Solutions to end–of–chapter problems
Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 2
Factors: How Time and Interest Affect Money
Determination of F, P and A
2.1 (1) (F/P, 10%, 7) = 1.9487
2.2 F = 1,200,000(F/P,7%,4)
2.3 F = 200,000(F/P,10%,3)
2.4 P = 7(120,000)(P/F,10%,2)
2.5 F = 100,000,000/30(F/A,10%,30)
2.6 P = 25,000(P/F,10%,8)
2.7 P = 8000(P/A,10%,10)
2.8 P = 100,000((P/A,12%,2)
2.9 F = 12,000(F/A,10%,30)
2.10 A = 50,000,000(A/F,20%,3)
2.11 F = 150,000(F/P,18%,5)
2.12 P = 75(P/F,18%,2)
2.13 A = 450,000(A/P,10%,3)
2.14 P = 30,000,000(P/F,10%,5) – 15,000,000
2.15 F = 280,000(F/P,12%,2)
2.16 F = (200 – 90)(F/A,10%,8)
2.17 F = 125,000(F/A,10%,4)
2.19 P = 90,000(P/A,20%,3)
2.20 A = 250,000(A/F,9%,5)
2.22 P = (110,000* 0.3)(P/A,12%,4)
Factor Values
2.25 (a) 1. Interpolate between i = 8% and i = 9% at n = 15:
(P/F,8.4%,15) = 0.3152 – 0.0163
= 0.2989
2. Interpolate between i = 16% and i = 18% at n = 10:
(b) 1. (P/F,8.4%,15) = 1/(1 + 0.084)15
= 0.2982
2.26 (a) 1. Interpolate between i = 18% and i = 20% at n = 20:
2. Interpolate between i = 25% and i = 30% at n = 15:
1/5 = x/0.5911
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(c) 1. = -FV(19%,20,1) displays 165.41802
2. = -PV(26%,15,1) displays 3.72607
2.27 (a) 1. Interpolate between n = 32 and n = 34:
2. Interpolate between n = 50 and n = 55:
2.28 Interpolated value: Interpolate between n = 40 and n = 45:
3/5 = x/(72.8905 – 45.2593)
Arithmetic Gradient
2.30 P0 = 500(P/A,10%,9) + 100(P/G,10%,9)
2.31 (a) Revenue = 390,000 + 2(15,000)
2.32 A = 9000 – 560(A/G,10%,5)
2.34 A = 100,000 + 10,000(A/G,10%,5)
2.36 In $ billion units,
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917,910,000 = 100,000,000(P/A,18%,5) + G(P/G,18%,5)
917,910,000 = 100,000,000(3.1272) + G(5.2312)
G = $115,688,561
2.37 95,000 = 55,000 + G(A/G,10%,5)
2.38 P in year 0 = 500,000(P/F,10%,10)
Geometric Gradient
2.39 Find (P/A,g,i,n) using Equation [2.32] and A1 = 1
2.40 Decrease deposit in year 4 by 7% per year for three years to get back to year 1.
2.43 First find Pg and then convert to F in year 15
2.44 (a) Pg = 260{1 – [(1 + 0.04)/(1 + 0.06)]20}/(0.06 – 0.04)
=$210,087,870
2.45 Solve for Pg in geometric gradient equation and then convert to A
2.46 First find Pg and then convert to F
Interest Rate and Rate of Return
2.47 1,000,000 = 290,000(P/A,i,5)
2.48 50,000 = 10,000(F/P,i,17)
2.49 F = A(F/A,i%,5)
2.50 Bonus/year = 6(3000)/0.05 = $360,000
2.51 Set future values equal to each other
Simple: F = P + Pni
2.52 100,000 = 190,325(P/F,i,30)
2.53 400,000 = 320,000 + 50,000(A/G,i,5)
Number of Years
2.54 160,000 = 30,000(P/A,15%,n)
2.55 (a) 2,000,000 = 100,000(P/A,5%,n)
(b) 2,000,000 = 150,000(P/A,5%,n)
(c) The reduction is impressive from forever (n is infinity) to n = 22.5 years for a 50%
annually when a fixed amount and a specific rate of return are involved.
2.56 10A = A(F/A,10%,n)
2.57 (a) 500,000 = 85,000(P/A,10%,n)
2.58 1,500,000 = 6,000,000(P/F,25%,n)
2.59 15,000 = 3000 + 2000(A/G,10%,n)
2.60 First set up equation to find present worth of $2,000,000 and set that equal to P in the
Exercises for Spreadsheets
2.61
Part Function Answer
a= –FV(10%,30,100000000/30) $548,313,409
b= –FV(10%,33,100000000/30) $740,838,481
c= –FV(10%,33,100000000/30) + FV(10%,3,(100000000/30)*2) $718,771,814
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2.62
2.63 Goal Seek template before and result after with solution for G = $115.69 million
13
2.64 Here is one approach to the solution using NPV and FV functions with results (left) and
formulas (right).
Year, Present worth Future worth
nDeposit in year 0 in year n
0
110,000 9,346 10,000
211,000
18,954
21,700
312,100 28,831
35,319
413,310 38,985 51,101
514,641 49,424 69,319
616,105
60,155 90,277
717,716 71,188
114,312
819,487 82,529 141,801
921,436 94,189 173,163
10 23,579
106,176 208,864
11 25,937 118,498 249,422
12 28,531 131,167
295,412
13 31,384 144,190 347,475
14 34,523 157,578
406,321
25 98,497 332,095 1,802,424
26 108,347 350,752 2,036,941
27 119,182 369,932 2,298,709
28 131,100 389,650 2,590,718
29 144,210 409,920 2,916,279
30 158,631 430,759 3,279,049
Year, Present worth Future worth
nDeposit in year 0 in year n
0
= $A3+1 10000 =NPV(7%,$B$4:$B4) = -FV(7%,$A4,,$C4)
= $A4+1 = $B4*1.1 =NPV(7%,$B$4:$B5) = -FV(7%,$A5,,$C5)
= $A5+1 = $B5*1.1 =NPV(7%,$B$4:$B6) = -FV(7%,$A6,,$C6)
= $A6+1 = $B6*1.1 =NPV(7%,$B$4:$B7) = -FV(7%,$A7,,$C7)
= $A7+1 = $B7*1.1 =NPV(7%,$B$4:$B8) = -FV(7%,$A8,,$C8)
= $A8+1 = $B8*1.1 =NPV(7%,$B$4:$B9) = -FV(7%,$A9,,$C9)
= $A9+1 = $B9*1.1 =NPV(7%,$B$4:$B10) = -FV(7%,$A10,,$C10)
= $A10+1 = $B10*1.1 =NPV(7%,$B$4:$B11) = -FV(7%,$A11,,$C11)
= $A11+1 = $B11*1.1 =NPV(7%,$B$4:$B12) = -FV(7%,$A12,,$C12)
= $A12+1 = $B12*1.1 =NPV(7%,$B$4:$B13) = -FV(7%,$A13,,$C13)
= $A13+1 = $B13*1.1 =NPV(7%,$B$4:$B14) = -FV(7%,$A14,,$C14)
= $A14+1 = $B14*1.1 =NPV(7%,$B$4:$B15) = -FV(7%,$A15,,$C15)
= $A15+1 = $B15*1.1 =NPV(7%,$B$4:$B16) = -FV(7%,$A16,,$C16)
= $A16+1 = $B16*1.1 =NPV(7%,$B$4:$B17) = -FV(7%,$A17,,$C17)
= $A27+1 = $B27*1.1 =NPV(7%,$B$4:$B28) = -FV(7%,$A28,,$C28)
= $A28+1 = $B28*1.1 =NPV(7%,$B$4:$B29) = -FV(7%,$A29,,$C29)
= $A29+1 = $B29*1.1 =NPV(7%,$B$4:$B30) = -FV(7%,$A30,,$C30)
= $A30+1 = $B30*1.1 =NPV(7%,$B$4:$B31) = -FV(7%,$A31,,$C31)
= $A31+1 = $B31*1.1 =NPV(7%,$B$4:$B32) = -FV(7%,$A32,,$C32)
2.65 (a) Present worth is the value of the savings for each bid
(b) and (c) Spreadsheet for A values and column chart
ADDITIONAL PROBLEMS AND FE REVIEW QUESTIONS
2.67 P = 840,000(P/F,10%,2)
2.69 F = 25,000(F/P,10%,25)
2.70 A = 10,000,000(A/F,10%,5)
2.71 A = 2,000,000(A/F,8%,30)
= 2,000,000(0.00883)
2.73 AW = 26,000 + 1500(A/G,8%,5)
2.74 30,000 = 4202(P/A,8%,n)
2.76 A = 800 – 100(A/G,8%,6)
2.77 P = 100,000(P/A,10%,5) – 5000(P/G,10%,5)
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From tables, i = 4%
Answer is (a)
2.79 28,800 = 7000(P/A,10%,5) + G(P/G,10%,5)
Solution to Case Study, Chapter 2
The Amazing Impact of Compound Interest
1. Ford Model T and a New Car
(a) Inflation rate is substituted for i = 3.10% per year
2. Manhattan Island
3. Pawn Shop Loan
(b) Beginning amount: P = $200
4. Capital Investment
(a) i = 15+% per year
1,000,000 = 150,000(P/A,i%,60)
5. Diamond Ring
(a) i = 4% per year
(b) Beginning price: P = $50
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= 179 years
F = 50(F/P,4%,179)
= 50(1119.35)
= $55,968