 §4.6 BASIC VARYING ANNUITIES (conclusion)

[1, Exercise 37, p. 125] “Annual deposits are made into a fund at the beginning ofeach year for 10 years. The first 5 deposits are 1,000 each, and deposits increase by5% per year thereafter. If the fund earns 8% effective, find the accumulated valueat the end of 10 years.”

Solution: The first 5 deposits are today worth

1000a5 8% = 4312.12684 .

I see 2 ways of interpreting the words “increase by 5% per year thereafter”: eitherthe increase is geometric, by a factor of 1.05 applied repeatedly; or the depositsincrease in arithmetic progression (before discounting). If the deposits increase ingeometric progression, then the present value will be

1000(v5(1.05) + v6(1.05)2 + . . . + v9(1.05)5

)

= 1000v5(1.05) ·(1− (v(1.05))5)

1− v(1.05)

= 36000(1.05)(1.08)−5 ·(

1−(

1.05

1.08

)5)

= 3379.996182.

The present value will then be 4312.1268+3379.9962 = 7692.1230; the accumulatedvalue at the end of 10 years will be 16606.72. This is the answer given by thetextbook, so, presumably, our interpretation is the one the author intended.

But the language is ambiguous, and the other interpretation is plausible also. Ifthe deposits increase in arithmetic progression, then the present value will be

1000(1.08)−5a5 8% + 50(1.08)−5(Ia)5 8% = 3510.09347

so the present value is 7822.2203, and the value after 10 years is 16,887.59.

[1, Exercise 38, p. 125] “Find the present value of a 20-year annuity with annualpayments which pays 600 immediately and each subsequent payment is 5% greaterthan the preceding payment. The annual effective rate of interest is 10.25%.”


Solution:

Present Value =19∑

n=0

600vn(1.05)n

= 600 · 1− (1.05

1.1025

)20

1− 1.051.1025

= 7851.1926 .

A.21.2 §4.8 CONTINUOUS VARYING ANNUITIES

Omit this section.

A.21.3 §4.9 SUMMARY OF RESULTS

Omit this section.

Textbook Chapter 5. Yield Rates.


A.21.5 §5.2 DISCOUNTED CASH FLOW ANALYSIS

Definition A.15 The yield rate is that rate of interest at which the present value ofreturns from the investment is equal to the present value of contributions into the in-vestment.

Finding the yield rate may require the use of various approximation methods, since theequations that have to be solved may be polynomial of high degree.

A.21.6 §5.3 UNIQUENESS OF THE YIELD RATE

Example A.7 [1, p. 133] A person makes payments of 100 immediately and 132 at theend of 2 years, in exchange for a payment in return of 230 at the end of 1 year. Theyield rate i can be shown to satisfy the equation

((1 + i)− 1.1)((1 + i)− 1.2) = 0

which has two distinct solutions.

Example A.8 [1, Example 5.3, p. 136] “A is able to borrow 1000 from B for 1 year at8% effective, and to lend it to C for 1 year at 10% effective. What is A’s yield rate onthis transaction.”


Solution: The equation of value at time 0 is

1000− v(1080) = 1000− v(1100) ,

which has no finite solution. We can say that the yield rate is infinite.

Read [1, Example 5.4, p. 136], in which the borrower cannot possibly earn sufficientinterest to cover her payments; in this case we might wish to speak of an “imaginary”yield rate.


A.22 Supplementary Notes for the Lecture of March 2nd, 2005

Distribution Date: Monday, March 2nd, 2005, subject to further revision

A.22.1 §5.4 REINVESTMENT RATES

Single payment with interest reinvested Suppose 1 is invested at time 0, withinterest being paid at rate i at the ends of n years, but where the interest can bereinvested only at rate j. At the end of n years the total accumulated value of theinvestment is

1 + isn j

Annuity-immediate at rate i, with reinvestment at rate j If an annuity-immediateof 1 per period for n periods pays interest at rate i, but the interest can be reinvestedonly at rate j, the accumulated value at time n is

n + i · (Is)n−1 j = n +i

j· (sn j − n)

Some exercises on reinvestment rates Some of these problems were not discussedin the lecture.

[1, Exercise 10, p. 161] “It is desired to accumulate a fund of 1,000 at the end of 10years by equal deposits at the beginning of each year. If the deposits earn interestat 8% effective, but the interest can only be reinvested at only 4% effective, showthat the deposit necessary is

1000

2s11 0.04 − 12.

Solution: Let x denote the necessary deposit. We will sum the principal payments(x×10) and treat the interest payments as forming an increasing annuity-immediatewith increments of 0.08x.

x(10 + 0.08 · (Is)10 4%

)= 1000

⇔ x

(10 + 0.08 · s10 4% − 10

0.04

)= 1000

⇔ x =1000

2s11 0.04 − 12. ¤



[1, Exercise 11, p. 161] “A loan of 10,000 is being repaid with payments of 1,000 atthe end of each year for 20 years. If each payment is immediately reinvested at 5%effective, find the effective annual rate of interest earned over the 20-year period.”

Solution: Let the effective yield rate be i. The payments do not become availableuntil the maturity date, after 20 years. Until that time they are locked into apayment-scheme that accumulates to value 1000s20 5%. We are asked for the interestrate that was earned. There are thus just two transactions: the loan at time 0, inthe amount of 10,000, and the repayment at time 20, in the amount given above.The equation of value at time t = 0 is

1000s20 5%(1 + i)−20 = 10000

⇔ (1 + i)−20 =10(0.05)

(1.05)20 − 1

⇔ i = 6.1619905%.

[1, Exercise 12, p. 161] “An investor purchases a 5-year financial instrument havingthe following features:

“(i) The investor receives payments of 1000 at the end of each year for 5 years.

“(ii) These payments earn interest at an effective rate of 4% per annum. At the endof the year, this interest is reinvested at the effective rate of 3% per annum.

“Find the purchase price to the investor to produce a yield rate of 4%.”

Solution: To determine the yield we need to consider when the payment is finallyreleased to the investor. The payments of 1000 are to be invested at 4%; they willgenerate an increasing annuity whose payments start at 40 at the end of year 2,up to 160 at the end of year 5. But they are not released to the investor; rather,they earn interest at 3%. At the end of 5 years — and only then — the investorreceives

5(1000) + 40(Is)4 3%

and these amounts have to be discounted to the present at 4%, giving a presentvalue of

(1.04)−5(5(1000) + 40(Is)4 3%

)= 5000 +

40

0.03

(s5 3% − 5

)

= (1.04)−5

(5000 +

40

0.03

((1.03)5 − 1

0.03− 5

))

= 4448.418326

if the yield is to be 4%.


[1, Exercise 13, p. 161] “An investor deposits 1,000 at the beginning of each yearfor five years in a fund earning 5% effective. The interest from this fund can bereinvested at only 4% effective. Show that the total accumulated value at the endof ten years is

1250(s11 0.04 − s6 0.04 − 1

).”

Solution: The 5 deposits of 1000 would be worth 1000(s10 4% − s5 4%

)at the end of

10 years if they were earning interest together with the reinvested annual interestpayments. But they are locked into a fund where they earn 5%, and are not releaseduntil time 10, still worth 5,000. The interest payments constitute an increasingannuity to the investor, beginning with 50 at time 1, increasing to 250 at time 5,and then remaining constant until time 10. They are available to the investor asthey are paid, but she reinvests them at 4%. Their value at time 10 is, therefore50

((Is)10 4% − (Is)5 4%

). Summing yields

5000 + 50((Is)10 4% − (Is)5 4%

)= 5000 +

50

0.04

(s11 4% − 11− s6 4% + 6

)

= (5000− 6250) + 1250(s11 4% − s6 4%

)

= 1250(s11 0.04 − s6 0.04 − 1

)

= 7316.719914

Using the tables in the textbook, we would find the answer to be

1250(s11 0.04 − s6 0.04 − 1

)= 1250(13.4864− 6.6330− 1)

= 7316.7500 .

[1, Exercise 14, p. 161] “A invests 2,000 at an effective interest rate of 17% for 10years. Interest is payable annually and is reinvested at an effective rate of 11%. Atthe end of 10 years the accumulated interest is 5,685.48. B invests 150 at the endof each year for 20 years at an effective interest rate of 14%. Interest is payableannually and is reinvested at an effective rate of 11%. Find B’s accumulated interestat the end of 20 years.”

Solution: A’s interest payments begin at the end of year 1, in the amount of17% × 2000 = 340 and continue at the same level for 10 years. The equation ofvalue at the end of year 10 is

340s10 11% = 5685.48 ,

implying that

s10 11% =5685.48

340= 16.722.



It follows that (1.11)10 = (16.722)(0.11) + 1 = 2.83942, and that

s20 11% =(1.11)20 − 1

0.11= s10 11% ·

((1.11)10 + 1

)

= s10 11% ·(s10 11% + 2

)

= 16.722× 3.83942 = 64.20278124.

B’s interest payments constitute an increasing annuity-immediate whose first pay-ment is at the end of year 2, in the amount of 21. The accumulated value of B’saccumulated interest is

21(Is)19 11% = 21s20 11% − 20

0.11

= 21 · 64.20278124− 20

0.11= 8438.712782.

A.22.2 §5.5 INTEREST MEASUREMENT OF A FUND

Omit this section.

A.22.3 §5.6 TIME-WEIGHTED RATES OF INTEREST

Omit this section.

A.22.4 §5.7 PORTFOLIO METHODS AND INVESTMENT YEAR METH-ODS

Omit this section.

A.22.5 §5.8 CAPITAL BUDGETING

Omit this section.

A.22.6 §5.9 MORE GENERAL BORROWING/LENDING MODELS

Omit this section.


A.23 Supplementary Notes for the Lecture of March 4th, 2005

Distribution Date: Friday, March 4th, 2005, subject to further revision

Textbook Chapter 6. Amortization schedules and sinkingfunds.


We consider methods of repaying a loan, in particular

The Amortization Method: In this method the borrower makes instalment paymentsto the lender. Usually these payments are a regularly spaced periodic intervals; theprogressive reduction of the amount owed is described as the amortization of theloan.

The Sinking Fund Method: In this method the loan will be repaid by a single lumpsum payment at the end of the term of the loan. However the borrow may preparehimself for the repayment by making deposits to a fund called a sinking fund toaccumulate the repayment amount. (Sometimes the lender may be aware of theexistence of the sinking fund; for example, an institutional borrower that issues aseries of bonds may let the public know that the accumulation of funds to redeemthe bonds may be disciplined by a sinking fund.)

A.23.2 §6.2 FINDING THE OUTSTANDING LOAN BALANCE

When a loan is being amortized the outstanding balance is being reduced by the amor-tization payments. Each payment may be analyzed an interpreted as consisting of aninterest component and a component for reduction of principal . An equation of valuecan be set up at any time during the amortization, equating

Current value of payments = Accumulated value of Loan

where “Payments” consists of both past and future payments. Decomposing the termand rearranging the equation gives

Present Value of Future Payments=Accumulated Value of Loan − Accumulated Value of Past Payments

Synonymous terms:

• outstanding loan balance

• outstanding principal


• unpaid balance

• remaining loan indebtedness

Not all of the following examples were discussed at the lecture.

[1, Exercise 1, p. 195] “A loan of 1,000 is being repaid with quarterly payments at theend of each quarter for 5 years, at 6% convertible quarterly. Find the outstandingloan balance at the end of the 2nd year.”

Solution: The level payments under this annuity-immediate will be1000

a20 1.5%

.

Retrospective method: The value of the payments already made is

1000

a20 1.5%

· s8 0.015 = 491.1769.

Subtracting this from 1000(1.015)8 yields 635.3157.

Prospective method: The value of the 12 remaining payments is

1000

a20 1.5%

· a12 1.5% = 635.3157 .

[1, Exercise 2, p. 195] “A loan of 10,000 is being repaid by instalments of 2,000 at theend of each year, and a smaller final payment made one year after the last regularpayment. Interest is at the effective rate of 12%. Find the amount of outstandingloan balance remaining when the borrower has made payments equal to the amountof the loan.”

Solution: The problem asks for the outstanding loan balance just after paymentstotalling 10,000 have been made; this will be immediately after the 5th payment.We shall use the retrospective method only here. The accumulated value of theloan at time t = 5 is 10000(1.12)5. The accumulated value of the payments madeis 2000s5. The outstanding loan balance will, therefore be

10000(1.12)5 − 2000s5 = 10000(1.12)5 − 2000

0.12

((1.12)5 − 1

)

= 4917.72212

Were we to use the prospective method, we would need to determine the valueof the last drop payment. This is interesting information, and we could have beenasked for it. But it has not been requested, and so we shall not bother finding it.(But you should know how to do that if it is necessary.)



[1, Exercise 3, p. 195] “A loan is being repaid by quarterly instalments of 1,500 atthe end of each quarter, at 10% convertible quarterly. If the loan balance at theend of the first year is 12,000, find the original loan balance.

Solution: Denote the original loan balance by L. Here the retrospective method isthe most appropriate, since we don’t know how many future payments have to bemade. The equation of value at time 1 year is

L(1.025)4 − 1500s4 2.5% = 12000

which we solve to yield

L = (1.025)−4(1200 + 1500

((1.025)4 − 1

))

= −48000(1.025)−4 + 60000 = 16514.36905.

[1, Exercise 4, p. 195] A loan is being repaid by annual payments at the ends of 15successive years. The first 5 instalments are 4,000 each, the next 5 are 3,000 each,and the final 5 are 2,000 each. Find expressions for the outstanding loan balanceimmediately after the second 3,000 instalment.

1. prospectively;

2. retrospectively.

Solution:

1. Prospective Method. The value at time 7 of the payments yet to be madeis 2000a8 + 1000a3.

2. Retrospective Method. We have to use the prospective method at sometime to obtain the initial value of the loan. This will be the value of allpayments at time t = 0, i.e., 2000a15 + 1000a10 + 1000a5. The value of allpayments made before and at time t = 7 is 4000s−2s2. The outstanding loanbalance is, therefore

(2000a15 + 1000a10 + 1000a5

)(1 + i)7 − 4000s7 + 1000s2

[1, Exercise 5, p. 196] “A loan is to be repaid with level instalments payable at theend of each half-year for 31

2years, at a nominal rate of interest of 8% convertible

semiannually. After the fourth payment the outstanding loan balance is 5,000.Find the initial amount of the loan.”

Solution: The effective semiannual rate is 4%. There are to be 7 level payments inall. The amount of the payments is, by the prospective method

5000

a3 4%

=5000× 0.04

1− (1.04)−3.


It follows that the amount of the loan, now by the prospective method, is

5000

a3 2%

· a7 2% = 5000

(1− (1.04)−7

1− (1.04)−3

)= 10814.15817 .

[1, Exercise 6, p. 196] “A 20,000 loan is to be repaid with annual payments at the endof each year for 12 years. If (1+ i)4 = 2, find the outstanding balance immediatelyafter the fourth payment.”

Solution: The annual payment is, by the prospective method,20000

a12 i

. Again by

the prospective method, the outstanding balance after the 4th payment is

20000

a12 i

· a8 i = 20000

(1− (1 + i)−8

1− (1 + i)−12

)=

6

7· 20000 = 17142.85714.

[1, Exercise 7, p. 196] “A 20,000 mortgage is being repaid with 20 annual instalmentsat the end of each year. The borrower makes 5 payments, and then is temporarilyunable to make payments for the next 2 years. Find an expression for the revisedpayment to start at the end of the 8th year if the loan is still to be repaid at theend of the original 20 years.”

Solution: The original payments are (by the prospective method)20000

a20

. The

outstanding balance at the end of the 7th year (with no payment then or at theend of the previous year) is the value then of all unpaid payments, i.e.

20000

a20

· (a13 + s2

).

It follows that the level payment needed to repay the loan in 13 payments (underan annuity-immediate) is

20000

a20

· (a13 + s2

) · 1

a13

=20000 · (1 + i)2 · a15

a20 · a13

.

[1, Exercise 8, p. 196] “A loan of 1 was originally scheduled to be repaid by 25 equalannual payments at the end of each year. An extra payment K with each of the 6ththrough the 10th scheduled payments will be sufficient to repay the loan 5 yearsearlier than under the original schedule. Show that

K =a20 − a15

a25 · a5

.”



Solution: The level payment of this loan of 1 is1

a25

. The extra payments are equal

in value to the value of the last 5 payments, so, at time t = 5,

K · a5 =a20 − a15

a25

which yields the desired value for K.

[1, Exercise 9, p. 196] “A loan is being repaid with level payments. If Bt, Bt+1, Bt+2

and Bt+3 are four successive outstanding loan balances, show that

“1. (Bt −Bt+1) (Bt+2 −Bt+3) = (Bt+1 −Bt+2)2

“2. Bt + Bt+3 < Bt+1 + Bt+2”

Solution:

1. Let P denote the level payment. By the prospective method, the unpaidbalance after payment #t, Bt = P · an−t. It follows that

(Bt −Bt+1) (Bt+2 −Bt+3) = P(an−t − an−t−1

) (an−t−2 − an−t−3

)

= vn−tP · vn−t−2P

=(vn−t−1P 1

)2

=(P · an−t−1 − P · an−t−2

)2

= (Bt+1 −Bt+2)2

2.

Bt + Bt+3 −Bt+1 −Bt+2 =(an−t − an−t−1

)− (an−t−2 − an−t−3

)

= P · vn−t − P · vn−t−2

= P (v2 − 1) · vn−t−2

in which product P , vn−t−2 are positive, and v2 − 1 is negative.




Distribution Date: Monday, March 7th, 2005, subject to further revision

A.24.1 §6.3 AMORTIZATION SCHEDULES

An amortization schedule is a chart showing, for each payment date, the informationshown in the columns of the following beginning of such a schedule with n payments ofsize x:

Year Payment Interest Principal Outstandingamount paid repaid loan balance

0 x · an i

1 x x(1− (1 + i)−n) x(1 + i)−n x · an−1 i

2 x x(1− (1 + i)−n+1) x(1 + i)1−n x · an−2 i

· · · · · · · · · · · · · · ·In practice there will be rounding errors as the table is generated line by line, and thelast line may not quite balance. “Standard practice is to adjust the last payment so thatit is exactly equal to the amount of interest for the final period plus the outstandingloan balance at the beginning of the final period,” in order to bring the outstanding loanbalance to 0. In order to determine the entries in one row of the table, we do not need togenerate the table line by line; the table is useful when a number of rows are of interest.

Some worked examples Few of the exercises in the textbook require extensive com-putation of schedules. Students should, however, be capable of completing a full tablefor a loan when all the information needed is available.

[1, Exercise 10, p. 196] “A loan is being repaid with quarterly instalments of 1,000 atthe end of each quarter for 5 years at 12% convertible quarterly. Find the amountof principal in the 6th instalment.”

Solution: The principal is

1000a20 3% =1000

0.03

(1− (1.03)−20

)= 14877.47486 .

After the 5th payment, the outstanding principal is (by the prospective method)1000a20−5 3%, so the interest component of the next payment is

30a20−5 3% =30

0.03

(1− (1.03)−15

)= 358.1380526 ,

and the amount of principal reduction is

1000− 358.1380526 = 641.8619474 .


[1, Exercise 11, p. 196] “Consider a loan which is being repaid with instalments of 1at the end of each period for n periods. Find an expression at issue for the presentvalue of the interest which will be paid over the life of the loan.”

Solution: We can use the information in [1, Table 6.1, p. 170]. The sum of the

interest portions of the payments is, as shown in the table,n∑

r=1

(1 − vr) = n − an.

This, however, is not what the problem requests. The present value of the interestpayments — presumably as of time t = 0, is

n∑r=1

vn−r+1(1− vr) = an − nvn+1 .

[1, Exercise 12, p. 196] “A loan of 10,000 is being repaid with 20 instalments at theend of each year, at 10% effective. Show that the amount of interest in the 11thinstalment is

1000

1 + v10.”

Solution: The amount of each payment is, by the prospective method,

10000

a20 10%

.

By the prospective method, the unpaid balance after the 10th payment is

10000

a20 10%

· a10 10% =10000

1 + (1.1)10

so the interest component of the 11th payment is

0.01

(10000

1 + (1.1)10

)=

100

1 + (1.1)10.

[1, Exercise 13, p. 196] “A loan is being repaid with 20 instalments at the end of eachyear at 9% effective. In what instalment are the principal and interest portionsmost nearly equal to each other?”

Solution: Without limiting generality, assume the payments are all of size 1, so thatthe loan is for a20 9%, and [1, Table 6.1, p. 170] applies. In that table we see that therepayments of principal range between v20 in the first payment to v1 in the last. Thetwo portions under consideration sum to a constant, so they cannot both exceed 1

2

at any time. Both sequences — interest payments and reductions of principal —


are monotone: there will be only one pair of values t = a, t = a + 1, where theyreverse their positions from being greater than/less than 1

2to the reverse. But it

is not clear which of those two will have the closest values! We see from [1, Table6.1, p. 170] that the difference between principal and interest components in thetth payment is |1− 2vn−t+1|. We shall begin by determining the largest value of t— if any — for which

v20−t+1 ≤ 1

2,

equivalently, the largest value of t such that

(1 + i)21−t ≥ 2 ,

equivalently, the largest value of t such that

t ≤ 21− ln 2.

ln 1.09= 12.95676827 ,

i.e., t = 12. This is the “best” choice where the difference 1 − 2vn−t+1 is positive.We need also to consider the “best” choice when the difference is negative, i.e.,t = 13. We compute the differences:

t = 12 1− 2(1.09)−9 = 0.0791444410t = 13 1− 2(1.09)−8 = −0.003732559

and see that the values are closest when t = 13.



Distribution Date: Friday, March 11th, 2005, subject to further revision

A.25.1 §6.3 AMORTIZATION SCHEDULES (continued)

[1, Exercise 14, p. 197] “A loan is being repaid with a series of payments at the endof each quarter, for 5 years. If the amount of principal in the 3rd payment is 100,find the amount of principal in the last 5 payments. Interest is at the rate of 10%convertible quarterly.”

Solution: The loan is being repaid in 5 × 4 = 20 quarterly payments, and theeffective interest rate per quarter is 1

4(10%) = 2.5%. Let’s assume that the amount

of each level payment is x, and begin by compiling the first part of the amortizationtable.

Payment Payment Interest Principal Outstandingamount paid repaid loan balance

0 x · a20 2.5%

1 x x(1− (1.025)−20) x(1.025)−20 x · a19 2.5%

2 x x(1− (1.025)−19) x(1.025)−19 x · a18 2.5%

3 x x(1− (1.025)−18) x(1.025)−18 x · a17 2.5%

from which we see that x(1.025)−18 = 100, so

x = 100(1.025)18 = 155.9658718 .

(We didn’t need to use the schedule here. By the Prospective Method, the unpaidbalance just after the 2nd payment is x · s18, so the interest component of the 3rdpayment is ix · s18 and the residue for reduction of principal is

x(1− (

1− v18))

= xv18 .)

We could now compile the last lines of the amortization table backwards. Alter-natively, if the author is requesting the total amount of principal in the last 5payments, that is

x(v + v2 + v3 + v4 + v5) = 155.9658718 · a5 0.025 = 724.5906916 .

[1, Exercise 15, p. 197] “A loan is being repaid with instalments of 1 at the end ofeach year for 20 years. Interest is at effective rate i for the first 10 years, andeffective rate j for the second 10 years. Find expressions for


“a) the amount of interest paid in the 5th instalment;

“b) the amount of principal repaid in the 15th instalment.”

Solution: The present value of the last 10 payments is (1 + i)−10a10 j; the principalof the loan is, therefore,

P = a10 i + (1 + i)−10a10 j .

We compile the first lines of the amortization table:

Year Payment Interest Principal Outstandingamount paid repaid loan balance

0 P1 1 iP 1− iP P (1 + i)− s1 i12 1 i(P (1 + i)− 1) (1 + i)(1− iP ) P (1 + i)2 − s2 i

3 1 i(1 + i)2((

P − a2 i

))s2 i − i(1 + i)2P P (1 + i)3 − s3 i

4 1 i(1 + i)3((

P − a3 i

))s3 i − i(1 + i)3P P (1 + i)4 − s4 i

5 1 i(1 + i)4((

P − a4 i

))s4 i − i(1 + i)4P P (1 + i)5 − s5 i

Thus the amount of interest in the 5th instalment is

i(1 + i)4((a10 i + (1 + i)−10a10 j − a4 i

)= i

(a6 i + (1 + i)−6a10 j

).

More simply, we can observe (using the Prospective Method) that the outstandingbalance just after the 4th instalment is

(1 + i)−6a10 j + a6 i .

The interest component of the 5th payment is obtained by multiplying this amountby i.

After the 10th instalment has been paid, we shift to the second interest rate. Theoutstanding balance after the 14th payment is, again by the Prospective Method,a6 j; interest one payment later will be

j · a6 j = 1− v6

so the payment will reduce principal by 1− (1− v6) = v6. the 5th of the j-portionof the loan — is (1 + j)11−5.

[1, Exercise 16, p. 197] (not discussed in the lectures) “A mortgage with originalprincipal A is being repaid with level payments of K at the end of each yearfor as long as necessary, plus a smaller final payment. The effective rate of interestis i.


“a) Find the amount of principal in the tth instalment.

“b) Is the principal repaid column in the amortization schedule in geometric pro-gression (excluding the irregular final payment)?”

Solution: Just after the (t − 1)th instalment the outstanding principal is, by theretrospective method,

A(1 + i)t−1 −K · st−1

and the interest earned during the next interval will be the product with i; hencethe amount of principal in the tth instalment will be

K − i(A · (1 + i)t−1 −K · st−1

)= K · st − i(1 + i)t−1A

= (K − Ai)(1 + i)t−1

which shows that the column entries are, indeed, in geometric progression, withcommon ratio 1 + i.

[1, Exercise 17, p. 197] (not discussed in the lectures) “A borrower has a mortgagewhich calls for level annual payments of 1 at the end of each year for 20 years. Atthe time of the 7th regular payment an additional payment is made equal to theamount of principal that, according to the original amortization schedule, wouldhave been repaid by the 8th regular payment. If payments of 1 continue to be madeat the end of the 8th and succeeding years until the mortgage is fully repaid, showthat the amount saved in interest payments over the full term of the mortgage is1− v13.”

Solution: At time 0 the amount owing is a20. By the 8th regular payment on theoriginal schedule, the principal repaid would have been

A− A(1 + i)8 + s8 = (Ai− 1)s8 .

It is intended that this amount is added to the 7th payment. Immediately afterthe original 7th payment the principal owing would have been

A(1 + i)7 − s7 .

The additional payment at time 7 reduces this amount to

A(1 + i)7 − s7 −(Ai− 1)s8

)= (1 + i)7 + A

(1− i(1 + i)7

)

[1, Exercise 18, p. 197] (not discussed in the lectures) “A loan of L is being amortizedwith payments at the end of each year for 10 years. If v5 = 2

3, find the following:


“a) The amount of principal repaid in the first 5 payments.

“b) The amount due at the end of 10 years if the final 5 payments are not madeas scheduled.”

Solution:

1. The annual level payments constitute an annuity-immediate with annual pay-ment of

L

a10

=Li

1− (23

)2 =9iL

5.

By the prospective method, the amount of principal remaining to be paidimmediately after the 5th annual payment is

9iL

5· a5 =

9L

5·(

1− 2

3

)=

3L

5.

Hence the amount of principal that has already been paid at that time is(1− 3

5

)L =

2L

5.

2. If no further payments are made, the amount repayable at the end of 10 yearsis

(1 + i)5 · 3L

5=

3

2· 3L

5=

9L

10.

[1, Exercise 19, p. 197] (not discussed in the lectures) “A 35-year loan is to be repaidwith equal instalments at the end of each year. The amount of interest paid in the8th instalment is 135. The amount of interest paid in the 22nd instalment is 108.Calculate the amount of interest paid in the 29th instalment.”

Solution: Suppose the amount of the loan at time 0 was L. The annual instalments

are eachL

a35

. By the prospective method the amount of principal outstanding just

after the 7th instalment is La35·a28. This will incur an interest payment of i · L

a35·a28

in the 8th instalment; we thus have the equation

iL · 1− v28

1− v35= 135 .

A similar computation involving instalments 21 and 22 gives

iL · 1− v14

1− v35= 108 .


Taking the ratio of the two equations, we obtain

1 + v14 =1− v28

1− v14=

135

108=

5

4

so v7 = 12; substitution in either equation above yields iL =

279

2. The amount of

interest in the 29th instalment is

iL · a7

a35

= iL · 1− v7

1− v35=

16iL

31= 72 .

A.25.2 §6.4 SINKING FUNDS

Differences between the Amortization and Sinking Fund Methods of repay-ment In the Amortization Method for repaying a loan, the borrower makes regularpayments — often level payments — directly to the lender. The Outstanding Loan Bal-ance at any time is then the net amount owing on the lender’s books — either the excessof the accumulated value of the loan minus the accumulated value of the payments madeRetrospective Method or the present value of the payments yet to be made ProspectiveMethod .

Where a loan is repaid by a Sinking Fund , portion’s of the borrower’s payments are isnot transmitted to the lender until a later date, usually when they have accumulated infund — the Sinking Fund — often at an interest rate different from the rate associatedwith the loan. Where the borrower is required or permitted to make regular paymentsdirectly to the lender in addition to those paid into the fund, those are described asservice on the loan; where those regular payments cover the interest costs on the loan,the principal outstanding will remain constant. This should not be confused with theNet Amount of the loan, which will be the excess of the outstanding principal over theaccumulated value of the sinking fund.

Where the interest rate associated with the Sinking Fund is the same rate as is beingpaid on the loan, the Sinking Fund Method is equivalent to the Amortization Method.

The identity1

an

=1

sn

+ i. While the identity can easily be proved algebraically,

it admits an interesting verbal proof in terms of a loan of 1 being repaid over n periods.1

an

is the regular payment necessary under an annuity-immediate.1

sn

is the portion of

that regular payment that will accumulate in a sinking fund to a value of 1 just afterthe nth payment; the complement — i — is the amount necessary to service the loanannually until the time that the sinking fund matures.

Omit pages 178-179


[1, Exercise 20, p. 197] “A has borrowed 10,000 on which interest is charged at 10%effective. A is accumulating a sinking fund at 8% effective to repay the loan. Atthe end of 10 years the balance in the sinking fund is 5000. At the end of the 11thyear A makes a total payment of 1500.

“a) How much of the 1500 pays interest currently on the loan?

“b) How much of the 1500 goes into the sinking fund?

“c) How much of the 1500 should be considered as interest?

“d) How much of the 1500 should be considered as principal?

“e) What is the sinking fund balance at the end of the 11th year?”

Solution: This problem requires attention to the terminology used.

1. Presumably we are to assume that the borrower is servicing the loan so thatthe outstanding balance remains constant. The interest payment necessary atthe end of the 11th year will, therefore, be 0.10× 10000 = 1000.

2. The remainder of the payment of 1500 is a contribution of 500 to the sinkingfund.

3. The net interest paid is the excess of the interest paid — 1000 — over the8%× 5000 = 400 interest earned by the sinking fund, or 600.

4. The excess of the contribution over the net interest payment can be assignedto reducing the principal of the loan. The amount is 1500− 600 = 900. Thiscan be considered as made up of two components: 500 which is paid intothe sinking fund, and 8% of 5000, which is 400 — the interest earned by thesinking fund, and which will ultimately be paid to the lender to retire theloan.

5. At the end of the 11th year, the sinking fund balance is

5000(1.08) + 500 = 5900 .

[1, Exercise 21, p. 198] “A loan of 1000 is being repaid with level annual payments of120 plus a smaller final payment made one year after the last regular payment. Theeffective rate of interest is 8%. Show algebraically and verbally that the outstandingloan balance after the 5th payment has been made is:

“a) 1000(1.08)5 − 120 · s5

“b) 1000− 40s5.”

Solution: I give only the verbal explanations.


1. Interpret the loan as being amortized by the regular payments and the finaldrop payment. The amount of 1000(1.08)5 − 120 · s5 is that given by theretrospective method.

2. Now interpret the loan as being repaid by the sinking fund method. As thereis no formal sinking fund, one may simply view a portion of the paymentsas being contributed to a sinking fund in the hands of the lender, where thesinking fund earns interest at the same rate as the principal. Each annualpayment of 120 may be interpreted as being the sum of the service cost of80, plus a contribution of 40 to the sinking fund which will repay the loan atmaturity. After the 5th payment the value of the sinking fund is 40 · s5. The“outstanding loan balance” will be the excess of the face value of the loan(which has been serviced annually, so there is no additional accumulation ofinterest) over the value of the sinking fund.

[1, Exercise 23, p. 198] “On a loan of 10,000, interest at 9% effective must be paidat the end of each year. The borrower also deposits X at the beginning of eachyear into a sinking fund earning 7% effective. At the end of 10 years the sinkingfund is exactly sufficient to pay off the loan. Calculate X.”

Solution: An equation of value at time t = 10 is X · s10 7% = 10000, so

X =10000× 0.07

((1.07)10 − 1) 1.07= 676.4252593 .

The interest rate of 9% is totally irrelevant.

[1, Exercise 24, p. 198] “A borrower is repaying a loan with 10 annual payments of1,000. Half of the loan is repaid by the amortization method at 5% effective. Theother half of the loan is repaid by the sinking fund method, in which the lenderreceives 5% effective on the investment and the sinking fund accumulates at 4%effective. Find the amount of the loan.”

Solution: Let the amount of the loan be L. The amortization of the loan of L2

entails an annual payment ofL

2a10 5%

=0.025L

1− (1.05)−10. For the other half of the

loan the borrower must pay interest annually in the amount of 0.05× L2

= 0.025L.These two expenses — the amortization of half the loan, and the servicing of theother half — leave from his annual payment a balance of

1000− 0.025L

1− (1.05)−10− 0.025L


which must accumulate at 4% in the sinking fund to produce a balance of L2

atmaturity. We have the equation of value

1000− 0.025L

1− (1.05)−10− 0.025L =

L

2s10 4%

=0.02L

(1.04)10 − 1

⇔ L =40000

11−(1.05)−10 + 1 + 0.8

(1.04)10−1

= 7610.479836



Distribution Date: Monday, March 14th, 2005, subject to further revision

The start of the lecture was delayed 15 minutes because of the expectedadministration of a course evaluation. (However, the person who was toadminister it did not arrive.)

A.26.1 §6.4 SINKING FUNDS (concluded)

Negative final payment? I begin this lecture with an example from an earlier chapterwhich can be relevant to sinking fund problems. In [1, Example 3.7, pp. 74–75] there isa fund that has a designated target value, where the payments are not designed so thatan integer number of them will exactly bring the fund to its desired value. This meansthat there can be a final payment which is less than a prescribed maximum value. But itcan also happen that, at the time of a payment, the outstanding balance is just slightlylarger than a prescribed maximum payment, so that, after the payment is made, a verysmall balance remains. One waits for the next payment, and, during that time, the valueof the fund grows by an accumulation factor, and may exceed the target value. So thenext payment is negative! Read the example. This type of situation can occur in otherways, so you should be prepared for it, even though it is not the most likely outcome.

[1, Exercise 25, p. 198] “A borrows 12,000 for 10 years, and agrees to make semian-nual payments of 1,000. The lender receives 12% convertible semiannually on theinvestment each year for the first 5 years and 10% convertible semiannually for thesecond 5 years. The balance of each payment is invested in a sinking fund earning8% convertible semiannually. Find the amount by which the sinking fund is shortof repaying the loan at the end of the 10 years.”

Solution: The interest payments for the first 10 half-years are 6% of 12,000, i.e.720 per half-year; and, for the second 10 half-years, 600 per half-year. This leaves280 at the end of each of the first 10 half-years, and 400 at the end of each of thesecond 10 half-years to accumulate in the sinking fund, which earns 4% effectiveevery half year. The accumulated balance in the sinking fund at maturity will be

120s10 4% + 280s20 4% =1

0.04

(120

((1.04)10 − 1

)+ 280

((1.04)20 − 1

))

= 25(120(1.04)10 + 280(1.04)20 − 400

)

= 9778.594855

implying that the shortfall to repay the loan will be 12, 000− 9778.59 = 2221.41.


[1, Exercise 26, p. 198] 1. “A borrower takes out a loan of 3000 for 10 years at 8%convertible semiannually. The borrower replaces one-third of the principalin a sinking fund earning 5% convertible semiannually, and the other two-thirds in a sinking fund earning 7% convertible semiannually. Find the totalsemiannual payment.

2. “Rework (a) if the borrower each year puts one-third of the total sinking funddeposit into the 5% sinking fund and the other two-thirds into the 7% sinkingfund.

3. “Justify from general reasoning the relative magnitude of the answers to (a)and (b).”

Solution:

1. The semiannual contribution to the sinking funds is

1000

s20 2.5%

+2000

s20 3.5%

and the semiannual interest payment is 4% of 3, 000, or 120. Hence the totalsemiannual payment is

1000

s20 2.5%

+2000

s20 3.5%

+ 120 =25

(1.025)20 − 1+

70

(1.035)20 − 1+ 120

= 229.8692824

2. Let the total sinking fund deposit be D. Then the equation of value at ma-turity is

D

3· s20 2.5% +

2D

3· s20 3.5% = 3000 ,

implying that

D =9000

s20 2.5% + 2s20 3.5%

=9000

(1.025)20−10.025

+ 2 · (1.035)20−10.035

= 109.6170427 ,

so the total semi-annual payment is 109.6170427+120=229.6170427.

3. In the original repayment scheme the portion of the payment contributed tothe 5% sinking fund grows more slowly than that to the 7% fund. Thus, whilethe final accumulations in the funds will be in the ratio of 1:2, the proportion


of the contribution to the 5% fund would have been more than 13. By reducing

that proportion to 13

we increased the interest earned by the fund, so a smallertotal contribution was required for the sinking fund.

[1, Exercise 27, p. 198] “A payment of 36,000 is made at the end of each year for 31years to repay a loan of 400,000. If the borrower replaces the capital by means of asinking fund earning 3% effective, find the effective rate paid to the lender on theloan.”

Solution: The annual contribution to the sinking fund is

400000

s31 3%

=12000

(1.03)31 − 1= 7999.571516.

Hence the annual interest payment is 36, 000−7, 999.57 = 28000.43, i.e., 7% of theprincipal of 400,000.

[1, Exercise 28, p. 198] “A 20-year annuity-immediate has a present value of 10,000,where interest is 8% effective for the first 10 years, and 7% effective for the second10 years. An investor buys this annuity at a price which, over the entire period,yields 9% on the purchase price; and, further, allows the replacement of capital bymeans of a sinking fund earning 6% for the first 10 years and 5% for the second 10years. Find an expression for the amount that is placed in the sinking fund eachyear.”

Solution: The level annual payments under the annuity will be

10000

a10 8% + (1.08)−10a10 7%

.

It appears to be intended that the sinking fund payments be level also. If theirvalue is S, and the purchase price is P , then

S((1.05)10s10 6% + s10 5%

)= P .

The purchase price satisfies the following equation of value at time 0:

(10000

a10 8% + (1.08)−10a10 7%

− S

)a20 9% + (1.09)−20P = P ,

which implies that

(10000

a10 8% + (1.08)−10a10 7%

− S

)= 0.09P .


Solving with the earlier equation yields

S =10000(

a10 8% + (1.08)−10a10 7%

) (1 + 0.09

((1.05)10s10 6% + s10 5%

)) .

[1, Exercise 29, p. 199] “A loan of 1 yields the lender rate i per period for n periods,while the borrower replaces the capital in a sinking fund earning rate j per period.Find expressions for the following if 1 ≤ t ≤ n:

1. Periodic interest paid to the lender.

2. Period sinking fund deposit.

3. Interest earned on sinking fund during the tth period.

4. Amount in sinking fund at end of the tth period.

5. Net amount of loan at the end of the tth period.

6. Net interest paid in period t.

7. Principal repaid in period t.

Solution:

1. The yield rate is i, so the lender must be receiving an amount of i each period.

2. The capital is 1, so the borrower is depositing s−1n j regularly into the sinking

fund. (I understand that interest will be paid regularly, and the capital willbe repaid at maturity.)

3. At the beginning of the tth period the balance in the sinking fund isst−1 j

sn j;

during the period it earns interest in the amount of j · st−1 j

sn j, payable at the

end of the period, i.e., at time t.

4. The amount in sinking fund at end of the tth period isst j

sn j.

5. The net amount of loan at the end of the tth period is the excess of 1 over

the balance in the sinking fund, i.e., 1− st j

sn j.

6. The net interest paid in the tth period is the excess of interest paid over

interest earned, i.e. i− j · st−1 j

sn j

7. By 5. above, the change in the amount of the loan between the t − 1th andthe tth payment is

(1− st j

sn j

)−

(1− st−1 j

sn j

)=

st j − st−1 j

sn j

=(1 + i)t−1

sn j

.


A.26.2 §6.5 DIFFERING PAYMENT PERIODS AND INTEREST CON-VERSION PERIODS

Omit this section.

A.26.3 §6.6 VARYING SERIES OF PAYMENTS

Omit this section.

A.26.4 §6.7 AMORTIZATION WITH CONTINUOUS PAYMENTS

Omit this section.

A.26.5 §6.8 STEP-RATE AMOUNTS OF PRINCIPAL

Omit this section.



Distribution Date: Wednesday, March 16th, 2005, subject to further revision

Textbook Chapter 7. Bonds and other securities.


The chapter is concerned with relations between the price of a security and its yield rate,and with the value of a security at any time after it has been purchased, even at a timethat is not an interest compounding date.

A.27.2 §7.2 TYPES OF SECURITIES

We shall confine our study to bonds , which are a commitment by the issuer to repaya loan at a particular time, with interest payments according to a prescribed rate andschedule. Many variations are possible, and this type of instrument is still evolving.

Read the book and become familiar with the following terms concerning bonds:

Definition A.16 • The word bond originally had a much more general meaning: weare using the word in the sense of a security that commits a borrower to pay oneor more specific sums at specific times, subject to detailed requirements of interestand/or bonuses.

• The term of the bond is the length of time from the date of issue until the date offinal payment, which date is the maturity date.

• The detailed conditions of the bond may permit the bond to be called at some dateprior to the maturity date, at which time the issuer (=the lender) will repay thecommitment, possibly with some additional amounts. Such a bond is callable.

• While the calling of a bond is at the initiative of the borrower, the lender (=thepurchaser) may possibly have the right to redeem the bond prior to the date ofmaturity. Or, he may have some other type of right, e.g., to exchange the bond forshares of the stock of the issuing company; this is a convertible bond.

• A bond may be supplied with coupons — portions of the paper bond that are to becut (=coupes) from the bond and exchanged for interest payments. Nowadays thecoupons may no longer be printed, but the purchaser may receive regular paymentsfrom the lender. This can be the case if the bond is fully registered , so that theissuer has the coordinates of the purchaser. (A bond could also be registered onlyas to principal , in which case the coupons are of the traditional type.)


• A coupon bond may be “stripped”, separating the coupons from the commitmentto repay the face value of the bond, following which the two parts may be sold toseparate purchasers.

The preceding is just a brief introduction: this is not the course to learn about the varietyof investment vehicles available in today’s financial markets.


A.28 Supplementary Notes for the Lecture of March 21st, 2005

Distribution Date: Monday, March 21st, 2005, subject to further revision

A.28.1 §7.2 TYPES OF SECURITIES (conclusion)

Unlike the problems we have been considering in earlier chapters, those here sometimesinvolve technical definitions that are not necessarily intuitive, and need simply to bememorized. I will eventually discuss with the class what definitions need to be absorbedfor examination purposes.

[1, Exercise 1, p. 240] “Find the price which should be paid for a zero coupon bondwhich matures for 1000 in 10 years to yield:

1. 10% effective

2. 9% effective

3. Thus a 10% reduction in the yield rate causes the price to increase by whatpercentage?”

Solution:

1. The bond is now worth 1000(1.10)−10 = 385.5432894.

2. When the interest rate is reduced to 9%, the present value of the bond in-creases to 422.4108069.

3. The 10% decrease in the interest rate thereby increases the price by

422.4108069

385.5432894− 1 = 9.5624846%.

[1, Exercise 2, p. 240] “A 10-year accumulation bond with an initial par value of 1000earns interest of 8% compounded semiannually. Find the price to yield an investor10% effective.”

Solution:

Definition A.17 [1, p. 205] An accumulation bond is one in which the redemptionprice includes the original loan plus all accumulated interest.

Solution: The only return payment is at maturity. The price to yield 10% interestwill therefore be

(1.10)−10(1000(1.04)20

)= 844.7728240.


[1, Exercise 3, p. 240] “A 26-week (U.S.) T(reasury)-bill is bought for 9,600 at issue,and will mature for 10,000. Find the yield rate computed as:

1. A discount24 rate, using the typical method for counting days on a T-bill25.

2. An annual effective rate of interest, assuming the investment period is exactlyhalf a year.”

Solution: Here is an example of a problem requiring some technical preparation.While we have encountered the use of discount rather than interest in isolatedproblems, this is the first time we have met it in a complex transaction.

1. The time is 26 weeks, i.e., 26×7 = 182 days, or, under the actual/360 system,182

360of a year. The discount rate will, therefore, be

360

182× 400

10000= 7.912087912% .

2. The effective interest rate for half a year is

400

9600=

1

24.

The effective rate for a full year will be

(1 +

1

24

)2

− 1 =1

12+

1

576= 8.506944444% .

A.28.2 §7.3 PRICE OF A BOND

Remember that a bond is essentially a contract to pay a large amount “at maturity”,and smaller amounts of interest periodically on “coupon dates” until maturity. It differsfrom the types of loans we have studied hitherto in that the contract is often (but notalways)26 transferable from one owner to another, and so it is reasonable to investigatethe value of the entire contract under conditions of varying yield.

24simple discount25i.e., “actual/360” [1, p. 39], using the exact number of days, but assuming 360 days in the year.26Of course, the issuer of a security cannot prevent the purchaser from arranging privately to transfer

the proceeds to another person, and to receive payment for that. What the issuer may be able to do is torestrict the establishment of a secondary market at which the security may be routinely bought and sold,and which would create a market value for the security. Some securities have restricted transferabilityconditions, e.g., only on the death of the owner. And no security is immune from a court order.


Further definitions Familiarize yourself with the following terms, defined in the text-book, and with the symbols usually used for them:

Definition A.18 1. The price P paid for a bond. In practice bond prices are usuallyquoted in terms of a bond with face value (see next item) of 100.

2. The par value or face value or face amount . This amount is usually printed inthe bond contract, but may not be the amount paid at maturity. Its functionis to determine, once the coupon rate r has been specified, the magnitude of thecoupons.

3. The redemption value C is the amount paid when the bond is redeemed. Whena bond is “redeemable at par”, C = F . Where the redemption value exceeds theface value, the word premium may be used for the excess; this word premium isalso used to denote the excess of the price paid for a bond over what would havebeen the value if the yield rate was the same as the coupon rate.

4. The coupon rate r is the effective rate per coupon payment period, based on whichthe amount of the coupon is calculated. The default payment period is a half-year.

5. The amount of a coupon is the product Fr.

6. The modified coupon rate g =Fr

Cis the coupon rate per unit of redemption value,

rather than per unit of par value.

7. The yield rate or yield to maturity i is the actual interest rate earned by theinvestor.

8. The number of coupon payment periods from the date of calculation until maturityis denoted by n.

9. The present value of the redemption value, discounted back to the present by theyield rate, is denoted by K; so K = C(1 + i)−n.

10. The base amount G isFr

i: the amount which, if invested at the yield rate i, would

produce periodic interest payments equal to the coupons.

11. A callable bond is one where the lender has the right to declare that interest pay-ments will stop and a bond may be redeemed at certain dates before the maturitydate; there could be a premium paid in addition to the redemption value, to en-courage lenders to cash in the bond.

12. The word discount is often used where we have been using the word premium ifthe premium is negative: the discount is the negative of the premium.


Four formulæ for price. We shall consider four different ways of determining thepresent value, or price of a bond. The formulae are derivable one from the other: thereare situations where one may be more useful than another for specific applications; aswith many of the other formula we have met, the relative advantages were often linked tothe number of times that tables had to be consulted in computing the bond value; suchdistinctions may no longer be significant. The most “basic” of the formulæ computes theprice of a bond as the present value of the coupons, interpreted as an annuity-immediate,to which is added the present value K of the redemption value C of the bond, the facevalue plus any premium that is payable upon redemption. This formula may be appliedeither at the maturity date, or, if the bond can be called (by the issuer) or redeemed (bythe purchaser), at some earlier date.

Theorem A.9 (The “Basic” Formula) P = Fr · an i + Cvn = Fr · an i + K

Theorem A.10 (The Premium/Discount Formula) P = C + (Fr − Ci) · an

Proof: This formula may be derived from the preceding by recalling that 1 = vn + i ·an i. ¤

The Base Amount G was defined above to beFr

i.

Theorem A.11 (The Base Amount Formula) P = G + (C −G)vn

Theorem A.12 (Makeham’s Formula) P = K +g

i(C −K)

There are interesting verbal explanations of the preceding formulæ, to which we mayreturn.

[1, Exercise 5, p. 241] “A 10-year 100 par value bond bearing a 10% coupon ratepayable semi-annually, and redeemable at 105, is bought to yield 8% convertiblesemiannually. Find the price. Verify that all four formulæ produce the sameanswer.”

Solution:F 100C 105r 5%

Fr 5g = Fr

C121

i 4%n 20

K = C(1 + i)−n 105(1.04)−20 = 47.92G = Fr

i125


Basic Formula P = 100(0.05)·a20 4%+105(1.04)−10 = 100(0.05)·a10+105(1.04)−20 =115.8722611.

Premium/Discount Formula P = 105+(100(0.05)−105(0.04))·a20 4% = 115.8722611.

Base Amount Formula P = 125 + (105− 125)(1.04)−20 = 115.8722611.

Makeham’s Formula P = 105(1.04)−20+121

0.04(105− 105(1.04)−20) = 115.8722611.

[1, Exercise 6, p. 241] “For the bond in [1, Example 7.3, pp. 212-213] determine thefollowing:

1. nominal yield based on the par value

2. nominal yield, based on the redemption value

3. current yield

4. yield to maturity.”

(The bond is described as “a 1000 par value 10-year bond with coupons at 8.4%convertible semiannually, which will be redeemed at 1050, purchased to yield 10%convertible semiannually.”.)

Solution:

1. The nominal yield “is simply the annualized coupon rate on the bond”. Sincethe coupon rate is 8.4% (convertible...), that is the nominal yield.

2. This usage of the term nominal differs slightly from that given in the textbook.The intention is that we should interpret the coupons as percentages of, notthe face value, but the redemption value, i.e, we should use the modifiedcoupon rate, which here is Fr

C= 42

1050= 4%. But that is the effective rate

per half-year, so the corresponding nominal annual rate compounded semi-annually is 8%.

3. The current yield “is the ratio of the annualized coupon to the original priceof the bond” [1, p. 221]. Here the coupon is 84, and the price has been shownin the example to be 919.15, so the current yield is 84

919.15= 9.139%.

4. The purchase price was calculated to yield 10% (convertible semi-annually).

[1, Exercise 7, p. 241] “Two 100 par value bonds, both with 8% coupon rates payablesemi-annually are currently selling at par. Bond A matures in 5 years at par, whileBond B matures in 10 years at par. If prevailing market rates of interest suddenlygo to 10% convertible semiannually, find the percentage change in the price of

1. Bond A


2. Bond B

3. Justify from general reasoning the relative magnitudes of the answers to (a)and (b).”

Solution:

1. For both of the bonds F = C = 100, r = 4%; for bond A n = 10. Using thePremium/Discount formula we find that

P = C + (Fr − Ci)an 4%

= 100 + 100(0.04− 0.05) · 1− (1.05)−10

0.05= 92.27826507

The percentage change in the price of this bond is −7.72173493100

= −7.72173493%.

2. For bond B n = 20. Using the Premium/Discount formula we find that

P = C + (Fr − Ci)an 4%

= 100 + 100(0.04− 0.05) · 1− (1.05)−20

0.05= 87.53778966


= −12.46221034%.

3. the last coupon payments under bond B are more affected by the rate changethan the first.

[1, Exercise 8, p. 241] “Two 1000 bonds redeemable at par at the end of the sameperiod are bought to yield 4% convertible semiannually. One bond costs 1136.78,and has a coupon rate of 5% payable semiannually. The other bond has a couponrate of 21

2% payable semiannually. Find the price of the second bond.”

Solution: For the first bond we have F1 = C1 = 1000, i1 = 2%, P1 = 1136.78,r1 = 2.5%. By the Premium/Discount formula,

1136.78 = 1000 + (25− 20) · an 2% ,

implying that an = 27.356. For the second bond we have F2 = C2 = 1000, i2 = 2%,r2 = 1.25%, n2 = n1,

P2 = 1000 + (12.5− 20) · an

= 794.83 .


[1, Exercise 9, p. 241] “A 1000 bond with a coupon rate of 9% payable semiannuallyis redeemable after an unspecified number of years at 1125. The bond is bought toyield 10% convertible semiannually. If the present value of the redemption value is225 at this yield rate, find the purchase price.”

Solution: We have F = 1000, C = 1125, r = 4.5%, i = 5%, K = 225. From this

last fact we have 225 = 1125(1.05)−n, so (1.05)n = 5, and an i =1− 1

5

0.05= 16. By the

Basic Formula, the price is Fr · an + K = 45(16) + 225 = 945.

[1, Exercise 10, p. 241] A 1000 par value n-year bond maturing at par with annualcoupons of 100 is purchased for 1110. If K = 450, find the base amount G.

Solution: We have F = C = 1000, Fr = 100 (so r = 10%), P = 1110, K = 450.By Makeham’s formula,

1110 = 450 +Fr

Ci(C −K) = 450 +

100

1000i(1000− 450)

so i = 112

, and

G =Fr

i= 1200 .

[1, Exercise 11, p. 241] “An investor owns a 1000 par value 10% bond with semian-nual coupons. The bond will mature at par at the end of 10 years. The investordecides that an 8-year bond would be preferable. Current yield rates are 7% con-vertible semiannually. The investor uses the proceeds from the sale of the 10%bond to purchase a 6% bond with semiannual coupons, maturing at par at the endof 8 years. Find the par value of the 8-year bond.”

Solution: We have C = F = 1000, r = 5%, n1 = 20, i = 3.5%. The price of thebond presently owned is, by the premium/discount formula,

P1 = C + (Fr − Ci)a20 3.5% = 1000 + (50− 35) · 1− (1.035)−20

0.035= 1213.186050 .

With these proceeds the investor buys a bond with par value F = C, n = 16,r = 3%; by the premium/discount formula,

1213.186050 = F + F (0.03− 0.035) · 1− (1.035)−16

0.035,

implying that F = 1291.269895.

[1, Exercise 12, p. 241] “An n-year 1000 par value bond matures at par and has acoupon rate of 12% convertible semiannually. It is bought at a price to yield 10%


convertible semiannually. If the term of the bound is doubled, the price will increaseby 50. Find the price of the n-year bond.”

Solution: We have F = C = 1000, r = 6%, i = 5%. Since we know that

a4n = 2 · a2n − i(a2n

)2,

we can eliminate P between the two equations

P = 1000 + (60− 50)a2n ,

P + 50 = 1000 + (60− 50)a4n ,

and deduce that (a2n

)2 − 20a2n + 100 = 0

so a2n = 10, and P = 1000 + 10(10) = 1100.


A.29 Supplementary Notes for the Lecture of March 21st, 2005

Distribution Date: Monday, March 21st, 2005, subject to further revision

A.29.1 §7.2 TYPES OF SECURITIES (conclusion)

[1, Exercise 1, p. 240] “Find the price which should be paid for a zero coupon bondwhich matures for 1000 in 10 years to yield:

1. 10% effective

2. 9% effective

3. Thus a 10% reduction in the yield rate causes the price to increase by whatpercentage?”

Solution:

1. The bond is now worth 1000(1.10)−10 = 385.5432894.

2. When the interest rate is reduced to 9%, the present value of the bond in-creases to 422.4108069.

3. The 10% decrease in the interest rate thereby increases the price by

422.4108069

385.5432894− 1 = 9.5624846%.

[1, Exercise 2, p. 240] “A 10-year accumulation bond with an initial par value of 1000earns interest of 8% compounded semiannually. Find the price to yield an investor10% effective.”

Solution:

Definition A.19 [1, p. 205] An accumulation bond is one in which the redemptionprice includes the original loan plus all accumulated interest.

Solution: The only return payment is at maturity. The price to yield 10% interestwill therefore be

(1.10)−10(1000(1.04)20

)= 844.7728240.

[1, Exercise 3, p. 240] “A 26-week (U.S.) T(reasury)-bill is bought for 9,600 at issue,and will mature for 10,000. Find the yield rate computed as:

1. A discount27 rate, using the typical method for counting days on a T-bill28.

27simple discount28i.e., “actual/360” [1, p. 39], using the exact number of days, but assuming 360 days in the year.


2. An annual effective rate of interest, assuming the investment period is exactlyhalf a year.”

Solution:

1. The time is 26 weeks, i.e. 26×7 = 182 days, or, under the actual/360 system,182

360of a year. The discount rate will, therefore, be

360

182× 400

10000= 7.912087912% .

2. The effective interest rate for half a year is

400

9600=

1

24.

The effective rate for a full year will be

(1 +

1

24

)2

− 1 =1

12+

1

576= 8.506944444% .

[1, Exercise 4, p. 241] “T-bills of all maturities yield 8% compounded on a discountbasis. Find the ratio of the annual effective rate of interest earned on a 52-weekT-bill to that earned on a 13-week T-bill. Use an approach which does not involvethe counting of days.”

Solution: The 52-week T-bill will earn interest at the rate

d

1− d=

0.08

0.92= 8.695652174%.

The 13-week T-bill earns interest quarterly at the rate

d

1− d=

0.02

0.92= 2.173913043% ,

so the effective annual rate is

(1.02173913043)4 − 1 = 8.9833377% ,

and the ratio to the 52-week rate to this is8.695652174

8.9833377= 0.9679756527. THIS IS

NOT THE AUTHOR’S ANSWER. HE GIVES 1.0332. Note that1

0.9679756527=

1.033083836. The effective rate for a shorter period should be higher than that fora longer period because of the compounding, since the rates of 8% given are simplediscount rates.


A.29.2 §7.3 PRICE OF A BOND

Remember that a bond is essentially a contract to pay a large amount “at maturity”,and smaller amounts of interest periodically on “coupon dates” until maturity. It differsfrom the types of loans we have studied hitherto in that the contract is often (but notalways) transferable from one owner to another, and so it is reasonable to investigate thevalue of the entire contract under conditions of varying yield.

Familiarize yourself with the following terms, defined in the textbook, and with thesymbols usually used for them:

• The price P paid for a bond. In practice bond prices are usually quoted in termsof a bond with face value (see next item) of 100.

• The par value or face value or face amount . This amount is usually printed inthe bond contract, but may not be the amount paid at maturity. Its functionis to determine, once the coupon rate r has been specified, the magnitude of thecoupons.

• The redemption value C is the amount paid when the bond is redeemed. Whena bond is “redeemable at par”, C = F . Where the redemption value exceeds theface value, the word premium may be used for the excess; this word premium isalso used to denote the excess of the price paid for a bond over what would havebeen the value if the yield rate was the same as the coupon rate.

• The coupon rate r is the effective rate per coupon payment period, based on whichthe amount of the coupon is calculated. The default payment period is a half-year.

• The amount of a coupon is the product Fr.

• The modified coupon rate g =Fr

Cis the coupon rate per unit of redemption value,

rather than per unit of par value.

• The yield rate or yield to maturity i is the actual interest rate earned by theinvestor.

• The number of coupon payment periods from the date of calculation until maturityis denoted by n.

• The present value of the redemption value, discounted back to the present by theyield rate, is denoted by K; so K = C(1 + i)−n.

• The base amount G isFr

i: the amount which, if invested at the yield rate i, would

produce periodic interest payments equal to the coupons.


• A callable bond is one where the lender has the right to declare that interest pay-ments will stop and a bond may be redeemed at certain dates before the maturitydate; there could be a premium paid in addition to the redemption value, to en-courage lenders to cash in the bond.

• The word discount is often used where we have been using the word premium ifthe premium is negative: the discount is the negative of the premium.

Four formulæ for price.

Theorem A.13 (The “Basic” Formula) P = Fr · an + Cvn = Fr · an + K

Theorem A.14 (The Premium/Discount Formula) P = C + (Fr − Ci) · an

Theorem A.15 (The Base Amount Formula) P = G + (C −G)vn

Theorem A.16 (Makeham’s Formula) P = K +g

i(C −K)

[1, Exercise 5, p. 241] “A 10-year 100 par value bond bearing a 10% coupon ratepayable semi-annually, and redeemable at 105, is bought to yield 8% convertiblesemiannually. Find the price. Verify that all four formulæ produce the sameanswer.”

Solution:F 100C 105r 5%

Fr 5g = Fr

C121

i 4%n 20

K = C(1 + i)−n 105(1.04)−20 = 47.92G = Fr

i125

Basic Formula P = 100(0.05)·a20 4%+105(1.04)−10 = 100(0.05)·a10+105(1.04)−20 =115.8722611.

Premium/Discount Formula P = 105+(100(0.05)−105(0.04))·a20 4% = 115.8722611.

Base Amount Formula P = 125 + (105− 125)(1.04)−20 = 115.8722611.

Makeham’s Formula P = 105(1.04)−20+121

0.04(105− 105(1.04)−20) = 115.8722611.


[1, Exercise 6, p. 241] “For the bond in [1, Example 7.3, pp. 212-213] determine thefollowing:

1. nominal yield based on the par value

2. nominal yield, based on the redemption value

3. current yield

4. yield to maturity.”

(The bond is described as “a 1000 par value 10-year bond with coupons at 8.4%convertible semiannually, which will be redeemed at 1050, purchased to yield 10%convertible semiannually.”.)

Solution:

1. The nominal yield “is simply the annualized coupon rate on the bond”. Sincethe coupon rate is 8.4% (convertible...), that is the nominal yield.

2. This usage of the term nominal differs slightly from that given in the textbook.The intention is that we should interpret the coupons as percentages of, notthe face value, but the redemption value, i.e, we should use the modifiedcoupon rate, which here is Fr

C= 42

1050= 4%. But that is the effective rate

per half-year, so the corresponding nominal annual rate compounded semi-annually is 8%.

3. The current yield “is the ratio of the annualized coupon to the original priceof the bond” [1, p. 221]. Here the coupon is 84, and the price has been shownin the example to be 919.15, so the current yield is 84

919.15= 9.139%.

4. The purchase price was calculated to yield 10% (convertible semi-annually).

[1, Exercise 7, p. 241] “Two 100 par value bonds, both with 8% coupon rates payablesemi-annually are currently selling at par. Bond A matures in 5 years at par, whileBond B matures in 10 years at par. If prevailing market rates of interest suddenlygo to 10% convertible semiannually, find the percentage change in the price of

1. Bond A

2. Bond B

3. Justify from general reasoning the relative magnitudes of the answers to (a)and (b).”

Solution:


1. For both of the bonds F = C = 100, r = 4%; for bond A n = 10. Using thePremium/Discount formula we find that

P = C + (Fr − Ci)an 4%

= 100 + 100(0.04− 0.05) · 1− (1.05)−10

0.05= 92.27826507


= −7.72173493%.

2. For bond B n = 20. Using the Premium/Discount formula we find that

P = C + (Fr − Ci)an 4%

= 100 + 100(0.04− 0.05) · 1− (1.05)−20

0.05= 87.53778966


= −12.46221034%.

3. the last coupon payments under bond B are more affected by the rate changethan the first.

[1, Exercise 8, p. 241] “Two 1000 bonds redeemable at par at the end of the sameperiod are bought to yield 4% convertible semiannually. One bond costs 1136.78,and has a coupon rate of 5% payable semiannually. The other bond has a couponrate of 21

2% payable semiannually. Find the price of the second bond.”

Solution: For the first bond we have F1 = C1 = 1000, i1 = 2%, P1 = 1136.78,r1 = 2.5%. By the Premium/Discount formula,

1136.78 = 1000 + (25− 20) · an 2% ,

implying that an = 27.356. For the second bond we have F2 = C2 = 1000, i2 = 2%,r2 = 1.25%, n2 = n1,

P2 = 1000 + (12.5− 20) · an

= 794.83 .


A.30 Supplementary Notes for the Lecture of March 23rd, 2005

Distribution Date: Wednesday, March 23rd, 2005, subject to further revision

The following problems are from the textbook, but were not discussed explicitly in thelecture.

[1, Exercise 9, p. 241] “A 1000 bond with a coupon rate of 9% payable semiannuallyis redeemable after an unspecified number of years at 1125. The bond is bought toyield 10% convertible semiannually. If the present value of the redemption value is225 at this yield rate, find the purchase price.”

Solution: We have F = 1000, C = 1125, r = 4.5%, i = 5%, K = 225. From this

last fact we have 225 = 1125(1.05)−n, so (1.05)n = 5, and an i =1− 1

5

0.05= 16. By the

Basic Formula, the price is Fr · an + K = 45(16) + 225 = 945.

[1, Exercise 10, p. 241] A 1000 par value n-year bond maturing at par with annualcoupons of 100 is purchased for 1110. If K = 450, find the base amount G.

Solution: We have F = C = 1000, Fr = 100 (so r = 10%), P = 1110, K = 450.By Makeham’s formula,

1110 = 450 +Fr

Ci(C −K) = 450 +

100

1000i(1000− 450)

so i = 112

, and

G =Fr

i= 1200 .

[1, Exercise 11, p. 241] “An investor owns a 1000 par value 10% bond with semian-nual coupons. The bond will mature at par at the end of 10 years. The investordecides that an 8-year bond would be preferable. Current yield rates are 7% con-vertible semiannually. The investor uses the proceeds from the sale of the 10%bond to purchase a 6% bond with semiannual coupons, maturing at par at the endof 8 years. Find the par value of the 8-year bond.”

Solution: We have C = F = 1000, r = 5%, n1 = 20, i = 3.5%. The price of thebond presently owned is, by the premium/discount formula,

P1 = C + (Fr − Ci)a20 3.5% = 1000 + (50− 35) · 1− (1.035)−20

0.035= 1213.186050 .

With these proceeds the investor buys a bond with par value F = C, n = 16,r = 3%; by the premium/discount formula,

1213.186050 = F + F (0.03− 0.035) · 1− (1.035)−16

0.035,

implying that F = 1291.269895.


[1, Exercise 12, p. 241] “An n-year 1000 par value bond matures at par and has acoupon rate of 12% convertible semiannually. It is bought at a price to yield 10%convertible semiannually. If the term of the bound is doubled, the price will increaseby 50. Find the price of the n-year bond.”

Solution: We have F = C = 1000, r = 6%, i = 5%. Since we know that

a4n = 2 · a2n − i(a2n

)2,

we can eliminate P between the two equations

P = 1000 + (60− 50)a2n ,

P + 50 = 1000 + (60− 50)a4n ,

and deduce that (a2n

)2 − 20a2n + 100 = 0

so a2n = 10, and P = 1000 + 10(10) = 1100.

Most of the lecture was devoted to a discussion of [5, Example 5.9, pp. 99-101]. Inthat example the bond with r = 4% was purchased at a premium because the yield ratewas i = 3%. Then we considered the same problem with the two rates reversed (notdiscussed explicitly in the cited source).



Distribution Date: Thursday, March 31st, 2005corrected Friday, April 1st, 2005,

subject to further revision

A.31.1 §7.4 PREMIUM AND DISCOUNT

Premium = P − C = (Fr − Ci)an i

Discount = C − P = −(Fr − Ci)an i

In practice the word premium is used when P − C > 0, and discount when P − C < 0;when P = C we speak of a purchase at par .

Book Value For accounting purposes it is necessary to show a gradual progression ofthe value of a bond from purchase to maturity that is in some reasonable relationshipwith its market value. There are a number of possible methods for doing this in practice,but we shall adhere to the method whereby the value is shown to be what would be theprice if the yield rate does not change after the purchase of the bond. Of course, the yieldrate could very well change after the purchase, and so this is not a completely realisticassumption. There are other methods for assigning a value to the bond. In this sectionwe shall consider the value only at coupon payment dates; generalization to other dateswill be considered in the next section.

When a bond is purchased at a premium29 the value of the redemption value andthe unpaid coupons just after the payment of the tth coupon — i.e. of the book valueBt — will decrease until the maturity date. Thus the coupons may be interpreted asconsisting of the interest which is earned at the yield rate i and an amount to amortizethe premium. The redemption value C is C = Bn, and the price is P = B0. We will usethis notation even when the bond is purchase with some of its coupons already paid (tothe vendor). The writing down of the premium may be shown on a schedule like

Coupon Coupon Interest Amount for Amortization BookNumber amount earned of Premium value

0 . . . . . .1 Fr

. . . Fr . . . . . . . . .

29mutatis mutandis, at a discount


[1, Exercise 16, p. 242] For a bond of face value 1 the coupon rate is 150% of theyield rate, and the premium is p. For another bond of 1 with the same number ofcoupons and the same yield rate, the coupon rate is 75% of the yield rate. Findthe price of the second bond.

Solution: We are assuming that C = F for both bonds, and that the commonvalue is 1. For the first bond we have

1 + p = C + (Fr − Ci)an i = 1 + (1.5− 1)ian i

so ian i = 2p. The price of the second bond is then, again by the Premium/Discountformula,

1 + (0.75− 1)ian i = 1− 0.25ian i = 1− p

2.

[1, Exercise 17, p. 242] “For a certain period a bond amortization schedule showsthat the amount for amortization of premium is 5, and that the required interestis 75% of the coupon. Find the amount of the coupon.”

Solution: The bond was purchased at a premium, because the coupon rate exceedsthe yield rate. In the question, the author means by “the amount of the coupon”,the total amount of all the coupons under discussion. We have

Frn− Fin = 5

andFin

Frn= 0.75 .

implying that Fr = 4(5) = 20.

[1, Exercise 18, p. 242] “A 10-year bond with semi-annual coupons is bought at adiscount to yield 9% convertible semiannually. If the amount for accumulation ofdiscount in the next-to-last coupon is 8, find the total amount for accumulation ofdiscount during the first four years in the bond amortization schedule.”

Solution: We have i = 4.5%, n = 20,

C(g − i)v2 = C

(Fr

C− i

)v2 = C · r − i

(1 + i)2= 8 ,

soC(g − i) = 8(1 + i)2 .


The total amount for accumulation of discount during the first four years in theschedule is

C(g − i)(v20 + v19 + . . . + v13) = 8(1 + i)2(v20 + v19 + . . . + v13)

= 8v14 · a4 i

= 8 · (1.045)−10 − (1.045)−18

0.045= 33.98

[1, Exercise 19, p. 242] “A 1000 par value 5-year bond with a coupon rate of 10%payable semi-annually and redeemable at par is bought to yield 12% convertiblesemiannually. Find the total of the interest paid column in the bond amortizationschedule.”

Solution: We have C = F = 1000, n = 10, r = 5%, i = 6%. The premium paid forthe bond is

1000(0.05− 0.06)a10 6% = −73.60

(i.e., a discount of 73.60.) The total of the coupons is 10Fr = 500., so the sumof the coupons and the discount is 500.00 + 73.60 = 573.60, and this is the trueinterest paid. While the problem did not ask us to actually set up the amortizationtable, let’s do it anyhow:

Coupon Coupon Interest Amount for Amortization BookNumber amount earned of Premium value

0 926.401 50.00 55.58 -5.58 931.982 50.00 55.92 -5.92 937.903 50.00 56.27 -6.27 944.174 50.00 56.65 -6.65 950.825 50.00 57.05 -7.05 957.876 50.00 57.47 -7.47 965.347 50.00 57.92 -7.92 973.268 50.00 58.39 -8.39 981.659 50.00 58.90 -8.90 990.5510 50.00 59.43 -9.43 999.98

TOTAL 500.00 573.58 -73.58

The following example illustrates a problem where the price P of a bond is not determinedexplicitly until an equation for P is determined and solved.

Example A.17 [1, Example 7.4, p. 219] “Find the price of a 1,000 par value 2-year8% bond with semi-annual coupons bought to yield 6% convertible semiannually, if the


investor plans to replace the premium by means of a sinking fund earning 5% convert-ible semi-annually. (Note: The intention is that this plan for a sinking fund has beenconsidered in calculating the yield rate of the bond.)Solution: Note that, following the usual convention, the 8% rate is interpreted as anominal interest rate compounded twice a year.

The unknown is the price, P . The coupons each have value 0.04 × 1000 = 40, butthe interest earned is 3% of the price P , which remains to be determined. The sinkingfund is to mature at value P − 1000 after 2 years, i.e., after 4 contributions. Thus anequation of value is

(40− .03P )s4 2.5% = P − 1000 . (102)

This can be solved for P :

P =1000 + 40s4 2.5%

1 + 0.03s4 2.5%

= 1, 036.93 .

Note the assumption used in equation (102): all of the coupon that exceeds the interest“earned” at the yield rate is contributed to the sinking fund; this was not explicitly statedin the problem, but is the author’s interpretation. If, for example, the purchaser haddecided that he would allocate only half of the excess to his sinking fund contributions,then the price of the bond would be about 1082.43.

A.31.2 §7.7 CALLABLE BONDS

[1, Exercise 31, p. 243] “A 1000 par value bond has 8% semiannual coupons, and iscallable at the end of the 10th through the 15th years at par.

1. “Find the price to yield 6% convertible semiannually.

2. “Find the price to yield 10% convertible semiannually.”

Solution:

1. Let n be the coupon number at whose date the bond is called matures. Then,by the Premium/Discount Formula

P = 1000 + (40− 30)an 3% , (n = 20, 22, 24, 26, 28, 30).

Without knowing which will be the date of call, we take the worst possibledate in order to minimize the price; since an 3% is an increasing function of n,and is multiplied by a positive number, 10, we minimize by making n as smallas possible, i.e., 2× 10 = 20:

P = 1000 + (40− 30)a20 3% = 1000.00 + 148.78 = 1148.78 .


2. When the semi-annual yield rate is 5%, the multiplier is negative, 40 − 50 =−10, and we must choose the largest value of n, i.e., n = 30, for a price of

P = 1000 + (40− 50)a30 5% = 1000.00− 153.72 = 846.28 .

[1, Exercise 32, p. 243] “If the bond in Exercise 31(b) were actually called at the endof 10 years, find the yield rate to the owner of the bond.”

Solution: The yield rate i satisfies the equation

846.28 = 1000 + 1000(0.04− i)a20 i ,

and can be found by iteration to be 10.52%%.

[1, Exercise 33, p. 243] “A 1,000 par value 8% bond with quarterly coupons is callablefive years after issue. The bond matures for 1,000 at the end of 10 years, and issold to yield a nominal rate of 6% convertible quarterly, under the assumption thatthe bond will not be called. Find the redemption value at the end of 5 years thatwill provide the purchaser the same yield rate.”

Solution: If the bond is certain not to be called, its present value (i.e., its price) is

1000 + (20− 15)a40 1.5% = 1000 + 5(29.9258) = 1149.63.

Let x denote the redemption value after 5 years that will provide the same yieldrate. Then

1149.63 = x + (20− 0.015(x))a20 1.5%

which implies that

x =1149.63− 20a20 1.5%

1− 0.05a20 1.5%

= 1085.91 .

[1, Exercise 34, p. 243] “A 1000 par value 4% bond with semiannual coupons maturesat the end of 10 years. The bond is callable at 1050 at the ends of years 4 through6, at 1025 at the ends of years 7 through 9, and at 1000 at the end of year 10. Findthe maximum price that an investor can pay and still be certain of a yield rate of5% convertible semiannually.”

Solution: We have to find the minimum of the following prices based on the givencall dates and premiums:

1050.00 + (20.000− 26.250)an 2.5% (n = 8, 10, 12) (103)

1025.00 + (20.000− 25.625)an 2.5% (n = 14, 16, 18) (104)

1000.00 + (20.000− 25.000)a20 2.5% . (105)


In each case the coefficient of an 2.5% is negative, so the lowest value will be whenn is as large as possible; that is, we have to compare the following three amounts

1050.00 + (20.000− 26.250)a12 2.5% = 1050− (6.250)10.2578 = 985.89

1025.00 + (20.000− 25.625)a18 2.5% = 1025− (5.625)14.3534 = 944.26

1000.00 + (20.000− 25.000)a20 2.5% = 1000− (5.000)15.5892 = 922.05 ,

whose minimum is the last, the price of the bond if not called before maturity.That is the highest price the investor may pay if she wishes to be sure that theyield will not be less than 5% convertible semiannually.

[1, Exercise 35, p. 243] “A 1,000 par value 6% bond with semiannual coupons iscallable at par 5 years after issue. It is sold to yield 7% under the assumptionthat the bond will be called. The bond is not called, and it matures at the endof 10 years. The bond issuer redeems the bond for 1000 + X without altering thebuyer’s yield rate of 7% convertible semiannually. Find X.”

Solution: Under the assumption that the bond will be called at par 5 years afterissue, its price, when yielding 3.5% effective semi-annually, would be

1000 + (30− 35)a10 3.5% = 958.42 .

The premium upon maturity will be given by the equation

958.42 = 1000 + X + (30− (0.035)(1000 + X))a20 3.5% .

which implies that

X =958.42− 30a20 3.5%

1− 0.035a20 3.5%

− 1000 = 58.66 .

Example A.18 We discussed in class the following problem: [1, Example 7.9, p. 231]“Consider a $100 par value 4% bond with semiannual coupons callable at $109 on anycoupon date starting 5 years after issue for the next 5 years, at $104.50 starting 10 yearsafter issue for the next 5 years, and maturing at $100 at the end of 15 years. What isthe highest price which an investor can pay and still be certain of a yield of: (1) 5%convertible semiannually, and (2) 3% convertible semiannually?”Solution: The solution was discussed in class. I recommended that you use the Pre-mium/Discount Formula, and remember that the function an i is an increasing functionof n; this means that, in a problem of this type, the values least favourable to the in-vestor will always be at the end of an interval of payment periods: at the lower end whenFr−Ci is positive, and at the upper end when Fr−Ci is negative. The intervals to be


considered will be consecutive payment dates when the call premium has the same value.Each of these intervals need to be treated separately, and then the least favourable fromeach interval taken together and the least favourable of them determined.

The textbook states, in connection with the part of the problem where the yieldrate is 3%, that “it is immediately clear that the latest possible redemption date is lessfavourable to an investor, since the bond will be selling at a discount.” This statementmay not be clear. The factor that determines which end of the interval is least favourableis Fr − Ci = C(g − i), where g is the modified coupon rate (cf. Definition A.18.4, p.2132). The given data lead to finding the minimum of the following three values:

100 + (2− 2.50)a30 0.025 = 89.53

109 + (2− 2.725)a29 0.025 = 94.17

104.50 + (2− 2.6125)a19 0.025 = 95.33

And, indeed, the minimum will be when n = 30. The situation is best viewed throughthe Base amount formula:

P = G + (C −G)vn

where G = Fri

(cf. Definition A.18.10). Since vn is a decreasing function of n, what iscritical is the sign of

C −G = C − Fr

i.


A.32 Supplementary Notes for the Lecture of April 1st, 2005

Distribution Date: Friday, April 1st, 2005, subject to further revision

Following is an example of a type of problem we have seen once on an assignment. Thisexample was not discussed in the lecture.

An amortization problem

Example A.19 [5, Exercise 4-9, p. 87] “A loan is being repaid with 30 equal annualinstallments, at i = 0.17. In what installment are the principal and interest portionsmost nearly equal to each other?”Solution: Let A be the amount of the loan, received at time n = 0, and let X be theannual payment. An equation of value at time n = 0 is Xa30 = A, yielding the value ofthe annual payment to be

X =A

a30

.

By the prospective method we see that the unpaid principal immediately after the (r −1)st payment is

Xa31−r = A · a31−r

a30

= A · 1− v31−r

1− v30. (106)

The unpaid principal is a decreasing function of r, and the interest earned on it duringthe next period and included in the next regular payment is its product with i, which willalso be a decreasing function of r. Since fixed payments are being made, and the portionof the payment that is interest will be decreasing, the balance of the rth payment, whichis applied to reduction of principal, is an increasing function of r, namely

A

a30

(1− i · a31−r

)= A · 1− a31−r

a30

; (107)

note that this amount is always positive until the loan is repaid. If, in the rth payment,the amount of interest is less than or equal to the amount for reduction of principal,then the same property will hold for all subsequent payments; if, in the rth payment,the amount of interest is greater than or equal to the amount for reduction of principal,then the same property will hold for all prior payments. The observations just madeabout functions increasing and decreasing hold even for non-integral values of r. So oneway of solving the present problem is to equate the interest and principal formulæ, andsolve for r. If r is not an integer — which will usually be the case — we know that the


payment for which the difference between interest and principal is minimal will be oneof payments ##brc and dre (the integers which are closest to r from below and above).

iA · a31−r

a30

= A · v31−r

a30

⇔ v31−r =1

2⇔ (1.17)31−r = 2

⇔ (31− r) ln 1.17 = ln 2

⇔ r = 31− ln 2

ln 1.17= 31− 4.41 = 26.58

so the candidates for closest interest and principal are payments ##26 and 27. Thedifference between the interest and principal in payments ##26 and 27 are, respectively,

A

∣∣∣∣1− 2v31−26

1− v30

∣∣∣∣ =

∣∣∣∣0.087777695

0.990996235

∣∣∣∣ = 0.0886

A

∣∣∣∣1− 2v31−27

1− v30

∣∣∣∣ =

∣∣∣∣−0.067300096

0.990996235

∣∣∣∣ = 0.0679

so the payment where these are closest is #27.The student should note that the formulæ we have derived above in (106) and (107)

are generally true when a loan is amortized in equal payments, and are found in [5, pp.78, 79]. But be cautioned — these formulæ are not necessary applicable under otherconditions, e.g. where payments are not level.

A sinking fund problem 30

Example A.20 “A borrower takes out a loan of 20000 for three years. Construct asinking fund schedule where the lender receives a nominal rate of 10% effective semi-annually and paid semi-annually on the loan, and the borrower replaces the amount ofthe loan with semi-annual deposits into a sinking fund earning 8% convertible quarterly.Use the following headings

Duration Contribution Interest Interest Earned Balance of(Months) to Sinking Fund on Loan in Sinking Fund Sinking Fund

0 0 0 0 03 . . . . . . . . . . . .6 . . . . . . . . . . . .

”

30Source of problem: modified from Final Examination in Math 329, April, 2000


Solution: Every six months the borrower pays the interest that has accrued on the loanat the rate of 10

2% = 5% per half-year; i.e. the amount of 5% of 20000, or 1000. In

order that the sinking fund accumulate an amount of 8% convertible quarterly, i.e. 2%quarterly, the borrower must make 6 semi-annual payments X which satisfy the equation

X · s6(1.02)2−1 = 20000

whence

X =20000

s61.022−1

=808

(1.02)12 − 1= 3012.21


0 0.00 0.00 0.00 0.003 0.00 0.00 0.00 0.006 3012.21 1000.00 0.00 3012.219 0.00 0.00 60.24 3072.4512 3012.21 1000.00 61.45 6146.1115 0.00 0.00 122.92 6269.0318 3012.21 1000.00 125.38 9406.6221 0.00 0.00 188.13 9594.7524 3012.21 1000.00 191.90 12798.8627 0.00 0.00 255.98 13054.8430 3012.21 1000.00 261.10 16328.1533 0.00 0.00 326.56 16654.7136 3012.20 1000.00 333.09 20000.00

Note that the last semi-annual payment has been reduced by 1 cent. An equivalenttable could have been compiled with 6 months intervals, with semi-annual interest rate(1.02)2 − 1 = .0404:


0 0.00 0.00 0.00 0.006 3012.21 1000.00 0.00 3012.2112 3012.21 1000.00 121.69 6146.1118 3012.21 1000.00 248.30 9406.6224 3012.21 1000.00 380.03 12798.8630 3012.21 1000.00 517.07 16328.1436 3012.20 1000.00 659.66 20000.00



Another final examination problem 31

Example A.21 “An investor plans to buy a twenty-year mortgage whose amount is150,000, having equal monthly payments; the nominal interest rate is 6%, compoundedmonthly. The investor plans to invest the monthly payments she receives in a savings ac-count earning a nominal annual rate of 3%, compounded monthly, in order to accumulatea retirement fund after 20 years.

1. “What is the monthly payment on the mortgage?

2. What should the investor pay for this mortgage, in order to receive an effectiveannual yield rate of 8%?

3. How much is in the retirement fund after 10 years.

4. How much is in the retirement fund after 20 years?”

Solution:

1. The mortgage is amortized over 20 years, i.e., 240 months. The monthly paymentX has the property that X · a240 6

12% = 150000, hence

X =150000

a2400.5%

= 1074.65

2. (a) Suppose that the intention of the problem is that the mortgagepayments are still to be placed in the sinking fund. The sinking fundbecomes available only at the end of 20 years. Discounting back to the presentat 8%, we find its present value, i.e. the price to be paid, to be

(1.08)−201074.65 · s2400.025%

= (1.08)−20 · 150, 000 · s2400.025%

a2400.05%

= 75, 694.46.

(b) Suppose that the intention of the problem is that the mortgagepayments are no longer to be placed in a sinking fund. Let i denotethe monthly interest rate that, when compounded 12 times, is equivalent to

31Source of problem: modified from Final Examination in Math 329, April, 1998.


an annual rate of 8%; i.e. i = (1.08)112 − 1. The present value of all future

payments is

X · a240i = X · 1− (1 + i)−240

(1.08)112 − 1

= X · 1− (1.08)−20

(1.08)112 − 1

= 1074.65× 122.0777 = 131190.83

3. After 10 years the payments have accumulated to X ·s1200.0025 = 1074.65×139.74 =150171.59.

4. After 20 years the payments have accumulated to X ·s2400.0025 = 1074.65×328.30 =352807.60.

The following is not examination material.

A.32.1 §7.6 DETERMINATION OF YIELD RATES

Omit this section.

A.32.2 §7.5 VALUATION BETWEEN COUPON PAYMENT DATES

In the preceding section we have considered the price of a bond on coupon paymentdates. We have chosen to use the term book value for the adjusted value of the bondon coupon payment dates after a coupon has been paid: the book value is the residualvalue of the instrument, where the yield rate is taken to be the rate in effect when thebond was purchased. These rates are related by the recurrence

Bt+1 = (1 + i)Bt − Fr . (108)

Do not confuse this with the equations of value that we computed earlier in the course,where we found that, under compound interest, sums of money could be moved forwardand backwards on in time by multiplication by the appropriate power of 1 + i. Thatprinciple is still in effect, but the sums we are considering — the book values at differenttimes — are not one single value, but a whole sequence of values; we normally areinterested in Bt precisely at time t, and it is not going to be the same as the book valueBt+1 at time t + 1 for 2 reasons:

• maturity is one year further away;

• the instrument guarantees one more coupon payment.

One way to prove equation (108) is by considering the value of Bt at time t + 1.


Market price and Flat price. The market price on a coupon payment date is whatwe earlier called simply the “price”, and will have an associated yield rate. It will bethe book value of a previously purchased bond when the yield rate has not changed sincepurchase, and we will normally make such an assumption. But bonds are not alwayspurchased on a coupon payment date; and, even when they are, there is normally a delaybetween the date of the formal order and the delivery of the bond. So we need to havea method for computing a “reasonable” price at times between coupon payment dates.The Flat Price is the actual sum of money that changes hands at the time of formalpurchase. It can be interpreted as being made up of two elements:

• the Market Price; and

• the Accrued Coupon

The Flat Price at time t + k between coupon dates ##t, t + 1 is obtained from the FlatPrice just after the payment of the coupon #t by one of two methods:

Theoretical Method: Compound interest, producing (1 + i)kBt;

Practical Method: Simple interest, producing (1 + ki)Bt.

The Accrued Coupon can also be calculated by both a Theoretical and a Practicalmethod:

Theoretical Method: Compound interest, producing Fr(

(1+i)k−1i

);

Practical Method: Simple interest, producing kFr.

(Even here there are still some conventions needed, in the way in which we determinethe time interval k as a fraction of a period.) In the Semi-theoretical Method the FlatPrice is calculated by the theoretical method, but the accrued coupon is calculated bysimple interest, i.e., by the Practical Method.

Note that it is the Flat Price that is being calculated by the choice of methods,not the Market Price. The Market Price can be obtained from the two elements bysubtraction:

Market Price = Flat Price - Accrued Coupon

In practice prices are quoted in the form

Flat Price = Market Price + Accrued Coupon


The graph of the Book Value (which we take to be the same as Market Price) will bemonotonely increasing when i > r, constant when i = r, and monotonely decreasingwhen i < r. Between the values at integer points the graph will be linear in the caseof the Practical Method, and exponential in the case of the theoretical method, butthe differences between the two are very small; in either case the function is continuous.However, the Flat Price will be discontinuous at coupon payment dates, because its valuedrops each time a coupon is paid, by the value of the coupon.

A.32.3 §7.8 SERIAL BONDS

Omit this section

A.32.4 §7.9 SOME GENERALIZATIONS

Omit this section.

A.32.5 §7.10 OTHER SECURITIES

Omit this section.

A.32.6 §7.11 VALUATION OF SECURITIES

Omit this section.


B Problem Assignments, Tests, and Examinations

from Previous Years

B.1 2002/2003

B.1.1 First 2002/2003 Problem Assignment, with Solutions

In all of the following problems students were are expected to show your work.

1. (a) What principal will earn interest of 100 in 7 years at a simple interest rate of6%?

(b) What simple interest rate is necessary for 10,000 to earn 100 interest in 15months?

(c) How long will it take for money to double at a simple interest rate of 8%?

(d) For the rate stated and the period of time computed in the previous part ofthe question, what would 1 grow to if interest were compounded annually?

Solution:

(a) Let P denote the unknown principal. Equating the interest earned, P ·(0.06)·7to 100 and solving for P , we obtain that P = 100

7×0.06= 238.10.

(b) If the interest rate is i, the amount of interest earned will be i· 1512·10000 = 100.

Solving yields i = 45· 100

10000= 0.008 or 0.8%.

(c) Let the number of years for money to double be denoted by t. We solve1 + (0.08)t = 2: t = 2−1

0.08= 12.5. Money doubles in 121

2years.

(d) (1.08)12.5 = 2.62.

2. The total amount of a loan to which interest has been added is 20,000. The termof the loan was four and one-half years.

(a) If money accumulated at simple interest at a rate of 6%, what was the amountof the loan?

(b) If the nominal annual rate of interest was 6% and interest was compoundedsemi-annually, what was the amount of the loan?

(c) If the rate of interest was 6%, interest was compounded annually for full years,but simple interest was paid for the last half-year, what was the amount ofthe loan?

(d) If the rate of interest was 6%, interest was compounded annually for full andpart years, what was the amount of the loan?


(e) If the effective annual rate of interest was 6%, but interest was compoundedsemiannually, what was the amount of the loan?

(f) If the nominal annual rate of interest was 6%, but interest was compoundedcontinuously, what was the amount of the loan?

(g) If interest was compounded continuously, and the force of interest was 6%,what was the amount of the loan?

Solution:

(a) If the amount of the loan is P , then 20000 = (1 + 0.06 × 4.5)P = 1.27P , soP = 20000

1.27= 15, 748.03.

(b) If the amount of the loan is P , then 20000 = (1 + 0.062

)2×4.5P ,

P = 20000(1.03)−9 = 15, 328.33 .

(c) If the amount of the loan is P , then 20000 = (1 + 0.06)4 · (1 + 0.062

)P ,

P = 20000(1.06)−4(1.03)−1 = 15, 380.46 . (109)

(d) If the amount of the loan is P , then 20000 = (1+0.06)4.5P , P = 20000(1.06)−4.5 =15, 386.99.

(e) If the semi-annual rate of interest is denoted by i, then (1 + i)2 = 1.06, so

1 + i = (1.06)12 . If the amount of the loan is P , then 20000 = (1 + i)9P =

(1.06)4.5P , so P = 20000× (1.06)−4.5 = 15, 386.99.

Note that this is exactly the same principal as in the preceding version of theproblem, since it is precisely the same problem! Note also that the amountof the principal is slightly more than in the version in part 2c, since simpleinterest for a fraction of a year in that case will have a higher yield thancompound interest in this case.

(f) We are told that the nominal rate of interest, compounded instantaneously,is 6%; thus, if i is the effective annual rate, 0.06 = δ = ln(1 + i); equivalently,e0.06 = 1 + i, so i = 6.18365%. In a full year an amount of 1 will grow by afactor 1.0618365; in a half year, by a factor

√1.0618365. In 41

2years we have

P (1.0618365)4.5 = 20000, so P = 20000(1.0618365)−4.5 = 20000(0.763379) =15267.59.

(g) The force of interest is the nominal rate of interest which is convertible contin-uously [5, p. 17]. We are told that interest was compounded continuously . Theeffective annual rate of interest, which we shall denote by i, has the property


that 0.06 = ln(1 + i), or that 1 + i = e0.06. If the amount of the loan is P ,then 20000 = (1 + i)4.5P = e4.5×0.06 = e0.27P , so P = 20000e−0.27 = 15267.59,the same result as in the preceding part.

3. (cf. [5, Exercise 1-13, p. 24]) Henry plans to have an investment of 10,000 onJanuary 1, 2006, at a compound annual rate of discount d = 0.11.

(a) Find the value that he would have to invest on January 1, 2003.

(b) Find the value of i corresponding to d.

(c) Using your answer to part (b), rework part (a) using i instead of d. Do youget the same answer?

Solution:

(a) The accumulation of 10,000 will have to be discounted by a factor of (1−0.11)three times to reduce it by compound discount to January 1, 2003. Theamount to be invested is, accordingly, 10000(1− 0.11)3 = 7049.69.

(b) The relationship between d and i is given, for example, by (1 + i)(1− d) = 1,which implies that i = d

1−d. Here

i =0.11

1− 0.11=

11

89= 0.1233596.. = 12.36..%.

(c) When i = 12.36%, v = 11+i

= 11.1236

= 0.89. The value on January 1, 2003 ofthe 10,000 expected on January 1, 2006 will then be 10000(0.89)3 = 7049.69,as before.

4. (cf. [5, Exercise 1-24, p. 26]) Recall that (cf. [5, (1.21)])[1 +

i(m)

m

]m

= 1 + i =1

1− d=

[1− d(m)

m

]−m

. (110)

(a) Determine whether there is an integer n such that

1 +i(n)

n=

1 + i(2)

2

1 + i(3)

3

(111)

and, if there is such an integer, find it.

(b) Replace the right member of (111) by a product(

1 +i(2)

2

)(1− d(3)

3

)

and then interpret this product verbally to show that it must be equal to1 + i(n)

nif a suitable n exists.


Solution:

(a) i(2) is the nominal annual interest rate which, when compounded semi-annually,

yields an effective annual rate of i. Thus 1+ i(2)

2=√

1 + i; similarly, 1+ i(3)

3=

3√

1 + i. The ratio1+ i(2)

2

1+ i(3)

3

is, therefore, equal to (1 + i)12− 1

3 = (1 + i)16 , which is

the accumulation of 1 after a period of 16

of a year; this is, by definition, equal

to 1 + i(6)

6.

(b) Under an effective annual interest rate of i, the factor(1 + i(2)

2

)is the value

of 1 after 122

= 6 months. If this amount is discounted back 123

= 4 months,

it decreases by a reduction factor of(1− d(3)

3

). The result is equivalent to a

net accumulation period of 6− 4 = 2 months, i.e. 16

of a year, under which it

would grow by a factor 1 + i(6)

6.

5. (cf. [5, Exercise 1-30, p. 27]) Show that f(t) = (1 + i)t − (1 + it) is minimized att = ln i−ln δ

δ.

Solution:

If only elementary calculus is used, this problem is more difficult thanit looks. Students were accorded a full grade for showing that the pointclaimed is, indeed, a local minimum; the proof that it is a global =absolute minimum, is more difficult; one possible solution is given below.No attempt has been made to produce a compact solution.

Applying elementary calculus, we find that

f ′ = (1 + i)t ln(1 + i)− i = (1 + i)tδ − i

f ′′ = (1 + i)tδ2

To find the critical points of the function, we solve for t the equation f ′(t) = 0.Taking natural logarithms yields

t ln(1 + i) + ln δ = ln i

which is satisfied only for

t0 =ln i− ln δ

δ=

ln iδ

δ. (112)

The second derivative, f ′′(t0) is positive everywhere, since it is the product of anexponential — always positive — and the square of a real number; this tells usthat the point t = t0 (112) is a local minimum.


Does this completely solve the problem? Not yet! To solve an extremum problemwe need to interpret local extremum information with reference to the domain ofthe function. For example, if the domain is infinite, then the function might noteven have a global or absolute minimum, even though it has a local minimum.32

And, if the domain of the function is a closed interval, we need to investigate thebehavior at the end points of that interval. The function f is meaningful for all realvalues of t. One interpretation would be to take the domain to be t ≥ 0; anotherinterpretation would be to take the domain to be −∞ ≤ t ≤ +∞.

What follows is just one possible way of completing this problem. We observe thatf(0) = f(1) = 0. Could f(t) = 0 for t different from 0, 1? Rolle’s theorem impliesthe existence of a point with zero slope between any two zeros of the function; as wehave seen that there is only one such point with zero slope, there cannot exist morethan two zeros of the function: and thus the point t0 is the only local extremum.Thus, by the Intermediate Value Theorem, f has the same sign throughout eachof the intervals −∞ < t < 0, 0 < t < 1, 1 < t. As t → ∞, lim f(t) → ∞; hencef(t) > 0 for all t > 1; as t → −∞, lim f(t) → ∞; so the function is positivein the interval −∞ < t < 0 also. Thus the global minimum is in the interval0 ≤ t ≤ 1; and, from our investigation of the critical point, we know that theminimum is attained at one (or more) of t = 0, t = 1 or t = t0. We can completethis investigation if we can argue that f(t0) < 0. As we know the sign of thefunction will be the same throughout the interval 0 < t < 1, we can take anyconvenient value of t in that interval.

f

(1

2

)=

√1 + i−

(1 +

i

2

)

=

(√1 + i− (1 + i

2)) · (√1 + i + (1 + i

2))

√1 + i + (1 + i

2)

=(1 + i)−

(1 + i + i2

4

)√

1 + i +(1 + i

2

)

= −i2

4· 1√

1 + i +(1 + i

2

) < 0

Thus the global minimum is attained at t0.

6. (cf. [5, Exercise 1-33, p. 27]) Find the accumulation function a(t) if it is knownthat δt = 0.04(1 + t)−1 for t > 0.

32Consider, for example, the function t3− t, which has a local minimum at t = 1, a local maximum att = −1, but has neither a global maximum nor a global minimum over its entire domain −∞ < t < +∞.


Solution: Applying definition [5, (1.28), p. 19], we solve the differential equation

d

dtln(a(t)) = 0.04(1 + t)−1 :

Integration gives

ln(a(t)) =

∫0.04

1 + tdt = 0.04 ln(1 + t) + C

where C is the constant of integration. Setting t = 0, where we know, by definition,that a(0) = 1, we have

0 = ln 1 = 0.04 ln 1 + C

so C = 0, and

a(t) = e0.04 ln(1+t) =(eln(1+t)

)0.04= (1 + t)0.04 .

7. Let φ(λ) denote the value of 1 at the end of 3 years, accumulated at an effectiverate of interest λ; let ψ(λ) denote the present value of 1, to be paid at the end of 3years at an effective rate of discount numerically equal to λ. Suppose it is knownthat φ(λ) + ψ(λ) = 2.0294. Determine λ.

Solution: (cf. [2, Exercise 52, p. 30]) φ(λ) = (1 + λ)3; ψ(λ) = (1 − λ)3. Summingyields φ(λ) + ψ(λ) = 2 + 6λ2, which we equate to 2.0294, and from which we infer

that λ =√

0.02946

= .07 = 7%.

8. Showing your work, determine a formula — in terms of the force of interest, δ, forthe number of years that are needed for a sum of money to double itself. Verifyyour answer by determining the value of δ when the annual interest rate is 100%.

Solution: Let the number of years needed be t, the interest rate be i, and the forceof interest δ. We solve the equation (1+ i)t = 2 by taking logarithms of both sides:t ln(1 + i) = ln 2, so

t =ln 2

ln(1 + i)=

ln 2

δ.

When the interest rate is 100% money doubles in one year; here δ = ln 2.

B.1.2 Second 2002/2003 Problem Assignment, with Solutions

1. (cf. Exercise 2-5, p. 36) A vendor has three offers for a house:

(a) three equal payments — one now, one 1 year from now, and the other 2 yearsfrom now;


(b) a single cash payment now of 120,000;

(c) two payments, 45,000 a year from now, and 90,000 two years from now.

He makes the remark that “one offer is just as good as another”. Determine theinterest rate and the sizes of the equal payments that will make this statementcorrect.

Solution: Let the interest rate be i, and the equal payments be k. Equating thepresent value of the payments of 45,000 and 90,000 to 120,000 yields

45000v + 90000v2 = 120000

which we solve for v, obtaining

v =−1

2±

√14

+ 163

2

=−1±√22.33333

4

The lower sign yields a negative value of v, and so that solution is extraneous. Weobtain v = 0.931454, so i = 0.0736 = 7.36%.

The present value of the equal payments is then k(1+v+v2) = 2.79906k = 120, 000,so the equal payments will each be 42,871.53.

2. (cf. [5, Exercise 2-9, p. 37]) Fund A accumulates at 9% effective, and Fund B at8% effective. At the end of 12 years the total of the two funds is 50,000. At theend of 6 years the amount in Fund B is 4 times that in Fund A. How much is inFund A after 15 years?

Solution: Let a and b denote the initial amounts in funds A and B. We have twoconstraints relating a and b:

a(1.09)12 + b(1.08)12 = 50000

b(1.08)6 = 4a(1.09)6

From the second of these we can determine the relative sizes of a and b; substitutingin the first equation and solving gives

a =50000

(1.09)12 + 4(

1.091.08

)6(1.08)12

=50000

(1.09)6 ((1.09)6 + 4(1.08)6)= 3, 715.25.

Hence b = 4(

109108

)6a = 15, 705.96. The value of Fund A after 15 years is, therefore,

3, 715.25(1.09)15 = 13, 532.73.


3. The initial balance in an investment fund was 100,000. At the end of 3 months ithad increased to 105,000; at that time 25,000 was added to the fund. Six monthslater the fund had increased to 143,000, and this time 30,000 was removed. Finally,at the end of a year, the fund had a balance of 120,000. What was the time-weightedrate of return?

Solution: [4, Example 2.3.2, p. 39] The balances and withdrawals are respectivelyB0 = 100, 000, B1 = 105, 000, B2 = 143, 000, B3 = 120, 000; W0 = 0, W1 = 25, 000,W2 = −30, 000. Hence the rates of interest in the successive time periods are givenby

1 + i1 =B1

B0 + W0

=105, 000

100, 000 + 0= 1.05

1 + i2 =B2

B1 + W1

=143, 000

105, 000 + 25, 000= 1.10

1 + i3 =B3

B2 + W2

=120, 000

143, 000− 30, 000= 1.062

The time-weighted rate of return i is, by definition, given by the product

1 + i = (1.05)(1.10)(1.062) = 1.227

so i = 22.7%.

4. (cf. [5, Exercise 2-15, p. 38]) A trust company pays 5% effective on deposits at theend of each year. At the end of every 3 years a 2% bonus is paid on the balance atthe time. Find the effective rate of interest earned by an investor if she leaves hermoney on deposit

(a) for 2 years;

(b) for 3 years (until after the bonus payment is made);

(c) for 4 years;

(d) forever — take a limit!

Solution:

(a) Since there are no bonus payments, the effective rate of interest is 5%.

(b) A deposit of 1 grows to 1.05 at the end of the first year, (1.05)2 at the endof the second year, and, after the bonus, (1.05)3(1.02) at the end of the 3rdyear. If the effective rate of interest is i, then we must solve the equation

(1 + i)3 = (1.05)3(1.02) .


Taking logarithms, we obtain 3 ln(1 + i) = 3 ln(1.05) + ln(1.02), so

ln(1 + i) =1

3(3 ln(1.05) + ln(1.02))

1 + i = e13(3 ln(1.05)+ln(1.02))

i = e13(3 ln(1.05)+ln(1.02)) − 1

= 1.053√

1.02− 1 = .056953846 = 5.70%.

(c) Analogously to the preceding,

i = e14(4 ln(1.05)+ln(1.02)) − 1

= 1.054√

1.02− 1 = .0552 = 5.52%.

(d) The effects of the bonuses depend on whether the remainder of the number ofyears is, upon division by 3, 0, 1, or 2. We have, for any non-negative integern,

(1 + i)3n = (1.05)3n(1.02)n

(1 + i)3n+1 = (1.05)3n+1(1.02)n

(1 + i)3n+2 = (1.05)3n+2(1.02)n

By taking logarithms and dividing, or, equivalently, by taking the appropriate(positive) root of both sides of the equation, we obtain

1 + i = (1.05)(1.02)1/3

1 + i = (1.05)(1.02)n

3n+1

1 + i = (1.05)(1.02)n

3n+2

As n → ∞, the exponent of 1.02 approaches 13, and so the interest rate

approaches the value of 5.70% we obtained in case of 3 years.

5. Alice borrows 5000 from The Friendly Finance Company, at an annual rate ofinterest of 18% per year, where the company compounds interest annually, butcharges simple interest for fractions of a year.

(a) She plans to pay the company 5000 at the end of 2 years.

i. How much will she continue to owe the company at that time?

ii. What is the present value of that residual amount, assuming the same18% interest rate?


(b) Alice discovers that she doesn’t need the loan, so she offers to lend the moneyto her brother, at an effective annual rate of 18%. When her brother pays offhis loan, Alice will pay off hers. How much will Alice still owe Friendly if

i. Her brother pays off his loan exactly 3 years from now?

ii. Her brother pays off his loan 3.5 years from now?

Solution:

(a) i. The residual amount immediately after the payment will be

5000((1.18)2 − 1

)= 1962 .

ii. The present value of the residual amount is 5000 ((1.18)2 − 1) (1.18)−2 =5000 (1− (1.18)−2) = 1409.08.

(b) i. If her brother pays off his loan after an integer number of years, and Aliceimmediately repays her loan, she will owe nothing to Friendly.

ii. After 3.5 years Alice will receive 5000(1.18)3.5, but will be owing5000(1.18)3(1.09); after making her payment, she will continue to owe

−5000(1.18)3.5 + 5000(1.18)3(1.09) = 5000(1.18)3(1.09−√

1.18) = 30.58 .

6. (cf. [5, Exercise 2-12, p. 37])

(a) Find an equation that gives information about the effective rate of interest iif payments of 200 at the present, 300 at the end of 1 year, and 400 at theend of 3 years, are to accumulate to 1000 at the end of 4 years.

(b) Use the Intermediate Value Theorem to argue that there exists a positive rateof interest less than 100% which can solve this problem. Then use the MeanValue Theorem to show that there is just one solution to the problem (byshowing that a certain derivative is positive).

(c) While there exist more efficient algorithms for solving problems like this, anaive solution could be found by successively subdividing an interval at whoseends a certain function would have values with opposite signs. Apply this ideato find the interest rate i to within an error of 0.1%.

Solution:


(a) The equation of value at the end of 4 years33 is

200(1 + i)4 + 300(1 + i)3 + 400(1 + i) = 1000

Define f(x) = 200x4 +300x3 +400x−1000. Then f(1) = −100 < 0 < 5400 =3200 + 2400 + 800− 1000 = f(2). By the Intermediate Value Theorem func-tion f , which, being a polynomial, is continuous, has a zero somewhere in1 < x < 2. If there were 2 or more zeroes, then, by Rolle’s Theorem, (since f ,being a polynomial, is differentiable), there would be a point between themwhere f ′ would be 0. But f ′(x) = 800x3+900x2+400 > 0 for 1 < x < 2. Fromthis contradiction we know that there is at most one zero for f , hence exactlyone solution i for our effective interest rate. We can apply the IntermediateValue Theorem between any two points in the domain. The most naive solu-tion would be to repeatedly halve the interval. We find that f(1.5) = 1625, sowe may confine ourselves to the interval 1 < x < 1.5; then f(1.25) = 574.22,f(1.125) = 197.51, f(1.0625) = 39.72, f(1.03125) = −32.29. We evalu-ate f at the midpoint of the interval [1.03125, 1.0625]: f(1.046875) = 3.17,so we next use the interval [1.046875, 1.0625], whose midpoint is 1.0546875,where f(1.0546875) = 21.30. As there will is a sign change in the interval[1.03125, 1.0546875], we next evaluate f at its mid-point: f(1.042968750) =−5.80.

In the course of these calculations we have not bothered to round the decimalexpansions of the midpoints. There is nothing to be gained by this persistence,as the procedure will work even if we do not take the precise midpoints.Having now confined the root to the interval [1.043, 1.055]. f(1.049) = 8.07,we try f

(1.043+1.049

2

)= f(1.046) = 1.15, f

(1.043+1.046

2

)= f(1.0445) = −2.29,

f(1.045) = −1.15, f(1.0455) = 0.002, f(1.04549913) = −0.0000013. Thusthe rate is approximately 4.55%.

B.1.3 Third 2002/2003 Problem Assignment, with Solutions

1. (a) Find the sum of the positive integers 1, 2, . . . , N .

(b) Find the sum of the odd integers 1, 3, 5, . . . , 2N + 1.

(c) In an arithmetic progression x1, x2, . . ., xn, . . . the third term is 4 times thefirst term, and the sixth term is 17. Find the general term xn.

33Non-trivial equations are never unique; also, we could have found an equation of value at anothertime. For example, an equation of value at the present could be

200 + 300v + 400v3 = 1000v4 .


(d) The sum of n terms of the arithmetic series 2, 5, 8, . . . is 950. Find n.

(e) The sum of the first 6 terms of a geometric progression is equal to 9 times thesum of the first 3 terms. Find the common ratio. Is it possible to determinethe sequence from this information?

(f) Use your knowledge of the sum of geometric series to determine a “vulgar”fraction of integers m

nwhich is equal to the repeating decimal number

3.157157157157...

Do not use a calculator for this problem.

Solution: These topics were once part of the standard high school curriculum.These problems were adapted from [7].

(a) The common difference is 1 and the first term is also 1; the sum of N terms

is, therefore, N2

(1 + N) = N(N+1)2

.34

(b) The common difference is 2, the first term is 1, and the N +1st term is 2N +1.The sum of N + 1 terms is N+1

2(2 · 1 + N · 2) = (N + 1)2.

[Many students failed to notice that the number of summands was N + 1 —not N . An error of this type might have been detected by checking one’scomputations for small values of N , e.g. N = 0 or N = 1. Carry out thesummation mechanically, then compare the sum that you obtain with thevalue of the formula you have derived; if the values are different, you need tocheck every step of your work carefully.]

(c) Let the first term be a and the common difference be d. We have to solve theequations:

x3 = 4x1 ⇔ a + 2d = 4a

x6 = 17 ⇔ a + 5d = 17

which yield a = 2, d = 3. Hence xn = 2 + 3(n− 1) = 3n− 1.

(d) We solve the equation n2

(2 · 2 + (n− 1) · 3) = 950, which reduces to 3n2 +n−1900 = 0, whose only positive solution is n = 25.

(e) If the first term is a and the common ratio is r, the given information impliesthat

a · r6 − 1

r − 1= 9a · r3 − 1

r − 1if r 6= 1 , (113)

6a = 9a if r = 1 . (114)

34This expression is known, from other considerations, to be the number of ways of choosing 2 objectsfrom a set of N distinct objects; it is often denoted by

(N2

).


When a = 0, both equations are satisfied: the sequence is 0, 0, 0, . . . ; thecommon ratio is indeterminate. When r 6= 1, (113) yields r6 − 1 = 9(r3 − 1),so r3 = 1 (which contradicts the hypothesis) or r3 = 8, hence r = 2 and thesequence is then

a, 2a, 4a, ..., 2n−1a, ...

But the sequence is not completely determined, since any value of a is ac-ceptable — including the value 0 which we already saw as the solution to(114).

(f)

3.157157157157... = 3 +

(1

10+

5

100+

7

1000

)+

1

1000

(1

10+

5

100+

7

1000

)

+1

10002

(1

10+

5

100+

7

1000

)+ . . .

= 3 +157

1000+

1

1000· 157

1000+

1

10002· 157

1000+ . . .

= 3 +157

1000· 1

1− 11000

= 3 +157

999=

3154

999.

2. (cf. [5, Exercise 3-6, p. 66] An annuity pays 1000 per year for 8 years. If i = 0.05,find each of the following

(a) The value of the annuity one year before the first payment.

(b) The value of the annuity one year after the last payment.

(c) The value of the annuity at the time of the 4th payment.

(d) If possible, the number of years an annuity-immediate would have to run inorder that its value, viewed one year before the first payment should be twicethat of the 8-payment annuity whose value at the same time was determinedabove.

(e) If possible, the number of years an annuity-immediate would have to runin order that its value, viewed one year before the first payment, should bethree times that of the 8-payment annuity whose value at the same time wasdetermined above.

(f) If possible, the number of years an annuity-immediate would have to runin order that its value, viewed one year before the first payment, should befour times that of the 8-payment annuity whose value at the same time wasdetermined above.35

35Note that the wording of the cited questions in the textbook required a number of assumptions thathave been made more explicit in the present questions.


(g) (cf. [5, Exercise 3-58, p. 73]) Redo part (a), assuming now that the annuityis continuous. (The effective annual interest rate remains 5%, and the time isstill 8 years.)

Solution:

(a) 1000a8 5% = 1000 · 1−(1.05)−8

0.05= 20000 (1− (1.05)−8) = 6463.21.

(b) 1000s8 5% = 1000 · (1.058−1)(1.05)

0.05= 10026.56.

(c)

1000s4 5% + 1000a4 5% = 1000(1.05)4a45%

= (1.05)4(6463.21) = 7856.07 .

(d) We have to solve for n:

an 5% = 2a8 5%

⇒ 1− vn = 2− 2v8

⇒ vn = 2v8 − 1

⇒ n =ln(2v8 − 1)

ln v= 21.30 years.

(e) We have to solve for n:

an 5% = 3a8 5%

⇒ 1− vn = 3− 3v8

⇒ vn = 3v8 − 2

⇒ n =ln(3v8 − 2)

ln v= 71.52 years.

(f) In this case we observe that the value of a perpetuity of 1000 per year at5% is only 1000

0.05= 20000 < 4(6463.21); so the problem will have no solution.

If we attempt to solve as in the preceding case, we will obtain the equation1− vn = 4− 4v8 ⇒ vn = 4v8 − 3 = −0.29, which has no solution.

(g)

1000a8 5% = 1000

8∫

0

vt dt

= 1000 · 1− (1.05)−8

ln(1.05)= 6623.48.


3. [5, Exercise 3.15] Prove each of the following identities

• algebraically; and

• verbally

(a) an = an + 1− vn

(b) sn = sn − 1 + (1 + i)n

Solution:

(a) an differs from an in that it has an immediate payment of 1 but lacks the finalpayment of 1 n years hence, whose value now is vn.

Algebraically,

an = an(1 + i)

= an + ian

= an + i · 1− vn

i= an + (1− vn)

(b) sn differs from sn in that is lacks a payment of 1 at time t = 0, but has apayment of 1 that has accumulated interest over n years, so that its presentvalue is (1 + i)n.

Algebraically,

sn = sn(1 + i)

= sn + isn

= sn + i · (1 + i)n − 1

i= sn + ((1 + i)n − 1)

4. [5, Exercise 3-45, p. 71] Wilbur leaves an inheritance to four charities: A, B, C,D. The total inheritance is a series of level payments at the end of each year,payable forever. During the first 20 years, A, B, C share each payment equally. Allpayments after 20 years are to revert to charity D. The present value of the sharesof A, B, C, and D are all equal. Showing all your work , prove that i = 0.07177.

Solution: It does not limit generality to assume that the level payments are all of1. The present value of the payments to each of A, B, C is 1

3a20 i%. The payments

to D constitute a perpetuity-immediate of 1 deferred 20 years; its value is v20 · 1i.

Accordingly we have to solve the following equation for i:

1

3· a20 i% = v20 · 1

i


⇔ 1− v20

3= v20

⇔ v20 =1

4

⇔ 1 + i = 4120 = 1.0717735

so i = 7.177%.

5. Find the present value at i effective of a perpetuity whose annual payments of 1000begin with a payment of 1000 after one year, with the property that each paymentthereafter is reduced by 10% from the preceding payment. In particular, determinethe present value when i = 2.5%.

Solution: [STUDENTS WERE ASKED NOT TO SUBMIT A SOLUTION TOTHIS PROBLEM.] The present value is

1000(v + 0.9v2 + 0.92v3 + 0.93v4 + ... + 0.9n−1vn + ...)

= 1000v∞∑

n=0

(0.9

1 + i

)n

=1000v

1− 0.9v

=1000

(1 + i)− 0.9=

1000

0.1 + i

When i = 2.5%, the present value is 10000.125

= 8000.

6. A fund of 10,000 is to be accumulated by means of deposits of 1000 made at theend of every year, as long as necessary. If the fund earns an effective rate of interestof 21

2%, find how many regular deposits will be necessary, and the size of a final

deposit to be made one year after the last regular deposit.

Solution: [2, Example 6.5, pp. 59-60] There are often tacit assumptions in inter-est problems; usually there is an “obvious” intended interpretation, while otherinterpretations might be justified by some unusual reading of the wording. In thepresent problem one is to assume that the deposits are not permitted to exceed1000, and that all deposits (the “regular” deposits) but the last are to be exactly1000. The last deposit can be smaller, but not larger.

Let n be the number of regular deposits required. Then n is the largest integerthat satisfies the “inequality of value”

1000sn 2.5% ≤ 10000 ;

equivalently, sn 2.5% ≤ 10 . Computing the values of sn 2.5%, we find that s8 2.5% =8.73612, s9 2.5% = 9.95452, s10 2.5% = 11.20338. Hence n = 9. The final partial de-posit will have to be the excess of 10,000 over the accumulated value of s9 2.5% after


one year, i.e., the excess of 10,000 over (1.025)s9 2.5% = 10, 000−(1.025)(9954.52) =−203.38. So rather than a final deposit, there will be a final refund of 203.38. Thissituation could also have been seen from the value of 10,000s10 2.5% = 11203.38,which would be the value after a 10th deposit; since this exceeds 11,000, no 10thdeposit would be required.

[ADDED March 10th, 2003] Another possible interpretation of the instructions inthis problem is to treat the 8th as the last “regular” deposit, and to reduce the 9thdeposit so that, when the time arrives for a possible 10th deposit, the balance inthe fund is exactly 10000. If we define the value of the 9th deposit to be x, then

(1 + i)((1 + i)s8 + x

)= 1000

⇔ x =10000

1.025− 1025s8

= 9756.0976− 8954.5188 = 801.58.

We can verify the correctness of this computation by observing that the excesspayment of 1000−801.58 = 198.42 accumulated at 2.5% to 1.025×198.42 = 203.38,which was computed earlier as the amount refunded one later.

[This assignment was intended as a learning exercise, rather than a testing exercise.Students were not expected to have seen an example of this type before.]

7. A deferred annuity is one that begins its payments later than might otherwise havebeen expected. We define

m |an = vman (115)

m |an = vman (116)

Prove, both algebraically and verbally, that, for non-negative integers m and n,

m |an = am+n − am (117)

(1 + i)m · sn = sm+n − sm (118)

1 |an = an (119)

Solution:

(a)

m |an = vm(v + v2 + . . . + vn)

= (vm+1 + vm+2 + . . . + vm+n

= (v1 + v2 + . . . + vm+n − (v1 + v2 + . . . + vm

= am+n − am



In deferring an n-payment annuity-immediate by m years we are planning forthe first payment to be made m + 1 years from now, and the last m + n yearsfrom now. These can be viewed as the last n payments of an m + n-paymentannuity-immediate whose first payment begins one year hence; thus we obtainthe value of m |an by subtracting am from am+n.

(b)

(1 + i)m · sn = (1 + i)m(1 + (1 + i)1 + . . . + (1 + i)n−1

)

= (1 + i)m + (1 + i)m+1 + . . . + (1 + i)m+n−1

=(1 + (1 + i)1 + . . . + (1 + i)m+n−1

)

− (1 + (1 + i)1 + . . . + (1 + i)m−1

)

= sm+n − sm

The payments associated with (1 + i)m · sn can be interpreted as the first mpayments of an m + n-payment annuity whose last payment has just beenmade. If we subtract from sm+n the value of the last n payments as viewedfrom the day of the last payment, we obtain the value of those first m pay-ments.

(c)

1 |an = v(1 + v + v2 + . . . + vn−1

)

= v + v2 + . . . + vn = an

When we defer an annuity-due one year it becomes an annuity-immediate.

B.1.4 Fourth 2002/2003 Problem Assignment, with Solutions

1. (cf. [5, Exercise 3-60, p. 73]) A student attending engineering school has increasingamounts of income as she advances through her programme. Accordingly she agreesto borrow a decreasing annual amount from her parents during her 5 training years,and to repay the loan with increasing amounts for 15 years after graduation. Shereceives amounts 5X, 4X, 3X, 2X and X at the beginning of each of 5 years,where the last payment is paid at the beginning of her final year. At the end ofher first year after graduation she pays 500, and then increases the amount by 200each year until a final payment of 3300.36 If the interest rate is 5%, determine X.

Solution: An equation of value at the time of graduation is

(1 + i)X(Ds)5 = 300a15 + 200(Ia)15

36The original version of this problem gave the final payment as 3500, which would have required 16years of payments. A correction was announced at the lecture of March 3rd, 2003.


implying that

X =300a15 + 200(Ia)15

(1 + i)(Ds)5

=300(1− v15) + 200((1 + i)a15 − 15v15)

5(1 + i)6 − (1 + i)6a5

= 993.11.

For students who corrected the error in the problem by increasing the number ofyears by 1, here is a solution:

Solution: An equation of value at the time of graduation is

(1 + i)X(Ds)5 = 300a16 + 200(Ia)16

implying that

X =300a16 + 200(Ia)16

(1 + i)(Ds)5

=300(1− v16) + 200((1 + i)a16 − 16v16)

5(1 + i)6 − (1 + i)6a5

= 1082.33.

2. (cf. [5, Exercise 4-2, p. 85]) A loan is being repaid by 36 monthly payments. Thefirst 12 installments are 250 each; the next 18 are 300 each; and the last 6 are 500each. Assuming a nominal annual interest rate of 12% compounded monthly,

(a) Find the principal, A(0), of the loan.

(b) Using the Prospective Method, find the loan balance immediately after the6th payment.

(c) Using the Retrospective Method, find the loan balance immediately after the6th payment.

(d) Divide the 7th and 8th payments into principal and interest.

Solution:

(a) Using the Prospective Method, the principal is seen to be

A(0) = 250a12 + 300v12 · a18 + 500v30 · a6

= 9, 329.46


(b) Immediately after the 6th payment, the value of the remaining 30 payments(at an interest rate of 1% per period) is

250a6 + 300v6 · a18 + 500v24 · a6

= 2501− v6

i+ 300v6 · 1− v18

i+ 500v24 · 1− v6

i

=1

i· ((250 + 500v24)(1− v6) + 300v6(1− v18)

)

=1

i· (250 + 50v6 + 200v24 − 500v30

)

= 8365.40

(c) The principal of the loan has been determined above. The outstanding prin-cipal is the accumulated value of this principal decreased by the accumulatedvalues of the payments that have been made, i.e.

(1.01)6A(0)− 250s6 = (1.016)(9, 329.46)− 250 · (1.01)6 − 1

0.01= 9, 805.38− 1, 538.00 = 8, 365.40.

(d) The principal owing immediately after the 6th payment is known to be 8,365.40.At the time of the 7th payment, this will have accumulated interest of 1%, or83.65; the balance of the payment, i.e. 250 − 83.65 = 166.35, will be appliedto reduction of principal. The reduced balance of 8365.40− 166.35 = 8199.05will accumulate interest in the amount of 0.01 × 8199.05 = 81.99 in the 8thmonth. The 8th payment will include, in addition to this amount of interest,an amount of 250 − 81.99 = 168.01 for the reduction of principle; the out-standing principal after the 8th payment will be 8199.05− 168.01 = 8031.04.

3. (a) [5, Exercise 4-16, p. 87] Harriet is repaying a car loan with payments of 2,000every three months and a final payment 3 months after the last full paymentof 2,000. If the amount of interest in the 4th installment (paid at the end ofthe first year) is 1,100, find the principal of the loan, the time and amountof the final payment, and the amounts of principal and interest in that finalpayment. Assume that interest is compounded monthly, at a nominal annualrate of 18%.

(b) Construct an amortization schedule for the first year of this loan.

Solution:

(a) The interest rate being charged monthly is 0.18/12 = 1.5%. Let the principalof the loan be A. The Retrospective Method shows that the amount owing


immediately after the 3rd installment (paid at 9 months) is

−2000((1.015)6 + (1.015)3 + (1.015)0

)+ (1.015)9A .

This unpaid balance will, in 3 months, earn the lender interest in the amountof

1100 =((1.015)3 − 1

) (−2000((1.015)6 + (1.015)3 + (1.015)0

)+ (1.015)9A

).

Thus

A =1100

(1.015)12 − (1.015)9

+2000((1.015)−9 + (1.015)−6 + (1.015)−3

)(120)

= 26552.32

We can consider the payments as constituting an annuity, with time interval 3months, and interest rate per 3 months of (1.015)3 − 1 = 0.045678375. Usingthe Prospective Method, we see that the value of n installments, as of the dayof the loan, is

2000an0.045678375 = 2000 · 1− (1.015)−3n

0.045678375.


1− (1.015)−3n ≥ 26552.32× 0.045678375

2000= 0.6064334

i.e., such that1.015−3n ≤ 0.3935666

or

n ≥ − ln 0.3935666

3 ln 1.015= 20.88

Thus there will be 20 full payments of 2,000, the last full payment being made5 years after the beginning of the loan. At that time the amount outstandingwill be

26552.32(1.015)60 − 2000s200.045678375

= 64873.15− 2000 · (1.0153)20 − 1

(1.015)3 − 1= 64873.15− 63190.50 = 1682.65

The last payment will be 1682.65(1.015)3 = 1759.51; of this, the interestcomponent will be 1682.65 ((1.015)3 − 1) = 76.86, and the balance will be theoutstanding principal of 1682.65.


(b)Duration Payment Interest Principal Repaid Outstanding Principal(Months)

0 26552.323 2000.00 1212.87 787.13 25765.196 2000.00 1176.91 823.09 24942.109 2000.00 1139.31 860.69 24081.4112 2000.00 1100.00 900.00 23181.41

4. John has borrowed 10000, on which he is paying interest at 10% effective peryear. He is required to pay the interest on the loan annually, and is permitted torepay only the entire loan, and only on an anniversary. He decides to accumulatea sinking fund to accumulate the funds to repay the loan. Suppose that Johnhas 2400 available at the end of each year, out of which to pay both the intereston the loan and an annual contribution to his sinking fund. If the sinking fundaccumulates at 6%, complete a table under the following headings to determinewhen John will be able to repay the loan.

Duration Contribution Interest Interest Earned Balance of(Years) to Sinking Fund on Loan in Sinking Fund Sinking Fund

0 0 0 0 0. . . . . . . . . . . . . . .

Solution: The instructions asked that the student “complete a table...to determinewhen John will be able to repay the loan”. The information could have beenobtained without the table, however, by finding the smallest value of n for which1400sn ≥ 10000; this can be seen to be n = 6, where

10000− 1400sn = 234.55 . (121)

From (121) we see that the shortfall in the balance of the sinking fund after thelast payment of 1400 is 234.55. The value of the sinking fund is not yet sufficientto repay the loan. Even without a 7th payment the sinking fund will exceed 10000by the time when that payment is due. It will, however, be necessary to pay theinterest charge of 1000 on the loan. If a full 6th payment of 1400 was made into thesinking fund, there would be a refund of (1.06)(1400)s6− 1000 = 351.38. However,a better solution would have been for John to make a smaller 6th deposit into thesinking fund — just sufficient to bring the fund up to the level of 10000 at the timeof the 7th interest payment. The balance just after such a 6th payment would needto be 10000

1.06, and the balance just prior to the 6th deposit would be (1.06)1400s5;


so the appropriate 6th deposit would be

10000

1.06− (1.06)1400s5 = 9433.963− 8365.446 = 1068.52 .

The first table below shows what would happen if John made a full 6th contribution:


0 0.00 0. 0.00 0.001 1400.00 1000. 0.00 1400.002 1400.00 1000. 84.00 2884.003 1400.00 1000. 173.04 4457.044 1400.00 1000. 267.42 6124.465 1400.00 1000. 367.47 7891.936 1400.00 1000. 473.52 9765.457 -351.38 1000. 585.93 10351.38

The following table shows the result of a reduced 6th contribution:


0 0.00 0. 0.00 0.001 1400.00 1000. 0.00 1400.002 1400.00 1000. 84.00 2884.003 1400.00 1000. 173.04 4457.044 1400.00 1000. 267.42 6124.465 1400.00 1000. 367.47 7891.936 1068.51 1000. 473.52 9433.967 0.00 1000. 566.04 10000.00

NOTE TO THE GRADER: PLEASE ACCEPT EITHER OF THESE TABLES.

5. (This is a complicated variant of [5, Exercise 4-6, p. 86]. It requires considerablepersistence, but is a very thorough exercise. Don’t panic! This is not a typicalexamination question.) Garfield is repaying a debt with 25 annual payments of1000 each, at an annual interest rate of i = 10%. The terms of his loan permithim to make additional payments on the date of any regular payment. After anysuch additional payment, the terms of the loan require the borrower to continuewith payments of 1000 until a last payment of 1000 or less which settles the debtcompletely.


(a) At the end of the 7th year Garfield proposes to make, in addition to hisregular annual payment of 1000, an extra payment of 5000. At that timehe also proposes to reduce his remaining payment period by 4 years, and tomake level payments over that time (replacing the originally agreed paymentsof 1000). Find the revised annual level payment, computed using the interestrate i = 10%.

(b) The lender is obliged to accept Garfield’s extra payment. But he is not obligedto accept Garfield’s proposed method to repay the loan in fewer payments.If, at the time of the change in the payment scheme, the lender insists oncharging an interest rate of i = 12% when the remainder of the loan will berepaid over 14 equal annual payments, what will be the revised annual levelpayment that will have to be paid at the end of the each of the next 14 years?

(c) Determine the premium Garfield is being asked to pay as a result of theincreased interest rate in part 5b. Express the amount as of the date of theproposed change in the payment scheme. Make two sets of calculations:

i. when the cost of money37 is 10% per annum;

ii. when the cost of money is 12% per annum.

(d) As the loan is repaid, the lender is able to put his money to work. Supposethat money now costs 12%, instead of the 10% that prevailed when the loanwas written. One might have expected the lender to encourage the borrower torepay the loan faster. Faced with the lender’s intransigence, Garfield makesthe supplementary payment of 5000, but decides to abandon his plans tochange the payment size; his payments will be 1000 per year until possiblythe last payment. Determine whether the lender has suffered from his ownstubbornness: express his loss (or gain) as of the time of the supplementarypayment made with the 7th payment.

(e) Suppose that, learning that the cost of money is 12% when he is about tomake his supplementary payment, Garfield changes his plans. He makes nochange to his loan, but invests his 5000 elsewhere in an annuity which willprovide him with payments of 1000 to apply to as many of the final paymentsunder his loan as possible. As of the beginning of the 8th year of the loan(immediately following the 7th payment and any supplementary payment)compare the cost of this scheme with

i. his commitment under the original loan contract;

37By the statement The cost of money is i we intend that Garfield is able — as of this particular date— to either borrow or lend money in any amount and for any period of time commencing immediately— at the interest rate i.


ii. his proposed scheme, whereby he would pay 5000 immediately and paythe rest of the loan over 14 years at 10%;

iii. the lender’s proposal, where an immediate payment of 5000 would befollowed by equal payments for 14 years, computed at a rate of 12%.

(f) Determine the yield earned by the lender under each of the following repay-ment schemes:

i. the loan as originally written — 25 annual payments of 1000;

ii. the repayment scheme proposed by Garfield: 1000 per year for 7 years,5000 additional at the end of the 7th year; level payments for 14 yearsthereafter, amount as computed in part 5a above;

iii. the repayment scheme proposed by the lender, in part 5b, where thelevel payments are recomputed at 12% charged from the time of the 7thpayment; (in this case it suffices to write down an equation that must besatisfied by the yield);

iv. the repayment scheme finally followed by Garfield in part 5d, where heinvests in an annuity to provide him with payments of 1000 for the finalpayments, and pays the rest annually from his savings.

Solution:

(a) Using the Prospective Method, we find that the unpaid balance immediatelyafter the 7th payment, but prior to the extra payment, is 1000a180.1; after theextra payment the amount owed is

1000a180.1 − 5000 = 8201.41− 5000 = 3, 201.41 .

The level payment to repay this principal in 18−4 = 14 years is (with i = 10%and v = 1

1.1)

1000a180.1 − 5000

a140.1

=1000(1− v18)− 5000i

1− v14(122)

= 434.58.

(b) We will have to evaluate the same ratio as in (122), but where numerator anddenominator involve different interest rates.

1000a180.1 − 5000

a140.12

=1000(1− (1.1)−18)− 5000(0.1)

1− (1.12)−14· 0.12

0.1

= 483.


(c) Let’s first determine the nature of Garfield’s commitment under the loan afterhe makes his supplementary payment. We have determined that the loanbalance is 3,201.41. We note that 1, 000a4.10 = 3, 169.87, while 1, 000a5.10 =3, 790.79. Garfield’s loan contract requires him to make 4 payments of 1000;and, at the end of the 5th year, to pay the balance of principal that would beowing at that time. That balance would be

(1.1)5(3201.41)− 1000(s5.1 − 1

)= 5155.90− 5105.10 = 50.80 .

These 5 payments are prescribed under his contract, and the calculation oftheir values is not affected by the cost of money today. What is affected is theway in which Garfield finances these payments; or, equivalently, the presentvalue of these payments, which may not be equal to the loan balance.

i. If the cost of money is 10%, the present value of the 14 payments Garfieldwould have to make would be 483a14.1 = 3558.11; the present value of thepayments required under the loan contract is the outstanding principal,3201.41; so the premium would be 356.70.

ii. If the cost of money is 12%, the present value of the 14 payments wouldbe the outstanding principal, 3,201.41. The value of the 4 payments of1000 and one final payment (i.e. the cost of financing them at 12%) is

1000a4.12 + (1.12)−550.80 = 3037.35 + 28.83 = 3066.18 .

In this case he would be paying a premium of

3201.41− 3066.18 = 135.23 .

(d) After his supplementary payment, Garfield owes an unpaid balance of 3201.41.Had he been permitted to repay this with 14 annual payments of 434.58, thepresent value of those payments would be 2880.47. But Garfield has now beendriven to repay the loan by continuing the planned payments of 1000 until afinal payment. In part 5c we have determined that the number of paymentsof 1000 is 4, and these are followed by a payment one year later of 50.80. Thevalue of these payments today, when money costs 12%, is

1000a40.12 + (1.12)−550.80 = 3037.35 + 28.83

= 3066.18 .

While neither of these repayment schemes yields the full amount owed —because interest rates are higher than at the outset — the lender was wiseto be unwilling to accept Garfield’s offer: he has reduced his losses under theloan by 3066.18− 2880.47 = 185.71.


(e) i. Garfield’s commitment under the original contract is for payments of 1000for 18 more years. At a rate of 12%, Garfield could buy an annuity to coverhis payments at a present cost of 1000a18.12 = 7249.67. We are asked tocompare this cost with the use of the 5000 to purchase a deferred annuityto cover the last payments due under the contract. We will answer thisquestion naively and then, when the answer looks “interesting”, observethat there is a much simpler solution.The present value of an annuity that will cover the payments due in years##k + 1, k + 2, ..., 18 is

1000(a18.12 − ak.12

).

Since

1000a18.12 = 7249.67− 5000 = 2249.67

1000a3.12 = 2401.83

1000a2.12 = 1690.05 ,

Garfield’s 5000 will buy him a deferred annuity paying 1000 per year,starting at the end of 4 years from now until 18 years from now, costinghim 1000s18.12−1000s3.12 = 7249.67−2401.83 = 4847.84 and he will have152.16 left over. The cost of the payments not covered by his 4847.84 is2401.83; the total of his commitments today is therefore 7249.67, preciselythe same as computed above. This should be no surprise, as both setsof computations are being made with an interest rate of 12%. Thus theexcess of one over the other is zero.

ii. An annuity to cover the payments of 434.58 per year for 14 years wouldcost Garfield today 434.58a14.12 = 2880.47; under this scheme he wouldalso be making a payment of 5000, for a total of 7880.47: the deferredannuity method would cost 7880.47− 7249.67 = 630.80 less.

iii. The payments of 483 per year for 14 years are worth today 483a14.12 =1000a180.1 − 5000 = 3201.41; the sum of the value of these payments andthe supplementary payment is 1000a180.1 = 8201.41 : the deferred annuitymethod would cost 8201.41− 7249.67 = 951.74 less.

(f) i. The loan was written to provide a yield of 10%. The fact that the costof money may have changed does not affect the yield, which is influencedonly by the lender’s payments and receipts under the loan.

ii. Since Garfield’s computation of the new level payment is based on aninterest rate of 10%, there has been no change in the yield to the lender:it remains 10%.


iii. When the lender demands that the computation of the replacement levelpayment be based on an interest rate of 12%, he effects a partial im-provement of the yield; but it cannot affect those funds that were alreadyrepaid. Setting up an equation of value at time 7, just after the supple-mentary payment and the 7th payment of 1000, we find that the yieldrate, i, will satisfy the equation:

−(1 + i)71000a2510% + 1000s7i + 5000 + 483a14i = 0

which is equivalent to

−9077.04(1+i)7+1000

((1 + i)7 − 1

i

)+5000+483

((1 + i)14 − 1

i(1 + i)14

)= 0 .

It can be shown that i = 0.10345 approximately. Thus even the increasein the interest rate for the final payments does not effect a marked increasein the yield rate.

iv. In this case the yield rate is 10%, as there are, from the lender’s perspec-tive, no changes.

B.1.5 Fifth 2002/2003 Problem Assignment, with Solutions

1. (a) (cf. [5, Exercise 5.1, p. 105]) A 15-year bond with face value 20000, redeemableat par, earns interest at 7.5%, convertible semiannually. Find the price to yieldan investor 8% convertible semiannually.

(b) What is the premium or discount at which the bond will be purchased?

(c) (cf. [5, Exercise 5-13, p. 109]) For the bond in part 1a find the market price38

and flat price at each of the following dates and times:

i. Just after the 7th coupon has been paid.

ii. 3 months after the 7th coupon has been paid. (Use simple interest forfractions of a period.)

iii. Just before the 8th coupon is paid.

iv. Just after the 8th coupon is paid.

(d) What would the market price and flat price have been just before and justafter payment of the 8th coupon if the bond had been purchased at par?

Solution:

38Market price=amortized value [5, p. 98] is obtained by interpolating linearly between book valueson coupon dates. Thus the market price is a continuous function of time.


(a) Both of the interest rates are nominal annual rates convertible semiannually;we must divide each by 2. Using the “general formula”, we find the price ofthe bond to be

(20000)(1.04)−30 + (0.0375× 20000)a30.04 (123)

= (20000)(1.04)−30 +

(750

.04

) (1− (1.04)−30

)(124)

= (20000)(1.04)−30 + 18750(1− (1.04)−30

)(125)

= 18750 + (20000− 18750)(1.04)−30 (126)

= 18750.00− 385.40 = 19135.40. (127)

Alternatively, using the ”alternate” formula, we find it to be

200 + (750− 800)a304%

= 2000− 50

0.04

(1− (1.04)−30

)= 19135.40.

(b) The bond is selling at a discount of 20000.00 − 19135.40 = 864.60 less thanits redemption value.

(c) The book value at the time of an interest payment is the present value ofthe unpaid portions of the bond; the coupon payments will enter into theaccounting in some other way, e.g., as income. These computations make useof the yield rate associated with the owner’s acquisition of the bond. Themarket price at these times will equal the book value.

i. The remaining 30− 7 coupons are worth 750a234% = 11142.53; the prin-cipal is worth 20000(1.04)−30+7 = 8114.53. The market price is the bookvalue, i.e., the sum, 19257.16.

ii. The book value immediately after the payment of the 8th coupon isthe sum of the value of the unpaid coupons, 750a224% = 10838.34 andthe present value of the principal, 20000(1.04)−30+8 = 8439.11; together,19277.45. The average of this book value and that after the payment ofthe 7th coupon is 19267.30, and this is what we define to be the marketprice.

iii. By our definition, market price is a continuous function of time: themarket price immediately before a coupon payment will be equal to thatafter the payment — here, 19277.45.

iv. As seen above, the book value is 19277.45. This could also have beencomputed by subtracting the value of the coupon from 1.04 times thebook value after the payment of the 7th coupon:

(1.04× 19257.16)− 750.00 = 20027.45− 750.00 = 19277.45 .


We compute the flat prices:

i. The flat price associated with yield rate 4% per interest period, just afterpayment of the 7th coupon, is the same as the market price, as there isno accrued interest. Here the value is, as above, 19257.16.

ii. The flat price is the book value at the time of the preceding couponpayment plus accrued simple interest. That is,

(1 +

1

2· 0.04

)19257.16 = 19642.30 .

In practice this is often quoted as the market price of 19267.30 plus ac-crued interest of 375.00, (half of the next coupon).

iii. The flat price just before the payment of the 8th coupon can be deter-mined in several different ways:

A. Viewed as book value plus accrued interest,

(1 + 0.04) 19257.16 = 20027.45 .

B. Viewed as the book value just after the payment of the coupon, plusthe value of the coupon, it is

19277.45 + 750.00 = 20027.45 .

iv. The flat price just after payment of a coupon is the book value, here19277.45. (The flat price is discontinuous at such points in time: thelimit as time approaches the point from the right is different from thelimit from the left: they differ by the value of the coupon.)

(d) The flat price just after payment of the 8th coupon would have been

20000(1.0375)−22 + 750a223.75%

= 8897.99 + 11102.01 = 20000.00 ;

are you surprised by this result? The flat price just before payment of thecoupon would have been 20000.00 increased by the interest that had beenearned but not paid, i.e. 20000.00 + 750.00 = 20750.00.

The market price would remain constant at 20000 throughout.

2. (a) [5, Exercise 5-3, p. 106] Prove the “Alternate Price Formula”:

P = C + (Fr − Ci)an

algebraically.


(b) (cf. [5, Exercise 5.5, p. 106]) Two bonds with face value 10000 each, redeemableat par at the end of the same period, are bought to yield 10%, convertiblesemiannually. The first bond costs 8246.56, and pays coupons at 7% per year,convertible semiannually. The second bond pays coupons at 6% per half-year.Find

i. the price of the second bond;

ii. the number of coupons remaining on each of the bonds.

Solution:

(a) We can derive the Alternate Price Formula from the “General Formula” [5,(5.1)] as follows:

P = (Fr)an + Cvn

= (Fr)an + C(1− ian) [5, (3.6), p. 45]

= C + (Fr − Ci)an ¤

(b) We apply the Alternate Price Formula proved above to the two bonds. Denotethe price of the second bond by P2, and the number of coupons remaining byn. Then

8246.56 = 10000 + (10000(0.035)− 10000(0.05))an (128)

P2 = 10000 + (10000(0.06)− 10000(0.05))an (129)

From (128) we find that

an5% =8246.56− 10000

10000(0.035− 0.05)= 11.6896 . (130)

i. Substituting in (129) yields P2 = 11168.96 as the price of the secondbond.

ii. We solve (130) for n:

1− vn

0.05= 11.6896

⇒ 1− vn = 0.58448

⇒ vn = 0.41552

⇒ −n ln(1.05) = ln(0.41552)

⇒ n = 18

There are 18 coupons remaining: the bonds mature in 9 years.



3. (cf. [5, Exercise 5-16, p. 108])

(a) Construct a bond amortization schedule for a 3 year bond of face amount5000, redeemable at 5250 with semiannual coupons, if the coupon rate is 5%and the yield rate is 6% — both converted semiannually. Use the format


0...

(b) Construct a bond amortization schedule for a 3 year bond of face amount5000, redeemable at 5250 with semiannual coupons, if the coupon rate is 6%and the yield rate is 5% — both converted semiannually.

Solution:

(a) The purchase price of the bond will be 5250(1.03)−6 + 125a63% = 4396.79 +677.15 = 5073.94.


0 5073.941 125.00 152.22 -27.22 5101.162 125.00 153.03 -28.03 5129.193 125.00 153.88 -28.88 5158.074 125.00 154.74 -29.74 5187.815 125.00 155.63 -30.63 5218.446 125.00 156.55 -31.55 5249.99

(b) The purchase price of the bond will be 5250(1.025)−6 +150a62.5% = 4527.06+826.22 = 5353.28.


0 5353.281 150.00 133.83 16.17 5337.112 150.00 133.43 16.57 5320.543 150.00 133.01 16.99 5303.554 150.00 132.59 17.41 5286.145 150.00 132.15 17.85 5268.296 150.00 131.71 18.29 5250.00


4. (cf. [5, Exercise 5-22, p. 109]) A 10-year bond of face value 12000 with semiannualcoupons, redeemable at par, is purchased at a premium to yield 10% convertiblesemiannually.

(a) If the book value (just after the payment of the coupon) six months before theredemption date is 11828.57, find the total amount of premium or discount inthe original purchase price.


(c) Give the amortization table for the last one and one-half years.

Solution:

(a) The book value just after the pænultimate39 coupon is

11828.57 = 12000v + Fr · a10.05

= 12000v + Fr · v =12000 + Fr

1.05

soFr = 1.05× 11828.57− 12000 = 420 .

Knowing the amount of each coupon we can now evaluate the purchase priceof the bond to have been

12000(1.05)−20 + 420a200.05 = 4522.67 + 5234.13

= 9756.80 .

The bond was purchased at a discount of 12000− 9756.80 = 2243.20.

(b) The rate per period was 42012000

= 3.5%; hence the nominal rate compoundedsemi-annually, is 2× 3.5% = 7%.

(c) For convenience we will compile this table backwards, beginning with Time=20.We were given that B19 = 11828.57. Hence the Principal Adjustment con-tained in the 20th coupon is

11, 828.57− 12, 000 = −171.43 .

book value at Time=18 will be (12000)(1.05)−2 + 420(1.05−1 + (1.05)−2) =11665.31; the book value at Time=17 will be (12000)(1.05)−3 + 420(1.05−1 +(1.05)−2 + (1.05)−3) = 11509.81.

392nd last



20 420.00 591.43 -171.43 12000.0019 420.00 583.26 -163.26 11828.5718 420.00 575.50 -155.50 11665.3117 420.00 . . . . . . 11509.81

5. A 4.5% bond40 with par value of 100 and semiannual coupons is issued on July 1,2003. It is callable at 110 on any coupon date from July 1, 2008 through January1, 2011; at 105 on any coupon date from July 1, 2011 through January 1, 2013;and at 102.50 on any coupon date from July 1, 2013 through January 1, 2015;thereafter it is callable without premium on any coupon date up to January 1,2018 inclusive; its maturity date is July 1, 2018. Determine the highest price thatan investor can pay and still be certain of a yield of

(a) 5% convertible semiannually;

(b) 4% convertible semiannually.

(c) 3% convertible semiannually.

[Hint: For each interest rate, and each range of payments for a given premium,express the price of the bond as a function of the payment number.]

Solution: As a first step towards organizing data, the student should determine thepayment numbers being referred to. If we label the payment dates with naturalnumbers, and define the issue date to be (non)-payment #0, the July dates willhave even numbers, and the January dates odd numbers. The premium of 10 ispayable when the calling date is ##10-15; the premium of 5 when the calling dateis ##16-19; the premium of 2.5 when the calling date is ##20-23; and no premiumis payable when the calling date is ##24-29 nor on the maturity date, which ispayment #30.

40The convention in bonds is that, lacking any indication to the contrary, the term an r% bond refersto a bond whose coupon rate is a nominal rate of r%; the rate is compounded (or converted) as often asindicated in the description of the bond, with the default being half-yearly if there is no indication tothe contrary. Under this convention the coupon rate for this bond is 2.25%. This convention is statedin the textbook [5, p. 93, 1st paragraph]; however, the author usually supplies additional, redundant,information in his problems.


(a) We tabulate the applicable price formulæ, based on the call or maturity date:

First Date Last Date Price10 15 110.00 · (1.025)−n + 2.25an2.5% = 90 + 20.00(1.025)−n

16 19 105.00 · (1.025)−n + 2.25an2.5% = 90 + 15.00(1.025)−n

20 23 102.50 · (1.025)−n + 2.25an2.5% = 90 + 12.50(1.025)−n

24 30 100.00 · (1.025)−n + 2.25an2.5% = 90 + 10.00(1.025)−n

One way to solve the problem would be to laboriously compute the price forevery possible call date, and then take the minimum. However, as the aboveformulæ express the value in terms of a decreasing function vn, it suffices toconsider the smallest value in each interval, i.e. the largest value of n. So wehave to compare the following four prices:

Call Date Price15 103.8119 99.3823 97.0830 94.77

Thus the highest price that the investor may safely pay is 94.77. Becausethe prices were expressible in the form 90 + A(1.025)−n, where A is a non-increasing function of n and (1.025)−n also a non-increasing function of n, wecould have stated immediately that the lowest price would be that for the bondheld to maturity: it was not necessary to carry out all these computations.The situation is not so clear when the yield rate is less than the coupon rate.

(b) Again we tabulate the applicable price formulæ, based on the call or maturitydate:

First Date Last Date Price10 15 110.00 · (1.02)−n + 2.25an2% = 112.50− 2.50(1.02)−n

16 19 105.00 · (1.02)−n + 2.25an2% = 112.50− 7.50(1.02)−n

20 23 102.50 · (1.02)−n + 2.25an2% = 112.50− 10.00(1.02)−n

24 30 100.00 · (1.02)−n + 2.25an2% = 112.50− 12.50(1.02)−n

These formulæ express the value in terms of an increasing function −vn, itsuffices to consider the largest value in each interval, i.e. the smallest value ofn. So we have to compare the following four prices:

Call Date Price10 110.4516 107.0420 105.7724 104.73


Thus the highest price that the investor may safely pay is 104.73.

(c) As before, we tabulate the applicable price formulæ, based on the call ormaturity date:

First Date Last Date Price10 15 110.00 · (1.015)−n + 2.25an1.5% = 150.00− 40.00(1.015)−n

16 19 105.00 · (1.015)−n + 2.25an1.5% = 150.00− 45.00(1.015)−n

20 23 102.50 · (1.015)−n + 2.25an1.5% = 150.00− 47.50(1.015)−n

24 30 100.00 · (1.015)−n + 2.25an1.5% = 150.00− 50.00(1.015)−n

As in the case of 4% we have to compare four prices:

Call Date Price10 115.5316 114.5420 114.7324 115.02

This time the highest price that the investor may safely pay is 114.54.

B.1.6 2002/2003 Class Tests, with Solutions

Versions 2 and 4 appear to have been slightly more difficult than Versions 1and 3, and the grades were adjusted to compensate for this.

Versions 1 (white) and 3 (yellow)

1. Showing your work, solve each of the following problems:

(a) [2 MARKS] Determine the nominal annual interest rate, i1, compounded every3 months, which is equivalent to a nominal annual interest rate of 12% com-pounded every 4 months.

(b) [4 MARKS] Determine the nominal annual interest rate compounded semi-annually, i2, which is equivalent to an effective annual discount rate of 6%.

(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded in-stantaneously (=convertible continuously), which is equivalent to an effectivemonthly discount rate of 1%.

Solution:


(a) In one year a sum of 1 will grow, under the first rate, to(1 + i1

4

)4, and under

the second to(1 + 0.12

3

)3. Equating these two yields 1 + i1

4= (1.04)

34 , so

i1 = 4((1.04)

34 − 1

)= 0.1194 = 11.94% .

(b) We know several relationships between i and the corresponding d. For exam-ple, d = iv = i

1+i= 1− 1

1+i. Solving these equations for i when d = 0.06, we

obtain, corresponding to an effective annual discount rate of 6%, an effectiveannual interest rate of 1

0.94− 1 = 6

94= 0.06383. The effective semi-annual

interest rate corresponding to this effective annual rate will be (1+ 694

)12 −1 =√

10094− 1 = 0.03142. Corresponding to this semi-annual rate, the nominal

annual interest rate compounded semi-annually will be twice this rate, i.e.6.284 %.

(c) Since d + v = 1, the value of v corresponding to d = 1% is 99100

, so 1 + i =10099

= 1 + 199

and i = 199

; hence the corresponding effective annual rate of

interest is(1 + 1

99

)12− 1 = 12.81781%. The “nominal annual interest rate, i3,compounded instantaneously (=convertible continuously)” will be the forceof interest

i3 = δ = ln

((1 +

1

99

)12)

= 12 ln

(1 +

1

99

)= 12 ln

100

99= 0.1206 = 12.06%.

2. (a) [2 MARKS] Define the sequence of payments whose value is represented bythe symbol sni, using a time diagram showing the payments, and indicatingthe point in time where the value of the various payments is being calculated.

(b) [4 MARKS] Derive a formula for sni by using formulæ known to you for thesummation of arithmetic or geometric progressions. Your final formula shouldbe expressed in closed form, i.e., without using summation symbols (

∑) or

dots (. . .),

(c) [4 MARKS] Define what is meant by sni. Give, without proof, a formulawhich expresses the value of sni in terms of i and n.

Solution:


(a)

1 1 1 1 1

0 21 3 · · · n

↑

(b)

sni = 1 + (1 + i) + (1 + i)2 + . . . + (1 + i)n−1

= 1 · (1 + i)n − 1

(1 + i)− 1

=(1 + i)n − 1

(1 + i)− 1

=(1 + i)n − 1

i

(c) sni is the value of the sum of n payments of 1 at the beginning of each year,

evaluated one year after the last payment. Its value is sn+1i−1 = (1+i)n+1−(1+i)i

;other formulæ would also have been acceptable.

3. A loan of 10,000 at i = 10% is to be repaid by ten equal annual payments.

(a) [5 MARKS] Determine the annual payment.

(b) [10 MARKS] Determine an amortization schedule for the first 5 payments,showing, for each payment, the interest portion and the portion for reductionof principal. Use the following format for your table.

Duration Payment Interest Principal Outstanding(Years) Repaid Principal

. . .

(c) [5 MARKS] If the loan is sold to an investor immediately after the 5th paymentat a price to yield 12% effective annual interest, determine the price paid bythe investor.

Solution: (Source = Deferred/Supplemental Examination in Math 329, August,2000, Problem 3.)


(a) If the annual payment is denoted by X, it must satisfy the equation X ·a1010% = 10, 000. Solving this equation yields X = 1000

1−(1.1)−10 = 10000.61445671

=1627.45 as the level annual payment.

(b) If we were interested only in the interest portion of the 5th payment, we mightrecall having proved that [5, p. 79] to be

X(1− (1.1)10−5+1) = 1627.45× (1− 0.56447393) = 708.80.

The portion for reduction of principal would then be

X − 708.80 = 1627.45− 708.80 = 918.65 .

However, the problem required the construction of an amortization table, sothese data can be used only to verify our computations in the table:

Duration Payment Interest Principal Outstanding(Years) Repaid Principal

0 10000.001 1627.45 1000.00 627.45 9372.552 1627.45 937.26 690.19 8682.363 1627.45 868.24 759.21 7923.154 1627.45 792.31 835.14 7088.015 1627.45 708.80 918.65 6169.36

(c) While the outstanding principal is shown as 6169.35, that will equal thepresent value of the remaining 5 payments of 1627.45 each only if the interestrate remains at 10%. If the interest rate changes to 12%, the present value

of the remaining 5 payments falls to 1627.45 · a512% = 1627.45 × 1−(1.12)−5

0.12=

5866.59. This will be the price paid by an investor who expects the 5 remain-ing payments to yield 12% effective interest.

Versions 2 (blue) and 4 (green)

1. [20 MARKS] A borrower takes out a loan of 2000 to be paid by one payment withfull interest at the end of two years. Construct a sinking fund schedule using theheadings

Duration Contribution to Interest Interest Earned Balance of Balance of(Years) Sinking Fund on Loan in Sinking Fund Sinking Fund Principal

. . .assuming that the lender receives 10% convertible semi-annually on the loan, andthe borrower replaces the amount of the loan with equal semi-annual deposits in a


sinking fund to mature when the loan becomes due, where the sinking fund earns8% convertible semi-annually.

Solution: (Source = Final Examination in Math 329, April, 2000, Problem 3; thepresent problem is simplified from that on the examination.)

(a) [7 MARKS] The sinking fund must attain the value of 2000(1.05)4; if wedenote the value of the semi-annual payments into this fund, than X · s40.04 =2000(1.05)4, so

X =2000× (1.05)4 × 0.04

(1.04)4 − 1=

97.2405

0.16986= 572.48.

(b) [13 MARKS] The schedule is as follows:Duration Contribution to Interest Interest Earned Balance of Balance of(Years) Sinking Fund on Loan in Sinking Fund Sinking Fund Principal

0.0 0.00 0.00 0.00 0.00 2000.000.5 572.48 100.00 0.00 572.48 2100.001.0 572.48 105.00 22.90 1167.86 2205.001.5 572.48 110.25 46.71 1787.05 2315.252.0 572.48 115.76 71.48 2431.01 2431.01

2. Showing your work, solve each of the following problems:

(a) [4 MARKS] Determine the nominal annual interest rate compounded semi-annually, i1, which is equivalent to an effective annual discount rate of 4%.

(b) [2 MARKS] Determine the nominal annual interest rate, i2, compounded every6 months, which is equivalent to a nominal annual interest rate of 24% com-pounded every 3 months.

(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded in-stantaneously (=convertible continuously), which is equivalent to an effectivequarterly discount rate of 2%.

Solution:

(a) We know several relationships between i and the corresponding d. For exam-ple, d = iv = i

1+i= 1− 1

1+i. Solving these equations for i when d = 0.04, we

obtain, corresponding to an effective annual discount rate of 4%, an effectiveannual interest rate of 1

0.96− 1 = 4

96= 0.04167. The effective semi-annual

interest rate corresponding to this effective annual rate will be (1+ 496

)12 −1 =√

10096− 1 = 0.02062. Corresponding to this semi-annual rate, the nominal


annual interest rate compounded semi-annually will be twice this rate, i.e.i1 = 4.124%.

(b) In one year a sum of 1 will grow, under the first rate, to(1 + i2

2

)2, and under

the second to(1 + 0.24

4

)4. Equating these two yields 1 + i1

2= (1.06)2, so

i2 = 2((1.06)2 − 1

)= 0.0472 = 24.72% .

(c) Since d+v = 1, the value of v corresponding to d = 2% is 98100

, so 1+ i = 10098

=1 + 2

98and i = 2

98; hence the corresponding effective annual rate of interest is(

1 + 298

)4− 1 = 8.4166%. The “nominal annual interest rate, i3, compoundedinstantaneously (=convertible continuously)” will be the force of interest

i3 = δ = ln

((1 +

2

98

)4)

= 4 ln

(1 +

2

98

)= 4 ln

100

98= 0.08081 = 8.081%.

3. (a) [2 MARKS] Define the sequence of payments whose value is represented bythe symbol ani, using a time diagram showing the payments, and indicatingthe point in time where the value of the various payments is being calculated.

(b) [4 MARKS] Derive a formula for ani by using formulæ known to you for thesummation of arithmetic or geometric progressions. Your final formula shouldbe expressed in closed form, i.e., without using summation symbols (

∑) or

dots (. . .).

(c) [4 MARKS] By allowing n to approach infinity, determine a (closed form)formula for the value of a∞i.

Solution:

(a)

1 1 1 1 1

0 21 3 · · · n

↑

(b)

an = v + v2 + . . . + vn


= v · 1− vn

1− v

= (1 + i)v · 1− vn

(1 + i)− (1 + i)v

=1− vn

(1 + i)− 1=

1− vn

i

(c) Since 1 + i > 1, 0 < 11+i

< 1, so a∞i = limn→∞

1−vn

i=

limn→∞ 1− lim

n→∞ vn

i= 1−0

i= 1

i.

B.1.7 Final Examination, 2002/2003

1. (a) [3 MARKS] The total amount of a loan to which interest has been added is5,000. The term of the loan was 4 years. If the nominal annual rate of interestwas 6% and interest was compounded semi-annually, determine the originalamount of the loan, showing all your work.

(b) [3 MARKS] Showing all your work, determine the simple interest rate underwhich a sum of money will double in 5 years.

(c) [4 MARKS] Showing all your work, determine the effective annual compounddiscount rate under which a sum of money will double in 8 years.

(d) [5 MARKS] Showing all your work, determine the rate of interest, convertiblecontinuously, that is equivalent to an effective interest rate of 1% per month.

2. (a) [8 MARKS] To repay a loan, X is obliged to pay Y 1,000 at the end of Decem-ber, 2004, and 1,200 at the end of December, 2006. He proposes to replacethese two payments by a single payment of 2,196 at the end of December,2005. If Y accepts this proposal, what yield rate will he be earning on hisloan? Show all your work.

(b) [7 MARKS] Showing all your work, determine the value at time t = 0 ofa continuous annuity that pays 10,000 per year for 2 years, at an effectiveannual interest rate of 5%.

3. Express each of the following only in terms of `x, and v.

(a) [2 MARKS] d27

(b) [2 MARKS] 4q24

(c) [2 MARKS] a20:25

(d) [2 MARKS] A120:25

(e) [2 MARKS] A20:25


(f) [2 MARKS] The probability that a 25-year old will survive 40 years, but willdie before reaching age 75.

(g) [3 MARKS] 12|a20:25

4. The Wallace Widget Company is planning to borrow 150,000 from the Bank ofAntigonish, and to undertake to pay interest annually at a rate of 12%; they planto contribute equal annual payments to a sinking fund that earns interest at therate of 9%. The sinking fund will repay the principal at the end of 10 years.Showing all your work, determine

(a) [2 MARKS] the annual interest payment,

(b) [3 MARKS] the annual payment into the sinking fund

At the end of 4 years, when Wallace has made its annual interest payment and its4th payment to the sinking fund, it proposes that this should be the last paymentto the sinking fund. It will apply the balance X accumulated to date in the sinkingfund to repay principal, and it will amortize the remainder of the principal by equalannual payments over the next 5 years, at a rate of 10%.

(c) [4 MARKS] Determine the annual level payment Y under this proposal.

(d) [6 MARKS] Construct an amortization table for this proposal, under thefollowing headings, beginning immediately after the 4th and last payment tothe sinking fund; assume also that all outstanding interest on the loan hasbeen made annually to date:

Duration Payment Interest Principal OutstandingRepaid Principal

4 0.00 0.00 0.00 150000.00−X5 Y =...

......

......

9 0.00

5. Consider a 100 par-value 15-year bond, with semi-annual coupons at the nominalannual interest rate of 4%, convertible every six months. Let t represent time inhalf-years; assume that the bond is callable at 109.00 on any coupon date fromt = 10 to t = 20 inclusive, at 104.50 from t = 21 to t = 29 inclusive, but maturesat 100.00 at t = 30. In each of the following cases, determine what price an investorshould pay to guarantee himself

(a) [7 MARKS] a nominal annual yield rate of 5%, convertible semi-annually;

(b) [8 MARKS] an effective annual yield rate of 3%.


6. In addition to her down payment, Mary’s purchase of her new home is financedby a mortgage of 60,000 payable to the vendor; the mortgage is amortized over 20years, with a level payment at the end of each month, at a nominal annual rate of6% compounded monthly.

(a) [3 MARKS] Determine the monthly payments under this mortgage.

(b) [2 MARKS] Divide the first payment into principal and interest.

(c) [3 MARKS] Determine the outstanding principal immediately after the 60thpayment.

(d) [4 MARKS] Divide the 60th payment into principal and interest.

(e) [3 MARKS] Determine the payment that Mary could make at the end of eachyear which would be equivalent to the year’s 12 monthly payments.

7. (a) [5 MARKS] Define what is meant by (Da)n and (Ia)n, and explain verballywhy

(Da)30 + (Ia)30 = 31a30 .

(b) [10 MARKS] Showing all your work, find the present value (using effectiveannual interest rate i = 6%) of a perpetuity which pays 100 after 1 year, 200after 2 years, increasing until a payment of 2000 is made, after which paymentsare level at 2000 per year forever. [For this problem you may assume that

(Ia)n =an − nvn

i(131)

(Ia)∞ =a∞i

(132)

(Is)n =sn − n

i.] (133)

8. In order to complete the sale of his home in Vancouver, John accepted, in partialpayment, a 200,000 mortgage amortized over 15 years with level semi-annual pay-ments at a nominal annual rate of 5% compounded semi-annually. Fred has cashavailable, and is prepared to buy the mortgage from John and to invest a fixedportion of the semi-annual payments he receives in a sinking fund that will replacehis purchase capital in 15 years. The sinking fund will earn interest at only 4%,compounded semi-annually. Showing all your work, determine the following:

(a) [3 MARKS] the amount of the semi-annual mortgage payments

(b) [4 MARKS] as a fraction of the purchase price Fred pays for the mortgage,the semi-annual payment into the sinking fund


(c) [8 MARKS] the amount that Fred should pay for the mortgage in order to ob-tain an overall yield rate of 6%, compounded semi-annually on his investment.(Note that the sinking fund earns 4% compounded semi-annually.)

B.2 2003/2004

B.2.1 First 2003/2004 Problem Assignment, with Solutions

Distribution Date: Solutions mounted on the Web on Wednesday, 4 February, 2004Assignment was mounted on the Web on Thursday, January 15th, 2005

Hard copy was distributed on Monday, January 19th, 2004Solutions were to be submitted by Monday, January 26th, 2004

(This is a short assignment. Subsequent assignments can be expected to belonger.)

1. It is known that the accumulation function a(t) is of the form b · (1.1)t + ct2, whereb and c are constants to be determined.

(a) If $100 invested at time t = 0 accumulates to $170 at time t = 3, find theaccumulated value at time t = 12 of $100 invested at time t = 1.

(b) Show that this function satisfies the requirement [1, p. 2, #2] that it be non-decreasing.

(c) Determine a general formula for in, and show that limn→∞

in = 10%. (Use

L’Hopital’s Rule.)

Solution: [1, Exercise 4, p. 30] Denote the corresponding amount function by A(t).

(a) An accumulation function must have the property that a(0) = 1; this impliesthat 1 = a(0) = b + 0, so b = 1.

The given data imply that

170 = 100(a(3)) = 100(1(1.331) + c · 32) (134)

which implies that c = 0.041. We conclude that

A(t)

A(0)= a(t) = (1.1)t + 0.041t2 , (135)

implying that a(1) = 1.141, a(12) = 9.042428377. Then

A(12) = A(1) · A(12)

A(1)= A(1) · a(12)

a(1)

= A(1) · 9.042428377

1.141= A(1)(7.925002959) = 792.5002959


so $100 at time t = 1 grows to $792.50 at time t = 12.

(b) It follows from (135) that a′(t) = (1.1)t ln 1.1 + 0.082t, which is positive forpositive t; thus a(t) is an increasing function of t for positive t.

(This property may also be proved “from first principles”. Let t1 ≥ t2. Then

a(t2)− a(t1) = (1.1)t2 + 0.041t22 − (1.1)t1 + 0.041t21= (1.1)t1

((1.1)t2−t1 − 1

)+ (0.041) (t2 − t1) (t2 + t1)

where both of the summands are non-negative for 0 ≤ t1 ≤ t1.

(c)

in =a(n)− a(n− 1)

a(n− 1)=

(0.1)(1.1)n−1 + (0.041)(2n− 1)

(1.1)n−1 − (0.041)(n− 1)2

=0.1 + 0.041

(2n−1

(1.1)n−1

)

1− 0.041(

(n−1)2

(1.1)n−1

)

By L’Hopital’s Rule

limx→∞

2x− 1

(1.1)x−1= lim

x→∞2

(1.1)x−1 ln 1.1= 0

limx→∞

(n− 1)2

(1.1)n−1= lim

x→∞2(x− 1)

(1.1)x−1 ln 1.1

= limx→∞

2

(1.1)x−1(ln 1.1)2= 0

Hence

limn→∞

in =0.1 + 0.041 (0)

1− 0.041 (0)= 0.1 = 10%.

2. It is known that 1000 invested for 4 years will earn 250.61 in interest, i.e., that thevalue of the fund after 4 years will be 1250.61. Determine the accumulated valueof 3500 invested at the same rate of compound interest for 13 years.

Solution: [1, Exercise 14, p. 30] Let i be the rate of compound interest. Then1000(1 + i)4 = 1250.61. The accumulated value of 3500 after 13 years will be

3500(1 + i)13 = 3500

(1250.61

1000

) 134

= 7239.57 .


3. It is known that an investment of 750 will increase to 2097.75 at the end of 25years. Find the sum of the present values of payments of 5000 each which willoccur at the ends of 10, 15, and 25 years.

Solution: [1, Exercise 21, p. 31] Let i be the interest rate. The known fact is that750(1+ i)25 = 2097.75. Hence (1+ i)25 = 2.797 , so v25 = 0.357535924. The presentvalue of three payments of 5000 after 10, 15, and 25 years will, therefore, be

5000(v10 + v15 + v25)

= 5000((0.357535924)

1025 + (0.357535924)

1525 + (0.357535924)

2525

)

= 5000(0.662709221 + 0.539500449 + 0.357535924)

= 7798.73.

4. Find the accumulated value of 1000 at the end of 10 years:

(a) if the nominal annual rate of interest is 6% convertible monthly;

(b) if the nominal annual rate of discount is 5% convertible every 2 years.

Solution: [1, Exercise 32, p. 31]

(a) The accumulation factor for each month is 1 + 6%12

= 1.005. After 10 years1000 grows to

1000(1.005)10×12 = 1819.40.

(b) The discount factor for each 2 years is 1−2×5% = 0.09 (moving backwards),corresponding to an accumulation factor of 1

0.9. After 10 years 1000 grows to

1000(0.09)−102 = 1693.51 .

5. Given that i(m) = 5√

66− 2 and d(m) = 2 − 8

√0.06, find m, the equivalent annual

compound interest rate, and the equivalent annual compound discount rate.

Solution: [1, Exercise 30, p. 32] For an mth of a year the relationship between i(m)

and d(m) is given by (1 +

i(m)

m

)(1− d(m)

m

)= 1

which is equivalent to(m + i(m))(m− d(m)) = m2

or

m =i(m)d(m)

i(m) − d(m).


Substituting the given values

i(m) = 0.041241452

d(m) = 0.040408205

gives m = 2. It follows that

i =

(1 +

i(2)

2

)2

− 1

= (1.020620726)2 − 1 = 4.1666667% = 41

6% =

1

24

d = 1−(

1− d(2)

2

)2

= 1− 0.96 = 4% =1

25

B.2.2 Second 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Friday, February 20th, 2004Assignment was mounted on the Web on January 19th, 2004.Hard copy was distributed on Wednesday, January 28th, 2004

Solutions were due by Monday, February 9th, 2004(Solutions presented subject to correction of errors and omissions.)

1. Find the present value of 1000, to be paid at the end of 37 months under each ofthe following scenarios:

(a) Assume compound interest throughout, and a (nominal) rate of discount of6% payable quarterly.

(b) Assume compound interest for whole years only at a (nominal) rate of discountof 6% payable quarterly, and simple discount at the rate of 1.5% per 3 monthsduring the final fractional period.

(c) Assume compound interest throughout, and a nominal rate of interest of 8%payable semi-annually.

Solution: (cf. [1, Exercise 2, p. 53])

(a) Present value = 1000

(1− 0.06

4

) 373

= 829.94.


(b) Present value = 1000

(1− 0.06

4

) 363

(1− 0.015

3

)= 1000×0.872823×0.9995 =

829.96.

(c) Present value = (1.04)−376 = 1000× 0.785165257 = 785.17.

2. The sum of 5,000 is invested for the months of April, May, and June at 7% simpleinterest. Find the amount of interest earned

(a) assuming exact simple interest in a non-leap year

(b) assuming exact simple interest in a leap year (with 366 days);

(c) assuming ordinary simple interest;

(d) assuming the “Bankers’ Rule”.


(a) The number of days is 30 + 31 + 30 = 91; exact simple interest is 91365· 5000 ·

(0.07) = 87.26.

(b) Exact simple interest is 91366· 5000 · (0.07) = 87.02

(c) Ordinary simple interest is 30+30+30360

· 5000 · (0.07) = 87.50

(d) Interest under the Banker’s Rule is 91360· 5000 · (0.07) = 88.47.

3. Find how long 4,000 should be left to accumulate at 5% effective in order that itwill amount to 1.25 times the accumulated value of another 4,000 deposited at thesame time at a nominal interest rate of 4% compounded quarterly.

Solution: (cf. [1, Exercise 13, p. 55]) The equation of value at n years is

4000(1.05)n = (1.25)(4000)(1.01)4n

so

n =ln 1.25

ln 105− 4 ln 101= 24.82450822 .

4. The present value of two payments of 100 each, to be made at the end of n yearsand 2n years is 63.57. If i = 6.25%, find n.

Solution: (cf. [1, Exercise 14, p. 55]) Solving the equation of value, 100v2n+100vn =63.57, we obtain

vn =−1±√3.5428

2,



in which only the + sign is acceptable, since vn > 0. Taking logarithms gives

n =0.818446587

ln 1.0625= 13.50023411.

We conclude, to the precision of the problem, that n = 13.5 years.

5. (a) Find the nominal rate of interest convertible quarterly at which the accumu-lated value of 1000 at the end of 12 years is 3000.

(b) Find the nominal rate of discount convertible semi-annually at which a pay-ment of 3000 12 years from now is presently worth 1000.

(c) Find the effective annual rate of interest at which the accumulated value of1000 at the end of 12 years is 3000.

Solution:

(a) (cf. [1, Exercise 19, p. 55]) The equation of value at time t = 12 is

1000

(1 +

i(4)

4

)4×12

= 3000 ,

implying that

i(4) = 4(3

148 − 1

)= 9.260676%.

(b) The equation of value at time t = 12 is

3000

(1− d(2)

2

)2×12

= 1000 ,

implying that

d(2) = 2(1− 3

124

)= 9.3678763%.

(c) The equation of value at time t = 12 is

1000 (1 + i)12 = 3000 ,

implying thati = 3

112 − 1 = 9.5872691%.

6. An investor deposits 20,000 in a bank. During the first 4 years the bank credits anannual effective rate of interest of i. During the next 4 years the bank credits anannual effective rate of interest of i−0.02. At the end of 8 years the balance in the


account is 22,081.10. What would the account balance have been at the end of 10years if the annual effective rate of interest were i + 0.01 for each of the 10 years?

Solution: (cf. [1, Exercise 32, p. 57]) The equation of value is

20000(1 + i)4(1 + (i− 0.02))4 = 22081.10 ,

which we interpret as a polynomial equation. The equation is of degree 8, and wedon’t have a simple algebraic method for solving such equations in general. Butthis equation has the left side a pure 4th power, so we can extract the 4th roots ofboth sides, obtaining

(1 + i)(1 + (i− 0.02)) = (1.104055)14 = 1.025056201 ,

which may be expressed as a quadratic equation in 1 + i:

(1 + i)2 − 0.02(1 + i)− 1.025056 = 0

whose only positive solution is

1 + i =0.02 +

√(0.02)2 + 4(1.025056)

2= 1.0225

from which we conclude that i = 2.25%, and that the account balance after 10years would be 20000(1.0225 + 0.01)10 = 20000(1.0325)10 = 27, 737.89.

7. A bill for 1000 is purchased for 950 4 months before it is due. Find

(a) the nominal rate of discount convertible monthly earned by the purchaser;

(b) the annual effective rate of interest earned by the purchaser.


(a) If d(12) be the nominal discount rate, then

1000

(1− d(12)

12

)4

= 950

implying that

1− d(12)

12= 0.9872585

so d(12) = 15.29%.


(b) Let i be the effective annual interest rate. Then

950(1 + i)13 = 1000

implies that

i =

(1000

950

)3

− 1 = 16.635% .

8. A signs a 2-year note for 4000, and receives 3168.40 from the bank. At the endof 6 months, a year, and 18 months A makes a payment of 1000. If interest iscompounded semi-annually, what is the amount outstanding on the note at thetime if falls due?

Solution: If i′ be the rate of interest charged semi-annually, then

3168.40(1 + i′)4 = 4000

so i′ = 6.00%; that is i(2) = 12.00%. The value of the 3 payments at the time thenote matures is

1000((1.06)3 + (1.06)2 + (1.06)1

)= 3374.62

so the amount outstanding before the final payment is 625.38.

9. The Intermediate Value Theorem for continuous functions tells us that such afunction f(x) whose value at x = a has the opposite sign from its value at x = bwill assume the value 0 somewhere between a and b. By computing the value off at the point 1

2(a + b), we can infer that there is a 0 of f in an interval half as

long as [a, b], and this procedure may be repeated indefinitely to determine a zeroof f to any desired accuracy. Assuming that polynomials are continuous, use thisidea to determine the nominal quarterly compound interest rate under which thefollowing payments will accumulate to 1000 at the end of 4 years:

• 300 today

• 200 at the end of 1 year

• 300 at the end of 2 years

Your answer should be accurate to 3 decimal places, i.e., expressed as a percentageto 1 decimal place.

Solution: (We will carry the computations to an accuracy greater than requestedin the problem.)


(a) Let the effective annual interest rate be i. The equation of value at the endof 4 years is

300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 . (136)

(b) We need a continuous function to which to apply the Intermediate ValueTheorem. Some choices may be better than others. We will choose

f(x) = 3x4 + 2x3 + 3x2 − 10 .

We observe that f(0) = −10, that f(2) = 48+16+12−10 = 66 > 0, and thatf(−2) = 48− 16 + 12− 10 = 34 > 0. This tells us that there is a solution toequation (136) for −2 ≤ x ≤ 0, equivalently for −3 ≤ i ≤ −1: such a solutionis of no interest to us, as it does not fit the constraints of this problem. Butthe function is a cubic polynomial, and has 2 other zeros. We see that it alsohas a solution in the interval 0 ≤ x ≤ 2, and we proceed to progressively halveintervals.

(c) The midpoint of interval [0, 2] is 1;

f(1) = 3 + 2 + 3− 10 = −2 < 0 < 66 = f(2) ,

so there must be a root in the interval [1, 2].

(d) The midpoint of [1, 2] is 1.5;

f(1.5) = 3(1.5)4 + 2(1.5)3 + 3(1.5)2 − 10

= 15.1875 + 6.75 + 6.75− 10 = 18.6875

> 0

so there must be a zero in the interval [1, 1.5], whose midpoint is 1.25.

(e)

f(1.25) = 3(1.25)4 + 2(1.25)3 + 3(1.25)2 − 10

= 7.3242 + 3.9063 + 4.6875− 10 = 5.918

> 0


(f)

f(1.125) = 3(1.125)4 + 2(1.125)3 + 3(1.125)2 − 10

= 4.8054 + 2.8477 + 3.7969− 10 = 1.45

> 0



(g)

f(1.0625) = 3(1.0625)4 + 2(1.0625)3 + 3(1.0625)2 − 10

= 3.8233 + 2.3989 + 3.3867− 10 = −0.3911

< 0

so there must be a zero in the interval [1.0625, 1.125], whose midpoint is1.09375.

(h)

f(1.09375) = 3(1.09375)4 + 2(1.09375)3 + 3(1.09375)2 − 10

= 4.2933 + 2.6169 + 3.5889− 10 = 0.4991

> 0


(i)

f(1.078125) = 3(1.078125)4 + 2(1.078125)3 + 3(1.078125)2 − 10

= 4.053197 + 2.506325 + 3.487061− 10 = 0.046582

> 0


(j)

f(1.0703125) = 3(1.0703125)4 + 2(1.0703125)3 + 3(1.0703125)2 − 10

= 3.936984 + 2.452233 + 3.436707− 10 = −0.174076

< 0

so there must be a zero in the interval [1.0703125, 1.078125], whose midpointis 1.07421875.

(k) f(1.07421875) = −.064207958 < 0 so there must be a zero in the interval[1.07421875, 1.078125], whose midpoint is 1.076172.

(l) f(1.076172) = −.008925 < 0 so there must be a zero in the interval[1.076172, 1.078125], whose midpoint is 1.077149.

(m) f(1.077149) = .018814 > 0 so there must be a zero in the interval[1.076172, 1.077149], whose midpoint is 1.076661.


(n) f(1.076661) = .004952 > 0 so there must be a zero in the interval[1.076172, 1.076661], whose midpoint is 1.076417.

(o) f(1.076417) = −.001974 < 0 so there must be a zero in the interval[1.076417, 1.076661], whose midpoint is 1.076539.

(p) f(1.076539) = .001488 > 0 so there must be a zero in the interval[1.076417, 1.076539], whose midpoint is 1.076478.

(q) f(1.076478) = −.000243 < 0 so there must be a zero in the interval[1.076478, 1.076539], whose midpoint is 1.076509.

(r) f(1.076509) = .000637 > 0 so there must be a zero in the interval[1.076478, 1.076509], whose midpoint is 1.076494.

(s) f(1.076494) = .000211 > 0 so there must be a zero in the interval[1.076478, 1.076494], whose midpoint is 1.076486.

(t) f(1.076486) = −.000016 > 0 so there must be a zero in the interval[1.076486, 1.076494], whose midpoint is 1.07649.

(u) f(1.07649) = .000097 > 0 so there must be a zero in the interval[1.076486, 1.07649], whose midpoint is 1.076488.

(v) f(1.076488) = .000041 > 0 so there must be a zero in the interval[1.076486, 1.076488], whose midpoint is 1.076487.

(w) f(1.076487) = .000012. One zero will be approximately x = 1.07649. Thusthe effective annual interest rate is approximately 7.649%. This, however, isnot what the problem asked for. The accumulation function for 3-months willthen be (1.0749)

14 = 1.01822, so the effective interest rate for a 3-month period

will be 1.822%, and the nominal annual interest rate, compounded quarterly,will be 7.288, or 7.3% to the accuracy requested.

THE FOLLOWING PROBLEM WAS CONSIDERED FOR INCLUSION IN THE AS-SIGNMENT, BUT WAS (FORTUNATELY) NOT INCLUDED.

10. [1, Exercise 6, p. 88]

(a) Show thatam−n = am − vmsn = (1 + i)nam − sn

where 0 < n < m.

(b) Show thatsm−n = sm − (1 + i)man = vnsm − an

where 0 < n < m.


(c) Interpret the results in (a) and (b) verbally.

Solution:

(a) We prove the first of these identities by technical substitutions in sums, anal-ogous to changes of variables in a definite integral. For the second identity wegive a less formal proof.

am−n =m−n∑r=1

vr

=m∑

r=1

vr −m∑

r=m−n+1

vr

=m∑

r=1

vr − vm

m∑r=m−n+1

vr−m

=m∑

r=1

vr − vm

1∑s=n

v1−s

under the change of variable s = m− r + 1

=m∑

r=1

vr − vm

n∑s=1

v1−s

reversing the order of the 2nd summation

=m∑

r=1

vr − vm

n∑s=1

(1 + i)s−1

= am − vm · sn

am−n = v + v2 + . . . + vm−n

= v−n+1 + v−n+2 + . . . + v0 + v1 + v2 + . . . + vm−n

− (v−n+1 + v−n+2 + . . . + v0

)

= v−n(v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

)

− ((1 + i)n−1 + (1 + i)n−2 + . . . + (1 + i)v0

)

= (1 + i)n(v1 + v2 + . . . + vn + vn+1 + vn+2 + . . . + vm

)

− ((1 + i)0 + (1 + i)1 + . . . + (1 + i)vn−1

)

= (1 + i)nam − sn

(b) These identities could be proved in similar ways to those used above. Instead,we shall show that these identities can be obtained from the preceding simply


by multiplying the equations by (1 + i)m−n:

sm−n = (1 + i)m−nam−n [1, (3.5), p. 60]

= (1 + i)m−n ((1 + i)nam − sn)

= (1 + i)mam − (1 + i)m · (1 + i)−nsn

= sm − (1 + i)man

sm−n = (1 + i)m−nam−n

= (1 + i)m−n (am − vmsn)

= (1 + i)−n · (1 + i)mam − (1 + i)−nsn

= (1 + i)−nsm − an

(c) i. An (m− n)-payment annuity-immediate of 1 has the same present valueas an annuity for a total term of m = (m− n) + n years minus a correc-tion paid today equal to the value of the deferred n payments. Those npayments are worth sm at time t = m, which amount can be discountedto the present by multiplying by vm.The preceding explanation was based on values at the commencement ofthe first year of an m-year annuity-immediate. Let us now interpret them− n payments as being the last payments of an m-year annuity whosemth payment has just been made. That m-payment annuity was wortham, n years ago — a year before its first payment; today it is worth(1 + i)nan, including the payments we attached at the beginning. Thosepayments are worth sn today, for a net value as claimed.

ii. Consider an value of the first m− n payments of an m-payment annuity-immediate of 1, just after the (m − n)th payment. Since these could beconsidered simply the accumulated value of an (m−n)-payment annuity,they are worth sm−n. But the payments of the m-payment annuity notyet made are worth an, and the entire annuity is worth sm at termination,hence vnsm today; hence vmsm − an is also the value today. This givesthe equality between the extreme members of the alleged inequality.Now let’s evaluate the same m−n payments, but this time consider themto be the last m − n payments of an m-payment annuity-certain; again,the m − nth payment has just been made. From first principles, theaccumulated value of the payments actually received is sm−n , and we areviewing them from the context of an annuity-certain of m payments thatwould be worth sm today: let’s determine the amount that would haveto be paid out today to correct for that expanded annuity. The value ofthe n payments we have tacked on in the past was an one year before


the payments began, and (1 + i)msn today; so we can also view the valuetoday as sm − (1 + i)man .

B.2.3 Third 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Wednesday, March 3rd, 2004.Assignment was mounted on the Web on February 8th, 2004,

hard copy of assignment was distributed on Wednesday, February 11th, 2004.Solutions were to be submitted by 9 a.m., Monday, March 3rd, 2004

SUBJECT TO CORRECTION OF TYPO’S AND OTHER ERRORS

Sketch a time diagram to accompany your solution of all problems except the last.

1. A skier wishes to accumulate 30,000 in a chalet purchase fund by the end of 8years. If she deposits 200 into the fund at the end of each month for the first 4years, and 200 + X at the end of each month for the next 4 years, find X if thefund earns a nominal (annual) rate of 6% compounded monthly.

Solution: The equation of value at the end of 8× 12 = 96 months is

200s96 + X · s48 = 30000 ,

which we may solve to yield

X =30000

s48 0.005

− 200s96 0.005

s48 0.005

=30000

s48 0.005

− 200(1.005)96 − 1

(1.005)48 − 1

= 30000s−1

48 0.005− 200((1.005)48 + 1)

= 30000(0.018485)− 200(2.27049) from the tables

= 100.452.

2. A fund of 2500 is to be accumulated by n annual payments of 50, followed by n+1annual payments of 75, plus a smaller final payment of not more than 75 made 1year after the last regular payment. If the effective annual rate of interest is 5%,find n and the amount of the final irregular payment.

Solution: We shall interpret the payments to be made under two annuities-due: thefirst, for 2n + 1 years, consists of an annual deposit of 50 in advance; the second,for n + 1 years, deferred n years after the first, consists of an annual deposit of 25in advance. It is at the end of year 2n + 1 that the final — drop — payment is to


be made, and it is to be under 75. (Note that this is the type of problem where thedrop payment could turn out to be negative.) We seek the smallest n for which

50 · s2n+1 + 25 · sn+1 > 2500− 75 = 2425

⇔ 2(1.05) ((1.05)n)2 + (1.05)n −(

3 +0.05(2425)

25(1.05)

)

⇔ ((1.05)n)2 +1

2.1(1.05)n − 1

2.1

(3 +

0.05(2425)

25(1.05)

)

⇔(

(1.05)n +1

4.2

)2

>1

(4.2)2+

1

2.1

(3 +

0.05(2425)

25(1.05)

)

⇒ (1.05)n +1

4.2>

(1

(4.2)2+

1

2.1

(3 +

0.05(2425)

25(1.05)

)) 12

= 1.919585178

⇒ n >ln 1.919585178

ln 1.05= 10.65133267

Thus the drop payment will be when t = 11 + 12, i.e., 23 years after the firstpayment under the annuity with payments of 50. Just before the drop paymentthe accumulated value of all previous payments is

50s23 + 25s12 =1.05

0.05

(50 · (1.05)23 + 25 · (1.05)12 − 75

)= 2592.924516

so the drop payment at time t = 23 is 2500− 2592.924516 = −92.92.

Thus we have an example here of a negative drop payment. Could this meanthat we should have taken n = 10? No. In that case we would find that thefinal payment would be larger than the permitted 75. (If tables like those in thetextbook were available, one could determine the value of n by inspecting the valueof 2s2n+1 + sn+1 . We observe from the 5% tables the following values:

n 2s2n+1 + sn

10 84.016511 97.067812 111.3713


50s2n+1 + 25sn > 2500− 75

i.e., such that2s2n+1 + sn > 97 ,


equivalently,

2s2n+1 + sn >97

1.05= 92.38 ,

and so can conclude that n = 11.)

3. On his 30th birthday, a teacher begins to accumulate a fund for early retirementby depositing 5,000 on that day and at the beginnings of the next 24 years aswell. Since he expects that his official pension will begin at age 65, he plans that,starting at age 55 he will make an annual level withdrawal at the beginning of eachof 10 years. Assuming that all payments are certain to be made, find the amountof these annual withdrawals, if the effective rate of interest is 6% during the first25 years, and 7% thereafter.

Solution: Let the constant amount of the withdrawals beginning at age 55 be X.The equation of value at age 55, just before the first withdrawal, is

5000 · s25 6% = X · a10 7%

⇒ Annual Withdrawal X =5000 · s25 6%

a10 7%

= 5000 · 1.06

1.07· 0.07

0.06· (1.06)25 − 1

1− (1.07)−10

= 33477.74

4. At an effective annual interest rate of i it is known that

(a) The present value of 5 at the end of each year for 2n years, plus an additional3 at the end of each of the first n years, is 64.6720.

(b) The present value of an n-year deferred annuity-immediate paying 10 per yearfor n years is 34.2642.

Find i.

Solution: It is convenient to distinguish two cases.

Case i 6= 0: From (4a) we have an equation of value

64.6720 = 5 · a2n + 3 · an ; (137)

from (4b) we have the equation of value

34.2642 = vn · 10 · an = 10(a2n − an

). (138)



Solving these equations, we obtain

a2n = 9.3689 (139)

an = 5.9425 , (140)

implying that

1− v2n

1− vn=

9.3689

5.9425⇒ 1 + vn = 1.5766

⇒ vn = 0.5766.

We can substitute in equation (140) to obtain

i = 0.07125 = 7.125%

which implies that

n = − ln 0.5766

ln 1.07125= 8.000 years.

Case i = 0: Here Equations (137) and (138) become

64.6720 = 5(2n) + 3n = 13n

34.2642 = 10n = 10(2n− n)

which are inconsistent. Thus this case is impossible.

5. (a) Find a12 if the effective rate of discount is 5%.

(b) Charles has inherited an annuity-due on which there remain 12 payments of10,000 per year at an effective discount rate of 5%; the first payment is dueimmediately. He wishes to convert this to a 25-year annuity-immediate at thesame effective rates of interest or discount, with first payment due one yearfrom now. What will be the size of the payments under the new annuity?

Solution:

(a) Since (1− d)(1 + i) = 1, v = 1− d = 0.95 when d = 0.05.

a12 =1− (0.95)12

0.05= 9.19279825.


(b) i = d(1 + i) = dv

= 0.050.95

= 119

. Let X be the size of the new payments. Wemust solve the equation of value,

91927.9825 = X · a25i

where v = 0.95. Hence

X =91927.9825

a25i

=91927.9825i

1− v25

=91927.9825

(1− (0.95)25) 19= 6695.606220

So the new annuity-immediate will pay 25 annual payments of 6695.61, be-ginning one year from now.

6. Give an algebraic proof and a verbal explanation for the formula

m |an = a∞ − am − vm+na∞ .

Solution:

(a)

a∞ − am − vm+na∞ =1

i− 1− vm

i− vm+n · 1

i

=1− (1− vm)− vm+n

i

=vm (1− vn)

i= vm · 1− vn

i= vm · an = m |an

(b) a∞ i is the present value of a perpetuity at rate i of 1 per year, paymentsstarting a year from now. am is the present value of the first m payments ofthat perpetuity; if we subtract this we have the present value of a perpetuity-immediate that starts m years from now, i.e., where the first payment is m+1years from now. vm+na∞ is the value of a perpetuity-immediate of 1 startingm + n years from now, i.e., where the first payment is m + n + 1 years fromnow; if we subtract this term as well, we are left with the present value ofpayments at the ends of years m + 1, m + 2, . . ., m + n, i.e., with the presentvalue of an n-payment annuity-certain of 1, deferred m years, i.e., of m |an


7. A level perpetuity-immediate is to be shared by A, B, C, and D. A receives thefirst n payments, B the next 2n payments, C payments ##3n + 1, . . . , 5n, and Dthe payments thereafter. It is known that the present values of B’s and D’s sharesare equal. Find the ratio of the present value of the shares of A, B, C, D.

Solution: The present values of the shares of A, B, C, D are, respectively, an,vna2n = a3n − an, v3na2n = a5n − a3n, and v5na∞ = a∞ − a5n. The fact that B’sand D’s shares are equal implies that

vn

i· (1− v2n) =

v5n

i

which is equivalent to (v2n)2 + v2n − 1 = 0, implying that

v2n =−1±√5

2.

Since v is positive, only the + sign is admissible:

v2n =−1 +

√5

2= 0.618033988...

sovn = 0.7861513777... .

Then the shares of A, B, C, D will be in the ratio

1− vn : vn − v3n : v3n − v5n : 1− v5n

i.e.0.2138486221 : 0.3002831059 : 0.1855851657 : 0.6997168937

8. (a) Find the present value of an annuity which pays 4,000 at the beginning ofeach 3-month period for 12 years, assuming an effective rate of 2% interestper 4-month period.

(b) Suppose that the owner of the annuity wishes to pay now so that paymentsunder his annuity will continue for an additional 10 years. How much shouldhe pay?

(c) How much should he pay now to extend the annuity from the present 12 yearsto a perpetuity?

(It is intended that you solve this problem “from first principles”, not by substitu-tion into formulæ in [1, Chapter 4].)

Solution:


(a) If i be the effective interest rate per 4-month period, the effective rate per

3-month period will be j = (1 + i)34 − 1. Accordingly the value of the desired

annuity is

4000a48 j = 4000(1 + j) · 1− (1 + j)−48

j

= 4000(1 + i)34 · 1− (1 + i)−36

(1 + i)34 − 1

= 40001− (1.02)−36

1− (1.02)−34

= 138, 317.4894.

(b) Repeating the calculations above for 48 + 40 = 88 payments, we obtain

4000a88 j = 4000(1 + j) · 1− (1 + j)−88

j

= 4000(1 + i)34 · 1− (1 + i)−66

(1 + i)34 − 1

= 40001− (1.02)−66

1− (1.02)−34

= 197, 897.4338,

so the additional payments will cost 197, 897.4338−138, 317.4894 = 59, 579.9444today.

(c) The cost of the perpetuity-due today would be

4000a88 j = 4000(1 + j) · 1

j

= 4000(1 + i)34 · 1

(1 + i)34 − 1

=4000

1− (1.02)−34

= 271, 329.4837,

so the additional payments will cost

271, 329.4837− 138, 317.4894 = 133, 011.9943

today.


9. (No time diagram is needed for the solution to this problem.) In Problem 9 ofAssignment 2 you were asked to apply the Bisection Method to determine thesolution to an interest problem to 3 decimal places. The equation in question was(136):

300(1 + i)4 + 200(1 + i)3 + 300(1 + i)2 = 1000 .

and the solution given began with the values of

f(x) = 3x4 + 2x3 + 3x2 − 10 .

at x = 0 (f(0) = −10), x = 2 (f(2) = 66 > 0), and x = −2 (f(−2) = 34 >0), and we were interested in the solution between 0 and 2 — a solution thatis unique because f ′ is positive in this interval. Apply Linear Interpolation 4times in an attempt to determine the solution we seek. (You are not expectedto know the general theory of error estimation.) The intention is that you applylinear interpolation unintelligently, using it to determine a point where you find thefunction value and thereby confine the zero to a smaller subinterval: the point thatyou find will replace the midpoint in the bisection method. In some situations, asin the present one, the procedure may not be better than the bisection method.Indeed, in the present example, it could take many more applications than thebisection method to obtain the accuracy you obtained with that method.

Solution: We take x1 = 0, x2 = 2. Then

x3 = 0 + (2− 0) · −10

−10− 66= 0.2631578947

f(x3) = −9.741407754

x4 = 0.2631578947 + (2− 0.2631578947) · −9.741407754

−9.741407754− 66= 0.4865401588

f(x4) = −8.891376200

x5 = 0.4865401588 + (2− 0.4865401588) · −8.891376200

−8.891376200− 66= 0.6662236083

f(x5) = −7.486007579

x6 = 0.6662236083 + (2− 0.6662236083) · −7.486007579

−7.486007579− 66= 0.8020951913

f(x6) = −5.796139773

x7 = 0.8020951913 + (2− 0.8020951913) · −5.796139773

−5.796139773− 66


= 0.8988026707

f(x7) = −4.166425912

x8 = 0.8988026707 + (2− 0.8988026707) · −4.166425912

−4.166425912− 66= 0.9641908820

f(x8) = −2.825434839

x9 = 0.9641908820 + (2− 0.9641908820) · −2.825434839

−2.825434839− 66= 1.006713115

f(x9) = −1.837664218

x10 = 1.006713115 + (2− 1.006713115) · −1.837664218

−1.837664218− 66= 1.033620406

f(x10) = −1.162055435

x11 = 1.033620406 + (2− 1.033620406) · −1.162055435

−1.162055435− 66= 1.050340958

f(x11) = −0.721587971

x12 = 1.050340958 + (2− 1.050340958) · −0.721587971

−0.721587971− 66= 1.060611435

f(x12) = −0.442976533

x13 = 1.060611435 + (2− 1.060611435) · −0.442976533

−0.442976533− 66= 1.066874356

f(x13) = −0.270019571

x14 = 1.066874356 + (2− 1.066874356) · −0.270019571

−0.270019571− 66= 1.070676410

f(x14) = −0.163879567

x15 = 1.070676410 + (2− 1.070676410) · −0.163879567

−0.163879567− 66= 1.072978227

f(x15) = −0.099198908

x16 = 1.072978227 + (2− 1.072978227) · −0.099198908

−0.099198908− 66= 1.074369462


f(x16) = −0.059950550

x17 = 1.074369462 + (2− 1.074369462) · −0.059950550

−0.059950550− 66= 1.075209488

f(x17) = −0.036195811

x18 = 1.075209488 + (2− 1.075209488) · −0.036195811

−0.036195811− 66= 1.075716385

f(x18) = −0.021840825

x19 = 1.075716385 + (2− 1.075716385) · −0.021840825

−0.021840825− 66= 1.076022149

f(x19) = −0.013174269

x20 = 1.076022149 + (2− 1.076022149) · −0.013174269

−0.013174269− 66= 1.076206548

f(x20) = −0.007944937

x21 = 1.076206548 + (2− 1.076206548) · −0.007944937

−0.007944937− 66= 1.076317739

f(x21) = −0.004790698

x22 = 1.076317739 + (2− 1.076317739) · −0.004790698

−0.004790698− 66= 1.076384781

f(x22) = −0.002888504

x23 = 1.076384781 + (2− 1.076384781) · −0.002888504

−0.002888504− 66= 1.076425201

f(x23) = −0.001741533

x24 = 1.076425201 + (2− 1.076425201) · −0.001741533

−0.001741533− 66= 1.076449571

f(x24) = −0.001049952

x25 = 1.076449571 + (2− 1.076449571) · −0.001049952

−0.001049952− 66= 1.076464263

f(x25) = −0.000632996


x26 = 1.076464263 + (2− 1.076464263) · −0.000632996

−0.000632996− 66= 1.076473120

f(x26) = −0.000381633

x27 = 1.076473120 + (2− 1.076473120) · −0.000381633

−0.000381633− 66= 1.076478460

f(x27) = −0.000230079

x28 = 1.076478460 + (2− 1.076478460) · −0.000230079

−0.000230079− 66= 1.076481679

f(x28) = −0.000138723

x29 = 1.076481679 + (2− 1.076481679) · −0.0001387233

−0.0001387233− 66= 1.076483620

f(x29) = −0.000083635

x30 = 1.076483620 + (2− 1.076483620) · −0.000083635

−0.000083635− 66= 1.076484790

f(x30) = −0.000050430

x31 = 1.076484790 + (2− 1.076484790) · −0.000050430

−0.000050430− 66= 1.076485496

f(x31) = −0.000030391

The reason that the method is not efficient here is that the graph of the functionis far from linear in the interval under consideration.41

B.2.4 Fourth 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Wednesday, March 31st, 2004Solutions were due by Wednesday, March 17th, 2004

Corrected as of April 29th, 2004.(Solutions presented subject to further correction of errors and omissions.)

1. (a) Find, to the nearest unit, the accumulated value 19 years after the first pay-ment is made of an annuity on which there are 7 payments of 3000 each made

41Issues of this type are beyond MATH 329, and are not covered adequately in the current textbook;if you are interested in Numerical Anaylsis, consider taking MATH 317.


at 112-year intervals. The nominal rate of interest convertible semiannually is

6%.

(b) Find, to the nearest unit, the present value of a 20-year annuity-due whichpays 200 at the beginning of each half-year for the first 8 years, increasing to250 per half-year thereafter. The effective annual rate of interest is 6%.

Solution:

(a) We will interpret the payments as being made under an annuity-immediatewith time-intervals of 11

2years. The clock starts ticking (i.e. t = 0) 11

2years

before the first payment; the last payment is made at time t = 7× 1.5 = 10.5.The evaluation is to be made at time t = 19−1.5 = 17.5, i.e., 7 years after thelast payment; this is 14

3intervals of length 11

2years; it is simpler to view this

as 14 intervals of length 12

year, for each of which the effective interest rate is3%. The effective interest rate — call it j — per 11

2years is j = (1.03)3 − 1 .

The accumulated value is, therefore,

3000(1.03)14 · s7 =3000

j· (1.03)14 · ((1 + j)7 − 1

)

=3000

(1.03)3 − 1· (1.03)14 · ((1.03)21 − 1

)

= 42100.12386

which is 42,100 to the nearest unit.

(b) With 1 + i = (1.06)12 , v = (1.06)−

12 ,

Present Value = 250a40 − 50a16

= (250− 50) + 250

(1− v39

i

)− 50

(1− v15

i

)

= 200 +250 (1− v39)− 50 (1− v15)

i

= 200 +200− 250v39 + 50v15

i= 5342.993032.

To the nearest unit the present value is 5343.

2. (a) The present value of a perpetuity-immediate paying 1 at the end of every 5years is 1.637975. Find i and d.

(b) The present value of a perpetuity-due paying 1 at the beginning of every 5years is 1.637975. Find d and i.

Solution:


(a) A payment of 1 at the end of every 5th year for 5 years is equivalent to apayment of s−1

5at the end of every year. The equation of value is

1.637975 = s−1

5· 1

i

⇒ 1.637975 =1

(1 + i)5 − 1

⇒ (1 + i)5 = 1.610500

⇒ i = 10.0000%,

⇒ d =0.1

1.1=

1

11.

(b) A payment of 1 at the beginning of every 5th year is equivalent to a paymentof a−1

5at the beginning of every year. The equation of value is

1.637975 = s−1

5· 1

d

⇒ 1.637975 =1

1− (1 + i)−5

⇒ (1 + i)−5 = 0.389490

⇒ i = 20.7538%,

⇒ d = 15.1703%.

3. Determine the present value, at a nominal interest rate of 6% compounded quar-terly, of the following payments made under an annuity: 120 at the end of the 3rdyear, 110 at the end of the 4th year, decreasing by 10 each year until nothing ispaid.

Solution: The payments decrease until a payment of 10 at the end of the 14thyear. We can think of a decreasing annuity-immediate of value 10 (Da)14 startingwith a payment of 140 at the end of the 1st year, and then make corrections. Theeffective annual interest rate i is given by

1 + i =

(1 +

0.06

4

)4

= (1.015)4 .

We need only subtract the present values of the payments due at the end of thefirst and second years.

Present Value = 10(Da)14 − 140v − 130v2


=10

(14− a14

)

i− 140(1.015)−4 − 130(1.015)−8

=10

(14− 1−(1.015)−56

(1.015)4−1

)

(1.015)4 − 1− 140(1.015)−4 − 130(1.015)−8

= 532.1438213

4. Find the present value, at an effective annual interest rate of 5.75%, of a perpetuity-immediate under which a payment of 100 is made at the end of the 1st year, 300 atthe end of the 2nd year, increasing until a payment of 2500 is made, which level ismaintained for exactly a total of 10 payments of 2500 (including the first of themin the count of 10), after which the payments fall by 400 each year until they reacha level of 100, which is maintained in perpetuity. (Note: You are expected to showexplicitly how you decompose the payments; it is not sufficient to simply show afew numbers and a sum.)

Solution: The payments reach the level of 2500 at the end of year 13, and continueat that level until the end of year 22, after which they fall by 400 annually untilthey reach 100 at the end of year 28.

Since the steady state is a constant perpetuity, we can begin with a perpetuity-

immediate of 100 per year, whose present value is100

0.0575. The remaining non-zero

portions of payments are finite in number: there are additional amounts of 200at the end of year 2, 400 at the end of year 3, . . . , 2400 at the ends of years 13through 22, 2000 at the end of year 23, . . . , and a final amount of 400 at the endof year 27.

The value at time t = 22 of the remaining amounts for years 23 through 27 is400 · (Da)5, so the value at time t = 0 is (1.0575)−22 ·400 · (Da)6. The present valueof the remainders of the payments at the ends of years 2, 3, ..., 12 is (1.0575)−1 ·200 · (Ia)11. Thus the sum

100

0.0575+ (1.0575)−22 · 400 · (Da)5 + (1.0575)−1 · 200 · (Ia)11

covers all the payments except for a level annuity of 2400 payable at the ends ofyears ##13. . . 22, whose present value is

2400(a22 − a12

).

Thus the present value of the entire scheme of payments is

100

0.0575+ (1.0575)−22 · 400 · (Da)5 + (1.0575)−1 · 200 · (Ia)11



+2400(a22 − a12

)

=100

0.0575+ (1.0575)−22 · 400 · 5− a5

0.0575+ (1.0575)−1 · 200 · a11 − 11v11

0.0575+2400

(a22 − a12

)

=100

0.0575+ (1.0575)−22 · 400 · 5− 1−(1.0575)−5

0.0575

0.0575

+(1.0575)−1 · 200 ·1.0575−(1.0575)−10

0.0575− 11(1.0575)−11

0.0575

+2400 · (1.0575)−12 − (1.0575)−22

0.0575= 1739.130435 + 1543.013920 + 8225.826240 + 9138.807264 = 20646.77786 .

5. Find, to the nearest unit, the present value of a 25-year annuity-due which pays100 immediately, 104 at the end of the 1st year, 108.16 at the end of the 2nd year,where each subsequent payment is obtained from its predecessor by multiplying bya factor of 1.04. The annual effective rate of interest is 8.%.

Solution:

Present Value =24∑

n=0

100vn(1.04)n

= 100 · 1− (1.041.08

)25

1− 1.041.08

= 1648.998444

or 1649 to the nearest unit.

6. (a) A loan of 15,000 is being repaid with payments of 1,500 at the end of eachyear for 20 years. If each payment is immediately reinvested at 6% effective,find the effective annual rate of interest earned over the 20-year period.

(b) A loan of 15,000 is being repaid with payments of 1,500 at the end of eachyear for 10 years. Determine the yield rate to the investor.

Solution:

(a) Let the effective yield rate be i. The payments do not become available untilthe maturity date, after 20 years. Until that time they are locked into apayment-scheme that accumulates to value 1500s20 6%. We are asked for theinterest rate that was earned. There are thus just two transactions: the loan



at time 0, in the amount of 15,000, and the repayment at time 20, in theamount given above. The equation of value at time t = 0 is

1500s20 6%(1 + i)−20 = 15000

⇔ (1 + i)−20 =10(0.06)

(1.06)20 − 1

⇔ i = 6.7293555%.

(b) We have to determine i such that 1500 · a10 i = 15000. This may appear to bea difficult problem. But remember that

a10 = v + v2 + v3 + . . . + v10

and that v ≤ 1. That means that the sum cannot be more than 10; in orderfor it to equal 10, each of the summands must equal 1, so v = 1. 1 + i = 1,and i = 0.

B.2.5 Fifth 2003/2004 Problem Assignment, with Solutions

Distribution Date: Mounted on the Web on Monday, April 7th, 2004.Assignment was mounted on the Web on Wednesday, March 17th, 2004,

hard copy of the assignment was distributed on Friday, March 19th, 2004.Solutions were to be submitted by Friday, April 2nd, 2004

(SUBJECT TO CORRECTION OF TYPO’S AND OTHER ERRORS)

1. A loan is being repaid with instalments of 1000 at the end of each year for 15 years,followed by payments of 2000 at the end of each year for 10 years. Interest is atan effective rate of 4% for the first 10 years, and an effective rate 6% for the next15 years.

(a) Showing all your work, find the numeric value of the amount of interest paidin the 4th instalment without making use of a schedule.

(b) Showing all your work, find the amount of principal repaid in the 20th instal-ment, without making use of a schedule.

(c) Then use the information you have computed to compile the lines of a schedulecorresponding to the payments at the ends of years 20, 21, . . . , 25.

(d) Now solve (a), (b), (c) again, this time assuming that all payments are at thebeginnings of the years: the interest rates remain precisely the same.

Solution:


(a) By the prospective method, the unpaid balance of the loan at time t = 0 is

1000 · a10 4% + (1.04)−10(2000 · a15 6% − 1000 · a5 6%

)

= 1000 · 1− (1.04)−10

0.04

+2000 · (1.04)−10 · 1− (1.06)−15

0.06− 1000 · (1.04)−10 · 1− (1.06)−5

0.06= 18387.66857 .

By the retrospective method, the unpaid balance just after the 3rd paymentof 1000 is, therefore,

18387.66857(1.04)3 − 1000 · s3 4% = 18387.66857(1.04)3 − 1000 · (1.04)3 − 1

0.04= 17562.02642 .

The interest component of the 4th instalment is, therefore,

0.04(17562.02642) = 702.48 .

(b) By the prospective method, the unpaid balance just after the 19th instalmentis

2000 · a6 6% = 2000 · 1− (1.06)−6

0.06= 9834.648653 .


0.06(9834.648653) = 590.0789192 ,

so the component for reduction of principal is

2000− 590.0789192 = 1409.921081 .

(c)

Payment Payment Interest Principal OutstandingNumber amount paid repaid loan balance

19 2000 . . . . . . 9834.6520 2000 590.08 1409.92 8424.7321 2000 505.48 1494.52 6930.2122 2000 415.81 1584.19 5346.0223 2000 320.76 1679.23 3666.7824 2000 220.01 1779.99 1886.7925 2000 113.21 1886.79 0


(d) Since the only interest rate affecting the last payments has not changed, therewill be no changes at all in (b) or (c).

By the prospective method, the unpaid balance of the loan at time t = 0, justbefore the first payment, is

1000 · a10 4% + (1.04)−10(2000 · a15 6% − 1000 · a5 6%

)

= 1000(1.04) · 1− (1.04)−10

0.04+ 2000(1.06) · (1.04)−10 · 1− (1.06)−15

0.06

−1000(1.06) · (1.04)−10 · 1− (1.06)−5

0.06= 19328.71077 .

By the retrospective method, the unpaid balance just after the 3rd paymentof 1000 is, therefore,

19328.71077(1.04)2 − 1000 · s3 4%

= 19328.71077(1.04)2 − 1000 · (1.04)3 − 1

0.04= 17784.33357 .


0.04(17784.33357) = 711.37 .

2. (This problem is modelled on [1, Exercise 23, p. 198].)

On a loan of 30,000, the borrower has agreed to pay interest at 7% effective at theend of each year until the loan is repaid. The borrower has decided to deposit afixed amount at the beginning of each year into a sinking fund earning 4% effective.At the end of 11 years the sinking fund is exactly sufficient to pay off exactly two-thirds of the loan. He plans to continue accumulating the sinking fund until a yearwhen a deposit of not more than this fixed amount will bring the fund balance upto 30,000 and the loan can be immediately repaid.

(a) Calculate the total amount the borrower has to pay out each year (at thebeginning, and at the end), except possibly in the year when the loan isrepaid.

(b) Complete the following table to show how the sinking fund attains the targetvalue of 30,000, and the net amount of the loan after payments ##10, 11, . . .until the loan is paid off.


Payment Interest Sinking Interest earned Amount in Net amountNumber paid fund deposit on sinking fund sinking fund of loan

101112. . .

Solution:

(a) The borrower is paying a constant amount each year of 7% of 30,000, or 2,100to cover the interest costs of servicing the loan; denote by X the constantamount the borrower spends each year; thus the amount she contributes tothe sinking fund is X − 2100. An equation of value at time t = 11 is (X −2100) · s11 4% = 20000, so

X = 2100 +20000× 0.04

((1.04)11 − 1) 1.04= 3525.943064 .

Thus the level contribution to the sinking fund at the beginning of each yearexcept the last is 3525.94− 2100.00 = 1425.94.

(b) The intention is that the column labelled “Amount in sinking fund” showsthe amount just after the payment with the given number.


10 2,100.00 1,425.94 603.62 17,120.03 12,879.9711 2,100.00 1,425.94 684.80 19,230.77 10,769.2312 2,100.00 1,425.94 769.23 21,425.64 8,574.3613 2,100.00 1,425.94 857.03 23,708.61 6,291.3914 2,100.00 1,425.94 948.34 26,082.89 3,917.1115 2,100.00 1,425.94 1,043.32 28,552.15 1,447.8516 2,100.00 305.76 1,142.09 30,000.00 0.00

3. A borrows 12,000 for 10 years, and agrees to make semiannual payments of 1,000,plus a final payment. The lender receives 12% convertible semiannually on theinvestment each year for the first 5 years and 10% convertible semiannually for thesecond 5 years. The balance of each payment is invested in a sinking fund earning8% convertible semiannually.

(a) Find the amount by which the sinking fund is short of repaying the loan atthe end of the 10 years.


(b) Complete the following table to show how the sinking fund attains its maxi-mum value, and the net amount of the loan after payments.


09101112

Solution:

(a) The interest payments for the first 10 half-years are 6% of 12,000, i.e. 720per half-year; and, for the second 10 half-years, 600 per half-year. This leaves280 at the end of each of the first 10 half-years, and 400 at the end of eachof the second 10 half-years to accumulate in the sinking fund, which earns4% effective every half year. The accumulated balance in the sinking fund atmaturity will be

120s10 4% + 280s20 4% =1

0.04

(120

((1.04)10 − 1

)+ 280

((1.04)20 − 1

))

= 25(120(1.04)10 + 280(1.04)20 − 400

)

= 9778.594855

implying that the shortfall to repay the loan will be 12, 000 − 9778.59 =2221.41.

(b) Just after the 8th payment the balance in the fund is

280 · s8 4% =280 ((1.04)8 − 1)

0.04= 2579.98 .

Interest earned by the 9th payments is 103.20. We can now begin to fill inthe schedule:


09 720.00 280.00 103.20 2963.18 9036.8210 720.00 280.00 118.53 3361.71 8638.2911 600.00 400.00 134.47 3896.18 8103.8212 600.00 400.00 155.85 4452.03 7547.97

4. (a) A borrower takes out a loan of 3000 for 10 years at 8% convertible semiannu-ally. The borrower replaces one-third of the principal in a sinking fund earning5% convertible semiannually, and the other two-thirds in a sinking fund earn-ing 7% convertible semiannually. Find the total semiannual payment.


(b) Rework (a) if the borrower each year puts one-third of the total sinking funddeposit into the 5% sinking fund and the other two-thirds into the 7% sinkingfund.

Solution:

(a) The semiannual contribution to the sinking funds is

1000

s20 2.5%

+2000

s20 3.5%

and the semiannual interest payment is 4% of 3, 000, or 120. Hence the totalsemiannual payment is

1000

s20 2.5%

+2000

s20 3.5%

+ 120 =25

(1.025)20 − 1+

70

(1.035)20 − 1+ 120

= 229.8692824

(b) Let the total sinking fund deposit be D. Then the equation of value at ma-turity is

D

3· s20 2.5% +

2D

3· s20 3.5% = 3000 ,

implying that

D =9000

s20 2.5% + 2s20 3.5%

=9000

(1.025)20−10.025

+ 2 · (1.035)20−10.035

= 109.6170427 ,

so the total semi-annual payment is 109.6170427+120=229.6170427.

5. A payment of 800 is made at the end of each month for 10 years to repay a loanof 30,000. The borrower replaces the capital by means of a sinking fund earning anominal annual rate of 6% compounded monthly.

(a) Find the effective annual rate i paid to the lender on the loan.

(b) Suppose that the lender had elected to amortize the loan by equal monthlypayments, at the same rate as he is now paying to the lender. What wouldbe the amount of those equal payments?

Solution:


(a) The monthly contribution to the sinking fund is

30000

s120 0.5%

=600

(1.005)120 − 1= 183.0615058 .

Hence the monthly interest payment is 800 − 183.0615058 = 616.9384942,i.e., 2.056461647% of the principal of 30,000. The effective annual rate is,therefore,

(1.02056461647)12 − 1 = 27.6691838%.

(b) The level monthly payment necessary to amortize a loan of 30000 over 120payments at an effective rate of 2.056461647% is

30000

a120 2.056461647%

=(30000)(0.02056461647)

1− (1.02056461647)−120= 675.6703969 .

(The amount is lower than the borrower is now paying because he is holdingfunds in his sinking fund and earning less there than he his paying the lenderfor the use of the capital.)

B.2.6 2003/2004 Class Test, Version 1

Instructions


• This test booklet consists of this cover, Pages 3080 through 3080 containing ques-tions together worth 60 marks; and Page 3080, which is blank.

• Show all your work. All solutions are to be written in the space provided on thepage where the question is printed. When that space is exhausted, you may writeon the facing page, on the blank page, or on the back cover of the booklet, but youmust indicate any continuation clearly on the page where the question is printed!(Please inform the instructor if you find that your booklet is defective.)


• Calculators. While you are permitted to use a calculator to perform arithmeticand/or exponential calculations, you must not use the calculator to calculate suchactuarial functions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without firststating a formula for the value of the function in terms of exponentials and/orpolynomials involving n and the interest rate. You must not use your calculator inany programmed calculations. If your calculator has memories, you are expectedto have cleared them before the test.


• In your solutions to problems on this test you are expected to show all your work .You are expected to simplify algebraic and numerical answers as much as you can.


1. (a) [5 MARKS] Suppose that the nominal annual rate of interest, compounded8 times per year, is 5%. Showing all your work, determine the equivalenteffective annual rate of discount.

(b) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 3times per year, is 7%. Showing all your work, determine the equivalent annualrate of interest.

(c) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 4%.

(d) [5 MARKS] Suppose that the effective interest rate for1

4year is 3%. Deter-

mine the equivalent nominal interest rate, compounded every 2 years.

2. [15 MARKS] The accumulated value just after the last payment under a 12-yearannuity of 1000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity at an interest rate of 6%, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the size of the payments under the perpetuity.

3. [25 MARKS] A loan of 5000 is to be repaid by annual payments of 250 to commenceat the end of the 6th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be larger thanthe regular payments. Assume i = 4%.




Instructions









1. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 100 to commenceat the end of the 5th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be NO largerthan the regular payments. Assume i = 4.5%.

2. (a) [5 MARKS] Suppose that the effective interest rate for1

5year is 0.02. Deter-


(b) [5 MARKS] Suppose that the nominal annual rate of interest, compounded9 times per year, is 6%. Showing all your work, determine the equivalenteffective annual rate of discount.

(c) [5 MARKS] Suppose that the nominal annual rate of discount, compounded 6times per year, is 5%. Showing all your work, determine the equivalent annualrate of interest.

(d) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 8%.


3. [15 MARKS] The accumulated value just after the last payment under a 12-yearannuity of 1000 per year, paying interest at the rate of 5% per annum effective,is to be used to purchase a perpetuity of 500 per annum forever, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the effective interest rate of the perpetuity, assuming it comes into effectjust after the last payment under the annuity.


Instructions








1. [15 MARKS] The accumulated value just after the last payment under a 9-yearannuity of 2000 per year, paying interest at the rate of 8% per annum effective,is to be used to purchase a perpetuity at an interest rate of 4%, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the size of the payments under the perpetuity.


2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commenceat the end of the 4th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be larger thanthe regular payments. Assume i = 5%.

3. (a) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 6%.

(b) [5 MARKS] Suppose that the effective interest rate for1

6year is 0.015. De-

termine the equivalent nominal interest rate, compounded every 4 years.

(c) [5 MARKS] Suppose that the nominal annual rate of interest, compounded3 times per year, is 8%. Showing all your work, determine the equivalenteffective annual rate of discount.

(d) [5 MARKS] Suppose that the nominal annual rate of discount, compounded12 times per year, is 6%. Showing all your work, determine the equivalentannual rate of interest.




Instructions









1. (a) [5 MARKS] Suppose that the nominal annual rate of discount, compounded6 times per year, is 0.5%. Showing all your work, determine the equivalentannual rate of interest.

(b) [5 MARKS] State the nominal annual interest rate, compounded instanta-neously, which is equivalent to an effective annual interest rate of 10%.

(c) [5 MARKS] Suppose that the effective interest rate for1

5year is 2%. Deter-


(d) [5 MARKS] Suppose that the nominal annual rate of interest, compounded7 times per year, is 2%. Showing all your work, determine the equivalenteffective annual rate of discount.

2. [25 MARKS] A loan of 1000 is to be repaid by annual payments of 200 to commenceat the end of the 4th year, and to continue thereafter for as long as necessary. Findthe time and amount of the final payment if the final payment is to be NO largerthan the regular payments. Assume i = 5%.

3. [15 MARKS] The accumulated value just after the last payment under a 10-yearannuity of 1000 per year, paying interest at the rate of 6% per annum effective,is to be used to purchase a perpetuity of 800 per annum forever, first payment tobe made 1 year after the last payment under the annuity. Showing all your work,determine the effective interest rate of the perpetuity, assuming it comes into effectjust after the last payment under the annuity.




B.2.10 Solutions to Problems on the 2003/2004 Class Tests

The first four problems listed are concerned with equivalent rates of interest and discount(each in 4 parts); the next four concern annuities and perpetuities; and the last four areconcerned with unknown type and final balloon or drop payments.

1. (a) [5 MARKS] [VERSION 1 #1(a)] Suppose that the nominal annual rate of in-terest, compounded 8 times per year, is 5%. Showing all your work, determinethe equivalent effective annual rate of discount.

Solution: We are given that i(8) = 0.05, and asked to determine d.

(1 +

i(8)

8

)8

= 1 + i

and (1− d)(1 + i) = 1

⇒ d = 1−(

1 +i(8)

8

)−8

= 0.0486225508 = 4.86%

(b) [5 MARKS] [VERSION 1 #1(b)] Suppose that the nominal annual rate of dis-count, compounded 3 times per year, is 7%. Showing all your work, determinethe equivalent annual rate of interest.

Solution: We are given that d(3) = 0.07, and asked to determine i.

(1− d(3)

3

)3

= 1− d

and (1− d)(1 + i) = 1

⇒ i =

(1− d(3)

3

)−3

− 1 = 0.073398300 = 7.34%

(c) [5 MARKS] [VERSION 1 #1(c)] State the nominal annual interest rate, com-pounded instantaneously, which is equivalent to an effective annual interestrate of 4%.

Solution: The rate we seek is the solution i to the equation

limn→∞

(1 +

i

n

)n

= 1.04

which is equivalent to ei = 1.04. Solving by taking logarithms, we obtaini = ln(1.04) = 0.039220713151 = 3.92%.


(d) [5 MARKS] [VERSION 1 #1(d)] Suppose that the effective interest rate for1

4year is 3%. Determine the equivalent nominal interest rate, compounded

every 2 years.

Solution: We are given the value ofi(4)

4= 3%, and asked to determine i(

12).

The equation we have to solve is

(1 +

i(4)

4

)4

= 1 + i =

(1 +

i(12)

12

) 12

.

This equation implies that

i(12) =

1

2

((1 +

i(4)

4

)8

− 1

)= 0.1333850405 = 13.3%.

2. (a) [5 MARKS] [VERSION 2 #2(b)] Suppose that the nominal annual rate of in-terest, compounded 9 times per year, is 6%. Showing all your work, determinethe equivalent effective annual rate of discount.


(1 +

i(9)

9

)9

= 1 + i

and (1− d)(1 + i) = 1

⇒ d = 1−(

1 +i(9)

9

)−9

= 0.0580479306 = 5.80%

(b) [5 MARKS] [VERSION 2 #2(c)] Suppose that the nominal annual rate of dis-count, compounded 6 times per year, is 5%. Showing all your work, determinethe equivalent annual rate of interest.


(1− d(6)

6

)6

= 1− d

and (1− d)(1 + i) = 1

⇒ i =

(1− d(6)

6

)−6

− 1 = 0.051491358 = 5.15%


(c) [5 MARKS] [VERSION 2 #2(d)] State the nominal annual interest rate, com-pounded instantaneously, which is equivalent to an effective annual interestrate of 8%.

Solution: The rate we seek is the solution to the equation

limn→∞

(1 +

i

n

)n

= 1.08


(d) [5 MARKS] [VERSION 2 #2(a)] Suppose that the effective interest rate for1

5year is 0.02. Determine the equivalent nominal interest rate, compounded

every 3 years.



13).


(1 +

i(5)

5

)5

= 1 + i =

(1 +

i(13)

13

) 13

.


i(13) =

1

3

((1 +

i(5)

5

)15

− 1

)= 0.1152894460 = 11.5%

3. (a) [5 MARKS] [VERSION 3 #3(c)] Suppose that the nominal annual rate of in-terest, compounded 3 times per year, is 8%. Showing all your work, determinethe equivalent effective annual rate of discount.


(1 +

i(3)

3

)3

= 1 + i

and (1− d)(1 + i) = 1

⇒ d = 1−(

1 +i(3)

3

)−3

= 0.0759156521 = 7.59%

(b) [5 MARKS] [VERSION 3 #3(d)] Suppose that the nominal annual rate ofdiscount, compounded 12 times per year, is 6%. Showing all your work,determine the equivalent annual rate of interest.



(1− d(12)

12

)12

= 1− d

and (1− d)(1 + i) = 1

⇒ i =

(1− d(12)

12

)−12

− 1 = 0.061996367 = 6.20%

(c) [5 MARKS] [VERSION 3 #3(a)] State the nominal annual interest rate, com-pounded instantaneously, which is equivalent to an effective annual interestrate of 6%.


limn→∞

(1 +

i

n

)n

= 1.06


(d) [5 MARKS] [VERSION 3 #3(b)] Suppose that the effective interest rate for1

6year is 0.015. Determine the equivalent nominal interest rate, compounded

every 4 years.


6= 1.5%, and asked to determine i(

14).


(1 +

i(6)

6

)6

= 1 + i =

(1 +

i(14)

14

) 14

.


i(14) =

1

4

((1 +

i(6)

6

)24

− 1

)= .1073757030 = 10.7%

4. (a) [5 MARKS] [VERSION 4 #1(d)] Suppose that the nominal annual rate of in-terest, compounded 7 times per year, is 2%. Showing all your work, determinethe equivalent effective annual rate of discount.


(1 +

i(7)

7

)7

= 1 + i


and (1− d)(1 + i) = 1

⇒ d = 1−(

1 +i(7)

7

)−7

= 0.197733746 = 1.98%

(b) [5 MARKS] [VERSION 4 #1(a)] Suppose that the nominal annual rate ofdiscount, compounded 6 times per year, is 0.5%. Showing all your work,determine the equivalent annual rate of interest.


(1− d(6)

6

)6

= 1− d

and (1− d)(1 + i) = 1

⇒ i =

(1− d(6)

6

)−6

− 1 = 0.005014616 = 0.5014616%

(c) [5 MARKS] [VERSION 4 #1(b)] State the nominal annual interest rate, com-pounded instantaneously, which is equivalent to an effective annual interestrate of 10%.


limn→∞

(1 +

i

n

)n

= 1.10


(d) [5 MARKS] [VERSION 4 #1(c)] Suppose that the effective interest rate for1

5year is 2%. Determine the equivalent nominal interest rate, compounded

every 6 years.



16).


(1 +

i(5)

5

)5

= 1 + i =

(1 +

i(16)

16

) 16

.


i(16) =

1

6

((1 +

i(5)

5

)30

− 1

)= 0.1352269307 = 13.5%.


5. [15 MARKS] [VERSION 1 #2] The accumulated value just after the last paymentunder a 12-year annuity of 1000 per year, paying interest at the rate of 5% perannum effective, is to be used to purchase a perpetuity at an interest rate of 6%,first payment to be made 1 year after the last payment under the annuity. Showingall your work, determine the size of the payments under the perpetuity.

Solution: Let X be the level payment under the perpetuity. The equation of valuejust after the last annuity payment is

1000 · s12 5% = X · a∞ 6%

implying that

X = 1000 · s12 5%

a∞ 6%

= 1000 · 6

5· ((1.05)12 − 1

)= 955.03.

6. [15 MARKS] [VERSION 2 #3] The accumulated value just after the last paymentunder a 12-year annuity of 1000 per year, paying interest at the rate of 5% perannum effective, is to be used to purchase a perpetuity of 500 per annum forever,first payment to be made 1 year after the last payment under the annuity. Showingall your work, determine the effective interest rate of the perpetuity, assuming itcomes into effect just after the last payment under the annuity.

Solution: Let i be the interest rate of the perpetuity. The equation of value justafter the last annuity payment is

1000 · s12 5% = 500 · a∞ i =500

i

implying that

i =500

1000· 1

s12 5%

=500

1000· 0.05

((1.05)12 − 1)

= 3.14%.

7. [15 MARKS] [VERSION 3 #1] The accumulated value just after the last paymentunder a 9-year annuity of 2000 per year, paying interest at the rate of 8% perannum effective, is to be used to purchase a perpetuity at an interest rate of 4%,first payment to be made 1 year after the last payment under the annuity. Showingall your work, determine the size of the payments under the perpetuity.


Solution: Let X be the level payment under the perpetuity. The equation of valuejust after the last annuity payment is

2000 · s9 8% = X · a∞ 4%

implying that

X = 2000 · s9 8%

a∞ 4%

= 2000 · 4

8· ((1.08)9 − 1

)= 999.00

8. [15 MARKS] [VERSION 4 #3] The accumulated value just after the last paymentunder a 10-year annuity of 1000 per year, paying interest at the rate of 6% perannum effective, is to be used to purchase a perpetuity of 800 per annum forever,first payment to be made 1 year after the last payment under the annuity. Showingall your work, determine the effective interest rate of the perpetuity, assuming itcomes into effect just after the last payment under the annuity.

Solution: Let i be the interest rate of the perpetuity. The equation of value justafter the last annuity payment is

1000 · s10 6% = 800 · a∞ i =800

i

implying that

i =800

1000· 1

s10 6%

=800

1000· 0.06

((1.06)10 − 1)

= 6.07%.

9. [25 MARKS] [VERSION 1 #3] A loan of 5000 is to be repaid by annual paymentsof 250 to commence at the end of the 6th year, and to continue thereafter foras long as necessary. Find the time and amount of the final payment if the finalpayment is to be larger than the regular payments. Assume i = 4%.

Solution: (cf. [1, Exercise 32, p. 91]) Let the time of the last — balloon — paymentbe n, and let the amount of the last payment be X. Then n is the largest integersolution to the inequality

5000 ≥ 250(1.04)−5an−5 = 250(1.04)−5 · 1− (1.04)−(n−5)

0.04


⇔ (1.04)−(n−5) ≥ 1− 5000× 0.04× (1.04)5

250⇔ −(n− 5) ln 1.04 ≥ ln

(1− 20× 0.04× (1.04)5

)

⇔ −(n− 5) ≥ ln (1− 0.8(1.04)5)

ln 1.04

⇔ n ≤ 5− ln (1− 0.8(1.04)5)

ln 1.04= 97.39832188.


5000(1.04)97 = 250s92 + (X − 250)

implying that

X = 250 + 5000(1.04)97 − 250

0.04

((1.04)92 − 1

)= 346.8818 .

10. [25 MARKS] [VERSION 2 #1] A loan of 1000 is to be repaid by annual paymentsof 100 to commence at the end of the 5th year, and to continue thereafter foras long as necessary. Find the time and amount of the final payment if the finalpayment is to be NO larger than the regular payments. Assume i = 4.5%.

Solution: Let the time of the last — drop — payment be n, and let the amount ofthe last payment be X. Then n is the smallest integer solution to the inequality

1000 ≤ 100(1.045)−4an−4 = 100(1.045)−4 · 1− (1.045)−(n−4)

0.045

⇔ (1.045)−(n−4) ≤ 1− 1000× 0.045× (1.045)4

100⇔ −(n− 4) ln 1.045 ≤ ln

(1− 10× 0.045× (1.045)4

)

⇔ −(n− 4) ≤ ln (1− 0.45(1.045)4)

ln 1.045

⇔ n ≥ 4− ln (1− 0.45(1.045)4)

ln 1.045= 21.47594530.

Thus we conclude that the drop payment is made at time t = 22. The equation ofvalue at time t = 22 is

1000(1.045)22 = 100s17 + X

implying that

X = 1000(1.045)22 − 100× 1.045

0.045

((1.045)17 − 1

)= 48.143638 .


11. [25 MARKS] [VERSION 3 #2] A loan of 1000 is to be repaid by annual paymentsof 200 to commence at the end of the 4th year, and to continue thereafter foras long as necessary. Find the time and amount of the final payment if the finalpayment is to be larger than the regular payments. Assume i = 5%.

Solution: (cf. [1, Exercise 32, p. 91]) Let the time of the last — balloon — paymentbe n, and let the amount of the last payment be X. Then n is the largest integersolution to the inequality

1000 ≥ 200(1.05)−3an−3 = 200(1.05)−3 · 1− (1.05)−(n−3)

0.05

⇔ (1.05)−(n−3) ≥ 1− 1000× 0.05× (1.05)3

200

⇔ −(n− 3) ln 1.05 ≥ ln

(1− 1000× 0.05× (1.05)3

200

)

⇔ −(n− 3) ≥ ln (1− 0.25(1.05)3)

ln 1.05

⇔ n ≤ 3− ln (1− 0.25(1.05)3)

ln 1.05= 10.00252595


1000(1.05)10 = 200s7 + (X − 200)

implying that

X = 200 + 1000(1.05)10 − 200

0.05

((1.05)7 − 1

)= 200.492935 .

12. [25 MARKS] [VERSION 4 #2] A loan of 1000 is to be repaid by annual paymentsof 200 to commence at the end of the 4th year, and to continue thereafter foras long as necessary. Find the time and amount of the final payment if the finalpayment is to be NO larger than the regular payments. Assume i = 5%.

Solution: Let the time of the last — drop — payment be n, and let the amount ofthe last payment be X. Then n is the smallest integer solution to the inequality

1000 ≤ 200(1.05)−3an−3 = 200(1.05)−3 · 1− (1.05)−(n−3)

0.05

⇔ (1.05)−(n−3) ≤ 1− 1000× 0.05× (1.05)3

200⇔ −(n− 3) ln 1.05 ≤ ln

(1− 5× 0.05× (1.05)3

)


⇔ −(n− 3) ≤ ln (1− 0.25(1.05)3)

ln 1.05

⇔ n ≥ 3− ln (1− 0.25(1.05)3)

ln 1.05= 10.00252595.

Thus we conclude that the drop payment is made at time t = 11. The equation ofvalue at time t = 11 is

1000(1.05)11 = 200s7 + X

implying that

X = 1000(1.05)11 − 200× 1.05

0.05

((1.05)7 − 1

)= 0.517581.

B.2.11 Final Examination, 2003/2004

1. In each of the following problems you are expected to show all your work.

(a) [3 MARKS] If v = 0.97, determine the value of d(4).

(b) [3 MARKS] If i(12) = 6%, determine the value of i(12).

(c) [3 MARKS] Showing all your work, determine the nominal interest rate, com-pounded semi-annually, under which a sum of money will triple in 12 years.

(d) [3 MARKS] Showing all your work, determine the rate of interest, convertiblecontinuously, that is equivalent to a nominal interest rate of 8% per annum,convertible monthly.

2. In each of the following problems, give a formula in terms of i alone; then evaluatethe formula and determine the numerical value.

(a) [5 MARKS] Determine the present value of a perpetuity-due of 100 payableevery three months, at an effective annual interest rate of 6%.

(b) [5 MARKS] The present value of a perpetuity-immediate paying 1000 at the

end of every 3 years is125, 000

91.Determine the effective annual interest rate.

3. In each of the following problems, give a formula in terms of i alone; then evaluatethe formula and determine its numerical value.

(a) [7 MARKS] At a nominal annual interest rate of 8% compounded quarterly,determine the value, 2 years after the last payment, of a decreasing annuitypaying 5,000 at the end of the first half-year, 4,500 at the end of the 2nd half-year, and continuing to decrease at 500 per half-year until the final paymentof 500.


Table 3: Several Useful Formulas that you were not expected to memorize

(Ia)n i =an i−nvn

i

(Is)n i =sn i−n

i

(Is)n i =sn+1 i

−(n+1)

i

(Da)n i =n−an i

i

(Ds)n i =n(1+i)n−sn i

i

(b) [8 MARKS] Three years before the first payment, determine the value of anannuity that pays 4,000 the first year, 3,900 the second year, with paymentscontinuing to decrease by 100 until it pays 2,000 per year, after which it pays2,000 forever. The interest rate is 5% effective per year until the payment of3,000, after which the interest rate becomes 4% effective forever.

4. One of the following equations is always true, and one is true only when i = 0.

I.1

an i

=1

sn i

+ i

II.1

sn i

=1

an i

+1

i

III.1

an i

=1

sn i

+1

i

IV.1

sn i

=1

an i

+ i

(a) Explain which is always true, and prove it

i. [4 MARKS] algebraically; and

ii. [4 MARKS] by a verbal argument, referring to a sinking fund.

(b) [4 MARKS] Prove algebraically that one of the other equations is true fori = 0.

5. The purchase of a new condominium is partially financed by a mortgage of 120,000payable to the vendor; the mortgage is amortized over 25 years, with a level pay-


ment at the end of each half-month, at a nominal annual rate of 6.6% compoundedevery half-month.

(a) [3 MARKS] Determine the half-monthly payments under this mortgage.

(b) [2 MARKS] Divide the 1st payment into principal and interest.


(d) [4 MARKS] Divide the 52nd payment into principal and interest.

(e) [3 MARKS] The amortization by half-monthly payments was designed to ac-commodate the purchase, whose salary was being deposited automatically tohis bank account every half-month. The purchase changes his profession, after2 years, and now would prefer to make a single payment once every half-year.Determine the amount of that payment if the interest rates are unchanged, butif the mortgage is now amortized to be paid off 5 years earlier than previously.

6. (a) [8 MARKS] Find the price of the following bond, which is purchased to yield6% convertible semi-annually: the bond has face value of 10,000, matures in15 years at a maturity value of 11,500, and has a nominal coupon rate of 9%per annum, compounded semi-annually; the investor is replacing the principalby means of a sinking fund earning 7% convertible semi-annually.

(b) [7 MARKS] Suppose that the bond is callable when t = 13 at a premium of1, 000 above the maturity value. Explain what price the investor should payif he is no longer plans to deposit any of the interest in a sinking fund.

7. A loan is being repaid with 15 annual payments of 1,000 each. At the time of the5th payment the borrower is permitted to pay an extra 2000, and then to repaythe balance over 5 years with a revised annual payment.

(a) [5 MARKS] If the effective annual rate of interest is 6%, find the amount ofthe revised annual payment.

(b) [8 MARKS] Complete an amortization table for the full 10 years of the loan,with the following columns:

Payment Payment Interest Principal Outstandingnumber amount paid repaid loan balance

01

. . . . . . . . . . . . . . .10


8. [8 MARKS] It was n years ago when James deposited 10,000 in a bank paying2.4% interest compounded monthly. If he had, instead, placed his deposit in asyndicate paying interest by cheque annually at the rate of 5% per annum, andhe had invested only this interest with the bank, how much more interest wouldhe have earned altogether? Show all your reasoning, and express your answer interms of n.

B.2.12 Supplemental/Deferred Examination, 2003/2004

1. In each of the following problems you are expected to show all your work.

(a) [3 MARKS] If v = 0.95, determine the value of d(3).

(b) [3 MARKS] Showing all your work, determine the nominal interest rate, com-pounded quarterly, under which a sum of money will double in 10 years.

(c) [3 MARKS] Showing all your work, determine the rate of discount, convertiblecontinuously, that is equivalent to a nominal discount rate of 8% per annum,convertible semi-annually.

(d) [3 MARKS] If i(12) = 1

30, determine the value of i(12).

2. You must show all your work in solving the following problems:

(a) [5 MARKS] Determine the present value of a perpetuity-immediate of 1000payable every three months, at an effective annual interest rate of 8%.

(b) [5 MARKS] The present value of a perpetuity-immediate paying 100 at the

end of every 4 years is129, 600

1, 105. Determine the effective annual interest rate.

3. Show detailed work in your solutions to each of these problems.

(a) [7 MARKS] At a nominal annual interest rate of 6% compounded quarterly,determine the value — 5 years after the last payment — of a decreasingannuity paying 6,000 at the end of the first half-year, 5,500 at the end of the2nd half-year, and continuing to decrease at 500 per half-year until the finalpayment of 500.

(b) [8 MARKS] Three years before the first payment, determine the present valueof an annuity that pays 6,000 the first year, 5,900 the second year, withpayments continuing to decrease by 100 until it pays 4,000 per year, afterwhich it pays 4,000 forever. The interest rate is 8% effective per year untilthe first payment of 4,000, after which the interest rate becomes 5% effectiveforever.


Table 4: Several Useful Formulas that you were not expected to memorize

(Ia)n i =an i−nvn

i

(Is)n i =sn i−n

i

(Is)n i =sn+1 i

−(n+1)

i

(Da)n i =n−an i

i

(Ds)n i =n(1+i)n−sn i

i

4. The purchase of a new condominium is partially financed by a mortgage of 120,000payable to the vendor; the mortgage is amortized over 35 years, with a level pay-ment at the end of each half-month, at a nominal annual rate of 9.6% compoundedevery half-month.

(a) [3 MARKS] Determine the half-monthly payments under this mortgage.

(b) [2 MARKS] Divide the 1st payment into principal and interest.


(d) [4 MARKS] Divide the 62nd payment into principal and interest.

(e) [3 MARKS] The amortization by half-monthly payments was designed to ac-commodate the purchaser, whose salary was being deposited automatically tohis bank account every half-month. The purchaser changes his profession 4years after the mortgage is executed, and now wishes to make a single paymentonce every half-year. Determine the amount of that payment if the interestrates are unchanged, but if the mortgage is now amortized to be paid off after25 years.

5. One of the following equations is always true, and one is true only when i = 0.

I.1

sn i

=1

an i

+ i

II.1

sn i

=1

an i

+1

i


III.1

an i

=1

sn i

+ i

IV.1

an i

=1

sn i

+1

i

(a) Explain which is always true, and prove it

i. [4 MARKS] algebraically; and

ii. [4 MARKS] by a verbal argument, referring to a sinking fund. (A detailedexplanation is expected.)

(b) [4 MARKS] Prove algebraically that one of the other equations is true fori = 0.

6. (a) [8 MARKS] Find the price of the following bond, which is purchased at apremium to yield 5% convertible semi-annually: the bond has face value of10,000, matures in 12 years at a maturity value of 11,500, and has a nominalcoupon rate of 8% per annum, compounded semi-annually; the investor isreplacing the premium by means of a sinking fund earning 4% convertiblesemi-annually.

(b) [7 MARKS] Suppose that the bond is callable at the end of 9, 10, or 11 yearsat a premium of 1, 000 above the maturity value. Explain what price theinvestor should pay if she is no longer plans to deposit any of the interest ina sinking fund.

7. [8 MARKS] It was n years ago when James deposited 10,000 in a bank paying1.8% interest compounded monthly. If he had, instead, placed his deposit in asyndicate paying interest by cheque annually at the rate of 6% per annum, andhe had invested only this interest with the bank, how much more interest wouldhe have earned altogether? Show all your reasoning, and express your answer interms of n.

8. A loan is being repaid with 16 annual payments of 1,000 each. At the time of the4th payment the borrower requests permission, and is permitted to pay an extra3000, and then to repay the balance over 8 years with a revised annual payment.

(a) [5 MARKS] If the effective annual rate of interest is 6%, find the amount ofthe revised annual payment.

(b) [8 MARKS] Complete an amortization table for the last 8 payments, with thefollowing columns:


Payment Payment Interest Principal Outstandingnumber amount paid repaid loan balance

5. . . . . . . . . . . . . . .12

information for students in math 329 2005 01

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