mcgill university faculty of science department … · mcgill university faculty of science...

McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATH 329 2003 01

THEORY OF INTEREST

Information for Students(Winter Term, 2002/2003)

Pages 1 - 7 of these notes may be considered theCourse Outline for this course.

W. G. Brown

April 23, 2003

Information for Students in MATH 329 2003 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 . . 2

1.3 Evaluation of Your Progress . . 31.3.1 Term Mark . . . . . . . 31.3.2 Assignments. . . . . . . 31.3.3 Class Test . . . . . . . . 31.3.4 Final Examination . . . 31.3.5 Supplemental Assessments 41.3.6 Machine Scoring . . . . 41.3.7 Plagiarism . . . . . . . . 4

1.4 Published Materials . . . . . . 51.4.1 Required Text-Book . . 51.4.2 Website . . . . . . . . . 5

1.5 Other information . . . . . . . 51.5.1 Prerequisites . . . . . . 51.5.2 Calculators . . . . . . . 61.5.3 Self-Supervision . . . . . 61.5.4 Escape Routes . . . . . 61.5.5 Showing your work; good

mathematical form; sim-plifying answers . . . . . 6

1.6 Timetable showing tentative lec-ture contents . . . . . . . . . . 8

2 First Problem Assignment 10

3 Second Problem Assignment 12

4 Third Problem Assignment 14

5 An amortization problem 17

6 Solutions, First Problem Assign-ment 19

7 Solutions, Second Problem Assign-ment 25

8 Fourth Problem Assignment 30

9 Two problems for classroom dis-cussion 33

10 Fifth Problem Assignment (Revised5 March 03) 36

11 Solutions, Third Problem Assign-ment 38

12 Solutions, Fourth Problem Assign-ment 45

13 Class Tests 5613.1 Version 1 . . . . . . . . . . . . 5613.2 Class Test, Version 2 . . . . . . 5813.3 Class Test, Version 3 . . . . . . 6013.4 Class Test, Version 4 . . . . . . 62

14 Solutions to Problems on the ClassTests 6414.1 Versions 1 (white) and 3 (yellow) 6414.2 Versions 2 (blue) and 4 (green) 67

15 Solutions, Fifth Problem Assign-ment 70

16 A problem on sinking funds 78

17 A problem on “reinvestment” 79

18 Textbook problems worked in thelectures 81

19 References 901

Information for Students in MATH 329 2003 01 1

1 General Information

Distribution Date: Monday, December 30th, 2002(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/Stat website (cf.§1.4.2 below), and will also be accessible via a link from WebCT.)

The Course Outline for MATH 329 2003 01 can be considered to be pages 1 through 7of 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]: RPHYS 115CLASS 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 most of the topics in the first five chapters ofthe textbook [4]1; section numbers, where shown, refer to that book:

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


Chapter 1. INTEREST: THE BASIC THEORY

Chapter 2. INTEREST: BASIC APPLICATIONS

Chapter 3. ANNUITIES

Chapter 4. AMORTIZATION AND SINKING FUNDS

Chapter 5. BONDS

In addition, some time will be spent at the end of the course introducing studentsto the mathematics of life insurance, through selections from the following chapters;the details, showing which portions of the chapters are to be considered examinationmaterial, will be be provided later in the term.

Chapter 6. PREPARATION FOR LIFE CONTINGENCIES

Chapter 7. LIFE TABLES AND POPULATION PROBLEMS

Chapter 8. LIFE ANNUITIES

Chapter 9. LIFE INSURANCE

[added March 25th, 2003] The following constitute the formal syllabus from Chap-ters 6, 7, 8, 9: Students should read all of Chapter 6, in order to study the following threesections, which could be examination material: §7.2 Life Tables, §8.1 Basic Concepts ofLife Annuities, §9.1 Basic Concepts of Life Insurance. Students should be able to solveproblems of the type included in the textbook for these three sections, and should read[4, pp. 133-135 of §7.3] in order to be able to work with analytic formulæ for `x, whenthese are presented; students are not expected to be familiar with the history or themotivation for the formulæ presented.

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 an

UPDATED TO April 23, 2003


explanation 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 on3 Wednesday, March 12th, 2003, at the regular class time,counting for 20 of the 30 marks in the Term Mark. There will be no “make-up” test forpersons who miss the test.

1.3.4 Final Examination

Examinations form an important part of the tradition of actuarial mathematics. Thefinal examination in MATH 329 2003 01 will count for either 70% or 100% of the nu-merical grade from which the submitted final letter grade will be computed. Where astudent’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 (Monday, April 14th, 2003 through Wednesday, April 30th,

2For a lexicon of actuarial terms in English/French, see

http://www.actuaries.ca/publications/other lexicon e.html

3Added 13 February: The date was chosen by the class on 10/12 February, to replace the tentativefirst date which had been announced.



2003). You are advised not to make any travel arrangements that would prevent you frombeing present on campus at any time during this period. Students who have religious orother constraints that could affect their ability to write examinations at particular timesshould watch for the Preliminary Examination Timetable, as their rights to apply forspecial consideration at their faculty may have expired by the time the final examinationtimetable 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, [2, §8.1, p. 364; or §8.1, p. 49].)

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)4 that

“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/

4http://ww2.mcgill.ca/students-handbook/chapter3secA.html


1.4 Published Materials

1.4.1 Required Text-Book

The textbook for the course this semester is [4] Michael M. Parmentier, Theory of Interestand Life Contingencies, with Pension Applications: A Problem-Solving Approach, 3rdedition. ACTEX Publications, Winstead, Conn. (1999), ISBN 0-56698-333-9.

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 5

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.

1.5 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. Whereadditional background is required in such areas as

• theory of infinite sequences and series (Math 222 material);

• probability (Math 323 material),

the textbook provides that background, and it will also be supplied in the lectures.

5At the time of this writing the current version is 5.1.


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 the needfor help.

Time Demands of your Other Courses. Be sure to budget enough time to attendlectures, for private study, and for the solution of many problems. Don’t be tempted todivert time required for this course to courses which offer instant gratification. Whilethe significance of the homework assignments and class test in the computation of yourgrade is minimal, these are important learning experiences, and can assist you in gaugingyour progress in the course. This is not a course that can be crammed for: you mustwork steadily through the term if you wish to develop the facilities needed for a strongperformance 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 beable to communicate your precise reasoning to others and to yourself. You are always


expected 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.


1.6 Timetable showing tentative lecture contents

Distribution Date: 0th version: Monday, January 6th, 2003(Subject to correction and change.)

[Section numbers refer to the text-book.]6

MONDAY WEDNESDAY FRIDAY

JANUARY6 Chapter 1 8 Chapter 1 10 §1.413 §1.4–§1.6 15 §2. 17 §2.

Course changes must be completed by Jan. 19, 200320 §3. 2¤ 22 §3. 24 §3.

Deadline for withdrawal with fee refund = Jan. 26, 200327 §3. 1

∑29 §3.4,§3.5 31 §3.6

FEBRUARYVerification Period: February 3–7, 2003

3 §3.6 2

∑3¤ 5 §4.1 7 §4.

10 §4. 12 §4. 14 §5.Deadline for withdrawal (with W) from course = Feb. 16

17 §5. 3

∑4¤ 19 §5. 21 X

Study Break: February 24–28, 2003No lectures, no office hours

24 NO LECTURE 26 NO LECTURE 28 NO LECTURE

(Page 9 of the timetable will be circulated later in the term.)

6Notation: #¤ = distribution date for assignment #n

∑= assignment #n due

R© = Read OnlyX = reserved for eXpansion or review


MONDAY WEDNESDAY FRIDAYMARCH

3 §5. 5 §5. 7 §5.10 §6. 12 CLASS TEST 14 §6.17 §6. 19 § 21 §24 § 26 § 2831 §

APRIL2 § 4 §

7 § 9 11


2 First Problem Assignment

Distribution Date: Monday, January 6th, 2003Solutions are to be submitted by Monday, January 27th, 2003.

In all of the following problems you are expected to show all 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?

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?

3. (cf. [4, 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?

4. (cf. [4, Exercise 1-24, p. 26]) Recall that (cf. [4, (1.21)])

[1 +

i(m)

m

]m

= 1 + i =1

1− d=

[1− d(m)

m

]−m

. (1)

(a) Determine whether there is an integer n such that

1 +i(n)

n=

1 + i(2)

2

1 + i(3)

3

(2)

and, if there is such an integer, find it.

(b) Replace the right member of (2) 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.

5. (cf. [4, Exercise 1-30, p. 27]) Show that f(t) = (1 + i)t − (1 + it) is minimized att = ln i−ln δ

δ.

6. (cf. [4, Exercise 1-33, p. 27]) Find the accumulation function a(t) if it is knownthat δt = 0.04(1 + t)−1 for t > 0.

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.

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%.


3 Second Problem Assignment

Distribution Date: Monday, January 19th, 2003Solutions are to be submitted by Monday, February 4th, 2003

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.

2. (cf. [4, 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?

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?

4. (cf. [4, 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!

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?

6. (cf. [4, 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%.


4 Third Problem Assignment

Distribution Date: Monday, February 3rd, 2003Solutions are to be submitted by Monday, February 17th, 2003

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.

(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.

2. (cf. [4, 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.7

7Note 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. [4, 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.)

3. [4, Exercise 3.15] Prove each of the following identities

• algebraically; and

• verbally

(a) an = an + 1− vn

(b) sn = sn − 1 + (1 + i)n

4. [4, 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.

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: 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.

7. A deferred annuity is one that begins its payments later than might otherwise havebeen expected. We define

m |an = vman (3)

m |an = vman (4)


Prove, both algebraically and verbally, that, for non-negative integers m and n,

m |an = am+n − am (5)

(1 + i)m · sn = sm+n − sm (6)

1 |an = an (7)

(8)


5 An amortization problem

Distribution Date: Mounted on the Web on Thursday, February 6th, 2003

[4, Exercise 4-9, p. 87] “A loan is being repaid with 30 equal annual install-ments, 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. (9)

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

; (10)

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 thepayment 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 (9) and (10)

are generally true when a loan is amortized in equal payments, and are found in [4, pp.78, 79]. But be cautioned — these formulæ are not necessary applicable under otherconditions, e.g. where payments are not level.


6 Solutions, First Problem Assignment

Distribution Date: Mounted on the Web on Friday, February 7th, 2003Solutions were to be submitted by Monday, January 27th, 2003.

Caveat lector! 8 There could be errors. If you don’t understand something, ask.

In all of the following problems you 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?

8Let the reader beware!


(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 . (11)

(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 [4, 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. [4, 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. [4, Exercise 1-24, p. 26]) Recall that (cf. [4, (1.21)])[1 +

i(m)

m

]m

= 1 + i =1

1− d=

[1− d(m)

m

]−m

. (12)

(a) Determine whether there is an integer n such that

1 +i(n)

n=

1 + i(2)

2

1 + i(3)

3

(13)

and, if there is such an integer, find it.

(b) Replace the right member of (13) 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. [4, 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. (14)

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 (14) 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.9

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. [4, Exercise 1-33, p. 27]) Find the accumulation function a(t) if it is knownthat δt = 0.04(1 + t)−1 for t > 0.

9Consider, 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 [4, (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. [1, 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.


7 Solutions, Second Problem Assignment

Distribution Date: Mounted on the Web on Monday, February 10th, 2003Solutions were to be submitted by Monday, February 3rd, 2003

Caveat lector! There could be errors. If you don’t understand something, ask.

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. [4, 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: [3, 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. [4, 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. [4, 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 years10 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%.

10Non-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 .


8 Fourth Problem Assignment

Distribution Date: Monday, February 17th, 2003Solutions are to be submitted by Monday, March 10th, 2003

1. (cf. [4, 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. If the interest rate is 5%, determine X.

2. (cf. [4, 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.

3. (a) [4, 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.

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 fund



accumulates 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. . . . . . . . . . . . . . .

5. (This is a complicated variant of [4, 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 bebe repaid over 14 equal annual payments, what will be the revised annuallevel payment that will have to be paid at the end of the each of the next 14years?

(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 money11 is 10% per annum;

ii. when the cost of money is 12% per annum.

11By 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.


(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;

ii. his proposed scheme, whereby he would pay 5000 immediately and paythe rest of the loan in 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.


9 Two problems for classroom discussion

Distribution Date: Discussed on 17 February, 2003Mounted on the Web on 15 February, 2003

(Corrected, 23 April, 2003)

1. 12 An investor plans to buy a twenty-year mortgage whose amount is 150,000,having equal monthly payments; the nominal interest rate is 6%, compoundedmonthly. The investor plans to invest the monthly payments she receives in asavings account earning a nominal annual rate of 3%, compounded monthly, inorder to accumulate a retirement fund after 20 years.

(a) What is the monthly payment on the mortgage?

(b) What should the investor pay for this mortgage, in order to receive an effectiveannual yield rate of 8%?

(c) How much is in the retirement fund after 10 years.

(d) How much is in the retirement fund after 20 years?

Solution:

(a) The mortgage is amortized over 20 years, i.e., 240 months. The monthlypayment X has the property that X · a240 6

12% = 150000, hence

X =150000

a2400.5%

= 1074.65

(b) i. Suppose that the intention of the problem is that the mortgagepayments are still to be placed in the sinking fund. The sinkingfund becomes available only at the end of 20 years. Discounting back tothe present at 8%, we find its present value, i.e. the price to be paid, tobe

(1.08)−201074.65 · s2400.025%

= (1.08)−20 · 150, 000 · s2400.025%

a2400.05%

= 75, 694.46.

ii. Suppose that the intention of the problem is that the mortgagepayments are no longer to be placed in a sinking fund. Let i

12Source of problem: modified from Final Examination in Math 329, April, 1998.



denote the monthly interest rate that, when compounded 12 times, isequivalent to 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

(c) After 10 years the payments have accumulated to X · s1200.0025 = 1074.65 ×139.74 = 150171.59.

(d) After 20 years the payments have accumulated to X · s2400.0025 = 1074.65 ×328.30 = 352807.60.

2. 13 A borrower takes out a loan of 20000 for three years. Construct a sinking fundschedule where the lender receives a nominal rate of 10% effective semi-annuallyand paid semi-annually on the loan, and the borrower replaces the amount of theloan 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 . . . . . . . . . . . .

Solution: Every six months the borrower pays the interest that has accrued onthe loan at 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% convertiblequarterly, i.e. 2% quarterly, the borrower must make 6 semi-annual payments Xwhich satisfy the equation

X · s6(1.02)2−1 = 20000

whence

X =20000

s61.022−1

=808

(1.02)12 − 1= 3012.21

13Source of problem: modified from Final Examination in Math 329, April, 2000




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 interestrate (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


10 Fifth Problem Assignment (Revised 5 March 03)

Distribution Date: Mounted on the Web on Monday, March 3rd, 2003;distributed in hard copy on Friday, March 7th, 2003.

Solutions are to be submitted by Monday, March 24th, 2003.

1. (a) (cf. [4, 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. [4, Exercise 5-13, p. 109]) For the bond in part 1a find the market price14

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?

2. (a) [4, Exercise 5-3, p. 106] Prove the “Alternate Price Formula”:

P = C + (Fr − Ci)an

algebraically.

(b) (cf. [4, 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.

3. (cf. [4, 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

14Market price=amortized value [4, p. 98] is obtained by interpolating linearly between book valueson coupon dates. Thus the market price is a continuous function of time.



Time Coupon Interest Principal BookValue Adjustment Value

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.

4. (cf. [4, 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.

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

(c) Give the amortization table for the last one and one-half years.

5. A 4.5% bond 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.]


11 Solutions, Third Problem Assignment

Distribution Date: Mounted on the Web on Monday, March 3rd, 2003Solutions were to be submitted by Monday, February 17th, 2003.


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.

(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 [6].

(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

.15

(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

15This 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

).


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 , (15)

6a = 9a if r = 1 . (16)

When a = 0, both equations are satisfied: the sequence is 0, 0, 0, . . . ; thecommon ratio is indeterminate. When r 6= 1, (15) 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(16).

(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. [4, 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.16

(g) (cf. [4, 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.

16Note that the wording of the cited questions in the textbook required a number of assumptions thathave been made more explicit in the present questions.


(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. [4, 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. [4, 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: [1, 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% afterone 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 (17)

m |an = vman (18)



Prove, both algebraically and verbally, that, for non-negative integers m and n,

m |an = am+n − am (19)

(1 + i)m · sn = sm+n − sm (20)

1 |an = an (21)

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.


12 Solutions, Fourth Problem Assignment

Distribution Date: Monday, March 10th, 2003Solutions were to be submitted by Monday, March 10th, 2003.


1. (cf. [4, 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 3300.17 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

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

= 1082.33.17The original version of this problem gave the final payment as 3500, which would have required 16

years of payments. A correction was announced at the lecture of March 3rd, 2003.


2. (cf. [4, 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 applied


to 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) [4, 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 owingimmediately 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

)(22)

= 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.


We seek the smallest n such that

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.



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 . (23)

From (23) we see that the shortfall in the balance of the sinking fund after the lastpayment of 1400 is 234.55. The value of the sinking fund is not yet sufficient torepay 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 [4, 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 bebe repaid over 14 equal annual payments, what will be the revised annuallevel payment that will have to be paid at the end of the each of the next 14years?

(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 money18 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;

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);

18By 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.


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(24)

= 434.58.

(b) We will have to evaluate the same ratio as in (24), 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.


This page has been left empty intentionally.


13 Class Tests

13.1 Version 1

McGILL UNIVERSITY FACULTY OF SCIENCE

CLASS TEST in MATH 329: THEORY OF INTEREST

EXAMINER: Professor W. G. Brown DATE: Wednesday, 12 March, 2003.ASSOCIATE EXAMINER: Professor N. Sancho TIME: 45 minutes, 14:35→15:20

FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions

• This test contains 3 questions, worth a total of 40 marks. The time available forwriting is about 45 minutes. This question sheet must be signed and handed in withyour solutions.

• Your rough work should be on the unruled pages of the examination book. All yourwriting — even rough work — must be handed in.

• Calculators. While you are permitted to use a calculator to perform arithmetic and/orexponential calculations, you must not use the calculator to calculate such actuarialfunctions as ani, sni, (Ia)ni, (Is)ni, (Da)ni, (Ds)ni, etc. without first stating a formulafor the value of the function in terms of exponentials and/or polynomials involving nand the interest rate. You must not use your calculator in any programmed calculations.If your calculator has memories, you are expected to have cleared them before the test.

• In your solutions to problems on this test you are expected to show all your work . Youare expected 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.


1. Showing your work, solve each of the following problems:

(a) [2 MARKS] Determine the nominal annual interest rate, i1, compounded every 3months, which is equivalent to a nominal annual interest rate of 12% compoundedevery 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 instan-taneously (=convertible continuously), which is equivalent to an effective monthlydiscount rate of 1%.

2. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol sni, using a time diagram showing the payments, and indicating the pointin 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 formula whichexpresses the value of sni in terms of i and n.

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, show-ing, for each payment, the interest portion and the portion for reduction of prin-cipal. 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 by theinvestor.


13.2 Class Test, Version 2




FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions







1. [20 MARKS] A borrower takes out a loan of 2000 to be paid by one payment with fullinterest at the end of two years. Construct a sinking fund schedule using the headings

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, and theborrower replaces the amount of the loan with equal semi-annual deposits in a sinkingfund to mature when the loan becomes due, where the sinking fund earns 8% convertiblesemi-annually.


(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 every 6months, which is equivalent to a nominal annual interest rate of 24% compoundedevery 3 months.

(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded instanta-neously (=convertible continuously), which is equivalent to an effective quarterlydiscount rate of 2%.

3. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol ani, using a time diagram showing the payments, and indicating the pointin 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) formulafor the value of a∞i.






FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions








(a) [2 MARKS] Determine the nominal annual interest rate, i1, compounded every 3months, which is equivalent to a nominal annual interest rate of 12% compoundedevery 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 instan-taneously (=convertible continuously), which is equivalent to an effective monthlydiscount rate of 1%.

2. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol sni, using a time diagram showing the payments, and indicating the pointin time where the value of the various payments is being calculated.


∑) or dots

(. . .),

(c) [4 MARKS] Define what is meant by sni. Give, without proof, a formula whichexpresses the value of sni in terms of i and n.



(b) [10 MARKS] Determine an amortization schedule for the first 5 payments, show-ing, for each payment, the interest portion and the portion for reduction of prin-cipal. Use the following format for your table.


. . .

(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 by theinvestor.






FAMILY NAME:

GIVEN NAMES:

STUDENT NUMBER:

Instructions







1. [20 MARKS] A borrower takes out a loan of 2000 to be paid by one payment with fullinterest at the end of two years. Construct a sinking fund schedule using the headings


. . .

assuming that the lender receives 10% convertible semi-annually on the loan, and theborrower replaces the amount of the loan with equal semi-annual deposits in a sinkingfund to mature when the loan becomes due, where the sinking fund earns 8% convertiblesemi-annually.


(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 every 6months, which is equivalent to a nominal annual interest rate of 24% compoundedevery 3 months.

(c) [4 MARKS] Determine the nominal annual interest rate, i3, compounded instanta-neously (=convertible continuously), which is equivalent to an effective quarterlydiscount rate of 2%.

3. (a) [2 MARKS] Define the sequence of payments whose value is represented by thesymbol ani, using a time diagram showing the payments, and indicating the pointin time where the value of the various payments is being calculated.


∑) or dots

(. . .),

(c) [4 MARKS] By allowing n to approach infinity, determine a (closed form) formulafor the value of a∞i.


14 Solutions to Problems on the Class Tests

Distribution Date: Mounted on the Web on Sunday, March 16th, 2003;distributed in hard copy on Wednesday, March 19th, 2003.

(Subject to correction of errors or omissions.)The tests were administered on Wednesday, March 12th, 2003

Versions 2 and 4 appear to have been slightly more difficult than Versions 1and 3, and the grades were adjusted to compensate for this.

14.1 Versions 1 (white) and 3 (yellow)


(a) [2 MARKS] Determine the nominal annual interest rate, i1, compounded ev-ery 3 months, which is equivalent to a nominal annual interest rate of 12%compounded 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.


∑) 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.



(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.


. . .

(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 [4, 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:



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.

14.2 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


. . .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 asinking 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:



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


(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 ev-ery 6 months, which is equivalent to a nominal annual interest rate of 24%compounded 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.


∑) 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.


15 Solutions, Fifth Problem Assignment

Distribution Date: Friday, March 28th, 2003; corrected Monday, March 31st, 2003Solutions were to be submitted by Monday, March 24th, 2003.


1. (a) (cf. [4, 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. [4, Exercise 5-13, p. 109]) For the bond in part 1a find the market price19

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:

(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 (25)

= (20000)(1.04)−30 +

(750

.04

) (1− (1.04)−30

)(26)

= (20000)(1.04)−30 + 18750(1− (1.04)−30

)(27)

= 18750 + (20000− 18750)(1.04)−30 (28)

= 18750.00− 385.40 = 19135.40. (29)

Alternatively, using the ”alternate” formula, we find it to be

200 + (750− 800)a304%

= 2000− 50

0.04

(1− (1.04)−30

)= 19135.40.

19Market price=amortized value [4, p. 98] is obtained by interpolating linearly between book valueson coupon dates. Thus the market price is a continuous function of time.


(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) [4, Exercise 5-3, p. 106] Prove the “Alternate Price Formula”:

P = C + (Fr − Ci)an

algebraically.

(b) (cf. [4, 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” [4,(5.1)] as follows:

P = (Fr)an + Cvn

= (Fr)an + C(1− ian) [4, (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 (30)

P2 = 10000 + (10000(0.06)− 10000(0.05))an (31)

From (30) we find that

an5% =8246.56− 10000

10000(0.035− 0.05)= 11.6896 . (32)

i. Substituting in (31) yields P2 = 11168.96 as the price of the second bond.

ii. We solve (32) 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. [4, 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. [4, 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.

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

(c) Give the amortization table for the last one and one-half years.

Solution:

(a) The book value just after the pænultimate20 coupon is

11828.57 = 12000v + Fr · a10.05

= 12000v + Fr · v =12000 + Fr

1.05202nd last


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.


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% bond21 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

21The 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 [4, p. 93, 1st paragraph]; however, the author usually supplies additional, redundant,information in his problems.


(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.

(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.


16 A problem on sinking funds

Distribution Date: On the Web: 30th March, 2003Discussed in the lecture of 28th March, 2003

[1, Example 5.10, p. 120] “An investor buys an n-year annuity with a presentvalue of 1000 at 5%, at a price which will allow him to accumulate a sinkingfund to replace his capital at 4%, and will produce an overall yield rate of6%. Find the purchase price of the annuity.”

Solution: The annual payments P received under the annuity will have the property that

Pan5% = 1000 ,

so P = 1000an5%

. Denote the amount paid for the annuity by X. The convention when

we speak of a sinking fund to accumulate a sum to repay a loan or replace capital isthat, unless information to the contrary is given, we are to assume that the annualcontributions are constant, and that the amount which changes hands annually is thesum of the contribution to the sinking fund and the interest on the borrowed capital.The annual deposit into the sinking fund will be X

sn4%, and it is the excess of the annual

annuity payment over this amount which is the interest on the capital, so

0.06X =1000

an5%

− X

sn4%

,

and we may solve this equation for X, obtaining

X =1000sn4%

an5%(1 + 0.06sn4%).


17 A problem on “reinvestment”

Distribution Date: On the Web: 30th March, 2003Discussed in the lectures of 28th and 31st March, 2003

[1, Exercise 5.37, p. 126] “It is desired to accumulate a fund of $1000 atthe end of 10 years, by equal deposits at the beginning of each year. If thedeposits earn interest at 2% effective, but the interest can be reinvested atonly 1% effective, show that the deposit necessary is

1000

2s110.01 − 12.”

Solution: Let X denote the amount of the regular deposit. We present two solutions.

1. First Solution: If we consider the annual deposits of X and the various accre-tions of interest, interest on the interest, etc., the only case where the interest isat a rate other than 1% is the annual interest earned by the 10 deposits of Xthemselves; let us split these payments into two parts — an earning of 1% and a“premium” earning of 1%. We can do this by interpreting the account as having afixed interest rate of 1%, and each of the 10 deposits generating a supplementarydeposit of 0.01X in advance, each year it is in the account. Viewed from t = 10these supplementary deposits can be interpreted as the accumulated value of anincreasing annuity 0.01X(Is)101%; the accumulation of the 10 deposits and halfof their interest payments, together with interest on the interest at 1%, will beXs101%. We equate the sum of these two accumulations to 1000:

0.01X(Is)101% + Xs101% = 1000

and solve for X:

X =1000

0.01(Is)101% + s101%

=1000

(s101% − 10) + s101%

=1000

2s101% − 10=

1000

2s111% − 12(33)

2. Second Solution: The deposit of X at time t = r generates an annuity of interestpayments of 0.02X at times t = r+1 . . . 10. Combining these payments as r ranges


from 0 to 9 we obtain an increasing annuity whose initial and incremental paymentsare 0.2X, first payment at time t = 1, whose value at time t = 10 is

0.02X(Is)101% = 2X(s101% − 10

)

= X(2s101% − 20

)

To this we must add the payments themselves, obtaining the equation

10X + X(2s101% − 20

)= 1000

which yields the same value of X obtained in (33).

(The student may ask, in puzzlement, why we have the right to add the paymentsthemselves without including a factor which is a power of 1 + i, as we have normallydone in calculations of this type, to represent the growth of principal in successive timeperiods. The reason is that we have already computed the growth separately in theincreasing annuity — only the payments themselves were not incorporated yet.)


18 Textbook problems worked in the lectures

Chapter Exercise Day Month/20031 6abc 10 11 7 10 11 8 10 12 3 15 12 8 15 13 3 20 13 4 20 13 7 24 13 9 27 13 11 24 13 12 24,27 13 24 27 13 27 27 13 38 29 13 47 29 13 51 29 13 63 3 24 1a 5,7 24 9 6 on Web 24 15 10 24 28 10,12 24 26 12 24 31 12 24 29 14 24 30 14 24 35 19 25 1 3 35 4 5 35 9 7 35 11 7 35 13 7,10 35 16 14 35 21 14 35 26 14 37 2 21 38 1 24 38 4 26, 28 38 11 3 26



• 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.

19 References

[1] S. G. Kellison, The Theory of Interest . Richard D. Irwin, Inc., Homewood, Ill. (1970).ISBN ???-083841.

[2] McGill Undergraduate Programs Calendar 2002/2003. Also accessible athttp://www.mcgill.ca/courses/#UGRAD .

[3] 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.

[4] 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.

[5] J. Stewart, Single Variable Calculus (Early Transcendentals), Fourth Edition.Brooks/Cole (1999). ISBN 0-534-35563-3.

[6] H. S. Hall and S. R. Knight, Higher Algebra, Fourth Edition, MacMillan & Co.(London, 1891).

mcgill university faculty of science department … · mcgill university faculty of science...

Documents