Basics of Engineering Economy - Every_third_solution

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. 1

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

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)

(c) NPER is n

2

(d) IRR is i

(e) PV is P

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)

3

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1

2.19

A = $4,470,884.66 (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 2

(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

3

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

4

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:

5

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)

6

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, 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) 7

(b) Spreadsheet

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)

8

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1

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

2

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)

3

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)

4

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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

4.7

XD

XE

XDE

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 1

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 B: Function is –PV(12,3,-8000,40000) – 140000 4.19

PW = $-107,360 PW = $-130,743

(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

2

(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. 3

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) 4

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) 5

+(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 6

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 F: = (-PMT(3%,32,-3000000,70000)+100000)/0.03 G: = -10000000+400000/0.03 4.61

Budget = $800,000

i = 10%

Display: $+4,775,295 Display: $-1,526,886 Display: $+3,333,333

6 viable bundles

Bundle

Projects

NCFj0

NCFjt

1 2 3 4 5 6

X Y Z XY XZ Do nothing

$-250,000 -300,000 -550,000 -550,000 -800,000 0

$ 50,000 90,000 150,000 140,000 200,000 0

S $ 45,000 -10,000 100,000 35,000 145,000 0

PW at 10% $-60,770 -21,539 -6,215 -82,309 -66,985 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 7

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)

8

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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)) 1

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:

2

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 3

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

Display: $270,187 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

4

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)

5

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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

too low 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 1

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

6.16

(spreadsheet)

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

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

6.22

(interpolation or spreadsheet)

(spreadsheet)

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

2

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

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 Product 4:

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 A to D: ∆i* = 12% < MARR B to D: ∆i* = 26% > MARR E to B: ∆i* = -16% < MARR C to B: ∆i* = 25% > MARR Select C

eliminate DN eliminate A eliminate D eliminate E eliminate B

3

(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) Year 0 1 2 3 4

Net Cash Flow, $ -40,000 32,000 18,000 -2000 -1000

Cumulative Cash Flow, $ -40,000 -8000 +10,000 +8000 +7000

There is one sign change in the cumulative cash flow series; only one positive i* value is indicated 6.49 Quarter Expenses 0 -20 1 -20 2 -10 3 -10 4 -10 5 -10 6 -15 7 -12 8 -15

Revenue 0 5 10 25 26 20 17 15 2

Net Cash Flow -20 -15 0 15 16 10 2 3 -13

4

Cumulative -20 -35 -35 -20 -4 +6 +8 +11 -2

(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) Year 0 1 2 3 4 5

Net CF, $ -17,000 -20,000 4,000 -11,000 32,000 47,000

Cumulative CF, $ -17,000 -37,000 -33,000 -44,000 -12,000 +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

Year 0 1 2 3 4 5 6 7 8 9 10

Cash Flow, $ 0 -5000 -5000 -5000 -5000 -5000 -5000 +9000 -5000 -5000 -5000 + 50,000

Cumulative Cash Flow, $ 0 -5,000 -10,000 -15,000 -20,000 -25,000 -30,000 -21,000 -26,000 -31,000 +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)

5

(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%

6

6.58 (a) Hand solution: ROIC procedure with ir = 14% F0 = 5000 F1 = 5000(1.14) –2000 = 3700 F2 = 3700(1.14) –1500 = 2718 F3 = 2718(1.14) – 7000 = -3901.48 F4 = -3901.48(1 + i″) + 4000

F0 > 0; use ir F1 > 0; use ir F2 > 0; use ir F3 < 0; use i″

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%. 7

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)

8

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 offset 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 1

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 2

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 Select site N

3

eliminate S

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 J to G = 1.07 I to J = 1.07 L to I = ?

eliminate DN eliminate G eliminate J

∆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)

4

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)

5

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1

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

2

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 . Cost, % change $/hour -15% 20.4 -5% 22.8 0% 24.0 +5% 25.2 +15% 27.6

VC1 2.04T 2.28T 2.40T 2.52T 2.76T

VC2 6.8T 7.6T 8.0T 8.4T 9.2T

AW1 = AW2 Breakeven relation T value, tons 4.76T = 5327 1119 5.32T = 5327 1001 5.60T = 5327 951 5.88T = 5327 906 6.44T = 5327 827

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

3

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.

4

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.

5

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 4% 6% 8%

AW1 $-2318 $-2444 $-2573

AW2 $-2234 $-2448 $-2673

Selection 2 1 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:

6

Percent Savings for A, Savings for B, variation $ per year AWA $ 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 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: 7

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: For np = 8:

0 = -10,000 +1700(5.2064) + 900(0.5835) = $-624 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

8

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)

9

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1 2 3 4 5 Economic service life is n = 4 years.

9.13

Total AW $-102,000 -84,334 -84,190 -83,857 -84,294

(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 1

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 2

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) 3

9.34

Answer is (b)

9.37 Years for Years for Defender Challenger A 0 3 B 1 2 C 2 1 Option A: AW = $-73,000 Option

AW, cash flows, $ per year Year 1 Year 2 Year 3 -73,000 -73,000 -73,000 -74,000 -84,000 -84,000 -74,000 -74,000 -84,000

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)

4

Option AW, $/year -73,000 -80,192 -76,880

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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%

1

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 2

(4.76%)

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) 3

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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

1

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 Rate Usage Annual cost (1) (2) (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 Dept

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.

2

(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. Last year; This year; flight basis baggage basis DFW $330,000 $343,620 YYZ 187,500 218,371 MEX 150,000 105,363 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)

3

Percent change + 4.1% +16.5 – 29.7

Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 Year, t Dt BVt 0 40,000 1 16,000 24,000 2 9,600 14,400 3 5,760 8,640 4 3,456 5,184 Spreadsheet solution with graphs follows.

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Book Depreciation Dt BVt 40,000 10,000 30,000 10,000 20,000 10,000 10,000 10,000 0

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 2

(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.

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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 dt

1 0.20

2 0.32

3 0.192

4 0.1152

5 0.1152

6__ 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 SL:

d4:

(selected)

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

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

Switch to SL occurs in year 4. 4

(select either)

d5:

Use the SL rate for n = 5 Accumulated D = 0.2 + 0.32 + 0.192+ 0.1152 = 0.8272 SL:

d6:

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

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 Volume, Cost depletion, Year 1000 tons $ 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)

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12.49 Salvage value is always $0 for the MACRS method. Answer is (a)

12.52 Percentage:

Cost:

GI = 65,000(40) = $2.6 million Depletion charge = 0.05(2.6 million) = $130,000 pt = $1.28 per ton Depletion charge = (65,000)(1.28) = $83,200

Answer is (b)

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Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1

Year GI E P and S CFBT 0 $-1900 $-1900 1 $800 $-100 0 700 2 950 -150 0 800 3 600 -200 0 400 4 300 -250 700 750

D

TI

$633 $ 67 845 -45 281 119 141 -91

Taxes $23 -16 42 -32

CFAT $-1900 677 816 358 782

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% 1- Te = 73.6% (0.736)

(0.264)

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

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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 0 1 2 3 4 5 6 7 8 Total

P and CFBT, $ -200,000 75,000 75,000 75,000 75,000 75,000 75,000 75,000 75,000

Rate

D, $

TI, $

Taxes, $

0.2000 0.3200 0.1920 0.1152 0.1152 0.0576 0 0

40,000 64,000 38,400 23,040 23,040 11,520 0 0

35,000 11,000 36,600 51,960 51,960 63,480 75,000 75,000

13,300 4,180 13,908 19,745 19,745 24,122 28,500 28,500 $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 3

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%

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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

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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)

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Solution to every third end-of-chapter problem Basics of Engineering Economy, 2nd edition Leland Blank and Anthony Tarquin

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 1 2 3 4 5 6

Importance 9.0 1.0 3.0 6.0 10.0 4.8 33.8

(b) Normalized weights, Wi = Score/33.8 Attribute 1 2 3 4 5 6 14.7

Wi______ 9/33.8 = 0.27 1/33.8 = 0.03 3/33.8 = 0.09 6/33.8 = 0.18 10/33.8 = 0.30 4.8/33.8 = 0.14

Estimates are for the future in engineering economic evaluations. Decision making under risk is, therefore, always present in any conclusion.

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14.10 Determine the probability values for N. N P(N)

0 0.12

1 0.56

2 0.26

3 0.032

4_ 0 .022

≥5_ 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, $ 600 800 1000 1200 1400 1600 1800 2000

P(Xi) 0.06 0.10 0.09 0.15 0.28 0.15 0.07 0.10 1.00

Xi × P(Xi), $ 36 80 90 180 392 240 126 200 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.

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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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