Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 2
Factors: How Time and Interest Affect Money
2.1 (1) (P/F, 6%, 8) = 0.6274
2.2 P = 21,300(P/A,10%,5)
2.3 Cost now = 142(0.60)
2.4 F = 100,000(F/P,10%,3) + 885,000
2.5 F = 50,000(F/P,6%,14)
2.6 F = 1,900,000(F/P,15%,3)
2.7 A = 220,000(A/P,10%,3)
2.8 P = 75,000(P/F,12%,4)
2.11 Gain in worth of building after repairs = (600,000/0.75 – 600,000) – 25,000 = 175,000
2.13 P = (110,000* 0.3)(P/A,12%,4)
2.14 P = 600,000(0.04)(P/A,10%,3)
2.15 A = 950,000(A/P,6%,20)
2.16 A = 434(A/P,8%,5)
2.17 F = (0.18 – 0.04)(100)(F/A,6%,8)
2.18 Fdifference = 10,500(F/P,7%,18) – 10,500(F/P,4%,18)
2.19 F = (200 – 90)(F/A,10%,8)
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2.20 A = 350,000(A/F,10%,3)
2.21 (a) 1. Interpolate between i = 12% and i = 14% at n = 15.
2.22 (a) 1. Interpolate between n = 60 and n = 65:
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2.23 Interpolated value: Interpolate between n = 40 and n = 45:
2.24 Interpolated value: Interpolate between n = 50 and n = 55:
2.25 (a) Profit
in year 5 = 6000 + 1100(4) = $10,400
2.26 (a) G = (241 – 7)/9 = $26 billion per year
2.27 A = 200 – 5(A/G,8%,8)
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2.29 (a) CF3 = 70 + 3(4) = $82 ($82,000)
2.30 601.17 = A + 30(A/G,10%,9)
2.32 75,000 = 15,000 + G(A/G,10%,5)
2.33 First find Pg (using equation) and then convert to A
For n = 1: Pg = {1 – [(1 + 0.04)/(1 + 0.10)]1}/(0.10 – 0.04)
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= $320,573
2.35 Pg1 = 10,000{1 – [(1 + 0.04)/(1 + 0.08)]10}/(0.08 – 0.04)
2.36 Pg = 260{1 – [(1 + 0.04)/(1 + 0.06)]20}/(0.06 – 0.04)
2.38 18,000,000 = 3,576,420(P/A,i,7)
2.39 813,000 = 170,000(F/P,i,15)
2.40 100,000 = 210,325(P/F,i,30)
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2.41 (1,000,000 – 1,900,000) = 200,000(F/P,i,4)
2.42 800,000 = 250,000(P/A,i,5)
2.43 87,360 = 24,000(F/A,i,3)
2.45 600,000 = 80,000(F/A,15%,n)
2.46 Starting amount = 1,600,000(0.55) = $880,000
2.47 200,000 = 29,000(P/A,10%,n)
2.48 1,500,000 = 18,000(F/A,12%,n)
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2.49 350,000 = 15,000(P/A,4%,n) + 21,700(P/G,4%,n)
2.50 16,000 = 13,000 + 400(A/G,8%,n)
2.51 140(0.06 – 0.03) = 12{1 – [(0.97170)]x}
2.52 135,300 = 35,000 + 19,000(A/G,10%,n)
2.54 P = 30,000(P/F,12%,3)
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2.55 30,000 = 4200(P/A,8%,n)
2.56 A = 22,000 + 1000(A/G,8%,5) = $23,847
2.58 A = 800 – 100(A/G,4%,6) = $561.43
2.60 F = 61,000(F/P,4%,4)
2.61 P = 90,000(P/A,10%,10)
2.62 A = 100,000(A/P,10%,7)
2.63 A = 1,500,000(A/F,10%,20)
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2.64 In $1 million units
A = 3(10)(A/P,10%,10)
2.65 75,000 = 20,000(P/A,10%,n)
2.66 50,000(F/A,6%,n) = 650,000
2.67 40,000 = 13,400(P/A,i,5)
(P/A,i,5) = 2.9851
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2.71 F = {5000[1 – (1.03/1.10) 20]/(0.10 – 0.03)}(F/P,10%,20)
Solution to Case Study, Chapter 2
There is no definitive answer to case study exercises. The following are examples only.
Time Marches On; So Does the Interest Rate
1. Situation A B C D
Interest rate 6% per year 6% per year 15% per year Simple: 780% per year
Comp’d: 143,213% per year
2. A: Start — $24
End — F = 24(1.06)385 = $132 billion
B: Start — $2000 per year or $20,000 total over 10 years
End — F32 = A(F/A,6%,10) = $26,361.60