978-0073523439 Chapter 14

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

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

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.

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)

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

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)

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

(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

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

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