Chapter 2: Time Value of Money
Chapter 2: Time Value of Money
2.1)
I = iPN = (0.09)($3,000)(5) = $1,350
2.2)
•
Simple interest:
F = P (1 + iN )
$4, 000 = $2, 000(1 + 0.08 N )
N = 12.5 years (or 13 years)
•
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
5
Principal
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
2.6)
P = $8, 000( P / F ,8%, 5) = $8, 000(0.6806) = $5, 444.8
2.7)
Remaining
Balance
$5,000.00
$4,164.54
$3,253.89
$2,261.28
$1,179.33
$0.00
F = $20,000( F / P,10%,2) = $20,000(1.21) = $24,200
2.8)
•
Alternative 1
P = $100
•
Alternative 2
P = $120( P / F ,8%,2) = $120(0.8573) = $102.88
•
Alternative 2 is preferred
(a)
F = $7,000( F / P,9%,8) = $7,000(1.9926) = $13,948.2
(b)
F = $1,250( F / P,4%,12) = $1,250(1.6010) = $2,001.25
(c)
F = $5,000( F / P,7%,31) = $5,000(8.1451) = $40,725.5
(d)
2.9)
F = $20,000( F / P,6%,7) = $20,000(1.5036) =
2.1)
I = iPN = (0.09)($3,000)(5) = $1,350
2.2)
•
Simple interest:
F = P (1 + iN )
$4, 000 = $2, 000(1 + 0.08 N )
N = 12.5 years (or 13 years)
•
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
5
Principal
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
2.6)
P = $8, 000( P / F ,8%, 5) = $8, 000(0.6806) = $5, 444.8
2.7)
Remaining
Balance
$5,000.00
$4,164.54
$3,253.89
$2,261.28
$1,179.33
$0.00
F = $20,000( F / P,10%,2) = $20,000(1.21) = $24,200
2.8)
•
Alternative 1
P = $100
•
Alternative 2
P = $120( P / F ,8%,2) = $120(0.8573) = $102.88
•
Alternative 2 is preferred
(a)
F = $7,000( F / P,9%,8) = $7,000(1.9926) = $13,948.2
(b)
F = $1,250( F / P,4%,12) = $1,250(1.6010) = $2,001.25
(c)
F = $5,000( F / P,7%,31) = $5,000(8.1451) = $40,725.5
(d)
2.9)
F = $20,000( F / P,6%,7) = $20,000(1.5036) =