Solutions to Chapter 9 Problems

Solutions to Chapter 8 Problems

8-1With no inflation in the U.S. economy, the GDP will grow at 3% per year. Inflation will add to the real rate of growth, so the actual GDP will exceed 3% per year by about the value of the inflation rate.

8-2A = $1,000; N = 10

(a)f = 6% per year; ir = 4% per year

In Part (a), the $1,000 is an A$ uniform cash flow (annuity)

im = 0.04 + 0.06 + (0.04)(0.06) = 0.1024, or 10.24% per year

PW(im) = $1,000 (P/A,10.24%,10) = $1,000(6.0817) = $6,082

(b)In Part (b), the $1,000 is a R$ uniform cash flow (annuity) because the A$ cash flow is $1,000 (1.06)k where 1  k  10; i.e.,

(R$)k = (A$)k= $1,000 (1.06)k= $1,000; 1  k  10

PW(ir) = $1,000 (P/A,4%,10) = $8,111

8-3f = 5% per year; N = ? (when does $1 equal $0.50 in today's purchasing power?)

From Equation 8-1, with k = 0, we have

(R$)N = $0.50 = ($1)

(1.05)N = 2

N (ln (1.05)) = ln (2)

N = 14 years

In 14 years, the dollar's purchasing power will be one-half of what it is now if the general price inflation rate is 5% per year.

8-4In ten years a service/commodity that increases at the 4% general inflation rate will cost (F/P, 4%, 10) = 1.4802 times its current cost. But health care costs will increase to (F/P, 12%, 10) = 3.1058 times their current value. The ratio of 3.1058 to 1.4802 is 2.10, which means that health care will cost 210% more than an inflation-indexed service/commodity in ten years.

8-5Situation a: FW5 (A$) = $2,500 (F/P, 8%, 5) = $2,500 (1.4693) = $3,673

Situation b: FW5 (A$) = $4,000 (given)

Choose situation b. (Note: The general inflation rate, 5%, is a distractor not needed in the solution.)

8-6f = 6% per year; ir = 9% per year; b = 0

Alternative A: Estimates are in actual dollars, so the combined (market) interest rate must be used to compute the present worth (PW).

im = ir + f + (ir)(f) = 0.09 + 0.06 + (0.09)(0.06) = 0.1554 or 15.54% per year

PW(15.54%)= − $120,000(P/F,15.54%,1) − $132,000(P/F,15.54%,2)

− $148,000(P/F,15.54%,3) − $160,000(P/F,15.54%,4)

= − $120,000(0.8655) − $132,000(0.7491) − $148,000(0.6483)

− $160,000(0.5611)

= − $388,466

Alternative B: Estimates are in real dollars, so the real interest rate must be used to compute the present worth (PW).

PW(9%)= − $100,000(P/A,9%,4) − $10,000(P/G,9%,4)

= − $100,000(3.2397) − $10,000(4.511)

= − $369,080

Alternative B has the least negative equivalent worth in the base time period (a PW value in this case since b = 0).

8-7(a)$6.00 = $50(P/F, i′, 50), or i′ = −4.33% (annual average loss in purchasing power).

(b)$2,000 = $50(F/P, i*, 50), or i* = 7.65% (which is the market interest rate). The real rate earned on the investment in stocks is (7.65% − 4.33%)/1.0433 = 3.18%.

8-8AlternativePresent Worth of Costs

I$10,000= $10,000

II$7,000 + $5,000(P/F,8%,6)= $10,151

III$5,000 + $5,000(P/F,8%,3) + $5,000(P/F,8%,6)= $12,120

Therefore, Alternative I is most economical with respect to the objective of

minimizing equivalent cost.

8-9The engineer's salary has increased by 6.47%, 7.18%, and 6.96% in years 2, 3, and 4, respectively. These are annual rates of change. By using Equation 8-1, but with each year's general price inflation taken into account separately, the R$ equivalents in year 0 dollars are calculated as follows.

8-10Tennessee index-adjusted salary = (95/132) × $70,000 = $50,379 per year

Tennessee actual salary = (1.00 − 0.11) × $70,000 = $62,300 per year

“Savings” = $62,300 − $50,379 = $11,921 per year

FW(10%) = $11,921(F/A, 10%, 5) = $72,779. Paul is not penalized at all!

8-11R$= $500M(1.60) = $1,300M

A$10 = $1,300M = $800M(1+f)10

1.625 = (1+f)10

f = − 1 = 0.0497 or 4.97%

8-12After-tax nominal return per year = 6% (1 – 0.33) = 4%

Approximate real return per year = 4% − 3% = 1% each year.

F (in today’s purchasing power) = $100,000(F/P, 1%, 10) = $110,460

8-13F = $100,000(F/P, 10%, 10) = $259,370Taxable Earnings = $159,370

After-tax F = $159,370(1 – 0.33) + $100,000 = $206,778

F (in today’s purchasing power) = $206,778 (P/F, 3%, 10) = $153,864

Your younger brother is ahead by about $43,400 with his concern over performance (total return) rather than risk avoidance (safety).

8-14Unit cost (8 years ago) = $89 / ft2SB = 80,000 ft2

X = 0.92SA = 125,000 ft2

eC = 5.4% per yearim = MARRc = 12% per year

eAE = 5.66% per yearf = 7.69% per year

(a)Using the power sizing technique (exponential cost estimating model) from Section3.4.1, with an adjustment for the price increase in construction costs, we have:

CA = CB (1 + eC)8

= ($89/ft2)(80,000 ft2) (1.054)8

= $16,350,060

Total Capital Investment = $16,350,060 (1 + 0.05 + 0.042 + 0.08 + 0.31)

= $24,230,790

(b)Note: The building is not being sold at the end of the 10 years. Therefore, working capital is not considered to be recovered at that time.

PW(12%) = −$24,230,790–($5)(125,000 ft2)

= −$24,230,790 –

= −$28,584,102

(c) or 4% per year

Assuming the base year to be the present (b = 0), we have:

AW(4%) = −$28,584,102 (A/P,4%,10) = −$3,524,420

8-15(a)$10,000(6.07) = $60,700

(b)R$20 = $60,700(P/F, 6%, 20) = $18,926.50

(c)$10,000(1 + ir)20 = $18,926.50

ir = 3.24%

(d)P2 = $10,000(1 – 0.18)(1 – 0.33) = $5,658

$5,658(1 + im)18 = $60,700

im = 14.1%

8-16In this problem, b = 2009 (i.e., the purchasing power of a real dollar is defined by the fiscal year 2009 dollar), and f = 6.0% per year.

For years 2007 and 2008, more 2009 dollars are required than the actual dollar amounts spent in those two years. This is because the 2009 dollar has less purchasing power than actual dollars in 2007 and 2008. Also, the entries in Column 4 indicate that in real dollars, the annual budget amounts decrease between 2008 and 2011. This contrasts with the actual dollar amounts in Column 2, which increase during this period. This difference reflects that the actual dollar amounts after 2008 increase at an annual rate less than the general price inflation rate and purchasing power is decreasing each year.

8-17In 10 years, the investor will receive the original $10,000 plus interest that has accumulated at 7.5% per year, in actual dollars. Therefore, the market rate of return (IRRc) is 7.5%.

Then, based on Equation 8-5, the real rate of return (IRRr) is:

ir' = = 0.0437, or 4.37% per year.

8-18(a)Lump sum interest in 2010= ($2.4 billion/5)(F/A, 10%, 5) – 2.4 billion

= $530,448,000

(b)C2005 = ($2.4 billion)= $3.12 billion

(c)C2015 = ($3.12 billion)(F/P, 9.2%, 10) = $7.52 billion

8-19(a)R$28 = $690(P/F, 3.2%, 28) = $285.64

(b)$850 = $690(1.032)N;N = 6.62 years

(c)$285.56 = $850(1 + ir)28; ir = −3.82%

This was not a good investment to have mad in January of 1980.

8-20(a) Cost in year 2020 = $15,000 (F/P,6%,15) = $35,949

Cost in year 2021 = $38,106

Cost in year 2022 = $40,392

Cost in year 2023 = $42,816

Total (un−discounted dollars) = $157,263

(b)Because im = , P2020 = 4($35,949) = $143,796

so A = $143,796 (A/P,0.5%,156 months)

= $143,796 (0.00925)

= $1,330 per month

8-21(a)R$ = $9.00(1.025)25 = $16.69 per thousand cubic feet

(b)A$ = $16.69(1.03)25 = $34.95 per thousand cubic feet

8-22(a)From Equation (8-1), we see that real dollars as of year k = 0 (today) is $1,107,706(P/F, 3%, 60) = $187,978 which is still a tidy sum of purchasing power.

(b)When f = 2% per year, we have R$0 = $1,107,706(P/F, 2%, 60) = $337,629. The impact of inflation is clear when you compare the results of Parts (a) and (b).

8-23(a)Cost in 10 years = = $3,700

(b)Left to student.

8-24You’ve got to be kidding! You have foregone 10% per year earnings on your money to save 5% per year on postage stamps? Give me a break. The U.S. Postal Service might just get away with this ruse. Only time will tell.

8-25Based on Equation (4-28), we have

PW = $2.5 billion = $48.88 billion and AW = $3.67 billion.

With inflation considered in this problem, the taxpayers can afford to increase the subsidy for the F-T technology.

8-26 $20,000 = $10,000(F/P,im,12) or im = 0.0595 (5.95%) per year.

ir= (im−f)/(1+f) = (0.0595−0.03)/(1.03) = 0.0286408 or 2.9% per year. This is a

reasonable return in real terms. Historically, real returns have been around

2−3% per year.

8-27MARR = im = 20% per year; Assume f = 6% per year; Let k = 0.

Note that the estimated cash flows are in R$ except for the contract maintenance agreement ($6,500 / year). However, the PW of $6,500 per year at im = 20% per year is equal to the PW of the R$ equivalent at ir. Therefore, the PW of the cash flows as a function of N is:

PW = −$60,000 + $23,000 (P/A, ir , N) − $6,500 (P/A, 20%, N)

and,

or 13.2075% per year

The life of the computer system must be at least 5 years for it to be economically justified.

8-28Device A:

Device B:

Device A should be selected to maximize after-tax present worth.

8-29 One approach would be to find the minimum value of N by trial and error.

At N = 15 years, the cost of an RA dose is $625.28 and the cost of a diabetes inhaler use is $97.64. The ratio of RA to diabetes is 6.40, so N = 15 years is correct.

8-30Option 1: Software with 3 year upgrade agreement.

Year

PW1(20%) = −$X + $0.1133X(P/A,20%,3)

Year

PW2(20%)= −$13,200(P/F,20%,1) − $14,520(P/F,20%,2) − $15,972(P/F,20%,3)

= −$30,326

Set PW1 = PW2 and solve for X.

−$X + $0.1133X(P/A,20%,3) = −$30,326

−$0.761X = −$30,326

X = $39,836

Therefore, $39,836 could be spent for software with a 3 year upgrade agreement (i.e., Option 1).

8-31$1 / 0.55 pound = $1.82 per pound

$1.82 per pound / 1.4 euro per pound = $1.30 per euro, or 1 U.S. dollor will buy 0.77 euro.

8-32In 2005 there was parity between the U.S. dollar and the Real. But in 2010 one Real is worth $0.33 U.S., so the investment is now worth $66.67 million and the bank has suffered a major loss. Conventional wisdom would be to cut the losses instead of chasing bad money with good money (i.e. sell out).

8-33 (a)In two years: $1 (1.026)2 = 6.4X

or, $1 = 6.4X / (1.026)2 = 6.08 units of X.

(b)In three years: $1 = (6.4X) (1.026)3

= 6.91 units of X.

8-34(a)The value of 0.5 pound Sterling is 90 cents, so 5 cents can be saved on each item purchased in U.S. dollars.

(b)100,000 items × $0.05 = $5,000 can be saved by purchasing in the U.S.

8-35iUS = 18% per year.

(a) fe = 10% per year.

ifm = 0.18 + 0.10 + (0.18)(0.10) = 0.2980, or 29.8% per year

(IRR on project in Country A currency)

(b)fe = −10% per year.

ifm = 0.18 + (− 0.10) + (0.18)(− 0.10) = 0.0620, or 6.20% per year

(IRR on project in Country B currency)

8-36Left to student.

8-37(a)

Project is not economically acceptable.

(b)IRRfc in terms of T−marks:

PW(i'%) = −3,600,000 + 450,000 (P/F, i'%, 1)

+ 1,500,000 (P/A, i'%, 6) (P/F, i'%, 1)

By linear interpolation, i'% = IRRfc = 0.2798, or 28.0% per year.

(c) From Equation 8-7, we have:

(IRR)US = = = 0.1429, or 14.29% < 18%

Note: This confirms our recommendation in part (a).

8-38500 euros × $1.40 = $700. The cost in U.S. dollars is $700 + $500 = $1200. Andrea may think she is paying too much for the jewelry, but she goes ahead with the purchase anyway. She could have converted her 500 euros into U.S. dollars, but there is a $.10 per euro transaction fee. Or she could have kept her 500 euros for the next trip she makes to Europe and simply charge the purchase to her credit card. The $25.00 premium ($1,200 vs. $1,175) is less costly than selling the euros and paying $50 for the transaction if there is a likelihood that she would not be wanting to use euros in the relatively near future and would be selling them and paying the commission anyway.

8-39ifm = 18% per year; fe = −2.5% per year.

Current exchange rate = $1 per 92 Z−Krons

= 21.03%

PW(21.03%) = −$175,000,000 − $37,000,000(P/F,21.03%,1)

+ $65,000,000(P/A,21.03%,10)

= −$175,000,000 − $37,000,000(0.82627) + $65,000,000(4.05063)

= $57,718,960 > 0

Yes, this project will meet the company's economic decision criteria.

8-40(a)

(b) im = (1.05)(1.09524) – 1 = 0.15 = 15% per year

PW = −$50,640(P/F,15%,1) − $38,904(P/F,15%,2) − $33,194(P/F,15%,3)

− $33,514(P/F,15%,4) − $33,865(P/F,15%,5) − $34,252(P/F,15%,6)

= −$146,084

EUAC = $146,084(A/P,15%,6) = $38,595

8-41 Demand = 400 million BTU/year; Efficiency = 85%, N = 15 years

f= 8% per year; b = 0

MARR = im = 15% per year

Annual gas demand = = 470,588 ft3 of gas

A1 = =

PW(18%)= − (470,588 ft3)

= − (470,588 ft3) (8.716237)

= −$39,869

8-42 (a) Cost of compressor replacement at EOY 8 (A$) = $500(1 + 0.06)8 = $797

Annual maintenance expense: A1(A$) = $100 (1.06) = $106

Annual electricity expense: A1(A$) = $680(1.10) = $748

PW(15%) = −$2,500 − $797 (P/F,15%,8)

−$106

−$748

= $2,500 − $797 (0.3269) −−

= −$10,871

A$: AW(15%) = −$10.871(A/P,15%,15) = −$1,859

(b) ir = = 0.085or 8.5% per year

= −$10,871 (0.1204) = −$1,309

8-43dk = ($150,000−0)/3 = $50,000

For discounting purposes, im = 26% would be used since the ATCFs are expressed in actual dollars.

8-44Annual revenues in year k (A$) = $360,000(1.025)k

Annual expenses in year k (A$) = −$239,000(1.056)k

(a)The values in the following table are expressed in A$.

PW(10%) = (P/F,10%,k) = $136,557

Total investment that can be afforded (including new equipment) =

$136,557 + $220,000 = $356,557

(b)ATCFk (R$) = ATCFk (A$)(P/F,4.9%,k)

8-45Purchase (A$ Analysis):

Notes:

a (MV)6 = $90,000 (1.02)6 = $101,355

b (O,I,OE)k = $26,000 (1.06)k

c (Maint)k = $32,000 (1.09)k

d Cost Basis = $600,000

im = 0.13208 + 0.06 + (0.13208)(0.06) = 0.20, or 20% per year

FW6 (A$) = (F/P, 20%, 6 − k) = − $1,823,920

8-45continued

Lease (A$ Analysis):

FW6 (A$) = (F/P, 20%, 6 − k) = − $1,952,551

Therefore choose Purchase Alternative due to smaller FW of costs.

8-46 Assuming that EOY 1 cost for purchased components is $85,000,000

i(r) = 0.178947368

(P/A,i(r),5) = 3.134641881

PW(w) = −20,000,000 – (85,000,000/(1−.05))[P/A,((0.12+0.05)/(1−0.05)),5]

= −300,467,957.81

PW(w/o) = −85,000,000(P/A,12%,5) = −306,405,977.20

PW(Difference) = $5,938,019.39

AW(Difference) = $1,647,264.37

Assuming that EOY 1 cost for purchased components is $85,000,000 (1−0.05)

i(r) = 0.178947368

(P/A,i(r),5) = 3.134641881

PW(w) = −20,000,000 – 85,000,000[P/A,((0.12+0.05)/(1−0.05)),5]

= −286,444,559.92

PW(w/o) = −85,000,000(P/A,12%,5) = −306,405,977.20

PW(Difference) = $19,961,417.28

AW(Difference) = $5,537,491.42

8-47Left to student.

8-48This is intended to be a tailor-made exercise (at the discretion of the instructor).

Assumptions:

 Salary and fringe benefits for the new analyst will be $28,000 (1.3) = $36,400 in year 1 purchasing power. This increases 6% per year thereafter.

 Staff retirements occur at the end of the year. Therefore, there are no realized savings in year 1. Savings of $16,200 in year 2, $32,400 in year 3, and $48,600 each year thereafter are expressed in real purchasing power keyed to year 0.

 First−year savings on purchases are 3% of $1,000,000 (1.10) = $33,000 and this increases by 10% per year thereafter.

 Contingency costs will not be considered as cash flows until they are spent (we assume they won’t be spent).

 The effective income tax rate is = 38%.

 There is no market value at the end of the 6−year project life.

PW(15%) = (P/F, 15%, k) = $10,263

In view of MARR of 15%, this investment should be undertaken. The instructor may wish to ask the class to explore various “what if” questions involving changes in the assumptions listed above. For example, how much change would occur in the PW value if we assume staff retirements occur at the beginning of the year?)

Solutions to Spreadsheet Exercises

8-49See P8-49.xls.

Typical solution for (*) is (P/F, 4%, 10) = 0.6756, so 32.44% of purchasing power has been lost due to inflation. This rounds to −32%.

8-50See P8-50.xls.

8-51 f = 4.5% per year; im (after−tax) = 12% per year; t = 40%; b = 0

increase rate = 6 % per year (applies to annual expenses, replacement costs, and market value)

Analysis period = 20 years; Useful life = 10 years

MACRS (GDS) 5−year property class

Capital investment (and cost basis, B)= −$260,000

Market value (at end of year 10) in year 0 dollars = $50,000

Annual expenses (in year 0 dollars) = −$6,000

Annual property tax = 4% of capital investment (does not inflate)

Assume like replacement at end of year 10.

* ATCF(R$) = ATCF(A$)  1/(1.045)k

ir = = 0.0718 or 7.18% per year

PW =(A$) (P/F, 12%, k) =(R$) (P/F, 7.18%, k) = −$359,665

Solutions to Case Study Exercises

8-52PW(maintenance costs)

where A1 = $1,000 for the induction motor and A1 = $1,250 for the synchronous motor.

PW(electricity costs)

where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor.

PWTC(induction motor) =$17,640 + $1,000(4.29785) + $50,552(4.37834)

=$17,640 + $4,298 + $221,334

=$243,272

PWTC(synchronous motor)=$24,500 + $1,250(4.29785) + $55,950(4.37834)

=$24,500 + $5,372 + $244,968

=$274,840

PWTC(A)=(3)($243,272) + [$17,640 + $4,298 + (350/400)($221,334)]

=$945,421

PWTC(B)=(3)($274,840) + [$24,500 + $5,372 + (50/500)($244,968)]

=$878,889

PWTC(C)=(3)($243,272) + [$24,500 + $5,372 + (350/500)($244,968)]

=$931,166

PWTC(D)=(3)($274,840) + [$17,640 + $4,298 + (50/400)($221,334)]

=$874,125

Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor.

8-53PW(electricity costs)

where A1 = $50,552 for the induction motor and A1 = $55,950 for the synchronous motor.

PWTC(induction motor) =$17,640 + $1,000(6.5136) + $50,552(6.9435)

=$17,640 + $6,514 + $351,008

=$375,162

PWTC(synchronous motor)=$24,500 + $1,250(6.5136) + $55,950(6.9435)

=$24,500 + $8,142 + $388,489

=$421,131

PWTC(A)=(3)($375,162) + [$17,640 + $6,514 + (350/400)($351,008)]

=$1,456,772

PWTC(B)=(3)($421,131) + [$24,500 + $8,142 + (50/500)($388,489)]

=$1,334,884

PWTC(C)=(3)($375,162) + [$24,500 + $8,142 + (350/500)($388,489)]

=$1,430,070

PWTC(D)=(3)($421,131) + [$17,640 + $6,514 + (50/400)($351,008)]

=$1,331,423

Option (D) has the lowest present worth of total costs. Thus, the recommendation is still to power the assembly line using three 500 hp synchronous motors operated at a power factor of 1.0 and one 400 hp induction motor.

8-54The following four options will be considered:

(A)Four induction motors (three at 400 hp, one at 100 hp)

(B)Three synchronous motors (two at 500 hp, one at 300 hp)

(C)Three induction motors at 400 hp plus one synchronous motor at 100 hp

(D)Two synchronous motors at 500 hp plus one induction motor at 300 hp.

PWTC(A)=(3)($364,045) + [$17,640 + $6,514 + (100/400)($339,891)]

=$1,201,262

PWTC(B)=(2)($408,827) + [$24,500 + $8,142 + (300/500)($376,185)]

=$1,076,007

PWTC(C)=(3)($364,045) + [$24,500 + $8,142 + (100/500)($376,185)]

=$1,200,014

PWTC(D)=(2)($408,827) + [$17,640 + $6,514 + (300/400)($339,891)]

=$1,096,726

Option (B) has the lowest present worth of total costs. Thus, the recommendation is to power the assembly line using three 500 hp synchronous motors: two operated at 500 hp and one at 300 hp.

Solutions to FE Practice Problems

8-55 Expected cost of Machine in 2008 = $3000(1.08)3= $3,779.14

True percentage increase in cost = x 100% = 19.07%

Select (c)

8-56 ir = 0.07; f = 0.09

im = 0.07 +0.09 + (0.07)(0.09) = 0.1663

Select (a)

8-57By using Equation (8-1), we have R$ = $1(P/F, 2.5%, 47) = $0.3133.

Select (a)

8-58 A$ Analysis: im – 0.098 + 0.02 + (0.098)(0.02) = 0.12 or 12%

PWA(12%) = −$27,000 + $4,000(P/A,12%,5) = −$2,581

PWB(12%) = −$19,000 + $5,000(P/A,12%,5) = −$976

Neither alternative is acceptable.

Select (c)

8-59 A1 = $35,000 (1.025) = $35,875

PW =

=

= $1,025,000

Select (b)

8-60($1.60 Canadian/Euro)($0.75 US/$1.00 Canadian) = $1.20 US/Euro

Select (c)

8-61$1 U.S. = 0.6767 Euro

$111 U.S. =111 (0.6767) = 75.11 Euro

Select (c)

1

Engineering Economy, Fourteenth Edition, by Sullivan, Wicks, and Koelling. ISBN 0-13-208342-6

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