Chapter 03

Chapter 3: Interest Rate and Economic Equivalence Types of Interest

• Simple interest:

yearsN

5.1208.01

2)08.01()08.01(000,5$000,10$

• Compound interest: yearsN

2.102)07.01(

)07.01(000,5$000,10$

• Simple interest: 400$)5)(000,1)($08.0( === iPNI

• Compound interest:

469$)14693.1(000,1$]1)1[( =−=−+= NiPI

3.3 • Option 1: Compound interest with 8%:

408,4$)4693.1(000,3$)08.01(000,3$ 5 ==+=F

• Option 2: Simple interest with 9% 350,4$)45.1(000,3$)509.01(000,3$ ==×+

• Option 1 is better

End of Year Principal Repayment

Interest payment

Remaining Balance

0 $10,000 1 $1,671 $900 $8,329 2 $1,821 $750 $6,508 3 $1,985 $586 $4,523 4 $2,164 $407 $2,359 5 $2,359 $212 $0

Contemporary Engineering Economics, Fourth Edition, By Chan S. Park. ISBN 0-13-187628-7.© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be

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Equivalence Concept

3.5 402,9$)7835.0(000,12$)5%,5,/(000,12$ === FPP

3.6 328,23$)1664.1(000,20$)2%,8,/(000,20$ === PFF

Single Payments (Use of F/P or P/F Factors)

3.7 (a) 388,7$)8%,5,/(000,5$ == PFF (b) 208,3$)12%,3,/(250,2$ == PFF (c) 161,65$)31%,7,/(000,8$ == PFF (d) 700,45$)7%,9,/(000,25$ == PFF

3.8 (a) 105,3$)6%,10,/(500,5$ == FPP (b) 338,3$)15%,6,/(000,8$ == FPP (c) 418,20$)5%,8,/(000,30$ == FPP (d) 851,3$)8%,12,/(000,15$ == FPP

3.9 (a) 428,5$)5%,13,/(000,10$ == FPP (b) 763,40$)4%,13,/(000,25$ == PFF

yearsNN

69.9)12.1(log3log

)12.01(3

• yearsN

NPPF N

96.4)15.1(log2log

)15.01(2

• Rule of 72: 72 /15 4.80 years=

Uneven Payment Series

3.12 (a) Single-payment compound amount factors for ),,/( NiPF

n 9% 10% 35 20.4140 28.1024 40 31.4094 45.2593

To find , first, interpolate for )38%,5.9,/( PF 38=n

n 9% 10% 38 27.0112 38.3965

Then, interpolate for %5.9=i

7039.32)38%,5.9,/( =PF As compared to formula determination

4584.31)38%,5.9,/( =PF

(b) Single-payment compound amount factors for ),,/( NiFPn 45 50 0.0313 0.0213

Then, interpolate for 47=n

0273.0)47%,8,/( =FP As compared with the result from formula

0269.0)47%,8,/( =FP

437,114$08.1

000,28$08.1

000,46$08.1000,43$

08.1000,32$

5432 =+++=P

3.14 231,11$)11%,6,/(000,2$)13%,6,/(800,1$)15%,6,/(500,1$ =++= PFPFPFF

$3,000,000 $2,400,000( / ,8%,1)$3,000,000( / ,8%,10)$20,734,618

P PP F

$3,000,000 $2,400,000( / ,8%,5)$3,000,000( / ,8%,5)( / ,8%,5)$20,734,618

P PP A P F

3.16 484,14$)7%,6,/(000,5$)5%,6,/(000,6$)2%,6,/(500,7$ =++= FPFPFPP

Equal Payment Series

3.17 (a) With deposits made at the end of each year

816,13$)10%,7,/(000,1$ == AFF

(b) With deposits made at the beginning of each year 783,14$)07.1)(10%,7,/(000,1$ == AFF

3.18 (a) 713,16$)5%,7,/(000,3$ == AFF (b) 043,77$)12%,25.8,/(000,4$ == AFF (c) 575,267$)20%,4.9,/(000,5$ == AFF (d) 236,134$)12%,75.10,/(000,6$ == AFF

3.19 (a) 166,1$)13%,6,/(000,22$ == FAA (b) 388,4$)8%,7,/(000,45$ == FAA (c) 5.479$)25%,8,/(000,35$ == FAA

(d) 361,1$)8%,14,/(000,18$ == FAA

3.20 $30,000 $1,500( / , 7%, )

( / , 7%, ) 2012.94 13

F A NF A N

N years

=== ≈

3.21 $15,000 ( / , 11%, 5)

$2,408.56A F A

3.22 (a) 310,2$)5%,5,/(000,10$ == PAA (b) 70.723,1$)4%,7.9,/(500,5$ == PAA (c) 85.975,2$)3%,5.2,/(500,8$ == PAA (d) 171,3$)20%,5.8,/(000,30$ == PAA

3.23 • Equal annual payment:

5.132,11$)3%,16,/(000,25$ == PAA

• Interest payment for the second year:

End of Year Principal Repayment

Interest payment

Remaining Balance

0 $25,000 1 $7,132.5 $4,000 $17,867.5 2 $8,273.7 $2,858.8 $9,593.8 3 $9,593.8 $1,535 -

3.24 (a) 2.781,6$)12%,8.5,/(800$ == APP (b) 25.403,16$)10%,5.8,/(500,2$ == APP

(c) 61.665,3$)5%,25.7,/(900$ == APP (d) 3.726,30$)8%,75.8,/(500,5$ == APP

(a) The capital recovery factor NiP for 5

),,/(An 6% 7% 35 0.0690 0.0772 40 0.0665 0.0750

To find , f nterpolate for

)38%,25.6,/( PA irst, i 38=n n 6% 7% 38 0.0675 0.0759

Then, interpolate for %25.6=i ; 6,/( PF 0696.0)38%,25. =

As compared with the result from fo ula rm0694.0)38%,25.6,/( =PF

(b) The equal payment series present-worth factor (P for )85,,/ iAi 9% 10% 11.1038 9.9970

Then, interpolate for

%25.9=i 8271.10)85%,25.9,/( =AP

As compared with the result from for ula m8049.10)85%,25.9,/( =AP

Linear Gradient Series

35.988,50$)5%,8,/)(5,%8,/(000,2$)5%,8,/(000,5$

)5%,8,/(000,2$)5%,8,/(000,5$21

AFGAAFGFAF

47.889,11$)5%,7,/)(5,%7,/(500$)5%,7,/(000,3$

)5%,7,/(500$)5%,7,/(000,3$

=−=−=

PFGPAFGFAFF

26.991$)7%,9,/)](2%,9,/(50$

)4%,9,/(50$)6%,9,/(50$)7%,9,/(100[$100$

+++=FPAF

AFAFAFP

3.404,10$)12%,8,/(000,1$000,15$

=−= GAA

0 (a 076,372,21$)7%,12%,10,/(000,000,6$ 1 =−= APP

Note that the oil price increases at the annual rate of 5% while the oil production decreases at the annual rate

(b) of 10%. Therefore, the annual

revenue can be expressed as follows:

)945.0(000,000,6$

)1.01(000,1000)05.01(60$−

−−

−+=n

ies is equivalent to a decreasing geometric gradient series g = -5.5%.

1)055.01(000,000,6$ −−= n

This revenue ser

with896,847,23$)7%,12%,5.5,/(000,000,6$ 1 =−= APP

(c) sent worth of the remaining series at the end of period 3 gives

Computing the pre 4 5 6 7( , , , )A A A A

652,269,14$)7%,12%,5.5,/(460,063,5=P $ =−AP

(2,000,000) (1.06) (1.06)

1.06(2,000,000 /1.06) ( )1.06

$396, 226, 415

− −

Note: if ,[1 ( 1)

(1 )When 6% and 8%,

1 0.98151$2,000,000 0.9815[1 (21)(0.6881) 20(0.6756)]

1.06 0.0003$334,935,843

i gx N x Nxnx

− + +=

−= =

+− +⎡ ⎤= ⎢ ⎥⎣ ⎦

3.32 (a) The withdrawal series would be

Period Withdrawal 11 $5,000 12 $5,000(1.08) 13 $5,000(1.08)(1.08) 14 $5,000(1.08)(1.08)(1.08) 15 $5,000(1.08)(1.08)(1.08)(1.08)

78.518,22$)5%,9%,8,/(000,5$ 110 == APP

Assuming that each deposit is made at the end of each year, then;

$22,518.78 ( / , 9%, 10) 15.1929$1482.19A F A A

(b) 85.491,24$)5%,6%,8,/(000,5$ 110 == APP

$24,491.85 ( / ,6%, 10) 13.1808$1858.15A F A A

Various Interest Factor Relationships

3.33 (a) 0058.0)2703.0)(0213.0()17%,8,/)(50%,8,/()67%,8,/( === FPFPFP

0058.0)08.01()67%,8,/( 67 =+=FP

(b) ),,/(1

),,/(NiFP

iNiPA−

0394.0)2%,8,/)(40%,8,/()42%,8,/( == FPFPFP

0833.00394.0108.0)42%,8,/( =

−=PA

0833.01)08.1(

)08.1(08.0)42%,8,/( 42

(c) 4996.1208.0

)35%,8,/)(100%,8,/(1),,/(1),,/( =−

iNiFPNiAP

4996.12)08.1(08.0

1)08.1()135%,8,/( 135

3.34 (a)

NiAFiNiPF

)1(11)1(

11)1()1(

1),,/(),,/(

+−+=

iNiAPNiFP

)1()1(

1)1()1()1(

)1(1)1(1)1(

),,/(1),,/(

1)1(]1)1[(

1)1()1(

),,/(),,/(

−+−+

−−+

iNiPANiFA

1)1()1(

)1()1(1)1(

)1()],,/(1[

iiNiFP

(f) & vid(e) (g) Di e the numerator and denominator by and take the

Equivalence Calculations

Ni)1( +li ∞→N

49.740$)10%,12,/)](5%,12,/(50$)7%,12,/(50$)9%,12,/(100[$

=++= FPAFAFAFP

92.373$)4%,8,/(300$

)3%,8,/(120$)2%,8,/(120$)1%,8,/(200$200$)08.1(

FPFPFPP

.37 Selecting the base period at n = 0, we find

.38 Selecting the base period at n =0, we find

75.782$)5%,6,/(75.782$

)5%,6,/(100$)4%,6,/(50$)1%,6,/(50$)5%,6,/(100$200$

$100( / ,13%,5) $20( / ,13%,3)( / ,13%, 2) ( / ,13%,5)$351.72 $36.98 (3.5172)

$110.51

P A P A P F A P AA

+ =+ =

82.185$

FPFPFPAPP

.40 Establish economic equivalent at

96.0$)12%,10,/(20$)5%,10,/(20$

=−= APGPP

0 1 2 3 4 5 6 7 8 9 10 11 12

$20$40

$60$80

3 8=n :

73.334,1$80.699,10$0164.8

)7833.1(000,6$)2597.1)(08.2(6366.10)2%,8,/(000,6$)3%,8,/)(2%,8,/()8%,8,/(

===−=−

APPFAFCAFC

3.41 The original cash flow series is

.42 Establishing equivalence at n = 8, we find

0 0 6 $9001 $800 7 $9202 $820 8 $3003 $840 9 $3004 $860 10 $300 $5005 $880

n nn A n A

13.297$)4872.9()1436.2(277.092,4$

)7%,10,/()8%,10,/(2)3%,10,/(200$)8%,10,/(300$

AFCPFCAFAF

.43 Establishing equivalence at 3 5=n

.44 Computing equivalence at

$200( / ,8%,5) $50( / ,8%,1)( / ,8%,5) ($200 )[( / ,8%,2) ( / ,8%,1)]

$1,119.32 (5.8666) ($200 )(2.2464)$185.09

F A F PX F A X F P F P

−= − + += − +=

2.623,29$)5%,9,/(000,3$)5%,9,/(000,3$ =+= APAFX

3.45 (2), (4), and (6)

3.46 (2), (4), and (5)

25.91$2638.1)5%,10,/(

32.115$)5%,10,/)](1%,10,/(50$50[$)5%,10,/(50$50($

=+−+=

AAPAAAA

PAFPGAA

3.48 (a)

3.49 (b)

.50 (b)

793,5$60.458,2$8137.6935,41$

)]6%,10,/)(6%,10,/(000,1$)12%,10,/()6%,10,/(000,30$000,25$

FPAPAPCFP

Solving for an Unknown Interest Rate of Unknown Interest Periods

3.52 Establishing equivalence at

52 (1 )log 2 5 log (1 )

14.87%

P P ii

= += +=

1 i )/(500,2$)6,,/(000,2$ APiAP 6,%,25,−=

Solving for I with Excel, we obtain %35.92=i

5$35,000 $10,000( / , ,5) $10,000(1 )28.47%

F P i ii= ==

$1,000,000 $2,000( / ,6%, )(1 0.06) 1500

0.0631 (1 0.06)

log30 log1.0658.37 years

+ −=

Short Case Studies

ST 3. uming that they are paid at the beginning of each year 1 Ass(a)

62.58$)3%,6,/(96.15$96.15$ =+ AP It is better to take the offer because of lower cost to renew.

ST 3 payment series at the end of

20 (or beginning of Year 21) is

$57.12 $15.96 $15.96( / , ,3)7.96%

P A ii= +=

.2 The equivalent future worth of the prize Year

F F A=

quivalent future worth of the lottery receipts is

F F P= −

resulting surplus at the end of Year 20 is

pute the equivalent present worth (in 2006) for each option at

P P F P FP F

= + + ++

(b) sing either Excel or Cash Flow Analyzer, both plans would be

$71,819,490=

$1,952,381( / ,6%,20)

$109,516,040=($36,100,000 $1,952,381)( / ,6%,20)

2 1 $109,516,040 $71,819,490$37,696,550

F F− = −=

ST 3.3 (a) Com %6=i .

Deferred $2,000,000 $566,000( / ,6%,1) $920,000( / ,6%, 2)$1, 260,000( / ,6%,11)

$8,574, 490

P P F PP F

= + ++

Non-Deferred $2,000,000 $900,000( / ,6%,1) $1,000,000( / ,6%, 2)$1,950,000( / ,6%,5)

$7, 431,560 ∴ At %6=i , the deferred plan is a better choice. Ueconomically equivalent at %72.15=i

ST 3 ximum amount to invest in the prevention program is

.4 The ma

467,50$)5%,12,/(000,14$ == APP

sing the

ST 3.5 geometric gradient series present worth factor, we can establish e equivalence between the loan amount $120,000 and the balloon payment

Uthserie

1) 1$120,000 (A= 1

1/ ,10%,9%,5) 4.6721$25,684.38

P A A==

2) Paym series

n Payment 1 2

$25,684.38 $28,252.82

3 $31,078.10 4 $34,185.91

ST 3.6 1) Compute the required annual net cash profit to pay off the investment and interest.

5 $37,604.50

$70,000,000 ( / ,1A P A 0%,5) 3.7908$18,465,759

) Decide the number of shoes, X

2$18,465,759 ($100)

184,657X

ST 3.7

$1,000( / ,9.4%,5) $500( / ,9.4%,5) $4,583.36P F F A$4,583.36( / ,9.4%,60) $1,005,132

ain question is whether or not the U.S. government will be able to vest the social security deposits at 9.4% interest over 60 years.

Stay with the original deferred plan.

The min

ST 3.8 (a)

Contract $3,875,000 $3,125,000( / ,6%,1)$5,525,000( / ,6%,2)$8,875,000( / ,6%,7)$39,547,242

P PP FP F

= ++ ++=

$1,375,000 $1,375,000( /P P= + ,6%,7)$9,050,775 $8,000,000

ST 3.9

e between two payment options. Selecting n = 0 as the base period, we can calculate the equivalent present worth for each option as follows:

Basically we are establishing an economic equivalenc

Option 1: $140,000Option 2: $32,639( / , %,9)

PP P A i==

P A i$140,000 $32,639( / , %,9)18.10%i

If Mrs. Setchfield can invest her money at a rate higher than 18

.10%, it is

better to go with Option 1. However, it may be difficult for her to find an investment opportunity that provides a return exceeding 18%.

Chapter 03

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