978-0073523439 Chapter 14

subject Type Homework Help
subject Pages 11
subject Words 1656
subject Authors Anthony Tarquin, Leland Blank

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Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 14
Effects of Inflation
Adjusting for Inflation
14.1 Inflated dollars are converted into constant-value dollars by dividing by one plus the
14.2 Cost will double in 10 years when the value of the money has decreased by exactly one
half. From Eq. [14.3]
14.3 (a) There is no difference.
14.4 Future amount = 10,000(1 + 0.07)10
14.7 (a) CV0, year 1 estimate = 13,000/(1 + 0.06)1 = $12,264
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14.8 (a) Freshman year cost = 19,548(1.07)3 = $23,947
14.9 (a) 5550(1 +f)2 = 5730
14.11 (a) CV purchasing power = 500,000/(1.03)27 = £225,095
14.12 (a) CV = 1,000,000/(1.04)40 = $208,289
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(b) Cost; year 10, 5% per year = 1000(1 + 0.05)10 = $1628.89
The cost is not the same because there are different inflation rates on
different costs in each year over the 10 years
14.14 23,930,909 = 1,000,000(1 + f)103
14.15 7984 = 10,000(1 + f)20
14.18 Increase is f = 100% per day
Present Worth Calculations with Inflation
14.19 Some examples of what Jake could do (there are many others):
1. Use savings and pay off the car loan immediately and do not apply for another loan for
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14.21 0.40 = i + 0.09 + i(0.09)
14.25 By hand:
(a) Inflation not considered uses i = 10%
(b) Inflation considered uses if
By spreadsheet: PW values and associated functions are shown.
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14.26 (a) Use if for future dollars and i for CV dollars
(b) Convert CF in year 7 to future dollars and develop NPV function
14.27 By hand:
Method 1: PW = -10,000 + 2000(P/F,20%,1) + 5000(P/A,20%,3)(P/F,20%,1)
= -10,000 + 2000(0.8333) + 5000(2.1065)(0.8333)
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By spreadsheet: Convert to CV; sample function for year 2: = $B7/(1 + $B$2)^$A7
14.28 (a) Use real i = 10%
= $ 66,938 < $68,000
Purchase later for $81,000
(b) Use if = 0.10 + 0.05 + (0.10)(0.05) = 0.155 (15.5% per year)
14.29 (a) PWA = -31,000 – 28,000(P/A,10%,5) + 5000(P/F,10%,5)
(b) if = 0.10 + 0.03 + (0.10)(0.03) = 0.133 (13.3%)
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= $-126,087
PWB = -48,000 – 19,000(P/A,13.3%,5) + 7000(P/F,13.3%,5)
= -48,000 – 19,000(3.4916) + 7000(0.5356)
= $-110,591
Select machine B
Spreadsheet function for PWA: = - PV(13.3%,5,-28000,5000) - 31000
PWB: = - PV(13.3%,5,-19000,7000) – 48000
(c) Maintain i = 10% for B; use Goal Seek to force PW difference to be 0 while changing
14.30 if = 0.12 + 0.03 + (0.12)(0.03) = 15.36%
Select alternative Y
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Deposit now: P = 4,929,431(P/F,8%,23)
= 4,929,431(0.1703)
= $839,482
14.32 (a) The $2.1 million are then-current dollars. Use if to find PW
14.33 if = 0.15 + 0.035 + (0.15)(0.035) = 0.19025 (19.025%)
14.34 (a) Find present worth of all three plans
(b) Rates are different for method 2 and 3, but they must both equal PW1 = $450,000. Use
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Future Worth and Other Calculations with Inflation
14.37 (a) Find F, then deflate the amount by dividing by (1 + f)n
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(b) Purchasing power in CV dollars = $61,762, which is less than the total amount
deposited of $73,000
14.38 740,000 = 625,000(F/P,f,5)
(F/P,f,5) = 1.184
14.39 (a) F = 10,000(F/P,10%,5)
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14.40 F = P[(1 + i)(1 + f)(1 + g)]n
14.41 (a) 1,030,000 = 653,000(1 + f)18
(b) The market rate is f + 5%.
14.42 Account will have to grow at rate of if
14.43 (a) In CV dollars, the cost will be the same as today, $96,000, for all inflation rates.
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14.44 Purchasing power = 1,800,000/(1 + 0.038)20
14.46 (a) Future amount is equal to the return at the market interest rate
14.47 (a) Required future amount is equal to a return of if on its investment
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(b) Set F relation equal to $2.5 million; find if; calculate MARR and then i
2,500,000 = 1,000,000(F/P,if,4)
economically justified.
Capital Recovery with Inflation
14.48 if = 0.15 + 0.06 + (0.15)(0.06) = 0.219 (21.9%)
14.49 (a) In constant value dollars, use i = 12% to recover the investment
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14.50 (a) To maintain purchasing power, use f to find future dollars.
F = 5,000,000(F/P,5%,4)
14.51 (a) if = 0.10 + 0.028 + (0.10)(0.028) = 0.1308 (13.08% per year)
displays A = $181,272 directly
14.52 Use market interest rate (if) to calculate AW in then-current dollars
14.53 Find amount needed at 2% inflation rate and then find A using market rate.
14.54 Use 15% rate to determine cost of expansion, then find A using market rate.
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= 76,045,000 (0.30211)
= $22,973,955 per year
Function: = - PMT(10%,3,,- FV(15%,3,,50000000)) displays A = $22,973,942
14.55 (a) Use if (market interest rate) to find AW
(b) For CV dollars, first find P using i (real interest rate); then find A using if
14.56 By hand:
(a) For CV dollars, use i = 12% per year
(b) For then-current dollars, use if
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Spreadsheet: Select A in both cases; the difference in AW values is larger for future dollars
Additional Problems and FE Exam Review Questions
14.57 if = 0.12 + 0.07 + (0.12)(0.07) = 0.1984 (19.84%)
Answer is (d)
14.62 Answer is (c)
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Nominal per year = 2.01(12) = 24.12%
Answer is (d)
14.66 Cost = 85,000(F/P,4%,3)
14.67 Find accumulated amount in future dollars, then divide by inflation rate

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