12/19/2015rd1 engineering economic analysis chapter 3 interest and equivalence

04/21/2304/21/23 rdrd 11

Engineering Economic Engineering Economic AnalysisAnalysis

Chapter 3 Interest and Equivalence

http://academic.udayton.edu/ronalddeep/enm530.htm

04/21/2304/21/23 rdrd 22

Irrelevant CharacteristicsIrrelevant Characteristics

Monetary Units Monetary Units DollarsDollarsPounds Pounds Yen Yen MarksMarks

Effective PeriodEffective PeriodDay Month Year Century

04/21/2304/21/23 rdrd 33

Interest and EquivalenceInterest and Equivalence

Computing Cash Flows

Time Value of Money

Equivalence

Single Payment Compound Interest Formulas

Why Engineering Economy?Why Engineering Economy?

Should I pay off my credit card balance with borrowed money?

What are the worth of graduate studies over my career?

Are tax deductions for my home mortgage a good deal or should I accelerate my mortgage payments?

Exactly what rate of return did we make on our stock investments?

Should I buy or lease my next car, or keep the one I have now and pay off the loan?

When should I replace my present car?

Which cash flow is preferable?

04/21/2304/21/23 rdrd 44

Time Value of MoneyTime Value of Money

The change in the amount of money over a given time period is call the the time value of money; and is the most important concept in engineering economy.

You borrow $10,000 and repay $10,700 a year later. Find the rate of interest.

i = (10.7K – 10K)/10K = 700/10000 = 7%

04/21/2304/21/23 rdrd 55

04/21/2304/21/23 rdrd 66

Cash Flow DiagramCash Flow Diagram

P ~ Present at time 0; F ~ Future A ~ Uniform or Equal

G ~ Gradient i ~ effective interest n ~ Number of pay periods

F

P

0 1 2 3 4 5 n

A

I = 7%

g

radient

uniform

Compounding ProcessCompounding Process

Given: i = 10% n = 7 years

P = $3000 F = P(1 + i)n

Find: F = 3000(1 + 0.10)7 compounding factor

= $5846.15

P = F(1 + i)-n

Discounting factor

F = P(F/P, i%, n)

Genie command is (FGP 3000 10 7) 5846.15

04/21/2304/21/23 rdrd 77

04/21/2304/21/23 rdrd 88

n Start Interest End1 P iP P(1 + i)1

2 P(1 + i)1 iP(1 + i)1 P(1 + i)2

3 P(1 + i)2 iP(1 + i)2 P(1 + i)3

.. …. …. …..n P(1 + i)n-1 iP(1 + i)n-1 P(1 + i)n

F/PF/P

04/21/2304/21/23 rdrd 99

F given P; F = P(1 + i)n = 1000(1 + 0.04)5 1000(1.2167) = 1216.652 1216.70

calculator table

F = 1216.65

i = 4% compounded annually

n = 5

P = $1000 (F/P 1000 4 5) 1216.65

(F/P, i%, n)(F/P, i%, n)

formula

04/21/2304/21/23 rdrd 1010

F = A(F/A, i%, n)

(F/A, i%, n) = A[(1 + i)n-1 + (1 + i)n-2 + … + (1 + i)1 + 1](Summing from right to left) = A[1 – (1 + i)(1 + i)n-1]/[1 – (1 + i)]

= A[(1 + i)n - 1]/i F

1 2 3 4 5 6 … n-2 n-1 n

A

(F/A, i%, n)

04/21/2304/21/23 rdrd 1111

F given A; F = A(F/A, i%, n); F = 500(5.4163, 4%, 5) = $2708.16

(F/A, i%, n) =

F = $2708.16 = (F/A 500 4 5)

i = 4%

0 1 2 3 4 5

A = $500

5(1 ) 1 (1 0.04) 15.4163

0.04

ni

i

(F/A, i%, n)(F/A, i%, n)

04/21/2304/21/23 rdrd 1212

P/AP/A

P/A =

Find the present worth of 5 yearly deposits of $1000 at 7% compounded annually.

P = A(P/A, 7%, 5) = 1000(4.100197) = $4100.20 = 1000(F/A, 7%, 5)(P/F, 7%, 5) = 1000 * 5.750749 * 0.712986 = $4100.20

n n

n n

F P (1+i) -1 1 (1+i) -1* = *

A F i (1+i) i(1+i)

F/A P/F

04/21/2304/21/23 rdrd 1313

A/F & A/PA/F & A/PnA[(1+i) -1]

F/Ai

n

FiA/F =

(1+i) -1

n

n

F P A[(1+i) -1]P/A * =

A F i(1+i)

n

n

A F Pi(1+i)A/P *

F P (1+i) -1

04/21/2304/21/23 rdrd 1414

Compound Interest Factors 7%Compound Interest Factors 7%

n F/P P/F A/F A/P F/A P/A A/G P/G 1 1.0700 0.9346 1.0000 1.0700 1.0000 0.9346 0.0000 0.0000

2 1.1449 0.8734 0.4831 0.5531 2.0700 1.8080 0.4831 0.8735 3 1.2250 0.8163 0.3111 0.3811 3.2149 2.6243 0.9549 2.5061 4 1.3108 0.7629 0.2252 0.2952 4.4399 3.3872 1.4155 4.7948 5 1.4026 0.7130 0.1739 0.2439 5.7507 4.1002 1.8650 7.6467 6 1.5007 0.6663 0.1398 0.2098 7.1533 4.7665 2.3032 10.9784 7 1.6058 0.6227 0.1156 0.1856 8.6540 5.3893 2.7304 14.7149 8 1.7182 0.5820 0.0975 0.1675 10.2598 5.9713 3.1466 18.7890 9 1.8385 0.5439 0.0835 0.1535 11.9780 6.5152 3.5517 23.1405 10 1.9672 0.5083 0.0724 0.1424 13.8165 7.0236 3.9461 27.7156 11 2.1049 0.4751 0.0634 0.1334 15.7836 7.4987 4.3296 32.4666 12 2.2522 0.4440 0.0559 0.1259 17.8885 7.9427 4.7025 37.3507

04/21/2304/21/23 rdrd CW3B-CW3B-1515

RelationshipsRelationshipsa. (F/P, i%, n) = i(F/A, i%, n) + 1

b. (P/F, i%, n) = 1 – i(P/A,i%, n)

c. (A/F, i%, n) = (A/P, i%, n) – i

d. (A/P i%, n) = i / [1 – (P/F, i%, n)]

e. Find (F/P, 10%, 37)(F/P, 10%, 37) = (F/P, 10%, 35)(F/P, 10%, 2) = 28.1024 *1.21 = 34.0039

f. (P/A, i%, n) – (P/A, i%, n-1) = (P/F, i%, n)

04/21/2304/21/23 rdrd 1616

Computing Cash FlowsComputing Cash Flows

You bought a machine for $30,000. You can either pay the full price now with a 3% discount, or pay $5000 now; at the end of 1 year pay $8000, then at the end of the next 4 years pay $6,000. i = 7% compound yearly.

Option 1 0.97 * 30K = $29,100

Option 2 $31, 470

0 1 2 3 4 5

-$5000

-$8000

-$6000 -$6000 -$6000 -$6000

Continued …

04/21/2304/21/23 rdrd 1717

Option 2Option 2

0 1 2 3 4 5

-$5000 -$6000 -$6000 -$6000 -$6000

-$8000

P = 6K(P/A, 7%, 4)(P/F, 7%, 1) + 8K(P/F, 7%, 1) + 5K

= 6K * 3.387 * 0.9346 + 8K(0.9346) + 5K

= 31,470

7% compounded annually

04/21/2304/21/23 rdrd 1818

Simple InterestSimple Interest

Simple interest is: P * i * n = Pin

You borrow $10,000 for 5 years at a simple interest rate of 6%. At the end of 5 years, you would repay: Principal plus simple interest

F = P + Pin = 10,000 + 10,000 * 0.06 * 10 = 10,000 + 600

= $10,600

04/21/2304/21/23 rdrd 1919

Compound InterestCompound Interest

Compound Interest is: P(1 + i)n

You borrow $10,000 for 5 years at 6% compounded annually. At the end of 5 years, you would repay: Principal plus compound interest

F = P(1 + 0.06)5 = $13,382.26versus the simple interest $10,600.00Difference $ 2,782.26

04/21/2304/21/23 rdrd 2020

Equivalence

04/21/2304/21/23 rdrd 2121

EquivalenceEquivalence

When comparing alternatives that provide the same service, equivalent basis depends on

Interest Rate

Amounts of money involved

Timing of the cash flow

Perspective (Point of View)

12 inches = 1 foot …continued

04/21/2304/21/23 rdrd 2222

EquivalenceEquivalenceYour borrow $5,000 at 8% compounded annually.You may return the $5,000 immediately, or pay according to Plan A or Plan B The 3 options are equivalent.n 1 2 3 4 5A -1400 -1320 -1240 -1160 -1080B -400 -400 -400 -400 -5400

PWA = 1400(P/A, 8%, 5) – 80(P/G, 8%, 5)= $5000

PWB = 400(P/A,8%,5) + 5000(P/F,8%,5) = 1597.08 + 3402.92 = $5000.

04/21/2304/21/23 rdrd 2323

Re-Payment PlansRe-Payment Plans

APR = 8%

n owed interest total principal total for year owed owed payment paid ( P+I)

1 $5000 $400 $5400 $1000 $14002 4000 320 4320 1000 13203 3000 240 3240 1000 12404 2000 160 2160 1000 11605 1000 80 1080 1000 1080 $1200 $6200

(IRR '(-5000 1400 1320 1240 1160 1080)) 8%

04/21/2304/21/23 rdrd 2424


APR = 8%

n owed interest total principal total for year owed owed payment paid1 $5000 $400 $5400 $0 $4002 5000 400 5400 0 4003 5000 400 5400 0 4004 5000 400 5400 0 4005 5000 400 5400 5000 5400 $2000 $7000

(IRR '(-5000 400 400 400 400 5400)) 8%

04/21/2304/21/23 rdrd 2525


APR = 8%

n owed interest total principal total for year owed owed payment paid1 $5000 $400 $5400 $ 852 $1252.282 4148 331 4479 921 1252.283 3227 258 3484 994 1252.284 2233 178 2411 1074 1252.285 1159 93 1252 1159 1252.28 $1260 $ 6261.40

(IRR '(-5000 1252.28 1252.28 1252.28 1252.28 1252.28)) 8%

04/21/2304/21/23 rdrd 2626


APR = 8%

n owed interest total principal total for year owed owed payment paid1 $5000 $400 $5400 $ 0 $ 02 5400 432 5832 0 03 5832 467 6299 0 04 6299 504 6803 0 05 6803 544 7347 5000 7347 $2347 $5000 7347

(IRR '(-5000 0 0 0 0 7347)) 8%

04/21/2304/21/23 rdrd 2727


-5000 -5000 -5000 -50001400 400 1252 .28 01320 400 1252 .28 01240 400 1252 .28 01160 400 1252 .28 01080 5400 1252 .28 7347

8 .000% 8.000% 8.000% 8.001%

(IRR '(-5000 1400 1320 1240 1160 1080)) 8%(IRR '(-5000 400 400 400 400 5400)) 8%(IRR '(-5000 1252.28 1252.28 1252.28 1252.28 1252.28)) 8%(IRR '(-5000 0 0 0 0 7347)) 8%

04/21/2304/21/23 rdrd 2828

Simple vs. Compound InterestSimple vs. Compound Interest

Fs = P(1 + ni) Simple interest

Fc = P(1 + i)n Compound interest

Given P = $24, i = 5%, n = 20 years vs. compounded annually.

Fs = 24(1 + 20 * 0.05) = $48

Fc = 24(1.05)20 = $63.68.

In 1626 Peter Minuit paid $24 for Manhattan Island.

In Year 2008: Fs = 24(1 + 382 * 0.05) = $482.40

Fc = 24(1.05)381 = $2,982,108,814.51

04/21/2304/21/23 rdrd 2929

Annual Percentage Rate (APR) r; for example, 12% per year

Effective interest rate ieff APY ~ annual percent yield

ieff = , where m is the number of pay periods

Example: APR is 12% compounded monthly.

ieff = = 12.68% effective yearly rate.

All interest rates is formulas are effective interest rates commensurate with the pay periods.

Effective Interest RateEffective Interest Rate

mr(1+ ) -1

m

120.12(1+ ) -1

12

04/21/2304/21/23 rdrd 3030


Annual Percentage Rate (APR) is 12%

If compounded monthly,

effective monthly rate is 1% 12/12

effective quarterly rate is 3.03% (1 + 0.03/3)3 - 1

effective yearly rate is 12.68% (1 + 0.12/12)12 - 1

If compounded quarterly, effective quarterly rate is 3% 12/4

effective yearly rate is 12.55% (1 + 0.12/4)4 - 1

ieff = mr

(1+ ) -1m

04/21/2304/21/23 rdrd 3131


$1000 is invested for 5 years at 12% APR compounded monthly. Compute its future worth using effective

a) annual rate b) quarterly rate c) monthly rate and d) 3-year rate.

F5 years = 1000(1 + 0.126825)5 = $1816.70F20 qtrs = 1000(1 + 0.030301)20 = $1816.70F60 mths = 1000(1 + 0.01)60 = $1816.70F2.5yrs = 1000(1 + 0.2697)2.5 = $1816.70

Effective 3 year rate is (1 + 0.36/36)36 – 1 = 43.07688%F3 yrs = 1000(1 + 0.4307688)5/3 = $1816.70

04/21/2304/21/23 rdrd 3232

Interest RatesInterest RatesYou borrow $1000 and agree to repay with 12 equal monthly payments of $90.30.

Find the a) effective monthly interest rate.

b) nominal annual interest rate

c) effective annual interest rate.

a) 1000 = 90.30(P/A, i%, 12) => P/A factor = 11.0742

=> imonth-eff = 1.25%

b) APR = 12 * 1.25% = 15%

c) Annual effective rate = (1 + 0.15/12)12 – 1 = 16.08%.

Continuous CompoundingContinuous Compounding

F = Pern where r is the APR and n is the number of years

(1 + 0.0001)10000 = 2.7181

The effective rate for continuous compounding is given by

er - 1.

Substitute er - 1 for i in the formulas.

For example, F = P(1 + i)n = P(1 + er – 1)n = Pern

04/21/2304/21/23 rdrd 3333

1/

0lim(1 ) x

xe x

04/21/2304/21/23 rdrd 3434

Continuous CompoundingContinuous Compounding

Traffic is currently 2000 cars per year growing at a rate of 5% per year for the next 4 years. How much traffic is expected at the end of 2 years?

F = 2000 e2*0.05 = 2210.34 cars

Investment is currently $2000 per year growing at a rate of 5% per year for the next 4 years. How much investment is expected at the end of 2 years?

F = 2000 e2*0.05 = $2210.34

04/21/2304/21/23 rdrd 3535

Problem Problem

$100 at time 0 is $110 at time pay period 1 and was $90 at time pay period –1. Find the APR for year –1 to 0 and for 0 to 1.

100 = 90(1 + i) => i = 11.11% 90 100 110

110 = 100(1 + i) => i = 10%;

If $90 is invested at time –1 returns $110 at time +1, find i. 110 = 90(1 + i)2 => i = 10.55%.

-1 0 1

04/21/2304/21/23 rdrd 3636

You plan to make 2 deposits, $25K now and $30K at the end of year 6. You draw out $C each year for the first 6 years and C + 1000 each year for the next 6 years. Find C at 10% interest compounded annually.

C(P/A, 10%, 12) + 1000(P/A, 10%, 6)(P/F, 10%, 6) = [25K + 30K(1.1)-6]

C = $5,793.60.

1 6 12

ProblemProblem

25K 30K

C + 1K

$C

04/21/2304/21/23 rdrd 3737

How many years until an investment doubles at 5% compounded annually?

F = 2P = P(1 + 0.05)n => 2 = 1.05n => Ln 2 = n Ln 1.05

n = Ln 2 / Ln 1.05 = 14.20669 years

CheckF = 1000 (1.05)14.21 = $2000

(NGPFI 1 2 5) 14.2067 years

Time to DoubleTime to Double

04/21/2304/21/23 rdrd 3838

F = P(1 + i)n

i = (F/P)1/n – 1

Example: What interest rate generates $3456 in 5 years by investing $1000 now?

i = 3.4561/5 – 1 = 28.14886%.

(IGPFN 1000 3456 5) 28.14886

Computing i given P, F and nComputing i given P, F and n

04/21/2304/21/23 rdrd 3939

Computing n given P, F and iComputing n given P, F and i

F = P(1 + i)n

n = Ln (F/P) / Ln (1 + i)

How many years for $1000 deposited now to accumulate to $3465 at an APR of 28.14886%?

n = Ln 3.465 / Ln 1.2814886

= 5 years

(NGPFI 1000 3465 28.1488) 5

04/21/2304/21/23 rdrd 4040

Arithmetic GradientArithmetic Gradient

0 1 2 3 . . . n -1 n

Gradient begins in Year 2

= A(P/A, i%, n) + G(P/G, i%, n)

G2G

(n -1)G

P

A

(n – 2)G

04/21/2304/21/23 rdrd 4141

11.Given the cash flow in the diagram below, find the present worth value at time 0 with interest rate 5% per year.

PW(5%) = [1000(P/A, 5%, 4) + 100(P/G, 5%, 4)](P/F, 5%, 2)

= $3679.12 (P/F (PGG 1000 100 5 4) 5 2) 3679.12

0 1 2 3 4 5 6

1000

1100

1200

1300

04/21/2304/21/23 rdrd 4242

Arithmetic GradientArithmetic Gradient

Find the present worth of the following cash flow at 7% compounded annually: A = 50, G = 20

n 1 2 3 4 5 6cf 50 70 90 110 130 150

P = A(P/A, i%, n) + G(P/G, i%, n)

= 50(P/A, 7%, 6) + 20(P/G, 7%, 6)

= 238.326 + 219.567 = $457.89

(PGG A G i n) (PGG 50 20 7 6) $457.89

04/21/2304/21/23 rdrd 4343

Geometric GradientGeometric Gradient

A1; A2 = A1 + gA1 = A1(1 + g);

A3 = A2 +gA2 = A1(1 + g) + gA1(1 + g) = A1(1 + g)2

An = A1(1 + g)n-1

Pn = An(P/F, i%, n) = An(1 + i)-n

= A1(1 + g)n-1(1 + i)-n

P = A1(1 + i)-n

(P/A1, i, g, n) =

1

1 (1 ) (1 )n ng iA

i g


Find the present worth of a cash flow beginning at $10K and increasing at 8% for 4 years at 6% per year interest.

(PGGG-table 10000 8 6 4)

n Cash-flow 8% PW-factor 6% PWorth

1 10000.00 0.9434 9433.96 2 10800.00 0.8900 9611.96 3 11664.00 0.8396 9793.32 4 12597.12 0.7921 9978.10

$38,817.54

PW = 10K[1 – (1.08)4/(1.06)4(0.06 – 0.08)] = $38,817.54

04/21/2304/21/23 rdrd 4444

04/21/2304/21/23 rdrd 4545


Example: Calculate the present worth of a contract awarded at $1000 per year and increasing at a uniform rate of 10% per year for 5 years at 7% APR compounded annually.

P = A1[1 – (1 + g)n(1 + i)-n]/(i – g)

= 1000[1 – 1(1 + 0.1)5(1 + 0.07)-5 ]/(0.07 - 0.10)

= $4942.38.

(PGGG A g i n) ~ (PGGG 1000 10 7 5) $4942.38

04/21/2304/21/23 rdrd 4646


A modification costs $8K, expected to last 6 years with a $1300 salvage value. Maintenance is $1700 the first year and increasing 11% per year thereafter. The interest rate is 8%. Find the present worth.

PW = -8K – 1700[1- (1.11/1.08)6/(0.08 – 0.11)] =1300(1.06)-6

= -$17,305.88

(+ 8000 (PGGG 1700 11 8 6) (PGF -1300 8 6)) 17305.88

04/21/2304/21/23 rdrd 4747

Problem 3-11Problem 3-11

n 0 1 2 3 4cf -100 25 45 45 30

Find the compound annual interest rate.

Guess and test or

(IRR '(-100 25 45 45 30)) 16.189%

04/21/2304/21/23 rdrd 4848


Compute the difference in borrowing $1E9 at 4.5% for 30 years versus at 5.25%.

Diff = 1E9[(F/P, 5.25%, 30) – (F/P, 4.5%, 30)]

= $896,232,956.47

04/21/2304/21/23 rdrd 4949


1903 painting valued at $600

1995 painting valued at $29,152,000

Find i.

29152000 = 600(F/P, i%, 92) = 600(1 + i)92

48586.7 = (1 + i)92

1.12445066 = 1 + i

12.45% = i

04/21/2304/21/23 rdrd 5050

Manhattan IslandManhattan Island

Manhattan Island was bought for $24 in 1626 from the native Americans. Compute the present day worth if they had invested the money in an 8% APR compounded yearly.

F = 24 (1 + 0.08)2008-1626 = $140,632,545,502,000

Today each of the 300 millions Americans could be given $468,775.15. F is considerably more than what Manhattan Island is now worth.

04/21/2304/21/23 rdrd 5151

Learning CurveLearning Curve

The time to make the first unit is $1000 and the learning curve is 90%. What is the cost to make the 2nd, 4th and 8th units?

T2 = 1000 * 2(log 0.90 2) = 1000 *2 -0.1520 = $900

T4 = 0.90 * $900 = $810

T8 = 0.90 * $810 = $729.

(sim-lc 1000 8 90) Unit Hours Cumulative 1 1000.00 1000.00 2 900.00 1900.00 3 846.21 2746.21 4 810.00 3556.21 5 782.99 4339.19 6 761.59 5100.78 7 743.95 5844.73 8 729.00 6573.73

04/21/2304/21/23 rdrd 5252

Cost of a MortgageCost of a Mortgage

Interest Rate

Loan Amount

Payment Frequency

Points (Prepaid interest of 1% of loan)

Fees (application, loan origination)

04/21/2304/21/23 rdrd 5353

Principal Reduction and Interest Paid for Principal Reduction and Interest Paid for each paymenteach payment

A = P(A/P, i%, N)

PRn = A(P/F, i%, N – n + 1); Interestn = A - PRn

PR5 = 1627.45(P/F, 10%, 10 – 5 + 1)

= $918.65

(P/F, i%, n) = (P/A, i%, n) – (P/A, i%, n - 1)

04/21/2304/21/23 rdrd 5454

Pay Num Payment Principal Interest Balance Total Interest 1 1627.45 627.45 1000.00 9372.55 1000.00 2 1627.45 690.19 937.26 8682.36 1937.26 3 1627.45 759.21 868.24 7923.15 2805.50 4 1627.45 835.13 792.32 7088.02 3597.82 5 1627.45 918.65 708.80 6169.37 4306.62 6 1627.45 1010.51 616.94 5158.86 4923.56 7 1627.45 1111.56 515.89 4047.30 5439.45 8 1627.45 1222.72 404.73 2824.58 5844.18 9 1627.45 1344.99 282.46 1479.59 6126.64 10 1627.45 1479.49 147.96 0.10 6274.60

Mortgage Payments Loan = $10,000 Interest 10%/year N = 10 years

A = 10K(A/P, 10%, 10) = $1627.45

Computing the BalanceComputing the Balance

On a loan of $10,000 for 10 years at 10% per year, determine the annual payment and the balance after 6 years.

1 2 3 4 5 6 7 8 9 10

A = P(A/P, i%, N) = 10K(A/P, 10%, 10) = $1627.45

B6 = 1627.45(P/A, 10%, 4) = $5158.81

04/21/2304/21/23 rdrd 5555

04/21/2304/21/23 rdrd 5656

Points or No PointsPoints or No Points

You finance a home for $100K at 15 year interest rate. You can either pay one point with 6.375% interest or no points with 6.75% interest compounded monthly.

The 1 point implies that you get $99K but get charged as if you borrowed $100K

A = 100K(A/P, 6.375/12 %, 180) = $864.2599K = 864.25(P/A, i%, 180) looking for 114.55 P/A factor for 180 pay periods. i = 0.545 % => APR = 6.54%

Thus go with paying the point as 6.54% < 6.75%.

04/21/2304/21/23 rdrd 5757

Monthly PaymentsMonthly Payments

What is the monthly payment for a 5-year car loan of $35,000 at 6% compounded monthly?Find the amount of the principal reduction of the 25th payment.

A = $35,000(A/P, 6% /12, 60) = $676.65.

PR25 = 676.65(P/F, 6%/12, 60 – 25 + 1) = $565.44.

04/21/2304/21/23 rdrd 5858

Mortgage PaymentsMortgage Payments

You take out a loan for $350,000 at 6% APR compounded monthly for 30 years.

a) Your monthly payment is _________.

b) The principal reduction PR225 is ________.

c) The interest paid Int225 at this payment is _______.

d) The total interest paid to date at this payment is _______.

e) The balance at this payment is __________.

f) The per cent of your loan paid by the 225th is ______.

Ans. a) $2098.43 b) $1064.90 c) $1033.53 d) $327,786.21

e) $205,640.17 f) 41.245%

04/21/2304/21/23 rdrd 5959

Compound ContinuouslyCompound Continuously

How long does it take for an investment to triple if interest rate is compounded continuously at i = 7%?

F = Pern

3 = 1e0.07n => n = 15.69 years

You deposit $1000 a year for 5 years at 7% compounded continuously. At the end of 5 years you can withdraw _______. Substitute er – 1 for i into the F/A factor.

ans. $5779.59

04/21/2304/21/23 rdrd 6060

Interest RateInterest Rate

A credit card company charges 1.5% interest on the unpaid balance each month.

Nominal annual interest rate is ________. ans. 18%

Effective annual interest rate is ________.

ans. 19.56%

04/21/2304/21/23 rdrd 6161

1 2 3 4 5

A = $500

Place an amount F3 at year 3 which is equivalent to the cash flow below at 7%.

F3 = 500(P/A, 7%,5)(F/P,7%,3) = $2511.46.


PreferencePreference

Which alternative would you choose at i = 10%? i = 9%?

1) Receive $100 now.

2) $120 two years from now.

04/21/2304/21/23 rdrd 6262

P/AP/A

Find the exact P/A factor at 9.35% compounded continuously for n = 10 pay periods.

P/A =

04/21/2304/21/23 rdrd 6363

0.935

0.0935 0.935

(1 ) 1 16.1974%

(1 ) ( 1)

n

n

i e

i i e e

Time Value of MoneyTime Value of Money

a) Joe wants to figure his annual car cost given the following cash flow at i = 7%:

n 1 2 3 4 5cf $45 90 135 180 225

(list-pgf '(0 45 90 135 180 225) 7) $528.61

(A/P 528.61 7 5) 128.92

b) He got the 3rd and 4th year mixed. Refigure.

(list-pgf '(0 45 90 180 135 225) 7) $531.01

(AGP 531.01 7 5) $129.51

04/21/2304/21/23 rdrd 6464


What sum of money is equivalent to $8250 2 years later if interest is 4% compounded semiannually?

8250(1.04)-4 = $7052.13

04/21/2304/21/23 rdrd 6565


A local bank pays 5% annual interest while an out of town bank pays 1.25% compounded quarterly. You have $3000 to deposit for 2 years. In which bank do you deposit your money?

3000(1.05)2 = $3307.50

3000(1.0125)8 = $3313.46

$5.96

04/21/2304/21/23 rdrd 6666


P1 = P2 at interest rate i. Which cash flow is better at 2i?

F1 F2

0 1 2 0 1 2 3

P1 = F1 (1 + i)-2 = F2 (1 + i)-3 = P2 => F2 = F1(1 + i)

P'1 = F1(1 + 2i)-2 P'2 = F2 (1 + 2i)-3 = F1(1 + i)(1 + 2i)-3

P'2 /P'1 = (1 + i)/(1 + 2i) < 1 => P'1 > P'2

For example, try F1 = $100 and F2 = $107.

04/21/2304/21/23 rdrd 6767


A sum of money Q will be received 6 years from now. At 5% annual interest the present worth of Q is $60. At the same interest rate, what would be the value of Q in 10 years?

F = 60(1.05)10 = $97.73

04/21/2304/21/23 rdrd 6868

04/21/2304/21/23 rdrd 6969

04/21/2304/21/23 rdrd CW3B-CW3B-7070

Find A to equate the series at 10% compounded annually. 120 120 120 100 100

1 2 3 4 5

P0 = 100(P/A,10%,2) + 120(P/A,10%,3)(P/F,10%,2)

A A A A

1 2 3 4 5

P0 = A(P/A,10%,4)(P/F,10%,1) = $420.18 => A = $145.81

12/19/2015rd1 engineering economic analysis chapter 3 interest and equivalence

Documents

n iap i

gradient i

n formulard

f1 in discounting factorf

p1 i1 ip1 i1 p1 i2

p ip p1 i1

p1 infind

p1 infprd