basics of engineering economy - every_third_solution

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Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition

Leland Blank and Anthony Tarquin

Chapter 1

Foundations of Engineering Economy

1.1 If the alternative that is actually the best one is not even recognized as an

alternative, it obviously will not be able to be selected by using any economic

analysis tools.

1.4 The analysis techniques that are used in engineering economic analysis are only

as good as the accuracy of the cash flow estimates.

1.7 In engineering economy, the evaluation criterion is financial units (dollars, pesos,

etc).

1.10 Time value of money means that there is a certain worth in having money and that

worth changes as a function of time.

1.13 The term that describes compensation for “renting” of money is time value of

money, which manifests itself as interest.

1.16 Minimum attractive rate of return is the lowest rate of return (interest rate) on a

project that companies or individuals consider to be high enough to induce them

to invest their money.

1.19 (a) Equivalent cost in 1 year = 38,000 + 38,000(0.10)

= $41,800

(b) Since $41,600 is less than $41,800, the firm should remodel 1ater (i.e. 1 year

from now).

1.22 Rate of return = (45/966)(100%)

= 4.7%

1.25 Rate of return = (2.3/6)(100%)

= 38.3%

1.28 Amount of earnings in year one = 400,000,000(0.25)

= $100,000,000

1.31 The engineer is wrong, unless the MARR is exactly equal to the cost of capital.

Usually, the inequality ROR ≥ MARR > cost of capital is used, and the MARR is

established higher than the cost of capital so that profit, risk and other factors are

considered.

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1.34 F = P + Pni

100,000 = 1000 + 1000(n)(0.1)

99,000 = 1000(n)(0.10)

n = 990 years

1.37 P(1.20)(1.20) = 20,000

P = $13,888.89

1.40 F = P + P(n)(i)

3P = P + P(n)(0.20)

n = 10 years

1.43 All engineering economy problems will involve i and n

1.46 P = $50,000; F = ?; i = 15%; n = 3

1.49 F = $400,000; n = 2; i = 20% per year; P = ?

1.52 P = $16,000,000; A = $3,800,000; i = 18% per year; n = ?

1.55 The difference between cash inflows and cash outflows is known as net cash

flow.

1.58 Assuming down is negative: down arrow of $40,000 in year 5; up arrow in year 0

identified as P =?; i = 15% per year.

1.61 (a) FV is F (b) PMT is A (c) NPER is n (d) IRR is i (e) PV is P

1.64 For built-in spreadsheet functions, a parameter that does not apply can be left

blank when it is not an interior one. For example, if no F is involved when using

the PMT function, it can be left blank because it is an end parameter. When the

parameter involved is an interior one (like P in the PMT function), a comma must

be put in its position.

1.67 (a) Assuming that Carol’s supervisor is a trustworthy and ethical person himself,

going to her supervisor and informing him of her suspicion is probably the

best of these options. This puts Carol on record (verbally) as questioning

something she heard at an informal gathering.

(b) Another good option is to go to Joe one-on-one and inform him of her

concern about what she heard him say at lunch. Joe may not be aware he is on

the bid evaluation team and the potential ethical consequences if he accepts

the free tickets from Dryer.

1.70 Answer is (a)

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1.73 F = P(1+i)n

16,000 = 8000(1 + i)9

21/9 = 1 + i

1.08 = 1 + i

i = 0.08 (8%)

Answer is (b)

1.76 2P = P + P(n)(0.05)

1 = 0.05n

n = 20

Answer is (d)

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Chapter 2

Factors: How Time and Interest Affect Money

2.1 (a) (F/P,10%,20) = 6.7275

(b) (A/F,4%,8) = 0.10853

(c) (P/A,8%,20) = 9.8181

(d) (A/P,20%,28) = 0.20122

(e) (F/A,30%,15) = 167.2863

2.4 (a) F = 885,000 + 100,000(F/P,10%,3)

= 885,000 + 100,000(1.3310)

= $1,018,000

(b) Spreadsheet function is = -FV(10%,3,,100000) + 885000

Display is $1,018,000

2.7 (a) F = 3000(F/P,10%,12) + 5000(F/P,10%,8)

= 3000(3.1384) + 5000(2.1436)

= $20,133.20

(b) Sum two calculator functions

FV(10,12,,-3000) + FV(10,8,-5000)

9,415.29 + 10,717.94 = $20,133.23

(c) If the spreadsheet function is = – FV(10%,12,,3000) – FV(10%,8,,5000), the

display is $20,133.23

2.10 A = 12,700,000(A/P,20%,8)

= 12,700,000(0.26061)

= $3,309,747

2.13 A = 20,000,000(A/P,10%,6)

= 20,000,000(0.22961)

= $4,592,200

2.16 (a) A = 3,000,000(10)(A/P,8%,10)

= 30,000,000(0.14903)

= $4,470,900

(b) If calculator function is PMT(8,10,-30000000,0), the answer is $4,470,884.66

(c) If the spreadsheet function is = -PMT(8%,10,30000000), display is

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A = $4,470,884.66

2.19 (a) A = 225,000(A/P,15%,4)

= 225,000(0.35027)

= $78,811

(b) Recall amount = 78,811/0.10

= $788,110 per year

2.22 P = (280,000-90,000)(P/A,10%,5)

= 190,000(3.7908)

= $720,252

2.25 P = 2,100,000(P/F,10%,2)

= 2,100,000(0.8264)

= $1,735,440

2.28 P = 95,000,000(P/F,12%,3)

= 95,000,000(0.7118)

= $67,621,000

2.31 (a) P = 7000(P/F,10%,2) + 9000(P/F,10%,4) + 15,000(P/F,10%,5)

= 7000(0.8264) + 9000(0.6830) + 15,000(0.6209)

= $21,245.30

(b) Three calculator functions are added.

-PV(10,2,0,7000) – PV(10,4,0,9000) – PV(10,5,0,15000)

Total is 5785.12 + 6147.12 + 9313,82 = $21,246.06

2.34 A = 10,000,000(A/P,10%,10)

= 10,000,000(0.16275)

= $1,627,500

2.37 A = 3,250,000(A/P,15%,6)

= 3,250,000(0.26424)

= $858,780

2.40 A = 5000(7)(A/P,10%,10)

= 35,000(0.16275)

= $5696.25

2.43 (a) Let CF4 be the amount in year 4

100,000(F/P,9%,3) + 75,000(F/P,9%,2) + CF4(F/P,9%,1) = 290,000

100,000(1.2950) + 75,000(1.1881) + CF4(1.0900) = 290,000

(1.09)CF4 = 71.392.50

CF4 = $65,497.71

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(b) F in year 5 for 2 known amounts

= -FV(9%,3,0,100000) - FV(9%,2,0,75000)

P in year 4 of $290,000 minus amount above (assume it’s in cell H9)

= -PV(9%,1,0,290000-H9)

Answer is $65,495.05

2.46 F = P(F/P,10%,n)

3P = P(F/P,10%,n)

(F/P,10%,n) = 3.000

From 10% interest tables, n is between 11 and 12 years. Therefore, n = 12 years

2.49 (a) P = 26,000(P/A,10%,5) + 2000(P/G,10%,5)

= 26,000(3.7908) + 2000(6.8618)

= $112,284

(b) Spreadsheet: enter each annual cost in adjacent cells and use the NPV

function to display P = $112,284

Calculators have no function for gradients; use the PV function on each cash

flow and add the five P values to get $112,284.55

2.52 A = 9000 – 560(A/G,10%,5)

= 9000 – 560(1.8101)

= $7986

2.55 A = 100,000 + 10,000(A/G,10%,5)

= 100,000 + 10,000(1.8101)

= $118,101

F = 118,101(F/A,10%,5)

= 118,101(6.1051)

= $721,018

2.58 475,000 = 25,000(P/A,10%,6) + G(P/G,10%,6)

475,000 = 25,000(4.3553) + G(9.6842)

9.6842G = 366,117.50

G = $37,805.65

2.61 A = 7,000,000 - 500,000(A/G,10%,5)

= 7,000,000 - 500,000(1.8101)

= $6,094,950

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2.64 P = (23,000) 1 – (1.02/1.10)5

(0.10 – 0.02)

= $90,405

2.67 First find P and then convert to A. (in million-people units)

P = 15,000(10)[1 – (1.15/1.08)5]/(0.08 – 0.15)

= $790,491,225,000

A = 790,491,225,000(A/P,8%,5)

= 790,491,225,000(0.25046)

= $197.986 billion (spreadsheet answer is $197,983,629,604)

2.70 Solve for P in geometric gradient equation and then convert to A

A1 = 5,000,000(0.01) = 50,000

P = 50,000[1 – (1.10/1.08)5]/(0.08 – 0.10)

= $240,215

A = 240,215(A/P,8%,5)

= 240,215(0.25046)

= $60,164

2.73 Solve for A1 in geometric gradient equation and then find cost in year 3

400,000 = A1[1 – (1.04/1.10)5]/(0.10 – 0.04)

4.0759 A1 = 400,000

A1 = $98,138

Cost in year 3 = 98,138(1.04)2

= $106,146

2.76 Since 4th deposit is known to be $1250, increase it by 5% each year to year one

A1 = 1250/(0.95)3

= $1457.94

2.79 F = 200,000(F/A,10%,6)

= 200,000(7.7156)

= $1,543,120

2.82 F in year 8 = 100(F/A,10%,3)(F/P,10%,6) + 200(F/A,10%,4)(F/P,10%,2)

= 100(3.3100)(1.7716) + 200(4.6410)(1.21)

= $1709.52

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2.85 Find the future worth Fpaid of 3 payments in year 4

Fpaid = 2,000,000(F/A,8%,3)(F/P,8%,1)

= 2,000,000(3.2464)(1.08)

= $7,012,224

Find total amount owed Fowed after 4 years

Fowed = 10,000,000(F/P,8%,4)

= 10,000,000(1.3605)

= $13,606,000

Due in year 4 = 13,606,000 - 7,012,224

= $6,593,776

2.88 Find P in year 0 then convert to F. In $ million units,

P0 = 450 – 40(P/F,10%,1) + 200(P/A,10%,6)(P/F,10%,1)

= 450 – 40(0.9091) + 200(4.3553)(0.9091)

= $1205.52

F7 = 1205.52(F/P,10%,7)

= 1205.52(1.9487)

= $2349.20

2.91 Factors: (a) P = 31,000(P/A,8%,3) + 20,000(P/A,8%,5)(P/F,8%,3)

= 31,000(2.5771) + 20,000(3.9927)(0.7938)

= $143,278

(b) A = 143,278(A/P,8%,8)

= 143,278(0.17401)

= $24,932

Spreadsheet:

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2.94 In $ billion units,

Gross revenue first 2 years = 5.8(0.701) = $4.0658

Gross revenue last 2 years = 6.2(0.701) = $4.3462

F = 4.0658(F/A,14%,2)(F/P,14%,2) + 4.3462(F/A,14%,2)

= 4.0658(2.1400)(1.2996) + 4.3462(2.1400)

= $20.6084 billion

2.97 First find F in year 8 and then solve for A

F8 = 15,000(F/A,8%,7) + 10,000(F/A,8%,4)

= 15,000(8.9228) + 10,000(4.5061)

= $178,903

A = 178,903(A/F,8%,8)

= 178,903(0.09401)

= $16,819

2.100 Find P in year -1for geometric gradient, than move to year 0 to find P

P-1 = (30,000) 1 – (1.05/1.10)8

(0.10 – 0.05)

= $186,454

F = P0 = 186,454(F/P,10%,1)

= 186,454(1.10)

= $205,099

2.103 Find P in year –6 using arithmetic gradient factor and then find F today

P-6 = 10,000(P/A,12%,6) + 1000(P/G,12%,6)

= 10,000(4.1114) + 1000(8.9302)

= 41,114 + 8930.20

= $50,044.20

F = 50,044.20(F/P,12%,6)

= 122,439(1.9738)

= $98,777

2.106 (a) Add and subtract $2400 and $2600 in periods 3 and 4, respectively, to use

gradient

30,000 = 2000 + 200(A/G,10%,8) – 2400(P/F,10%,3)(A/P,10%,8)

-2600(P/F,10%,4)(A/P,10%,8) + x(P/F,10%,3)(A/P,10%,8)

+ 2x(P/F,10%,4)(A/P,10%,8)

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30,000 = 2000 + 200(3.0045) – 2400(0.7513)( 0.18744)

-2600(0.6830)( 0.18744) + x(0.7513)(0.18744)

+ 2x(0.6830)( 0.18744)

30,000 = 2000 + 600.90 – 337.98 – 332.86 + 0.14082x + 0.25604x

0.39686x = 28,069.94

x = $70,730

(b) Spreadsheet uses Goal Seek to find x = $70,726

2.109 (a) Find P in year 4 for the geometric gradient, (b) Spreadsheet

then move all cash flows to future

P4 = 500,000[1 – (1.15/1.12)16]/(0.12 – 0.15)

= $8,773,844

F = 500,000(F/A,12%,4)(F/P,12%,16) + P4(F/P,12%,16)

= 500,000(4.7793)(6.1304) + 8,773,844(6.1304)

= $68,436,684

2.112 Answer is (a)

2.115 A = 10,000,000((A/P,15%,7)

= $2,403,600

Answer is (a)

2.118 F = 50,000(F/P,18%,7)

= 50,000(3.1855)

= $159,275

Answer is (b)

2.121 10,000 = 2x(P/F,10%,2) + x(P/F,10%,4)

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10,000 = 2x(0.8264) + x(0.6830)

2.3358x = 10,000

x = $4281

Answer is (a)

2.124 1000(F/P,10%,20) + 1000(F/P,10%,n) = 8870

1000(6.7275) + 1000(F/P,10%,n) = 8870

1000(F/P,10%,n) = 2142.5

(F/P,10%,n) = 2.1425

n = 8

Deposit year = 20 - 8 = 12

Answer is (d)

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Chapter 3

Nominal and Effective Interest Rates

3.1 (a) Nom i/semi = 0.02*2 = 4% (b) Nom i/year = 0.02*4 = 8%

(c) Nom i/2 years = 0.02*8 = 16%

3.4 (a) year; (b) quarter; (c) day; (d) continuous (period length is zero); (e) hour

3.7 (a) r = 2% per quarter

(b) r = 2(2/3 quarter) = 1.33% per 2 months

(c) r = 2(8 quarters) = 16% per 2 years

(d) r = 2(4 quarters) = 8% per year

(e) r = 2(2 quarters) = 4% per semiannual period

3.10 (a) i/week = 6.8/26 = 0.262%; (b) effective

3.13 PP = day; CP = quarter

3.16 (a) Need effective i/quarter: i/quarter = 12%/4 = 3%

(b) Need effective i/semi: i/semi = (1 + 0.06/2)2 -1

= 0.0609 (6.09%)

(c) Need effective i/year: i/year = (1 + 0.12/4)4 -1

= 0.1255 (12.55%)

3.19 In $ million units,

F = (54 +14 + 10)(F/P,1.5%,12)

= 78(1.1956)

= $93.2568 ($93,256,800)

3.22 F = 242,000(F/P,1.5%,48)

= 242,000(2.0435)

= $494,527

3.25 P = 190,000(P/F,2%,8) + 120,000(P/F,2%,16)

= 190,000(0.8535) + 120,000(0.7284)

= $249,573

3.28 F = 18,000(F/P,8%,6) + 26,000(F/P,8%,5) + 42,000(F/P,8%,4)

= 18,000(1.5869) + 26,000(1.4693) + 42,000(1.3605)

= $123,907

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3.31 P = 51(100,000)(0.25)(P/A,0.5%,60)

= (1,275,000)(51.7256)

= $65,950,140

3.34 First find P; then convert to A

P = 40,000 + 40,000(P/A,12%,2) + 50,000(P/A,12%,3)(P/F,12%,2)

= 40,000 + 40,000(1.6901) + 50,000(2.4018)(0.7972)

= $203,340

A = 203,340(A/P,12%,5)

= 203,340(0.27741)

= $56,409

3.37 i/week = 0.25%

P = 2.99(P/A,0.25%,40)

= 2.99(38.0199)

= $113.68

3.40 P = 90(P/A,3%,12) + 2.50(P/G,3%,12)

= 90(9.9540) + 2.50(51.2482)

= $1023.98

3.43 Find cost of treatments after one year, then monthly equivalent A over 5 years

Cost of treatment = 10,000(F/A,1%,12)

= 10,000(12.6825)

= $126,825

A = 126,825(A/P,1%,60)

= 126,825(0.02224)

= $2821 per month

3.46 i/quarter = (1 + 0.01)3 –1 = 3.03%

P = [140,000 + 140,000(0.20)](P/A,3.03%,12)

= 168,000(9.9362)

= $1,669,282

3.49 F = 9000(F/A,1%,24)

= 9000(26.9735)

= $242,762

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3.52 (a) A = 48,000 – 2000(P/G,1.5%,7)(P/F,1.5%,5)(A/P,1.5%,12)

= 48,000 – 2000(19.4018)(0.9283)(0.09168)

= $44,698

(b)

3.55 290,000(P/F,0.5%,48) = 4000(P/A,0.5%,48) + G(P/G,0.5%,48)

290,000(0.7871) = 4000(42.5803) + G(959.9188)

959.9188G = 57,938

G = $60.36

3.58 i = e0.0125 – 1 = 1.26% per month

F = 100,000(F/A,1.26%,24)

= 100,000{[1 + 0.0126)24 –1]/0.0126}

= 100,000(27.8213)

= $2,782,130

3.61 A/3 months = 3(1500) = $4500

F = 4500(F/A,1.5%,20)

= 4500(23.1237)

= $104,057

3.64 Move chemical cost to end of interest period; find A

Chemical cost/month = 11(30) = $330

A = 1200(A/P,1%,48) + 330

= 1200(0.02633) + 330

= $361.60

3.67 Move cash flow to end of interest period (year); find P

Fuel savings = 800(0.50)(12) = $4800 per year

P = 4800(P/A,12%,3)

= 4800(2.4018)

= $11,529

3.70 Answer is (a)

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3.73 P = 30(P/A,0.5%,60)

= $1552

Answer is (b)

3.76 PP > CP; must use i over PP of 1 year. Therefore, n = 7

Answer is (a)

1



Chapter 4

Present Worth Analysis

4.1 The do-nothing alternative is not an option (1) when it is absolutely required that

one of the defined alternatives be selected (such as for legal purposes), and (2)

when each alternative has only cost cash flow estimates.

4.4 (a) A, B and C are mutually exclusive; D and E are independent.

(b) X is in all bundles as the mutually exclusive selection. The two independent

projects have 22 = 4 bundles. The 4 viable options are:

X only XD XE XDE

4.7 PW solar = -14,000 - 1500(P/A,10%,4) + 0.25(14,000)(P/F,10%,4)

= -14,000 - 1500(3.1699) + 3500(0.6830)

= $-16,364

PW line = -12,000 - 600(P/A,10%,4)

= -12,000 - 600(3.1699)

= $-13,902

Install power line; its cost is lower.

4.10 PWmanual = -425,000 – [90,000(P/A,8%,5) + 7000(P/G,8%5)] + 80,000(P/F,8%,5)

= -425,000 – [90,000(3.9927) + 7000(7.3724)] + 80,000(0.6806)

= $-781,502

PWrobotic = -850,000 –[10,000(P/A,8%,5) + 1000(P/G,8%5)] + 300,000(P/F,8%,5)

= -850,000 – [10,000(3.9927) + 1000(7.3724)] + 300,000(0.6806)

= $-693,119

Select the robotic system

4.13 Interest rate of 12% per year compounded monthly is 1% per month.

PW = -950 + 70(P/A,1%,36)

= -950 + 70(30.1075)

= $+1157.53

Since PW > 0, the service is financially justified

4.16 (a) PWA = -70,000 – 20,000(P/A,12%,3) + 15,000(P/F,12%,3)

= -70,000 – 20,000(2.4018) + 15,000(0.7118)

= $-107,359

2

PWB = -140,000 – 8,000(P/A,12%,3) + 40,000(P/F,12%,3)

= -140,000 – 8,000(2.4018) + 40,000(0.7118)

= $-130,742

Select method A

(b) A: Function is –PV(12,3,-20000,15000) – 70000 PW = $-107,360

B: Function is –PV(12,3,-8000,40000) – 140000 PW = $-130,743

4.19 (a) Monetary units are in $1000. Calculate PW values to select the pull system.

PWpull = -1500 – 700(P/A,10%,8) + 100(P/F,10%,8)

= -1500 – 700(5.3349) + 100(0.4665)

= $-5187.780 ($-5,187,780)

PWpush = -2250 – 600(P/A,10%,8) + 50(P/F,10%,8) – 500(P/F,10%,3)

= -2250 – 600(5.3349) + 50(0.4665) – 500(0.7513)

= $-5803.265 ($-5,803,265)

(b) By spreadsheet, enter the following into single cells to display the PW values.

PWpull : = - PV(10%,8,-700000,100000)-1500000

PWpush : = - PV(10%,8,-600000,50000)-2250000-PV(10%,3,,-500000)

4.22 (a) For Allison (A), use i = 1% per month and n = 60 months to calculate PW.

PWA = -40,000 - 5000(P/A,1%,60) + 10,000(P/F,1%,60)

= -40,000 - 5000(44.9550) + 10,000(0.5504)

= $-259,271

For Joshua (J), use effective semiannual i and n = 10 to calculate PW.

effective i = (1.01)6 -1 = 6.152%

PWJ = -60,000 – 13,000(P/A,6.152%,10) + 8,000(P/F,6.152%,10)

= -60,000 – 13,000(7.30737) + 8,000(0.55045)

= $-150,592

Select Joshua’s plan

3

(b) A spreadsheet solution follows; select Joshua’s plan

4.25 (a) Semiannual bond dividend is 1000(0.05)/2 = $25 per 6 months.

Semiannual interest rate is 5%/2 = 2.5%.

PW = -925 + 25(P/A,2.5%,16) + 800(P/F,2.5%,16)

= -925 + 25(13.0550) + 800(0.6736)

= $-59.74

No, the bond investment did not make the target rate since PW < 0.

(b) A spreadsheet solution follows to obtain PW = $-59.72

4.28 Semiannual bond dividend is 10,000(0.05)/2 = $250 per 6 months.

4

Semiannual interest rate expected is 6%/2 = 3%.

PW = -9000 + 250(P/A,3%,40) + 10,000(P/F,3%,40)

= -9000 + 250(23.1148) + 10,000(0.3066)

= $-155

No, the bond investment does not make the target rate since PW < 0.

A spreadsheet solution follows.

4.31 LCM is 48 months; repurchase P after 24 months

PWP = -13,650 - 200(P/A,1%,48) - 13,650(P/F,1%,24)

= -13,650 - 200(37.9740) - 13,650(0.7876)

= $-31,996

PWF = -22,900 - 50(P/A,1%,48) + 2000(P/F,1%,48)

= -22,900 - 50(37.9740) + 2000(0.6203)

= $-23,558

Select the fiber optic sensors

4.34 (a) LCM is 4 years; repurchased CFRP after 2 years

PWCFRP = -205,000 – 29,000(P/A,10%,4) – 203,000(P/F,10%,2)

+ 2000(P/F,10%,4)

= -205,000 – 29,000(3.1699) – 203,000(0.8264) + 2000(0.6830)

= $-463,320

PWFRC = -235,000 – 27,000(P/A,10%,4) + 20,000(P/F,10%,4)

= -235,000 – 27,000(3.1699) + 20,000(0.6830)

= $-306,927

Select material FRC

(b) PWCFRP function: = - 205000 - PV(10%,4,-29000,2000)+ PV(10%,2,,203000)

5

PWFRC function: = - 235000 - PV(10%,4,-27000,20000)

4.37 (a) PWC = -375,000 -200(P/A,8%,13)(P/F,8%,7)

= -375,000 – 200(7.9038)(0.5835)

= $-375,922

For asphalt, repave after 10 years and re-start maintenance charge in year 12.

PWA = -250,000[1+(P/F,8%,10)] – 2500(P/A,8%,9)[(P/F,8%,1)+(P/F,8%,11)]

= -250,000[1.4632] – 2500(6.2469)[0.9259 + 0.4289]

= $-386,958

Select the concrete option with a marginal advantage.

(b) Maintenance costs are incurred over 5 years only; there are none for concrete.

Now, select the asphalt option by a large margin.

PWC = $-375,000

PWA = -250,000 – 2500(P/A,8%,4)(P/F,8%,1)

= -250,000 – 2500(3.3121)(0.9259)

= $-257,667

A spreadsheet solution for parts (a) and (b) follows.

4.40 FWC = -80,000[(F/P,12%,15)+(F/P,12%,10)+(F/P,12%,5)+1]

= -80,000[5.4736 + 3.1058 + 1.7623 + 1]

= $-907,336

FWM = -200,000(F/P,12%,15) – 500[(F/P,12%,12)+(F/P,12%,9)+(F/P,12%,6)

6

+(F/P,12%,3)+1] +25,000

= -200,000(5.4736) - 500[3.8969 + 2.7731 + 1.9738 + 1.4049 + 1] + 25,000

= $-1,075,244

Select the concrete exterior by a future worth amount of approximately $168,000.

4.43 LCCA = -750,000 – (6000 + 2000)(P/A,0.5%,240) – 150,000[(P/F,0.5%,60)

+ (P/F,0.5%,120) + (P/F,0.5%,180)]

= -750,000 – (8000)(139.5808) – 150,000[(0.7414) + (0.5496) + (0.4075)]

= $-2,121,421

LCCB = -1.1 – (3000 + 1000)(P/A,0.5%,240)

= -1.1 – (4000)(139.5808)

= $-1,658,323

Select proposal B.

4.46 CC = - [38(120,000) + 17(150,000)]/0.08

= $-88,875,000

4.49 CC = -1,700,000 – 350,000(A/F,6%,3)/0.06

= - 1,700,000 – 350,000(0.31411)/0.06

= $-3,532,308

4.52 CC = 100,000 + 100,000/0.08

= $1,350,000

4.55 Monetary terms are in $1000 units.

CC = -200 - 300(P/F,8%,4) - 50(A/F,8%,5)/0.08 - (8/0.08)(P/F,8%,14)

= -200 - 300(0.7350) - 50(0.17046)/0.08 - (8/0.08)(0.3405)

= $-561.088 ($-561,088)

4.58 Quarterly interest rate is 12/4 = 3% with 4 quarters per year

7

CC = AW/i; select alternative E. Monetary values are in $1000 units.

CCE = [-2000(A/P,3%,16) + 300 + 50(A/F,3%,16)]/0.03

= [-2000(0.07961) + 300 + 50(0.04961)]/0.03

= $+4775.35 ($+4,775,350)

CCF = [-3000(A/P,3%,32) + 100 + 70(A/F,3%,32)]/0.03

= [-3000(0.04905) + 100 + 70(0.01905)]/0.03

= $-1527.217 ($-1,527,217)

CCG = -10,000 + 400/0.03

= $+3333.333 ($+3,333,333)

By spreadsheet, enter the following into single cells to display the CC values

Answer: Select alternative E.

E: = (-PMT(3%,16,-2000000,50000)+300000)/0.03 Display: $+4,775,295

F: = (-PMT(3%,32,-3000000,70000)+100000)/0.03 Display: $-1,526,886

G: = -10000000+400000/0.03 Display: $+3,333,333

4.61 Budget = $800,000 i = 10% 6 viable bundles

Bundle Projects NCFj0 NCFjt S PW at 10%

1 X $-250,000 $ 50,000 $ 45,000 $-60,770

2 Y -300,000 90,000 -10,000 -21,539

3 Z -550,000 150,000 100,000 -6,215

4 XY -550,000 140,000 35,000 -82,309

5 XZ -800,000 200,000 145,000 -66,985

6 Do nothing 0 0 0 0

PWj = NCFj(P/A,10%,4) + S(P/F,10%,4) - NCFj0

Since no bundle has PW > 0. Select ‘Do nothing’ project.

4.64 P = -35,000 - 20,000(P/A,10%,4) - 25,000(P/F,10%,2) + 10,000(P/F,10%,4)

= -35,000 - 20,000(3.1699) - 25,000(0.8264) + 10,000(0.6830)

= $-112,228

Answer is (d)

4.67 CC = A/i

8

A = 10,000(A/F,10%,5)

= 10,000(0.16380)

= $-1638

CC = -1638/0.10

= $-16,380

Answer is (b)

4.70 Answer is (d)

1



Chapter 5

Annual Worth Analysis

5.1 AW = -7,000,000(A/P,15%,3) - 860,000

= -7,000,000(0.43798) - 860,000

= $-3,925,860

Therefore, required revenue is $3,925,860 per year

5.4 In $ millions,

AW = [-13 - 10(P/F,15%,1)](A/P,15%,10) – 1.2

= [-13 - 10(0.8696)](0.19925) – 1.2

= $-5.5229

Revenue required is $5,522,900 per year

5.7 First find PW, then annualize over three years

PW = -150,000[(400)(P/F,6%,1) + 300(P/F,6%,2) + 600(P/F,6%,3)]

= -60,000,000(0.9434) - 45,000,000(0.8900) - 90,000,000(0.8396)

= $-172,218,000

AW = -172,218,000(A/P,6%,3)

= -172,218,000(0.37411)

= $-64,428,476

5.10 (a) Use Equation [5.3] for CR per year.

CR = -3,800,000(A/P,12%,12) + 250,000(A/F,12%,12)

= -3,800,000(0.16144) + 250,000(0.04144)

= $-603,112

(b) AW = CR + A of AOC

= -603,112 – 350,000 – 25,000(A/G,12%,12)

= -603,112 – 350,000 - 25,000(4.1897)

= $-1,057,855

(c) One way to use a spreadsheet follows:

CR: single cell entry = -PMT(12%,12,-3800000,250000)

AW: Enter increasing gradient into B3:B14 and combine PMT functions

= -PMT(12%,12,-3800000,250000) - PMT(12%,12,NPV(12%,B3:B14))

2

5.13 (a) AWplastic = -0.90(110)(43,560)(A/P,8%,20)

= -4,312,440(0.10185)

= $-439,222

AWrubber = -2.20(110)(43,560)(A/P,8%,30)

= -10,541,520(0.08883)

= $-936,403

Select the plastic liner

(b) Plastic: liner cost; 0.9(110)43560) = $4,312,440

Function: -PMT(8,20,0,-4312440) Display: AW = $-439,232

Rubber liner cost; 2.20(110)43560) = $10,541,520

Function: -PMT(8,30,0,-10541520) Display: AW = $-936,376

Select the plastic liner

5.16 Tabulated factor solution:

(a) Monetary terms in $ million. From AW, project clearly makes 10% per year.

AW = -170(A/P,10%,20) + 85(1-0.225)

= -170(0.11746) + 65.875

= $+45.90 ($+45.9 million)

(b) AW goes negative between 30% and 40% per year.

@ 20%: AW = -170(A/P,20%,20) + 65.875 = $+30.96 million

@ 30%: AW = -170(A/P,30%,20) + 65.875 = $+14.60 million

@ 40%: AW = -170(A/P,40%,20) + 65.875 = $-2.21 million

Spreadsheet solution with x-y scatter chart:

3

5.19 (a) AW evaluation indicates chamber 490G to be more economic.

AWD103 = -400,000(A/P,10%,3) + 40,000(A/F,10%,3) – 4000

= -400,000(0.40211) + 40,000(0.30211) – 4000

= $-152,760

AW490G = -250,000(A/P,10%,2) + 25,000(A/F,10%,2) – 3000

= -250,000(0.57619) + 25,000(0.47619) – 3000

= $-135,143

(b) At P = $-300,000 and S = $30,000:

AWD103 = -300,000(A/P,10%,3) + 30,000(A/F,10%,3) – 4000

= -300,000(0.40211) + 30,000(0.30211) – 4000

= $-115,570

At P = $-500,000 and S = $50,000:

AWD103 = -500,000(A/P,10%,3) + 50,000(A/F,10%,3) – 4000

= -500,000(0.40211) + 50,000(0.30211) – 4000

= $-189,950

Cheaper model, P = $-300,000 will change the decision to D103.

(c) By spreadsheet for part (b) use the PMT functions at different P values.

490G: = -PMT(10%,2,-250000,25000)-3000

Display: $-135,143

D103 @ P=-300,000: = -PMT(10%,3,-300000,30000)-4000

Display: $-115,571

D103 @ P=-500,000: = -PMT(10%,3,-500000,50000)-4000

Display: $-189,952

5.22 AW = PW(i)

AW = [5,000,000 + 2,000,000(P/F,10%,10) + 100,000/(0.10)(P/F,10%,10)](0.10)

= [5,000,000 + 2,000,000(0.3855) + 1,000,000(0.3855)](0.10)

= $615,650

5.25 (a) Determine amount needed at end of year 20, followed by A to accumulate

this future amount.

CC = P = A/i = 24,000/0.08 = $300,000

F = A(F/A,8%,20)

300,000 = A(45.7620)

A = $6556 per year

4

By spreadsheet, enter = -PMT(8%,20,,300000) to display $6556.

(b) P = 24,000(P/A,8%,30)

= 24,000(11.2578)

= $270,187

270,187 = A(F/A,8%,20)

= A(45.7620)

A = $5904 per year

By spreadsheet, enter two functions.

P after 20 years: = -PV(8%,30,24000) Display: $270,187

A over 20 years: = -PMT(8%,20,,270187) Display: $5904

(c) For 30 year payout, difference = 6556 – 5904

= $652 per year less

5.28 Factors: Perpetual AW is equal to AW over one life cycle.

AW = {[-150,000(P/A,10%,4) - 25,000(P/G,10%,4)](P/F,10%,2)

- [225,000(P/A,10%,4)(P/F,10%,6)]}(A/P,10%,10)

= {[-150,000(3.1699) – 25,000(4.3781)](0.8264)

-[225,000(3.1699)(0.5645)]}(0.16275)

= $-144,198

Spreadsheet

5

5.31 In $ million units. Effective annual i = (1.025)4 – 1 = 10.381%.

AWA = -10(A/P,10.381%,5) + 0.7(A/F,10.381%,5) – 1.8

= -10(0.26637) + 0.7(0.16256) – 1.8

= $-4.35 ($-4.35 million)

AWB = -35(0.10381) – 0.6

= $-4.23 ($-4.233 million)

Select B

5.34 Answer is (b)

5.37 Answer is (d)

5.40 Perpetual AWA = -50,000(A/P,10%,3) – 20,000 + 10,000(A/F,10%,3)

= -50,000(0.40211) – 20,000 + 10,000(0.30211)

= $-37,084

AW is same for all years, including an infinite life.

Answer is (d)

1



Chapter 6 Rate of Return Analysis

6.1 (a) The highest possible is infinity

(b) The lowest possible is -100%

6.4 In $ million units

Annual payment = 6,000,000(A/P,10%,10)

= 6,000,000(0.16275)

= $976,500

Principal remaining after year 1 = 6,000,000(1.10) – 976,500

= $5,623,500

Interest, year 2 = 5,623,500(0.10)

= $562,350

6.7 (a) Move all cash flows to year 1

0 = -80,000 + 9000(P/F,i*,1) + 70,000(P/F,i*,2) + 30,000(P/F,*i,3)

By trial and error, i* = 15.32%

(b) Enter cash flows for years 1 to 4 in cells B1:B4

Function = IRR(B1:B4) displays i* = 15.32%

6.10 (a) Tabulated factors

0 = -400,000(F/A,i*,10) + 270,000/i*

Try 5%: -400,000(12.5779) + 270,000/0.05 = $368,840 too low

Try 6%: -400,000(13.1808) + 270,000/0.06 = $-772,320 too high

Interpolation yields i* = 5.32%

(b) Solve for i* by spreadsheet using the FV function and GOAL SEEK. For

example, place a guess i* in cell B1and set up the single-cell function

= -FV($B$1,10,-400000) + 270000/$B$1

Display is i* = 5.29% per year

2

6.13 0 = -3,000,000 + 1,500,000(P/A,i*,3)

(P/A,i*,3) = 2.0000

Solve for i* = 23.38% per year (spreadsheet)

6.16 0 = -65,220(P/A,i*,4) + (57,925 – 35,220)(P/A,i*,31)(P/F,i*,4)

= -65,220(P/A,i*,4) + (22,705)(P/A,i*,31)(P/F,i*,4)

Solve by trial and error:

Try 6%: 0 = -225,994 + 250,510 = $24,516 i too low

Try 7%: 0 = -220,913 + 217,071 = $-3842 i too high

i* = 6.86% per year (interpolation or spreadsheet)

6.19 0 = 2,000,000 – 200,000(P/A,i*,2) – 2,200,000(P/F,i*,3)

Solve for i* = 10% per year (spreadsheet)

6.22 0 = -130,000 + (78,000 – 49,000)(P/A,i*,8) + 1000(P/G,i*,8) + 23,000(P/F,i*,8)

Solve for i* = 19.17 % (spreadsheet)

6.25 (a) The rate of return on the increment has to be larger than 18%

(b) The rate of return on the increment has to be smaller than 10%

(c) Two samples follow with approximate ROR values of 10% for X

and 18% for Y

6.28 (a) Incremental CF, year 0: -25,000 –(-15,000) = $-10,000

(b) Incremental CF, year 3: -400 - (-1600 – 15,000 + 3000) = $+13,200

(c) Incremental CF, year 6: (-400+6000) – (-1600+3000) = $+4200

3

6.31 This is an incremental ROR analysis to find ∆i*

0 = -250,000(F/P,∆i*,8) + 50,000(F/A,∆i*,8) + 30,000

Solve for ∆i* = 13.14% < MARR = 15% (spreadsheet)

Select P

6.34 Incremental CF amounts for (Y-X)

Incremental first cost = $-62,000

Incremental M&O = $3,000

Incremental revenue = $23,000

Incremental salvage = $7,000

0 = -62,000 + 3,000(P/A,∆i*,3) + 23,000(P/A ∆i*,3) + 7000(P/F,∆i*,3)

Solve for ∆i* by trial and error or spreadsheet

∆i* = 16.83% > MARR = 15% (spreadsheet)

Select robot Y

6.37 Alternatives are independent; compare each against DN

Product 1: 0 = -340,000 + (180,000 – 70,000)(P/A,i*,5)

i* = 18.52% > MARR = 15% Accept

Product 2: 0 = -500,000 + (190,000 – 64,000)(P/A,i*,5)

i* = 8.23% < MARR = 15% Reject

Product 3: 0 = -570,000 + (220,000 –48,000)(P/A,i*,5)

i* = 15.49% > MARR = 15% Accept

Product 4: 0 = -620,000 + (205,000 – 40,000)(P/A,i*,5)

i* = 10.35% < MARR = 15% Reject

Company should manufacture product lines 1 and 3

6.40 Rank revenue alternatives by increasing initial investment: DN, D, A, B, E, C

(a) MARR = 13.5%

DN to D: ∆i* = 15% > MARR eliminate DN

A to D: ∆i* = 12% < MARR eliminate A

B to D: ∆i* = 26% > MARR eliminate D

E to B: ∆i* = -16% < MARR eliminate E

C to B: ∆i* = 25% > MARR eliminate B

Select C

4

(b) MARR = 16%

D to DN: ∆i* = 15% < MARR eliminate D

A to DN: ∆i* = 13% < MARR eliminate A

B to DN: ∆i* = 23% > MARR eliminate DN

E to B: ∆i* = -16% < MARR eliminate E

C to B: ∆i* = 25% > MARR eliminate B

Select C

6.43 (a) Select A and C with project ROR > MARR = 15.5%

(b) Rank proposals: DN, A, B, C, D; MARR = 10%

A to DN: ∆i* = 29% < MARR eliminate DN

B to A: ∆i* = 1% < MARR eliminate B

C to A: ∆i* = 7% < MARR eliminate C

D to A: ∆i* = 10% = MARR eliminate A Select D

(c) Same as part (b) above, except last comparison against MARR = 14%

D to A: ∆i* = 10% < MARR eliminate D Select A

6.46 (a) Two sign changes; maximum number of i* values is two

(b) Cumulative

Year Net Cash Flow, $ Cash Flow, $

0 -40,000 -40,000

1 32,000 -8000

2 18,000 +10,000

3 -2000 +8000

4 -1000 +7000

There is one sign change in the cumulative cash flow series; only one positive i*

value is indicated

6.49 Quarter Expenses Revenue Net Cash Flow Cumulative

0 -20 0 -20 -20

1 -20 5 -15 -35

2 -10 10 0 -35

3 -10 25 15 -20

4 -10 26 16 -4

5 -10 20 10 +6

6 -15 17 2 +8

7 -12 15 3 +11

8 -15 2 -13 -2

5

(a) From net cash flows, there are two possible i* values

(b) From cumulative cash flow, sign starts negative and changes twice. Norstrom’s

criterion is not satisfied, there may be up to two i* values.

6.52 (a) 3 changes in sign of net cash flow; 3 possible i* values

(b) Cumulative

Year Net CF, $ CF, $

0 -17,000 -17,000

1 -20,000 -37,000

2 4,000 -33,000

3 -11,000 -44,000

4 32,000 -12,000

5 47,000 +35,000

Per Norstrom’s criterion, there is only one positive i* value

0 = -17 - 20(P/F,i*,1) + 4(P/F,i*,2) -11(P/F,i*,3) + 32(P/F,i*,4) + 47(P/F,i*,5)

i* = 17.39% (spreadsheet)

6.55 Tabulate net cash flow and cumulative cash flow values

Cumulative

Year Cash Flow, $ Cash Flow, $

0 0 0

1 -5000 -5,000

2 -5000 -10,000

3 -5000 -15,000

4 -5000 -20,000

5 -5000 -25,000

6 -5000 -30,000

7 +9000 -21,000

8 -5000 -26,000

9 -5000 -31,000

10 -5000 + 50,000 +14,000

(a) Cash flow rule of signs test (Descartes’ rule): three possible ROR values

Cumulative cash flow test (Norstrom’s criterion): one positive ROR value

(b) Move all cash flows to year 10 and solve for i*

0 = -5000(F/A,i*,10) + 14,000(F/P,i*,3) + 50,000

Solve for i* = 6.29% (spreadsheet)

6

(c) Hand solution: MIRR with ir = 20% and ib = 8%

PW0 = -5000(P/A,ib,6) – 5000(P/A,ib,3)(P/F,ib,7)

= -5000(P/A,8%,6) – 5000(P/A,8%,3)(P/F,8%,7)

= -5000(4.6229) – 5000(2.5771)(0.5835)

= $-30,631

FW10 = 9000(F/P,ir,3) + 50,000

= 9000(F/P,20%,3) + 50,000

= 9000(1.7280) + 50,000

= $65,552

Find ROR at which PW0 is equivalent to FW10

PW0(F/P,i′,10) + FW10 = 0

-30,631(1 + i′)10 + 65,552 = 0

(1 + i′)10 = 2.1400

i′ = 7.91%

Spreadsheet solution: Enter cash flows for years 0 through 10 into cells

B2 through B12, The function = MIRR(B1:B11,8%,20%) displays

i′ = 7.90%

(d) Hand solution: In applying the ROIC procedure, all F values are negative

except the last one. Therefore, i″ is used in all equations. The ROIC EROR

(i″) is the same as the internal ROR value (i*) of 6.29% per year.

Spreadsheet solution: Use Figure 6.7 as a model and apply GOAL SEEK to

obtain i″ = 6.29%

7

6.58 (a) Hand solution: ROIC procedure with ir = 14%

F0 = 5000 F0 > 0; use ir

F1 = 5000(1.14) –2000 = 3700 F1 > 0; use ir

F2 = 3700(1.14) –1500 = 2718 F2 > 0; use ir

F3 = 2718(1.14) – 7000 = -3901.48 F3 < 0; use i″

F4 = -3901.48(1 + i″) + 4000

Set F4 = 0 and solve for i″

i″ = 2.53%

Spreadsheet solution: ROIC with ir = 14% results in i″ = 2.53%

Before GOAL SEEK After GOAL SEEK

(b) Hand solution: Apply MIRR procedure with ib = 7% and ir = 14%

PW0 = -2000(P/F,ib,1) – 1500(P/F,ib,2) - 7000(P/F,ib,3)

= -2000(P/F,7%,1) – 1500(P/F,7%,2) - 7000(P/F,7%,3)

= -2000(0.9346) – 1500(0.8734) – 7000(0.8163)

= $-8893

FW4 = 5000(F/P,ir,4) + 4000

= 5000(F/P,14%,4) + 4000

= 5000(1.6890) + 4000

= $12,445

Find EROR at which PW0 is equivalent to FW4

PW0(F/P,i′,4) + FW4 = 0

-8893(1 + i′)4 + 12,445 = 0

(1 + i′)4 = 1.3994

i′ = 8.76%

Spreadsheet solution:

Enter cash flows in cells B2:B6 and apply function =MIRR(B2:B6,7%,14%) to

display i′ = 8.76%.

8

6.61 Answer is (b)

6.64 0 = -60,000 + 10,000(P/A,i*10)

(P/A,i*,10) = 6.0000

From tables, i* is between 10% and 11%

Answer is (a)

6.67 Answer is (c)

6.70 Answer is (c)

1



Chapter 7

Benefit/Cost Analysis and Public Sector Projects

7.1 It is best to take a limited viewpoint in determining benefits and disbenefits

Because, in the broadest sense, benefits and disbenefits will usually exactly off-

set each other.

7.4 Under a DBOMF contract, the contractor is responsible for managing the cash

flows for the project and the government unit remains responsible for obtaining

capital and operating funds for the project.

7.7 B/C = (2,740,000 – 380,000)/2,000,000

= 1.18

7.10 (a) AW of Costs = 13,000(A/P,10%,20) + 400

= 13,000(0.11746) + 400

= $1927

AW of B – D = 3800 – 6750(A/F,10%,20)

= 3800 – 6750(0.01746)

= $3682

B/C = 3682/1927

= 1.91

(b) Let P = minimum first cost

AW of costs = P(A/P,10%,20) + 400 = 3682

P = (3682 – 400)/0.11746

= $27,941

The first cost must > $27,941,000 to force B/C < 1.0

7.13 AW of initial cost, C = 35,000,000(A/P,5%,30)

= 35,000,000(0.06505)

= $2,276,750

AW of B = AW of Benefits – AW of disbenefits – AW of M&O costs

= 6,500,000 - 1,700,000 - 900,000

= $3,900,000

Modified B/C = 3,900,000/2,276,750

= 1.71

2

7.16 Let P = initial cost

1.3 = 500,000/[P(A/P,7%,50) + 200,000]

1.3 = 500,000/[P(0.07246) + 200,000]

500,000 = 1.3[P(0.07246) + 200,000]

240,000 = (0.094198)P

P = $2,547,825

7.19 (a) B = $600,000

D = $190,000

C = 650,000(A/P,6%,20) + 150,000

= 650,000(0.08718) + 150,000

= $206,667

B/C = (600,000 – 190,000)/206,667

= 1.98

Project is justified since B/C > 1.0

(b) Modified B/C = (600,000 – 190,000 – 150,000)/650,000(A/P.6%,20)

= 260,000/56,667

= 4.59

Project is justified since modified B/C > 1.0

7.22 B = 41,000(33 –18) = $615,000

D = 1100(85) = $93,500

C = 750,000(A/P,0.5%,36)

= 750,000(0.03042)

= $22,815

B/C = (615,000 – 93,500)/22,815

= 22.86

7.25 Calculate AW of total cost and rank according to increasing cost.

AWpond = 58(A/P,6%,40) + 5.5

= 58(0.06646) + 5.5

= $9.35

AWexpand = 76(A/P,6%,40) + 5.3

= 76(0.06646) + 5.3

= $10.35

3

AWprimary = 2(A/P,6%,40) + 2.1

= 2(0.06646) + 2.1

= $2.23

AWpartial = 48(A/P,6%,40) + 4.4

= 48(0.06646) + 4.4

= $7.59

Benefits are directly estimated; DN is first alternative

Ranking is as follows: DN, Primary, Partial, Pond, Expand

Primary to DN: ∆B/C = 2.7/2.23

= 1.21 eliminate DN

Partial to Primary: ∆B/C = (8.3-2.7)/(7.59 – 2.23)

= 1.04 eliminate Primary

Pond to Partial: ∆B/C = (11.1 - 8.3)/(9.35 - 7.59)

= 1.59 eliminate Partial

Expand to Pond: ∆B/C = (12.0 - 11.1)/(10.35 - 9.35)

= 0.90 eliminate Expand

Select the Pond System

7.28 Benefits are direct; determine AW of costs; order is DN, N, S

CN = 11,000,000(0.06) + 100,000 = $760,000 per year

CS = 27,000,000(0.06) + 90,000 = $1,710,000 per year

N to DN: B/C = (990,000-120,000)/760,000

= 1.14 eliminate DN

S to N: Δ(B-D) = (2,100,000 – 300,000) – (990,000 – 120,000) = $930,000

ΔC = 1,710,000 – 760,000 = $950,000

∆B/C = 930,000/950,000 = 0.98 eliminate S

Select site N

4

7.31 Rank by increasing AW of total costs; order is: DN, EC, NS

AW of costs, EC = 38,000(A/P,7%,10) + 49,000

= 38,000(0.14238) + 49,000

= $54,410

AW of costs, NS = 87,000(A/P,7%,10) + 74,000

= 87,000(0.14238) + 74,000

= $86,387

EC to DN: B/C = (B-D)/C = (110,000 – 26,000)/54,410

= 1.54 eliminate DN

NS to EC:

∆B = 130,000 – 110,000 = $20,000

∆D = 18,000 – 26,000 = $-8000

∆C = 86,387 - 54,410 = $31,977

∆B/C = [20,000 -(-8000)]/31,977

= 0.88 eliminate NS

Select method EC

7.34 Rank alternatives by increasing cost: DN, G, J, H, I, L, K

Eliminate H and K based on B/C to DN < 1.0

G to DN = 1.15 eliminate DN

J to G = 1.07 eliminate G

I to J = 1.07 eliminate J

L to I = ?

∆B/C for L-to- I comparison is not shown. Must compare L to I incrementally;

survivor is selected alternative.

7.37 Answer is (c)

7.40 Can use either PW, AW, or FW values; For PW,

B/C = (245,784 – 30,723)/(100,000 + 68,798) = 1.27

Answer is (a)

5

7.43 B/CX = (110,000 – 20,000)/(60,000 + 45,000)

= 0.86 reject X

B/CY = (150,000 – 45,000)/(90,000 + 35,000)

= 0.84 reject Y

Answer is (a)

7.46 In $1000 units,

∆B/C = [(220 – 140) – (30 – 10)]/(450 - 300) = 0.40

Answer is (b)

7.49 B/C = (360,000 – 42,000)/[2,000,000(0.06)] = 2.65

Answer is (d)

1



Chapter 8 Breakeven, Sensitivity, and Payback Analysis

8.1 (a) QBE = 1,000,000/(9.90 – 4.50)

= 185,185 units

(b) Profit = R – TC

= 9.90(150,000) – 1,000,000 – 4.50(150,000)

= $-190,000 (loss)

(c) Profit = R – TC

= 9.90(480,000) – 1,000,000 – 4.50(480,000)

= $1,592,000

8.4 Let x = hours per month billed to realize a profit of $15,000

15,000 = -900,000(A/P,1%,120) – 1,100,000 + 1,500,000(A/F,1%,120) + 10(90)x

15,000 = -900,000(0.01435) – 1,100,000 + 1,500,000(0.00435) + 10(90)x

900x = 1,121,390

x = 1246 hours/month

8.7 (a) Plot shows maximum quantity at about 1350 units. Profit is about $20,000.

(b) Profit = R – TC = (-0.007 - 0.004) Q2 + (32-2.2)Q - 8

= -0.011Q2 + 29.8Q - 8

Qp = -b/2a = -29.8/2(-0.011)

= 1355 units

2

Max profit = -b2/4a + c = -29.82 / 4(-.011) - 8

= $20,175 per month

8.10 (a) Using the relation PW = 0, select different i values and solve for n. Details for

i = 8% and 15% are shown.

0 = -3,150,000 + 500,000(P/A,i%,n) + 400,000(P/F,i%,n)

i = 8%, n = 8: PW = -3,150,000 + 500,000(5.7466) + 400,000(0.5403)

= $-60,580

i = 8%, n = 9: PW = -3,150,000 + 500,000(6.2469) + 400,000(0.5002)

= $173,530

n = 8.2 (interpolation)

i = 15%, n = 19: PW = -3,150,000 + 500,000(6.1982) + 400,000(0.0703)

= $-22,780

i = 15%, n = 20: PW = -3,150,000 + 500,000(6.2593) + 400,000(0.0611)

= $4090

n = 19.8 years (interpolation)

(b) Retention ranges from 8 to 20 years for varying i values. This is a perfect

example where the spreadsheet is easier. Use the NPER function.

8.13 Let Pasp = maximum $/mile for asphalt

-2,300,000(A/P,8%,20) - 483 = -Pasp(A/P,8%,10) – 774

-2,300,000(0.10185) - 483 = -Pasp(0.14903) – 774

Pasp = $1,569,912

3

8.16 Let x = square yards per year to break even

-109,000 – 2.75x = -225,000(A/P,8%,15) – 13x

-109,000 – 2.75x = -225,000(0.11683) – 13x

10.25x = 82,713

x = 8070 square yards/year

8.19 Let x = ads per year

-21 x = -12,000(A/P,10%,3) -55,000 + 2000(A/F,10%,3) – 5x

-21x = -12,000(0.40211) -55,000 + 2000(0.30211) – 5x

16x = 59,221

x = 3701 ads

8.22 Let x = production in year 3

-40,000 – 50x = -70,000 – 12x

38x = 30,000

x = 789.5 or 790 units

8.25 (a) Let T = number of tons. Solve relation AW1 = AW2 for T.

Variable costs (VC) for each machine

VC1: 24T/10 = 2.4T VC2: 2(24)T/6 = 8T

-123,000(A/P,7%,10) – 5000 -2.4T = -70,000(A/P,7%.6) – 2500 – 8T

(8-2.4)T = -70,000(0.20980) – 2500 + 123,000(0.14238) + 5000

5.6T = 5327

T = 951.2 tons

If tonnage is less than breakeven, select machine 2 since the slope is steeper.

At 1000 tons, select machine 1.

(b) Set up the VC relation for each machine and solve for T in AW1 = AW2 .

SOLVER can be used to find breakeven with the constraint AW1 = AW2.

% change

Cost,

$/hour

VC1

VC2

AW1 = AW2

relation

Breakeven

T value, tons

-15% 20.4 2.04T 6.8T 4.76T = 5327 1119

-5% 22.8 2.28T 7.6T 5.32T = 5327 1001

0% 24.0 2.40T 8.0T 5.60T = 5327 951

+5% 25.2 2.52T 8.4T 5.88T = 5327 906

+15% 27.6 2.76T 9.2T 6.44T = 5327 827

4

8.28 (a) Solve relation Revenue - Cost = 0 for Q = number of filters per year

50Q - [200,000(A/P,6%,5) + 25,000 + 20Q] = 0

30Q = 200,000(0.23740) +25,000

Q = 72,480/30

= 2416 filters per year

At 5000 units, make the filters inhouse

(b) Solve the relation AWbuy = AWmake

(50-30)Q = -72,480 + (50-20)Q

Q = 7248 filters per year

Since 5000 < 7248, outsourcing is the correct choice; slope of 30Q is larger

(c) Inhouse: make 5000 at $20 each

Profit = 5000(50-20) - 72,480 = $77,520

Outsource: buy 5000 at $30 each

Profit = 5000(50-30) = $100,000

A spreadsheet can be used to answer all three questions.

5

8.31 AWcont = -130,000(A/P,15%,5) -30,000 + 40,000(A/F,15%,5)

= -130,000(0.29832) -30,000 + 40,000(0.14832)

= $-62,849

Lowest cost for batch will occur when the interest rate is the lowest (5%)

and life is longest (10 years)

AWbatch = -80,000(A/P,5%,10) – 55,000 + 10,000(A/F,5%,10)

= -80,000(0.12950) – 55,000 + 10,000(0.07950)

= $-64,565

Since batch AW is more costly, the batch system will never be less expensive

than the continuous option.

8.34 AWcurrent = $-62,000

AW10,000 = - 64,000(A/P,15%,3) – 38,000 + 10,000(A/F,15%,3)

= - 64,000(0.43798) – 38,000 + 10,000(0.28798)

= $-63,151

AW13,000 = - 64,000(A/P,15%,3) – 38,000 + 13,000(A/F,15%,3)

= - 64,000(0.43798) – 38,000 + 13,000(0.28798)

= $-62,287

AW18,000 = - 64,000(A/P,15%,3) – 38,000 + 18,000(A/F,15%,3)

= - 64,000(0.43798) – 38,000 + 18,000(0.28798)

= $-60,847

The decision is sensitive; replace the existing process only if the estimate of

$18,000 is reliable.

6

8.37 (a) AW1 = -10,000(A/P,i%,8) - 600 - 100(A/F,i%,8) - 1750(P/F,i%,4)(A/P,i%,8)

AW2 = -17,000(A/P,i%,12) - 150 - 300(A/F,i%,12)

- 3000(P/F,i%,6)(A/P,i%,12)

Calculate AW at each MARR value. The decision is sensitive to MARR,

changing at i = 6%.

MARR AW1 AW2 Selection

4% $-2318 $-2234 2

6% $-2444 $-2448 1

8% $-2573 $-2673 1

(b) Spreadsheet solution requires that the PW value is first determined using the

NPV function over the LCM of 24 years and then converting it to an AW value

using the PMT function. Spreadsheet follows.

8.40 Hand solution: Determine AW values at different savings, s.

AWA = -50,000(A/P,10%,5) – 7500 + 5,000(A/F,10%,5) + s

= -50,000(0.2638) – 7500 + 5000(0.1638) + s

= -19,871 + s

AWB = -37,500(A/P,10%,5) – 8000 + 3700(A/F,10%,5) + s

= -37,500(0.2638) – 8000 + 3700(0.1638) + s

= -17,286 + s

Selection changes when s is +40% of best estimate. Table follows:

7

Spreadsheet solution: PMT functions display AW values with savings variation

added to end of function.

8.43 (a) Set up the general AW relation for D103 and determine AW for the three

scenarios.

AWD103 = -P(A/P,10%,n) + (0.1P)(A/F,10%,n) – 4000

Strategy P S n AW

Pess -500,000 50,000 1 $-504,000

ML -400,000 40,000 3 -152,761

Opt -300,000 30,000 5 -78,226

The AW490G = $-135,143 is known. D103 is selected only if the optimistic

scenario is correct, i.e., P = $-300,000 and n = 5 years.

(b) Spreadsheet solution.

8.46 (a) Machine 1:

Percent

variation

Savings for A,

$ per year

AWA

Savings for B,

$ per year

AWB

Selection

-40% 9,000 $-10,871 7,800 $-9,486 B

-20 12,000 -7,871 10,400 -6,886 B

0 15,000 -4,871 13,000 -4,286 B

20 18,000 -1,871 15,600 -1,686 B

40 21,000 1,129 18,200 914 A

8

np = 40,000/10,000

= 4 years

Machine 2:

np = 90,000/15,000

= 6 years

Select only machine 1.

(b) Machine 1:

-40,000 + 10,000(P/A,10%,np) = 0

(P/A,10%,np) = 4.0000

From 10% interest tables, np is between 5 and 6 years; np > 5 years.

Machine 2:

-90,000 + 15,000(P/A,10%,np) = 0

(P/A,10%,np) = 6.0000

From 10% interest tables, np is between 9 and 10 years’ np > 9 years

Select neither alternative for further analysis

8.49 (a) Let np = number of months at 0.5% per month

0 = -90,000(A/P,0.5%,np) – 20,000 + 22,000

(A/P,0.5%,np) = 2000/90,000

= 0.02222

From 0.5% interest factor table, np is approximately 51 months

(b) Calculator function n(0.5,2000,-90000,0) displays np = 51.1 months

(c) Spreadsheet function =NPER(0.5%,2000,-90000) displays np = 51.1 months

8.52 Set up the PW relation and use trial and error or spreadsheet for np.

0 = -10,000 +1700(P/A,8%,np) + 900(P/F,8%,np)

For np = 7: 0 = -10,000 +1700(5.2064) + 900(0.5835) = $-624

For np = 8: 0 = -10,000 +1700(5.7466) + 900(0.5403) = $+255

np = 7.7 years (interpolation)

Lease; don’t purchase

Spreadsheet: NPER(8%,1700,-10000,900) displays 7.7 years

9

8.55 (a) Determine i* from AW relations

AW1 = -50,000(A/P,i*,5) + 24,000

(A/P,i*,5) = 0.4800

i1* = 38.6% (interpolated)

AW2 = -120,000(A/P,i*,10) + 42,000 – 2500(A/G,i*,10)

i2* = 26.55% (trial and error or spreadsheet)

Now, select process 1

(b) First of all, an ROR analysis always requires an incremental analysis over the

LCM of 10 years. The i2* = 26.55% is not correctly used in the fashion

shown here. Furthermore, this analysis assumes a return of 26.55% is made

from years 6 through 10, which may not be correct for these funds.

8.58 Both the fixed cost and variable cost of alternative X are higher than those of

alternative Y. Therefore, it can never be favored.

Answer is (a)

8.61 Answer is (a)

8.64 Answer is (c)

8.67 -100Q = -250,000(A/P,15%,4) – 80,000 – 40Q

60Q = 250,000(0.35027) + 80,000

Q = 2793

Larger than 2793 since slope of make item inhouse is lower

Answer is (b)

8.70 Select largest PW.

Answer is (d)

1



Chapter 9

Replacement and Retention Decisions

9.1 The defender refers to the currently-owned, in-place asset while the challenger

refers to the equipment/process that is under consideration as its replacement.

9.4 P = market value = $39,000

AOC = $17,000 per year

n = 4 years

S = $25,000

(Note: P and AOC will carry – signs in an evaluation)

9.7 The ESL of the defender is 3 years with the lowest AW of $-85,000.

9.10 Add AW values for first cost, operating cost, and salvage value; select lowest AW.

Years Retained Total AW

1 $-102,000

2 -84,334

3 -84,190

4 -83,857

5 -84,294

Economic service life is n = 4 years.

9.13 (a) Tabulated factors

AW1 = -65,000(A/P,10%,1) – 50,000 + 30,000(A/F,10%,1) = $-91,500

AW2 = -65,000(A/P,10%,2) – [50,000 + 10,000(A/G,10%,2)]

+ 30,000(A/F,10%,2)

= $-77,929

AW3 = -65,000(A/P,10%,3) – [50,000 + 10,000(A/G,10%,3)]

+ 20,000(A/F,10%,3)

= $-79,461

AW4 = -65,000(A/P,10%,4) – [50,000 + 10,000(A/G,10%,4)]

+ 20,000(A/F,10%,4)

= $-80,008

AW5 = -65,000(A/P,10%,5) – [50,000 + 10,000(A/G,10%,5)]

+ 20,000(A/F,10%,5)

= $-81,972

2

AW6 = -65,000(A/P,10%,6) – [50,000 + 10,000(A/G,10%,6)]

+ 20,000(A/F,10%,6)

= $-84,568

AW7 = -65,000(A/P,10%,7) – [50,000 + 10,000(A/G,10%,7)]

+ 20,000(A/F,10%,7)

= $-87,459

ESL is n = 2 years with AW = $-77,929

(b) Spreadsheet solution shows ESL is n = 2 years with AW = $77,929

9.16 AWC = -80,000(A/P,12%,3) - 19,000 + 10,000(A/F,12%,3)

= -80,000(0.41635) - 19,000 + 10,000(0.29635)

= $-49,345

AWD = -10,000(A/P,12%,3) -15,000 – 31,000 + 9000(A/F,12%,3)

= -10,000(0.41635) -15,000 – 31,000 + 9000(0.29635)

= $-47,496

AWD = -20,000(A/P,12%,3) -15,000 – 31,000 + 9000(A/F,12%,3)

= -20,000(0.41635) -15,000 – 31,000 + 9000(0.29635)

= $-51,660

Trade-in of $10,000: Select the defender

Trade-in of $20,000: Select the challenger

9.19 No option of retention of defender D more than one year

3

AWD = -9000(A/P,10%,1) – 192,000

= -9000(1.1000) -192,000

= $-201,900

AWC = -320,000(A/P,10%,4) – 68,000 + 50,000(A/F,10%,4)

= -320,000(0.31547) – 68,000 + 50,000(0.21547)

= $-158,177

Replace the defender now with System C

9.22 -RV(A/P,8%,3) – 60,000 + 15,000(A/F,8%,3) = -80,000(A/P,8%,5)

- [40,000 + 2000(A/G,8%,5)] + 20,000(A/F,8%,5)

-RV(0.38803) – 60,000 + 15,000(0.30803) = -80,000(0.25046)

- [40,000 + 2000(1.8465)] + 20,000(0.17046)

-0.38803 RV = 4941.05

RV = $12,734

9.25 -RV(A/P,12%,4) – [40,000 + 2000(A/G,12%,4)] = -150,000(A/P,12%,10)

- [10,000 + 500(A/G,12%,10)] + 50,000(A/F,12%,10)

-RV(0.32923) – [40,000 + 2000(1.3589)] = -150,000(0.17698)

- [10,000 + 500(3.5847)] + 50,000(0.05698)

-0.32923RV = -35,490 + 42,718

RV = $21,954

9.28 Only 2 options; replace defender, buy challenger now or retain defender for 3 years.

AWD = - (40,000 + 70,000)(A/P,15%,3) – 85,000 + 30,000(A/F,15%,3)

= -110,000(0.43798) – 85,000 + 30,000(0.28798)

= $-124,538

AWC = -220,000(A/P,15%,3) – 65,000 + 50,000(A/F,15%,3)

= -220,000(0.43798) – 65,000 + 50,000(0.28798)

= $-146,957

Keep the defender

9.31 Answer is (a)

4

9.34 Answer is (b)

9.37

Option Years for

Defender

Years for

Challenger

AW, cash flows, $ per year Option

AW, $/year Year 1 Year 2 Year 3

A 0 3 -73,000 -73,000 -73,000 -73,000

B 1 2 -74,000 -84,000 -84,000 -80,192

C 2 1 -74,000 -74,000 -84,000 -76,880

Option A: AW = $-73,000

Option B: AW = -84,000 + 10,000(P/F,15%,1)(A/P,15%,3)

= -84,000 + 10,000(0.8696)(0.43798)

= $-80,192

Option C: AW = -74,000 - 10,000(A/F,15%,3)

= -74,000 - 10,000(0.28798)

= $-76,880

Select option A: Replace now.

Answer is (a)

1



Chapter 10 Effects of Inflation

10.1 Inflated dollars are converted into constant value dollars by dividing by one plus

the inflation rate per period for however many periods are involved.

10.4 (a) Cost = 17,131(1 + 0.07)3

= $20,986 per year

(b) Cost = (17,131+ 4000)(1 + 0.07)3

= $25,886 per year

10.7 185(1 + f)5 = 225

(1 + f)5 = 1.2162

1 + f = 1.21620.2

f = 3.99% per year

10.10 (a) Salary in 2017 = 61,872(1.02)6

= $69,678

(b) Salary in 2017 = 61,872(1.07)6

= $92,853

10.13 Transportation cost in 5 years = 7.5(F/P,10%,5)

= 7.5(1.6105)

= 12.08¢ or 12.1¢

Labor cost in 5 years = 38.5(F/P,4%,5)

= 38.5(1.2167)

= 46.84¢ or 46.8¢

2012: Total cost = 38.5¢ + 19.5¢ + 12¢ + 7.5¢ + 7¢ + 6¢ + 5¢ + 4.5¢

= 100¢

% profit = (4.5/100)(100%)

= 4.50%

2017: Total cost = 46.8¢ + 19.5¢ + 12¢ + 12.1¢ + 7¢ + 6¢ + 5¢ + 4.5¢

= 112.9¢

% profit = (4.5/112.9)(100%)

= 3.99%

2

10.16 if per month = 24/12 = 2%. Use inflated rate equation to solve for real rate i.

0.020 = i + 0.005 + (i)(0.005)

1.005i = 0.015

i = 0.0149 (1.49% per month)

10.19 if = 0.10 + 0.07 + (0.10)(0.07)

= 17.7%

PW = -150,000 – 60,000(P/A,17.7%,5) + 0.20(150,000)(P/F,17.7%,5)

= -150,000 - 60,000(3.1485) + 30,000(0.44271)

= $-325,630

10.22 if = 0.12 + 0.04 + (0.12)(0.04) = 16.48%

PWA = 2,100,000(P/F,16.48%,2)

= 2,100,000(0.73705)

= $1,547,806

PWB = 2,100,000 - 400,000 = $1,700,000

The offer from vendor A is better.

10.25 Find F, then deflate the amount by dividing by (1 + f)n

F = 5000(F/P,15%,17) + 8000(F/P,15%,14) + 9000(F/P,15%,13)

+ 15,000(F/P,15%,10) + 16,000(F/P,15%,6) + 20,000

= 5000(10.7613) + 8000(7.0757) + 9000(6.1528)

+ 15,000(4.0456) + 16,000(2.3131) + 20,000

= $283,481

Purchasing power = 283,481/(1.03)17 = $171,511

10.28 (a) F = 10,000(F/P,10%,5)

= 10,000(1.6105)

= $16,105

(b) Purchasing Power = 16,105/(1 + 0.05)5 = $12,619

(c) if = i + 0.05 + (i)(0.05)

0.10 = i + 0.05 + (i)(0.05)

1.05i = 0.05

i = 0.0476 (4.76%)

or

i = 0.10 – 0.05/(1 + 0.05) = 0.0476 (4.76%)

3

10.31 Buying power = 1,800,000/(1 + 0.038)20 = $853,740

10.34 Cost in terms of constant-value dollars at real i = 5%

F = 40,000(F/P,5%,3)

= 40,000(1.1576)

= $46,304

10.37 Buying power is P value with inflation removed

P6000 = 6000/(1 + 0.04)2 = $5547

P9000 = 9000/(1 + 0.04)3 = $8001

P5000 = 5000/(1 + 0.04)6 = $3952

10.40 if = 0.15 + 0.05 + (0.15)(0.05)

= 20.75%

AWX = -65,000(A/P,20.75%,5) – 40,000

= -65,000(0.33991) – 40,000

= $-62,094

AWY = -90,000(A/P,20.75%,5) – 34,000 + 10,000(A/F,20.75%,5)

= -90,000(0.33991) – 34,000 + 10,000(0.13241)

= $-63,268

Select process X

10.43 if = 0.10 + 0.04 + (0.10)(0.04) = 14.4% per year

AW = -40,000(A/P,14.4%,3) - 24,000 + 6000(A/F,14.4%,3)

= -40,000(0.43363) - 24,000 + 6000(0.28963)

= $-39,607 per year

10.46 Answer is (a)

10.49 Answer is (c)

10.52 Answer is (c)

10.55 if per month = 0.01 + 0.01 + (0.01)(0.01)

= 2.01%

Nominal per year = 2.01(12) = 24.12%

Answer is (c)

1



Chapter 11

Estimating Costs

11.1 Elements of first cost include equipment cost, delivery charges, installation cost,

insurance coverage, and training of personnel for equipment use.

11.4 Cost = 98.23(50,000) = $4,911,500

11.7 Cost = 1,350,000(1.70/0.93) = $2,467,742

11.10 Cost = 185,000(1634.9/1375.4) = $219,904

11.13 From Table, index value in 2003 = 6695; index value in 2012 = 9290

Ct = 30,000,000(9290/6695)

= $41,628,000

11.16 Cost2011 = Cost2002(1490.2/1104.2)

328 = Cost2002(1.3496)

Cost2002 = $243

11.19 31.39 = Rate1913(16,520.53/100)

Rate1913 = $0.19 per hour

11.22 C2 = 9500(450/250)0.32

= 9500(1.207)

= $11,466

11.25 If C2 = 1.04 C1 and Q2/Q1 = 0.50, the cost-capacity relation is

1.04C1 = C1 (Q2/Q1)x

1.04 = (0.50)x

log 1.04 = x log 0.50

x = -0.0566

11.28 Cost = 3750(2)0.89(1520.6/1104.2) = $9570

11.31 1,320,000 = h(225,000)

h = 5.87

2

11.34 T12 = T1Ns

s = log 0.85/log 2

= -0.234

T12 = (56)12-0.234

= 31.3 months

11.37 Total direct labor hours = 3000 + 9000 + 5000

= 17,000 hours

Indirect cost/hour = 34,000/17,000

= $2.00

Allocation to Dept A = 3000(2.00)

= $6000

Allocation to Dept B = 9000(2.00)

= $18,000

Allocation to Dept C = 5000(2.00)

= $10,000

11.40 (a) Hand solution: Make alternative indirect cost computation

Dept Rate

(1)

Usage

(2)

Annual cost

(3) = (1)(2)

M $2.40 450,000 $1.08 million

P $0.50 850,000 425,000

Q $25 4800 120,000

$/year $1,625,000

Determine AW for make and buy alternatives.

AWmake = -3,000,000(A/P,12%,6) + 500,000(A/F,12%,6)

– 1,500,000 – 1,625,000

= -3,000,000(0.24323) + 500,000(0.12323) – 3,125,000

= $-3,793,075

AWbuy = -4,500,000 – 200,000(A/G,12%,6)

= -4,500,000 – 200,000(2.1720)

= $-4,934,400

Select make alternative.

3

(b) Spreadsheet solution: Select to make inhouse.

11.43 Compare last year’s allocation based on flight traffic with this year’s based on

baggage traffic. Significant change took place.

11.46 The answer is (c)

11.49 C2 = 22,000(300/200)0.64 = $28,518

Answer is (b)

11.52 CT = 2.45(390,000) = $955,500

Answer is (a)

11.55 3,000,000 = 550,000(100,000/6000)x

5.4545 = (16.67)x

log 5.4545 = xlog16.67

x = 0.60

Answer is (d)

11.58 Allocation = (900 + 1300)(2000) = $4.4 million

Percent allocated = 4.4/8.0 million = 55%

Answer is (c)

Last year;

flight basis

This year;

baggage basis

Percent

change

DFW $330,000 $343,620 + 4.1%

YYZ 187,500 218,371 +16.5

MEX 150,000 105,363 – 29.7

1



Chapter 12 Depreciation Methods

12.1 Depreciation is a tax-allowed deduction in the equation Taxes = (income –

expenses – depreciation)(tax rate)

12.4 Productive life – Time the asset is actually expected to provide useful service.

Tax recovery period – Time allowed by tax laws to depreciate the asset’s value to

salvage (or zero).

Book recovery period – Time used on company accounting books for depreciation

to salvage (or zero)

12.7 Tax depreciation: Dt = Rate(BVt-1)

Book depreciation: Dt = Rate(40,000)

Tax Depreciation Book Depreciation

Year, t Dt BVt Dt BVt

0 40,000 40,000

1 16,000 24,000 10,000 30,000

2 9,600 14,400 10,000 20,000

3 5,760 8,640 10,000 10,000

4 3,456 5,184 10,000 0

Spreadsheet solution with graphs follows.

2

12.10 (a) Depreciation charge is determined from the change in book value

D = 296,000 - 224,000 = 72,000/year

Since the asset has a 5-year life 2 more years of depreciation will reduce

BV to salvage value.

S = 224,000 – 2(72,000) = $80,000

(b) (B – 80,000)/5 = 72,000

B = $440,000

12.13 Spreadsheet solution uses SL depreciation.

Dt = (350,000 – 50,000)/5 = $60,000 per year

12.16 Substitute 0.25B for BV. Depreciation rate is d = 2/5 = 0.4

0.25B = B(1 – 0.4)t

log 0.25 = t(log 0.6)

t = 2.71years

Salvage value will be reached by end of 3rd year

12.19 (a) SL: BV10 = $10,000 by definition

DDB: Use Equation [12.9] to determine if the implied S < $10,000

with d = 2/7 = 0.2857

BV10 = BV7 = B(1-d)7 = 100,000(0.7143)7

= $9486

Both salvages are less than the market value of $12,500

3

(b) SL: D10 = (100,000 - 12,500)/10 = $8750

DDB: D10 = 0, since n = 7 years

A spreadsheet solution for both parts follows.

12.22 BV3 = 400,000 - 400,000(0.20 + 0.32 + 0.192)

= $115,200

12.25 SL: D4 = [80,000 - 0.25(80,000)]/5

= $12,000 per year

BV4 = 80,000 – 4(12,000)

= $32,000

MACRS: BV4 = 80,000 – 80,000(0.20 + 0.32 + 0.192 + 0.1152)

= $13,824

Difference = 32,000 – 13,824 = $18,176

MACRS has a lower BV after 4 years

12.28 MACRS: D1 = 0.01391(3,400,000) = $47,294

D2 to D10 = 0.02564(3,400,000) = $87,176

Total depreciation for 10 years = 47,294 + 9(87,176) = $831,878

BV10 = 3,400,000 – 831,878 = $2,568,122

Anticipated selling price is 1.5(BV10) = $3,852,183

Fairfield hopes to sell it for $452,183, or 13.3%, more than they paid for it.

4

12.31 d = 1/5 = 0.20, For year 2, by DB method

DB : D2 = 0.20(100,000)(1 – 0.20)1 = $16,000

For SL method, need BV1 from DB method with d = 0.20

BV1 = 100,000(1 – 0.2)1 = $80,000

New SL depreciation

D2 = (80,000 – 10,000)/4 = $17,500

SL has larger depreciation by 17,500 – 16,000 = $1500; switch to SL is advisable

12.34 Verify the following MACRS rates using the modified DDB-to-SL switching.

t 1 2 3 4 5 6__

dt 0.20 0.32 0.192 0.1152 0.1152 0.0576

The DDB rate to start is d = 2/5 = 0.40.

d1: DDB: d1 = ½d = 0.20

d2: Accumulated depreciation = Accumulated D = 0.20

DDB: By Equation [12.14]

d2 = 0.4(1 – 0.2) = 0.32 (selected)

SL: Modify Equation [12.15] to have denominator (n-t+1.5)

d2 = 0.8/(5-2+1.5) = 0.8/4.5 = 0.178

d3: Accumulated D = 0.2 + 0.32 = 0.52

DDB: d3 = 0.4(1 – 0.52) = 0.192 (selected)

SL: d3 = 0.48/(5-3+1.5) = 0.137

d4: Accumulated D = 0.2 + 0.32 + 0.192 = 0.712

DDB: d4 = 0.4(1 – 0.712) = 0.1152

SL: d4 = 0.288/2.5 = 0.1152 (select either)

Switch to SL occurs in year 4.

5

d5: Use the SL rate for n = 5

Accumulated D = 0.2 + 0.32 + 0.192+ 0.1152 = 0.8272

SL: d5 = 0.1728/(5-5+1.5) = 0.1728/1.5 = 0.1152

d6: d6 is the remainder or 1/2 the d5 rate. Accumulated D = 0.9424

d6 = d5 /2 = 0.0576

or

d6 = 1 – 0.9424 = 0.0576

12.37 Income = 6000(6) + 7000(9) = $99,000

Percentage depletion charge = 99,000(0.05) = $4950

12.40 (a) Determine the cost depletion factor in $/1000 tons and multiply by yearly

tonnage.

pt = 2,900,000/100 = $29,000 per 1000 tons

Annual cost depletion = volume × $29,000

Year

Volume,

1000 tons

Cost depletion,

$ per year

1 10 290,000

2 12 348,000

3 15 435,000

4 15 435,000

5 18 522,000

Total 70 $2,030,000

(b) Cost depletion is limited by the total first cost, which is $2.9 million. Since

only $2.03M has been depleted, all 5 years’ depletion is acceptable.

12.43 Depreciation is same for all years in straight line method.

D3 = [100,000 – 0.20(100,000)]/5 = $16,000

Answer is (a)

12.46 d = 2/5 = 0.4

BV2 = 30,000(1 – 0.4)2 = $10,800

Answer is (b)

6

12.49 Salvage value is always $0 for the MACRS method.

Answer is (a)

12.52 Percentage: GI = 65,000(40) = $2.6 million

Depletion charge = 0.05(2.6 million)

= $130,000

Cost: pt = $1.28 per ton

Depletion charge = (65,000)(1.28)

= $83,200

Answer is (b)

1



Chapter 13

After-Tax Economic Analysis

13.1 (a) From Table 13.1, marginal tax rate = 39%

(b) Taxes = 0.15(50,000) + 0.25(75,000 – 50,000) + 0.34(100,000 - 75,000) +

0.39(250,000 – 100,000)

= $80,750

(c) Average tax rate = (80,750/250,000)(100%) = 32.3%

13.4 TI = 450,000 – 230,000 – 48,000 = $172,000

Approximate taxes = 172,000(0.38) = $65,360

13.7 (a) TI = 320,000 – 149,000 – 95,000 = $76,000

(b) Use Table 13-1

Taxes = 13,750 + 0.34(76,000-75,000) = $14,090

(c) Te = 10.5 + (1 – 0.105)(18.5) = 27.06%

Tax estimate = 76,000(0.2706) = $20,566

Percent of GI = 20,566/320,000 × 100% = 6.43%

13.10 Before-tax ROR: 0 = -500,000 + 230,000(P/A,i*,5) + 100,000(P/F,i*,5)

i* = 38.48% (spreadsheet)

After-tax ROR = 38.48(1 – 0.35) = 25.01%

13.13 Missing values are shown in bold red in the table

CFBT2 = 950 – 150 = $800

D2 = 0.4445(1900) = $845

D4 = 0.0741(1900) = $141

TI3 = 600 – 200 – 281 = $119

Taxes2 = -45(0.35) = $-16

CFAT3 = 400 – 42 = $358

2

13.16 GI can be determined from the data provided.

CFBT = CFAT + taxes

GI – E = CFAT + (GI – E – D)(Te)

Solve for GI to obtain a general relation for each year t.

GIt = [CFAT + Et(1- Te) – DtTe]/ (1- Te)

where CFAT = $2.5 million

Te = 8% + (1-0.08)(20%) = 26.4% (0.264)

1- Te = 73.6% (0.736)

Year 1: GI1 = [2.5 million + 650,000(0.736) – 650,000(0.264)]/0.736

= $3,813,587

Year 2: GI2 = [2.5 million + 900,000(0.736) – 900,000(0.264)]/0.736

= $3,973,913

Year 3: GI3 = [2.5 million + 1,150,000(0.736) – 1,150,000(0.264)]/0.736

= $4,134,239

13.19 Total depreciation was 20% + 32% + 19.2% + 11.52% = 82.72%

BV4 = 80,000 – 80,000(0.8272)

= $13,824 < selling price of $15,000. Depreciation recapture is present

DR = 15,000 – 13,824

= $1176

Year GI E P and S CFBT D TI Taxes CFAT

0 $-1900 $-1900 $-1900

1 $800 $-100 0 700 $633 $ 67 $23 677

2 950 -150 0 800 845 -45 -16 816

3 600 -200 0 400 281 119 42 358

4 300 -250 700 750 141 -91 -32 782

3

13.22 Calculate taxes = GI – E – (TI + DR)(Te) for each year and add. System 2 has

lower total taxes by $26,600. Spreadsheet solution follows.

13.25 (a) MACRS depreciation

Year P and

CFBT, $ Rate D, $ TI, $ Taxes, $

0 -200,000

1 75,000 0.2000 40,000 35,000 13,300

2 75,000 0.3200 64,000 11,000 4,180

3 75,000 0.1920 38,400 36,600 13,908

4 75,000 0.1152 23,040 51,960 19,745

5 75,000 0.1152 23,040 51,960 19,745

6 75,000 0.0576 11,520 63,480 24,122

7 75,000 0 0 75,000 28,500

8 75,000 0 0 75,000 28,500

Total $152,000

PWtax = 13,300(P/F,8%,1) + 4180(P/F,8%,2) + … + 28,500(P/F,8%,8)

= $102,119 (spreadsheet)

Straight line depreciation D = 200,000/8 = $25,000 per year

Taxes = (75,000 – 25,000)(0.38) = $19,000 per year

Total taxes = 8(19,000) = $152,000

PWtax = 19,000(P/A,8%,8) = 19,000(5.7466)

= $109,185

4

MACRS is preferable with the lower PWtax value.

(b) Total taxes are $152,000 for both methods.

13.28 (a) Defender: Capital loss = BV - Sales price = [300,000 – 2(60,000)] - 100,000

= $-80,000

Taxes = -80,000(0.35) = $-28,000

This is a tax savings for the challenger in year 0.

CFATC in year 0 = $-420,000 + 28,000 = $-392,000

CFATD in year 0 = $-100,000

(b) Defender: TI = -120,000 – 60,000 = $-180,000

Taxes = 180,000(0.35) = $-63,000

CFAT = -120,000 - (-63,000) = $-57,000

Challenger: TI = -30,000 -140,000 = $-170,000

Taxes = -170,000(0.35) = $-59,500

CFAT = -30,000 - (-59,500) = $29,500

(c) AWD = -100,000(A/P,15%,3) – 57,000

= -100,000(0.43798) – 57,000

= $-100,798

AWC = -392,000(A/P,15%,3) + 29,500

= -392,000(0.43798) + 29,500

= $-142,188

Keep the defender

13.31 The debt portion of $15 million represents 40% of the total.

Total amount of financing = 15,000,000/0.40 = $37,500,000

13.34 Company’s equity = 30(0.35) = $10.5 million

Return on Equity = (4/10.5)(100%) = 38.1%

5

13.37 (a) WACCA = 0.5(9%) + 0.5(6%) = 7.5%

WACCB = 0.2(9%) + 0.8(8%) = 8.2%

Plan A has a lower WACC

(b) Let x = cost of debt capital

WACCA = 8.2% = 0.5(9%) + 0.5x

x = 7.4%

WACCB = 8.2% = 0.2(9%) + 0.8x

x = 8.0%

Plan A cost of debt goes up from 6% to 7.4%; Plan B maintains the same cost.

13.40 (a) Determine the after-tax cost of debt capital and WACC.

After-tax cost of debt capital = 10(1-0.36) = 6.4%

After-tax WACC = 0.65(14.5%) + 0.35(6.4%) = 11.665%

Interest charged to revenue for the project:

14 million × 0.11665 = $1,633,100

(b) After-tax WACC = 0.25(14.5%) + 0.75(6.4%)

= 8.425%

Interest charged to revenue for the project:

14 million × 0.08425 = $1,179,500

As more and more of the capital is borrowed, the company risks

higher loan rates and owns less and less of itself. Debt capital (loans) will

get more expensive and harder to acquire.

13.43 Find BVt -1 = BV1; solve for NOPAT; find TI from NOPAT; solve for E from TI.

BV1 = 550,000 – 550,000(0.3333) = $366,685

EVA2 = 28,000 = NOPAT – (0.14)(366,685)

NOPAT = $79,336

NOPAT = 79,336 = TI(1 – 0.35)

TI = $122,055

6

D2 = 550,000(0.4445)

= $244,475

122,055 = GI - E - D

= 700,000 - E - 244,475

E = $333,470

13.46 Te = 0.08 + (1 – 0.08)(0.34) = 39.3%

Answer is (b)

13.49 Answer is (d)

13.52 CFAT = GI – E – TI(Te)

26,000 = 30,000 – TI(0.40)

TI = (30,000 – 26,000)/0.40 = $10,000

Taxes = TI(Te) = 10,000(0.40) = $4000

TI = GI - E – D

10,000 = 30,000 – D

D = $20,000

Answer is (d)

13.55 BV5 = 100,000(0.0576) = $5760

DR = 22,000 – 5760 = $16,240

Tax on DR = 16,240(0.30) = $4872

Cash flow = 22,000 – 4872

= $17,128

Answer is (b)

1



Chapter 14 Alternative Evaluation Considering

Multiple Attributes and Risk

14.1 Wi = 1/6 = 0.1667

14.4 (a) Score by attribute with 10 for most important

Logic: #1: 0.90(#5); W1 = 0.90(10) = 9

#2: 0.10(#5); W2 = 0.10(10) = 1

#3: 0.3(#5); W3 = 0.30(10) = 3

#4: 2(#3); W4 = 2(3) = 6

#5: highest score: W5 = 10

#6: 0.8(#4); W6 = 0.80(6) = 4.8

Attribute Importance

1 9.0

2 1.0

3 3.0

4 6.0

5 10.0

6 4.8

33.8

(b) Normalized weights, Wi = Score/33.8

Attribute Wi______

1 9/33.8 = 0.27

2 1/33.8 = 0.03

3 3/33.8 = 0.09

4 6/33.8 = 0.18

5 10/33.8 = 0.30

6 4.8/33.8 = 0.14

14.7 Estimates are for the future in engineering economic evaluations. Decision making under

risk is, therefore, always present in any conclusion.

2

14.10 Determine the probability values for N.

N 0 1 2 3 4_ ≥5_

P(N) 0.12 0.56 0.26 0.032 0 .022 0.006

(a) P(N=0 or 1) = P(N=0) + P(N=1) = 0.12 + 0.56 = 0.68

Percentage is 68% of the households

(b) P(N = 1or 2) =0.56 + 0.26

= 0.82 (82%)

(c) P(N > 3) = P(N=4) + P(N≥5) = 0.022 + 0.006

= 0.028 (2.8%)

14.13 (a) Use Equation [14.5] to find E(X).

Cell,

Xi, $ P(Xi) Xi × P(Xi), $

600 0.06 36

800 0.10 80

1000 0.09 90

1200 0.15 180

1400 0.28 392

1600 0.15 240

1800 0.07 126

2000 0.10 200

1.00 1344

Sample expected value: E(X) = $1344 (b) Construct a graph with bars at $600, $800, ..., $2000 each at the height of the P(Xi) value. Show the E(X) = $1344 value on the graph.

14.16 As an example, a sample of 100 values generated the following results:

(a) = AVERAGE(A1:A100) resulted in 49.2532; very close to 50

(b) = STDEV(A1:A100) resulted in 28.08; very close to 28.87

14.19 A simulation similar to Example 14.6 is performed. MARR varies from 7% to 10% with

a 0.25 probability each. Use a lookup table for MARR (columns E and F) coupled with

the RAND() function. For CFAT, use RANDBETWEEN(4000,7000). PW values

(column M) for the simulation shown are positive 23 of the 30 trials.

Conclusion: The project appears economically viable under both certainty and risk.

3

14.22 Wenv = 50/(100 + 75 + 50) = 50/225 = 0.22

Answer is (c)

14.25 Answer is (c)

14.28 Reading #5 is missing

X = 79 = (81 + 74 + 83 + 66 + x5)/5

395 = 304 + x5

x5 = 91

2 values (66 and 91) are outside the limits

Answer is (b)

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