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Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 14
Effects of Inflation
14.1 (a) There is no difference.
(b) Today’s dollars are inflated compared to dollars of 2 years ago. Therefore, in order
14.2 (a) During periods of inflation
14.5 if per month = 0.30/12 + 0.015 + (0.30/12)(0.015)
14.6 0.35 = 0.25 + f + 0.25f
14.7 if = 0.04 + 0.01 + (0.04)(0.01)
14.8 if per month = 18/12 = 1.5%
Use inflation–adjusted interest rate equation to solve for i.
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14.9 Let CV = constant–value dollars
CV1 = 45,000/(1 + 0.05)1 = $42,857
14.12 Assume C1 is the cost today
2C1 = C1(1 + 0.07)n
14.13 0.28 = i + 0.06 + i(0.06)
14.15 Buying power = 250,000/(1 + 0.04)5
14.16 (a) Constant-value dollars have to increase by only the real interest rate of 5% per year.
(b) if = 0.05 + 0.04 + (0.05)(0.04)
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14.17 Find f using F/P or P/F factor
5400 = 4050(F/P,f,5)
14.18 Price next year = 28,000(1 + 0.021)1
14.19 (a) Cost in today’s dollars = $120,000
14.20 If price had increased only by inflation rate,
14.21 (a) Cost of T & F = 0.28(52,000) = $14,560
14.22 (a) At a 58% increase, $1 would increase to $1.58. Let x = annual percentage increase
1.58 = (1 + x)5
14.23 Pg = 350{1- [(1+0.03/1 + 0)31]/0 – 0.03}
14.24 The two ways to account for inflation in PW calculations are:
14.28 Convert all cash flows into CV dollars and then use i.
PW = 3000(P/F,8%,1) + [6000/(1 + 0.06)2](P/F,8%,2)
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14.29 The $1.9 million are then–current dollars. Use if to find PW
if = 0.15 + 0.03 + (0.15)(0.03) = 18.45%
14.30 (a) Use i = 10%
F = 68,000(F/P,10%,2)
14.31 Use the real i for salesman A and inflated if for Salesman B.
if = 0.20 + 0.04 + (0.20)(0.04) = 24.8%
14.32 if = 0.12 + 0.04 + (0.12)(0.04)
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14.33 if per month = 0.01 + 0.004 + (0.01)(0.004) = 1.4%
14.34 Find present worth of all three plans.
Method 1: PW1 = $480,000
Method 2: if = 0.10 + 0.06 + (0.10)(0.06) = 16.6%
14.35 if = 0.10 + 0.06 + (0.10)(0.06)
= 16.6% per year
14.36 Find F in future dollars using f = -3.0%
14.37 Purchasing power = 100,000(F/P,10%,15)/(1 – 0.01)15
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14.38 Buying power = 60,000(F/A,10%,5)/(1 + 0.04)5
14.39 8,000,000(1 + f)4 = 7,000,000(F/P,7%,4)
14.40 (a) 25,000 = 10,000(F/P,i,5)
14.42 (a) 1,400,000 = 653,000(1 + f)13
(b) The market rate is f + 5%.
14.43 if = 0.15 + 0.028 + (0.15)(0.028)
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14.44 (a) Cost, year 20: machine A = 10,000(1.10)(1.10)(1.02)(1.02)…(1.02)
= $31,617.58
Cost, year 20: machine B = 10,000(1.02)(1.02)(1.10)(1.10)…(1.10)
14.45 F = P[(1 + i)(1 + f)(1 + g)]n
14.46 if
= 0.07 + 0.04 + (0.07)(0.04)
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14.47 Calculate amount needed at 5% inflation rate and then find A using market rate.
F = 72,000(1 + 0.05)3
14.49 if = 0.15 + 0.05 + (0.15)(0.05)
14.50 if = 0.12 + 0.03 + (0.12)(0.03)
14.51 if = 0.10 + 0.04 + (0.10)(0.04)
14.52 if = 0.09 + 0.03 + (0.09)(0.03)
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A = -180,000(A/P,12.27%,5) – 70,000(P/F,12.27%,3)(A/P,12.27%,5)
14.53 if = 0.20 + 0.05 + (0.20)(0.05)
(a) CR = A = 2,500,000(A/P,26%,5)
14.57 0.16 = i + 0.09 + i(0.09)
14.58 0.06 = i + 0.02 + (i)(0.02)
14.59 Cost = 40,000/(1 + 0.06)10
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14.60 F = 1000(F/P,5%,25)
14.61 if = 0.06 + 0.04 + (0.06)(0.04)
14.62 if = 0.04 + 0.03 + (0.04)(0.03)
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Solution to Case Study, Chapter 14
Sometimes, there is not a definitive answer to a case study exercise. Here are example responses.
INFLATION VERSUS STOCK AND BOND INVESTMENTS
1. Stocks: Overall i* = 6.6% per year
2. if = 0.07 + 0.04 + 0.04(0.07) = 11.28%
4. Subtract the future value of each payment from the bond face value 5 years from now.
Both amounts take purchasing power into account.
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5. Stocks: F = 50,000(P/F,11.28%,12) – 1,000(F/A,11.28%,12)
(Note: Goal Seek will find the answers, also. Target cells are row 17, the i* values set to
Do the answers seem reasonable?