Ingeneria Economica

Chapter 2: Time Value of Money
2.1) 2.2) • Simple interest:
F = P (1 + iN ) $4, 000 = $2, 000(1 + 0.08 N ) N = 12.5 years (or 13 years)
I = iPN = (0.09)($3,000)(5) = $1,350

•

Compound interest:

$4, 000 = $2, 000(1 + 0.07) N 2 = 1.07 N log 2 = N log 1.07 N = 10.24 years (or 11 years) 2.3) •

Simple interest:
I = iPN = (0.07)($10, 000)(20) = $14, 000

•

Compound interest:

I =P ⎡(1 + i) N − 1⎤ = $10,000 ⎡(1.07)20 − 1⎤ ⎣ ⎦ ⎣ ⎦ = $28,696.84
2.4) •

Compound interest:
F = $1, 000(1 + 0.06)5

= $1,338.23 • Simple interest:

F = $1, 000(1 + 0.07(5)) = $1,350 The simple interest option is better.
Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. © 2008 Pearson Education, Inc.,Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

2 2.5)

•

Loan balance calculation:
End of period 0 1 2 3 4 5Principal Payment $0.00 $835.46 $910.65 $992.61 $1,081.94 $1,179.32 Interest Payment $0.00 $450.00 $374.81 $292.85 $203.52 $106.14 Remaining Balance $5,000.00 $4,164.54 $3,253.89 $2,261.28 $1,179.33 $0.00

2.6) 2.7) 2.8)

P = $8, 000( P / F ,8%, 5) = $8, 000(0.6806) = $5, 444.8 F = $20,000( F / P,10%,2) = $20,000(1.21) = $24,200

•

Alternative 1

P = $100
• Alternative 2
P = $120( P / F,8%,2) = $120(0.8573) = $102.88

• 2.9) (a) (b) (c) (d) 2.10) (a) (b) (c) (d)

Alternative 2 is preferred
F = $7,000( F / P,9%,8) = $7,000(1.9926) = $13,948.2 F = $1,250( F / P,4%,12) = $1,250(1.6010) = $2,001.25 F = $5,000( F / P,7%,31) = $5,000(8.1451) = $40,725.5 F = $20,000( F / P,6%,7) = $20,000(1.5036) = $30,072 P = $4,500( P / F ,7%,6) = $4,500(0.6663) = $2,998.35 P = $6,000( P / F ,8%,15)= $6,000(0.3152) = $1,891.2 P = $20,000( P / F ,9%,5) = $20,000(0.6499) = $12,998 P = $12,000( P / F ,10%,8) = $12,000(0.4665) = $5,598

Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright andwritten permission should be obtained from the publisher prior to any prohibited reproduction, storage 2 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

3 2.11) (a) (b) 2.12)
F = 3P = P(1 + 0.07) N log 3 = N log 1.07 N = 16.24 years (or 17 years)
P = $6,000( P / F ,8%,5) = $6,000(0.6806) = $4,083.6 F = $15,000( F/ P,8%,4) = $15,000(1.3605) = $20,407.5

2.13) •
F = 2 P = P(1 + 0.12) N log 2 = N log 1.12 N = 6.12 years

• 2.14)

Rule of 72:

72 /12 = 6 years

P = $35,000(P / F,9%,4) + $10,000( P / F,9%,2) = $35,000(0.7084) + $10,000(0.8417) = $33,211

2.15)

•

Simple interest:
I = iPN = (0.1)($1, 000)(3) = $300

•

Compound interest:
I = P ⎡(1 + i) N − 1⎤ = $1,000 ⎡(1 + .095)3 − 1⎤ ⎣⎦ ⎣ ⎦ = $312.93

• 2.16) 2.17)
P=

Susan’s balance will be greater by $12.93.

$3,000 $3,500 $4,000 $6,000 + + + = $13,260.58 1.06 2 1.063 1.064 1.065

F = $1, 000( F / P,8%,10) + $1,500( F / P,8%,8) + $2, 000( F / P,8%, 6) = $8,109.05

Instructor Solutions Manual to accompany Fundamentals of Engineering Economics, Second Edition, by Chan S. Park. ISBN-13: 9780132209618. © 2008 PearsonEducation, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage 3 in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

4

2.18)
P = $3, 000, 000 + $2, 400, 000( P /...