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Engineering Economy 16th Edition Sullivan Test Bank

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Exam Name___________________________________

SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.

Answer the question.

1) Consider the three mutually exclusive alternatives below. At the end of their useful lives, alternatives X and Z will be replaced with identical replacements so that a 10-year service requirement is met. If the MARR is 3% per year, which alternative (if any) should be chosen?

Alternative Capital Investment Annual savings Salvage value Life, years

X $300,000 $68,750 $90,000 10

Y $425,000 $108,750 $125,000 20

1)

Z $500,000 $188,750 $140,000 5

Answer: AW(X) = $41,438.00 AW(Y) = $84,840.00 AW(Z) = $105,926.00 Select the alternative with the largest positive annual worth.

Explanation: AW(X) = -300,000(A/P, 3%, 10) + 68,750 + 90,000(A/F, 3%, 10) = -300,000(0.1172) + 68,750 + 90,000(0.0872) = 41,438.00 AW(Y) = -425,000(A/P, 3%, 20) + 108,750 + 125,000(A/F, 3%, 20) = -425,000(0.0672) + 108,750 + 125,000(0.0372) = 84,840.00 AW(Z) = -500,000(A/P, 3%, 5) + 188,750 + 140,000(A/F, 3%, 5) = -500,000(0.2184) + 188,750 + 140,000(0.1884) = 105,926.00 Select the alternative with the largest positive annual worth.

2) Nadia recently moved to Celebration, Florida, an unincorporated master-planned community in Osceola County that connects directly to the Walt Disney World parks. To support the planned community's environmentally friendly transit system and to save on transportation costs, she wants to buy a new Neighborhood Electric Vehicle. She is looking into three models of the NEVs and has been provided with the following information.

Compare the alternatives shown below on the basis of rate of return and determine which model should be purchased if her personal MARR is 9% per year.

Alternative Initial costs Annual expenses Annual Saving Salvage value Life, years IRR (%)

Dynasty $13,000 $4781 $6781 $1300 10 9.64

Might-E $13,550 $5281 $7581 $1600 10 11.93

1

Zenn $14,000 $5081 $7581 $2000 10 13.26

2)

Answer: Incremental IRR (M-D) = 54.22% > 9%; Discard Dynasty Incremental IRR (Z-M) = 44.32% > 9%; Discard Mighty-E Zenn should be selected.

Explanation: Rank D, M, and Z. PW(M-D) = 0 = -13,550+ 13,000 + (7581 - 6781 - 5281 + 4781)(P/A, i%, 10)+ (1600-1300)(P/F, i%, 10) = -550 + 300 (P/A, i%, 10) +300 (P/F, i%, 10) Solve for i by interpolation and incremental IRR = 54.22% > 9%; therefore, discard Dynasty.

Or, PW(M-D, 9%) = -550 + 300 (P/A, 9%, 10) + 300 (P/F, 9%, 10) = -550 + 300(6.4177) + 300(0.4224) = 1502.03 PW(M-D, 9%) > 0; therefore, discard Dynasty.

PW(Z-M) = 0 = -14,000 + 13,550 - (5081 - 5281)(P/A, i%, 10) + (2,000-1600)(P/F, i%, 10) = -450 + 200(P/A, i%, 10) + 400(P/F, i%, 10) Solve for i by interpolation and incremental IRR = 44.32% > 9%; therefore, discard Mighty-E. Or, PW(Z-M, 9%) = -450 + 200(P/A, 9%, 10) + 400(P/F, 9%, 10) = -450 + 200(6.4177)+ 400(0.4224) = 1002.50 > 0; therefore, discard Mighty-E Zenn should be selected.

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3) A textile company is considering opening a production and shipping facility in Dallas to keep up with demand for its pillows. The 105,000-square-foot facility, if purchased, will require an initial investment of $255,000 and an annual operating cost of $68,500. It will have a $80,000 salvage value after 8 years. Alternatively, the facility can be leased with annual rent of $51,000 in year 1 and increasing by $1000 per year. If the company's minimum attractive rate of return is 6% per year, compounded quarterly, should the facility be purchased or leased?

3)

Answer: AW (Purchase) = -$101,744.50 AW (Lease) = -$54,188.40 Select the alternative with the least negative annual worth.

Explanation: Effective interest rate = (1 + r/M)M - 1 = (1+ 0.06/4)4 -1 = 0.0614 or 6.14% per year

AW (Purchase) = -255,000(A/P, 6.14%, 8) - 68,500 + 80,000(A/F, 6.14%, 8) = -255,000(0.1619) - 68,500 + 80,000(0.1005) = -101,744.50 AW (Lease) = -51,000 - 1000(A/G, 6.14%, 8) = -51,000 - 1000(3.1884) = -54,188.40 Select the alternative with the least negative annual worth.

4) Consider the three mutually exclusive alternatives below. Determine which alternative is preferable at an interest rate of 9% per year.

Alternative Capital investment Annual expense Annual revenue Salvage value Life, years

M $400,000 $189,000 $309,000 $65,000 24

N $4000 $94,500 $194,500 -

P $150,000 $134,000 $234,000 $130,000 12

Answer: AW(M) = $79,645.00 AW(N) = $99,640.00 AW(P) = $85,506.00 The alternative with the largest positive annual worth should be selected.

Explanation: AW(M) = -400,000(A/P, 9%, 24) - 189,000 + 309,000 + 65,000(A/F, 9%, 24) = -400,000(0.103) + 120,000 + 65,000(0.013) = 79,645.00 AW(N) = -4000 x 0.09 - 94,500 + 194,500 = 99,640.00 AW(P) = -150,000(A/P, 9%, 12) - 134,000 + 234,000 + 130,000(A/F, 9%, 12) = -150,000(0.1397) + 100,000 + 130,000(0.0497) = 85,506.00 Select the alternative with the largest positive annual worth.

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5) Two flash vaporizer machines are considered for the upgrade of biodiesel production. An engineer is asked to perform analyses to select the best machine. He prepares the following information for the evaluation. Machine X has a useful life of 8 years and machine Y has a useful life of 11 years. Compute the market value of Machine Y at the end of year 8 and determine which machine should be selected based on annual worth method using an interest rate of 13% per year and a study period of 8 years.

Machine First costs Net annual revenue Market value at the end of the useful life Life, years

X $29,000 $10,500

Y $32,000 $12,000

$6900 8

$6000 11

Answer: MV at the end of year 8 = $16,792.91 AW(X) = $4997.36 AW(Y) = $6647.76 Select machine Y. Explanation: Compute the PW at the end of year 8 of the remaining CR amounts of Y. PW (CR) = [32,000(A/P, 13%, 11) - 6000(A/F, 13%, 11)](P/A, 13%, 3) = [32,000(0.1758) - 6000(0.0458)](2.3612) = 12,634.31 Compute the PW at the end of year 8 of the original MV at the end of useful life.

PW (MV) = 6000(P/F, 13%, 3) = 6000(0.6931) = 4158.60 Thus, MV at the end of year 8 = PW(CR)+ PW(MV) = [(5625.60) - (6000*0.0458)] * 2.3612 + (4158.60) = 16,792.91 AW (X) = -29,000(A/P, 13%, 8) + 10,500 + 6900(A/F, 13%, 8) = -29,000(0.2084) + 10,500 + 6900(0.0784) = 4997.36 AW(Y) = -32,000(A/P, 13%, 8) + 12,000 + 16,792.91(A/F, 13%, 8) = -32,000(0.2084) + 12,000 + 16,792.91(0.0784) = 6647.76 AW(Y) > AW(X); therefore, select machine Y.

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6) A large textile company is trying to decide among three alternatives of sludge dewatering processes. The costs associated with these alternatives are shown below. Alternative Y will need an upgrade of $9700 at the end of year 2. At the end of year 2, alternative Z would be replaced with another alternative Z having the same installed and operating costs. If the MARR is 14% per year, which alternative should be chosen?

Alternative Installed costs Annual operating costs Overhaul cost in year 2 Salvage value Useful life, years

X $68,500 $6000 $33,250 8

Y $48,500 $4000 $9700 $28,250 4

Z $33,500 $5000 $15,750 2

Answer: AW (X) = -$18,254.90 AW(Y) = -$17,466.50 AW (Z) = -$17,984.57 Select the alternative with the least negative annual worth.

Explanation: AW(X) = -68,500(A/P, 14%, 8) - 6000 + 33,250(A/F, 14%, 8) = -68,500(0.2156) - 6000 + 33,250(0.0756) = -18,254.90 AW(Y) = -48,500(A/P, 14%, 4) - 4000 - 9700(P/F, 14%, 2)(A/P, 14%, 4) + 28,250(A/F,14%, 4) = -48,500(0.3432) - 4000 - 9700(0.7695)(0.3432) + 28,250(0.2032) = -17,466.50 AW(Z) = -33,500(A/P, 14%, 2) - 5000 + 15,750(A/F, 14%, 2) = -33,500(0.6073) -5000 + 15,750(0.4673) = -17,984.57 Select the alternative with the least negative annual worth.

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7) A leading cancer research institution received a gift for cancer research. The institution is deciding whether to use the gift to create an endowment fund or to build a new research facility. If the gift is used to establish an endowment fund, the fund would award research funding and scholarships totaling $9.75 million per year with the first awards to be granted at the end of next year and continue each year perpetually. Alternatively, if the gift is used to building an integrative cancer research facility, the construction of the facility will cost $16.5 million and the building is expected to be renovated every 10 years at a cost of $3.3 million. The annual maintenance costs are expected to be $1 million. If the facility is expected to last forever, which alternative should be selected at an interest rate of 2% per year?

Answer: CW (Fund) = -$487,500,000.00 CW (Building) = -$81,564,500.00 Select the alternative with the least negative capitalized worth.

Explanation: CW (Fund) = -9.750000000 × 106/0.02 = -487,500,000.00

CW (Building) = -16,500,000 - 1,000,000/0.02 - [3,300,000(A/F, 2%, 10]/0.02 = -16,500,000 - 50,000,000 - 3,300,000(0.0913)/0.02 = -81,564,500.00 Select the alternative with the least negative capitalized worth.

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8) An insulated shipping container supplier is considering expanding its product line. Three materials with different insulation properties are under consideration. The following information is prepared for the economic evaluation of the best material. If the company's MARR is 4% per year and the study period is 5 years, use an AW-based incremental rate of return equation to determine which alternative is preferred. Assume the salvage value is negligible.

Material First costs Net annual revenue

IRR (%) Incremental IRR (%) Q R S

Q $46,000 $10,000 in year 1, increasing by $100 per year thereafter 5.93

R $58,000 $11,900, increasing by $450 per year thereafter 5.15

2.50% 2.79%

5.09%

S $61,000 $13,450

5.14

-

Answer: AW (R-Q) = 0 = -12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5) IRR (R-Q) = 2.50% < 4%; therefore, discard material R. AW(S-Q) = 0 = -15,000(A/P, i%, 5) + 3450 - 100 (A/G, i%, 5) IRR (S-Q) = 2.79% < 4%; therefore, discard material S. Select material Q. Explanation: Rank Q, R and S AW (R-Q) = 0 = [-58,000 - (-46,000)](A/P, i%, 5) + 1900 + (450-100) (A/G, i%, 5) = -12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5) IRR (R-Q) = 2.50% < 4%; therefore, discard material R. AW(S-Q) = 0 = [-61,000 - (-46,000)](A/P, i%, 5) + 3450 - 100 (A/G, i%, 5) = 0 = -15,000(A/P, i%, 5) + 3450 - 100 (A/G, i%, 5) IRR (S-Q) = 2.79% < 4%; therefore, discard material S. Select material Q.

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9) A dentist is deciding between two X-ray machines for his new office. Estimated costs for each machine are the following: Machine Installed cost Annual maintenance cost Market value at year 6 Life, years

A $36,000 $1250 $5750 6

B $38,000 $950 $6250 6

Which machine should be recommended based on the annual worth method? Use a MARR of 15% and a study period of 6 years.

Answer: AW(A) = -$10,104.55 AW(B) = -$10,275.85 Therefore machine A should be recommended.

Explanation: AW(A) = -36,000(A/P, 15%, 6) - 1250 + 5750(A/F, 15%, 6) = -36,000(0.2642) - 1250 + 5750(0.1142) = -10,104.55

AW(B) = -38,000(A/P, 15%, 6) - 950 + 6250(A/F, 15%, 6) = -38,000(0.2642) - 950 + 6250(0.1142) = -10,275.85 AW(A) > AW(B) ;therefore, machine A should be recommended.

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10) The Pentagon is deciding between two models of light-duty support vehicles proposed by two high-technology defense contractors. The costs associated with each model are shown below. Which model should be selected if the MARR is 16% per year based on the future worth method?

Alternative Initial R&D costs Non-recurring investment costs (Years 1, 3, and 5) Recurring costs

Annual maintenance costs Disposal costs Life, years

A $481,000 $47,000

B $496,000 $72,000

$15,000 in year 4, increasing by $500 until year 10 $3000

$16,500 in year 4, increasing by $400 until year 10 $5000

$3000 14

$3500 14

Answer: FW(A) = -$5,053,855.29 FW(B) = -$5,682,668.31 The alternative with the least negative future worth should be selected.

Explanation: FW(A) = -481,000(F/P, 16%, 14) - 47,000(F/P, 16%, 13) - 47,000(F/P, 16%, 11) - 47,000(F/P, 16%, 9) - 15,000(F/A, 16%, 7)(F/P, 16%, 4) - 500(P/G, 16%, 7)(F/P, 16%, 11) - 3000(F/A, 16%, 14) - 3000 = -481,000(7.9875) - 47,000(6.8858) - 47,000(5.1173) - 47,000(3.803) - 15,000(11.4139)1.8106 - 500(9.761)(5.1173) - 3000(43.672) - 3000 = -5,053,855.29 FW(B) = -496,000 (F/P, 16%, 14) - 72,000(F/P, 16%, 13) - 72,000(F/P, 16%, 11) - 72,000(F/P, 16%, 9) - 16,500(F/A, 16%, 7)(F/P, 16%, 4) - 400(P/G, 16%, 7)(F/P, 16%, 11) - 5000(F/A, 16%, 14) - 3500 = (-496,000 * 7.9875) - 72,000(6.8858) - 72,000(5.1173) - 72,000(3.803) - 16,500(11.4139)1.8106 - 400(9.761)(5.1173) - 5000(43.672) - 3500 = -5,682,668.31 Select the alternative with the least negative future worth.

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11) Cassandra sets up a savings plan for her retirement. She plans to make a quarterly payment of $3250 into a savings account that earns 11% per year, compounded quarterly.

After 3 years, she will have an option to continue making $3250 quarterly payments or to switch to an annual savings plan that earns higher interest of 11.25% per year and requires annual payments of $13,000. If she plans to retire 13 years from now, which option will offer more money in the savings plan at that time?

Answer: FW(quarterly payment) = $231,621.33 FW(annual payment) = $220,021.10 Continuing to make quarterly payment offers more money at the time of retirement.

Explanation: Effective interest rate = 0.11/4 = 0.0275 or 2.75% per quarter FW(quarterly payment) = 3250(F/A, 2.75%, 40) = 3250(71.2681) = 231,621.33 FW (annual payment)

= 13,000(F/A, 11.25%, 10) = 13,000(16.9247) = 220,021.10

FW (quarterly payment) > FW(annual payment); therefore, continuing to make quarterly payments offers more money at the time of retirement.

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12) Compare the alternatives shown below on the basis of their future worth, using an interest rate of 18% per year. Which alternative should be selected?

Alternative Initial costs Annual revenues

L $238,000 $69,000 in year 1, increasing by $190 each year

M $213,000 $114,000

Annual expenses Salvage value Life, years

$20,000 $6000 16

$25,000 $9000 8

12)

N $293,000 $69,000 in years 1 to 5, $69,570 in years 6 to 16 $20,000 $6000 16

Answer: FW(L) = $277,411.15 FW(M) = $2,724,278.40 FW(N) = -$543,414.65 The alternative with the largest positive future worth should be selected.

Explanation: FW(L) = -238,000(F/P, 18%, 16)+[69,000 - 20,000](F/A, 18%, 16) + 190(P/G, 18%, 16)(F/P, 18%, 16) + 6000 = -238,000(14.129) + [69,000 - 20,000](72.939) + 190(22.3885)(14.129) + 6000 = 277,411.15 FW(M) = -213,000(F/P, 18%, 16)+[114,000 - 25,000](F/A, 18%, 16)+ [9000 213,000](F/P, 18%, 8) + 9000 = -213,000(14.129) + [89,000](72.939) + [-204,000](3.7589) + 9000 = 2,724,278.40 FW(N) = -293,000(F/P, 18%, 16) + 69,000(F/A, 18%, 5)(F/P, 18%, 11) + 69,570(F/A, 18%, 11) - 20,000(F/A, 18%, 16) + 6000 = -293,000(14.129) + 69,000(7.1542)(6.1759) + 69,570(28.7551) - 20,000( 72.9390) + 6000 = -543,414.65 The alternative with the largest positive future worth should be selected.

13) A thermoplastic film manufacturer is trying to decide between four types of thermoforming molding processes to be added to its molding operation. The estimated costs and revenue are shown below. Compare them on the basis of rate of return and determine which process should be selected if the company's MARR is 6% per year.

Alternative Initial costs Annual expenses Annual revenue Salvage value Life, years IRR (%)

Vacuum forming $31,000 $3300 $7300 $3100 8 2.66

Pressure forming $51,000 $3000 $11,000 $5100 8 6.87

11

Drape forming $43,000 $3200 $9700 $4300 8 6.03

Free blowing $45,000 $3800 $10,800 $4500 8 6.68

13)

Answer: IRR (V) = 2.66% < 6%; Discard vacuum forming.

Incremental IRR (F-D) = 19.43% > 6%; Discard drape forming.

Incremental IRR (Z-M) = 8.30% > 6%; Discard free blowing.

Pressure forming should be selected.

Explanation: Rank V, D, F and P IRR (V) = 2.66% < 6%; therefore, vacuum forming is discarded.

PW (F-D) = 0 = -45,000 + 43,000 + (10,800 - 3800 - 9700 + 3200)(P/A, 6%, 8) + ( 4500 - 4300) (P/F, 6%, 8) = -2,000 + 500 (P/A, i%, 8) + 200 (P/F, i%, 8) Solve for i by interpolation and incremental IRR = 19.43% > 6%; therefore, discard drape forming.

Or PW (F-D) = -2,000 + 500 (P/A, 6%, 8) + 200 (P/F, 6%, 8) = -2,000 + 500(6.2098) + 200(P/F, 6%, 8) = -2,000 + 500(6.2098) + 200(0.6274) = 1230.38 > 0 PW(F-D, 6%) > 0; therefore, discard drape forming.

PW (P-F) = 0 = -51,000 + 45,000 + (11,000 - 3000 - 10,800 + 3800)(P/A, 6%, 8) + ( 5100- 4500) (P/F, 6%, 8) = -6,000 + 1,000 (P/A, i%, 8) + 600 (P/F, i%, 8) Solve for i by interpolation and incremental IRR = 8.30% > 6%; therefore, discard free blowing.

Or PW (P-F) = -6,000 + 1,000 (P/A, 6%, 8) + 600 (P/F, 6%, 8) = -6,000 + 1,000(6.2098) + 600(P/F, 6%, 8) = -6,000 + 1,000(6.2098) + 600(0.6274) = 586.24 > 0 PW(P-F, 6%) > 0; therefore, discard free blowing.

Select pressure forming.

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14) HealthRx, a major manufacturer of non-invasive breast and cervical cancer detection products is planning to expand its market in Asia. Three single-patient-use disposable devices based on its proprietary technology to identify cancers and precancers painlessly and noninvasively by scanning the cervix with light are being considered. The costs to manufacture these devices are the following: Alternative Initial costs Annual operating costs Salvage Value Sell Price per unit

Soft Touch $500,000 $15,000 $70,000 $3500

Light Touch $520,000 $14,500 $80,000 $3600

Gentle Touch $535,000 $14,000 $85,000 $3650

If the company sells 1000 units per year, which device should be selected based on the present worth method? Assume the company uses a MARR of 11% and wants to recover its investment in 7 years.

Answer: PW(ST) = $15,955,736.00 PW(LT) = $16,414,129.10 PW(GT) = $16,639,503.70 Therefore, Gentle Touch should be selected.

Explanation: PW(ST) = -500,000 - 15,000(P/A, 11%, 7) + (3500)(1000)(P/A, 11%, 7) + 70,000(P/F, 11%, 7) = -500,000 + (-15,000 + 3,500,000)(4.7122) + 70,000(0.4817) = 15,955,736.00 PW(LT) = -520,000 - 14,500(P/A, 11%, 7) + (3600)(1000)(P/A, 11%, 7) + 80,000(P/F, 11%, 7) = -520,000 + (-14,500 + 3,600,000)(4.7122) + 80,000(0.4817) = 16,414,129.10 PW(GT) = -535,000 - 14,000(P/A, 11%, 7) + (3650)(1000)(P/A, 11%, 7) + 85,000(P/F, 11%, 7) = -535,000 + (-14,000 + 3,650,000)(4.7122) + 85,000(0.4817) = 16,639,503.70 PW(ST) < PW (LT) < PW (GT); therefore, Gentle Touch should be selected.

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15) Two grandparents are considering purchasing a baby bond for their first grandson. They are deciding between two bonds with the same face value of $42,000. Both bonds are offered at the same price of $37,000. Bond A has interest of 2.5% per year, payable quarterly, and matures in 6 years. Bond B, issued 2 years ago, has interest of 2.75% per year, payable semiannually, and a 8-years maturity date. If the current market rate is 3% per year, compounded quarterly, which bond should be purchased?

15)

Answer: PW(A) = $40,849.49 PW(B) = $41,381.95 Bond B should be purchased.

Explanation: From VN = C(P/F, i%, N) + rZ(P/A, i%, N) Bond A i = 0.03/4= 0.01 or 0.75% per quarter r = (2.5%)/4 = (0.625%) N = 6*4 periods remaining in the life of the bond.

PW(A) = 42,000(P/F, 0.75%, 24) + 262.5(P/A, 0.75%, 24) = 42,000(0.8358) + 262.5(21.8891) = 40,849.49 Bond B i = (1 + r/M)M - 1 = (1 + 0.015/2)2 -1 = 0.0001 or 1.51% per semiannually r = (2.75%)/2 = (1.375%) N = 6*2 periods remaining in the life of the bond.

PW(B) = 42,000(P/F, 1.51%, 12) + 577.50(P/A, 1.51%, 12) = 42,000(0.8354) + 577.50(10.9007) = 41,381.95 PW(B) >PW(A) and PW(B) > 37,000; therefore, bond B should be purchased.

16) You are deciding between three types of water heaters. The associated costs are shown below. At the end of year 5, the electricity and oil water heaters need to be replaced with the same model at the purchased prices. Which alternative should be selected on the basis of their future worth at an interest rate of 12% per year?

Alternative Price of water heater EF Fuel cost Annual Maintenance costs Life, years

Electricity $32,000

Gas $27,000

Oil $29,000

2.0 $0.075 /kWh $1000

0.57 $0.000009/Btu $200

0.75 $0.000009/Btu $500

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10

5

Hint: The estimated annual cost of operation for gas and oil heaters equals 365 x 41045/EF x Fuel Cost per Btu and the estimated annual cost of operation for electric water heaters equals 365 x 12.03/EF x Electricity Cost per kWh.

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Answer: FW(Electric) = -$176,217.48 FW(Gas) = -$91,517.46 FW(Oil) = -$153,104.10 The alternative with the least negative future worth should be selected.

Explanation: FW(Electric) = -32,000(F/P, 12%, 10)- [1000 + 164.660625](F/A, 12%, 10) - 32,000( F/P, 12%, 5) = -32,000(3.1058) - [1000 + 164.660625](17.5487) - 32,000(1.7623) = -176,217.48 FW(Gas) = -27,000(F/P, 12%, 10) - [200 + 236.548816](F/A, 12%, 10) = -27,000(3.1058) - [436.548816](17.5487) = -91,517.46 FW(Oil) = -29,000(F/P, 12%, 10)- [500 + 179.7771](F/A, 12%, 10) - 29,000(F/P, 12%, 5) = -29,000(3.1058) - [500 + 179.7771](17.5487) - 29,000(1.7623) = -153,104.10 Select the alternative with the least negative future worth.

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17) Two delivery methods have been proposed for a standard delivery method for new bridge constructions in Oregon to overcome traditional design-bid-build models that do not adequately account for site conditions and constructability, and often lead to additional expenses. One alternative must be selected to represent the delivery method specified in the RFP. Based on data from recent bridge construction projects of comparable size, we can estimate the savings from reduced redesign costs, risk management costs, and overhead costs during the life of the construction contract over the traditional delivery model. The estimated costs of a 18-month contract are the following: Alternative Preconstruction and design costs Monthly savings Project life, months

Construction manager /General contractor $80,000 $4000 18

Design-Build $90,000 $6500 18

Which alternative should be selected based on the present worth method? Use a MARR of 4%, compounded monthly. Assume other benefits and costs are negligible.

Answer: PW(CM/GC) = -$10,208.40 PW(DB) = $23,411.35 Therefore, the Design-Build method should be selected.

Explanation: Effective interest rate = 0.04/12 = 0.0033 or 0.33%

PW(CM/GC) = -80,000 + 4000(P/A, 0.33%, 18) = -80,000 + 4000(17.4479) = -10,208.40 PW(DB) = -90,000 + 6500(P/A, 0.33%, 18) = -90,000 + 6500(17.4479) = 23,411.35 PW(CM/GC) < PW(DB); therefore, the Design-Build method should be selected.

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18) A manufacturing company is deciding between three maintenance plans for a new waste management system. Plan A needs a single prepayment of $59,000 at the beginning of the year and the contract needs to be renewed every 3 years. Plan B is a two-year contract and requires a payment of $19,470 at the end of year 1 and another payment of $20,060 at the end of year 2. Plan C provides a three-year services with two payments of $29,500 made at the end of years 1 and 3. Which maintenance plan should be selected based on the present worth method? Assume the company uses a MARR of 11% and a study period of 6 years.

Answer: PW(A) = -$102,140.80 PW(B) = -$83,550.73 PW(C) = -$83,349.30 Therefore, plan C should be selected.

Explanation: Compare alternatives over a study period of 6 years.

PW(A) = -59,000 - 59,000(P/F, 11%, 3) = -59,000 - 59,000(0.7312) = -102,140.80 PW (B) = -19,470(P/F, 11%, 1) - 20,060(P/F, 11%, 2) - 19,470(P/F, 11%, 3) 20,060(P/F, 11%, 4) - 19,470(P/F, 11%, 5) - 20,060(P/F, 11%, 6) = -19,470(0.9009) - 20,060(0.8116) - 19,470(0.7312) - 20,060(0.6587) - 19,470(0.5935) - 20,060(0.5346) = -83,550.73 PW(C) = -29,500(P/F, 11%, 1) - 29,500(P/F, 11%, 3)- 29,500(P/F, 11%, 4) 29,500(P/F, 11%, 6) = -29,500(0.9009) - 29,500(0.7312) - 29,500(0.6587) - 29,500(0.5346) = -83,349.30 PW(A) < PW (B) < PW (C); therefore, plan C should be selected.

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19) Two advanced thermal insulating and anti-condensation protection alternatives have been proposed for new Antarctica marine vessels subject to the harsh marine environment. One alternative must be selected. Estimated savings from reduced total installation and maintenance costs over conventional insulation are the following: Alternative Installed cost Annual savings Life, years

Delta-T $13,000 $4750 8

Alpha-B $15,000 $6250 8

Which alternative should be recommended based on the present worth method? Use a MARR of 18 percent and a study period of 8 years. Assume negligible salvage values.

Answer: PW(Delta-T) = $6370.50 PW(Alpha-B) = $10,487.50 Therefore Alpha-B should be recommended.

Explanation: PW(Delta-T) = -13,000 + 4750(P/A, 18%, 8) = -13,000 + 4750(4.078) = 6370.50

PW(Alpha-B) = -15,000 + 6250(P/A, 18%, 8) = -15,000 + 6250(4.078) = 10,487.50 PW(Alpha-B) > PW(Delta-T); therefore, Alpha-B should be recommended.

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20) Two structural designs for a large public monument in Indonesia are under evaluation.

Use the repeatability assumption and the CW method to determine which design should be selected if the service period of the monument is indefinite and the interest rate is 2% per year.

Alternative Initial costs Annual maintenance costs Usual life, years

A $310,000 $23,000 8

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B $325,000 $24,500 11

Answer: CW (A) = -$57,715.00 CW (B) = -$2,885,750.00 Design B should be selected.

Explanation: AW(A) = -310,000(A/P, 2%, 8) - 23,000 = -310,000(0.1365) - 23,000 = -65,315.00 CW (A) = AW/0.02 = (-65,315.00)/(0.02) = -3,265,750.00 AW(B) = -325,000(A/P, 2%, 11) - 24,500 = -325,000(0.1022) - 24,500 = -57,715.00 CW (B) = AW/0.02 = (-57,715.00)/(0.02) = -2,885,750.00 CW(B) > CW(A); therefore, the design B should be selected.

21) A medical mobility equipment manufacturer is considering two alternatives as part of an upgrade of its power wheelchairs assembly. Alternative A has an installed cost of $10,000, net annual revenue of $6000, and a useful life of 3 years. Alternative B has an installed cost of $20,000, net annual revenue of $6350 and a useful life of 6 years. At the end of year 3, alternative A would be replaced with another alternative A having the same installed cost and net annual revenues. If the MARR is 8% per year, which alternative (if any) should be selected based on the present worth method? Assume negligible salvage value.

Answer: PW(A) = $9799.40 PW(B) = $9355.42 Therefore, alternative A should be selected.

Explanation: PW(A) = -10,000 - 10,000(P/F, 8%, 3) + 6000(P/A, 8%, 6) = -10,000 - 10,000(0.7938) + 6000(4.6229) = 9799.40

PW (B) = -20,000 + 6350 (P/A, 8%, 6) = -20,000 + 6350(4.6229) = 9355.42 PW(A) > 0 and PW(A) > PW (B) ; therefore, alternative A should be selected.

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22) A slip sheet manufacturer is considering two machines. An engineer is asked to perform analyses to select the best machine. She prepares the following information for the evaluation. All machines have a useful life of 5 years. If the company's MARR is 4% per year, which machine should be selected.

Machine First costs Annual expenses Annual revenue Salvage value IRR (%)

X $52,000 $5750 $15,250 $23,400 9.10

Y $37,000 $8050 $15,750 $5550 5.62

Answer: Incremental IRR (X-Y) = 14.83% > 4%; discard machine Y Machine X should be selected.

Explanation: Rank Y and X. PW(X-Y) = 0 = -52,000 + 37,000 + (-5750 + 8050 + 15,250 - 15,750)(P/A, i%, 5) + (23,400 - 5550)(P/F, i%, 5) = -15,000 + 1,800 (P/A, i%, 5) + (17,850) (P/F, i%, 5) Solve for i by interpolation and incremental IRR = 14.83% > 4%; therefore, discard machine Y. Or, PW(X-Y, 4%) = -15,000 + 1,800 (P/A, 4%, 5) + (17,850)(P/F, 4%, 5) = -15,000 + 1,800 (4.4518) + (17,850)(0.8219) = 7684.16 > 0 Select machine X.

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23) A manufacturer of automated optical inspection (AOI) devices is deciding on a project to increase the productivity of the manufacturing processes. The estimated costs for the two feasible alternatives being compared are shown below. Use the ERR method to determine which alternative should be selected if the analysis period is 8 years and the reinvestment rate equals the company's MARR of 4% per year.

Alternative Initial costs Net annual cash flow Life, years

M $30,000 $4,500 8

N $45,000 $7,000 8

Answer: ERR (M) = 4.11% > 4% ERR (N-M) = 5.55% > 4% Select project N. Explanation: Rank M and N. Find the ERR of project M. N N Ek (P/F, %, k)(F/P,i %, N) = Rk (F/P, %, N - k) k=0 k=0 30,000(F/P, i %, 8) = 4,500(F/A, 4%, 8) (1+ i )8 = 4,500(9.2142)/30,000 (1+ i ) = (1.38)1/8 i = 0.0411

Using a MARR of 4%, project M is justified.

Next, find the incremental ERR of N-M. 15000(F/P, i %, 8) = 2,500(F/A, 4%, 8) (1+ i )8 = 2,500(9.2142)/ 15,000 (1+ i ) = (1.54)1/8 i = 0.0555 i > 4%; therefore, project N is justified.

Select Project N.

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24) A major bakery-cafe chain is evaluating whether they should consolidate its two offices into one location when the two leases expire. In addition, the company also needs to decide if they want to purchase or lease the new location. The estimated costs for these three alternatives are as follows: Alternative

Initial costs Annual lease Annual maintenance costs Lease period Market value at the end of year 12

Continue Leasing Separate Offices $53,550 4 -

Lease Combined Office $63,000 12 -

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Purchase Combined Office $270,000 $6000 $216,000

Which alternative should be selected based on the annual worth method? Use a MARR of 11% and a study period of 12 year(s).

Answer: AW(Separate Lease) = -53,550 AW(Combined Lease) = -63,000 AW(Purchase) = -$38,076.00 The alternative with the least negative annual worth should be selected.

Explanation: AW(Separate Lease) = -53,550 AW(Combined Lease) = -63,000 AW(Purchase) = -270,000(A/P, 11%, 12) - 6000 + 216,000(A/F, 11%, 12) = -270,000(0.154) - 6000 + 216,000(0.044) = -38,076.00 Select the alternative with the least negative annual worth.

25) A major defense supplier is planning for the manufacturing of its Desert Hawk VII Unmanned Aerial Systems, a hand-launched air vehicle that provides intelligence, reconnaissance, and surveillance capabilities. Three manufacturing facilities are under consideration. Each site provides the same manufacturing capability but has different costs to manufacture and transport the system to the customers. The estimated costs are provided in the table below.

Alternative Set up costs Annual operating costs Salvage Value, at end of year8 Transportation costs per unit Useful lives, years

Texas Plant $720,000 $27,500 $420,000 $2000 8

Utah Plant $770,000 $22,500 $430,000 $2250 9

Minnesota Plant $755,000 $24,500 $445,000 $2300 10

If the company has to deliver 300 units per year for 8 years, which facility should be selected based on the present worth method? Assume the company uses a MARR of 5% and a study period of 8 years.

Answer: PW(TX) = -$4,491,402.00 PW(UT) = -$4,987,058.00 PW(MN) = -$5,071,780.40 Therefore, the Texas plant should be selected.

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Explanation: Compare alternatives over a study period of 8 years.

PW(TX) = -720,000 - 27,500(P/A, 5%, 8) - (2000)(300)(P/A, 5%, 8) + 420,000(P/F, 5%, 8) = -720,000 - (27,500 + 600,000)(6.4632) + 420,000(0.6768) = -4,491,402.00 PW(UT) = -770,000 - 22,500(P/A, 5%, 8) - (2250)(300)(P/A, 5%, 8) + 430,000(P/F, 5%, 8) = -770,000 - (22,500 + 675,000)(6.4632) + 430,000(0.6768) = -4,987,058.00 PW(MN) = -755,000 - 24,500(P/A, 5%, 8) - (2300)(300)(P/A, 5%, 8) + 445,000(P/F, 5%, 8) = -755,000 - (24,500 + 690,000)(6.4632) + 445,000(0.6768) = -5,071,780.40 PW(TX) > PW (UT) > PW (MN); therefore, the Texas plant should be selected.

26) VB Flantronics Inc. is a company that designs, makes, and sells computer and home entertainment sound systems, along with a line of headphones and microphones for personal digital media. The company is trying to decide whether it should purchase or lease a building for its manufacturing and research-and-development operation in China.

If the building is leased, a payment will have to be made at the beginning of each year.

The estimated costs are the following: Alternative Initial Cost Lease Annual Operating Costs Salvage Value Life, years

Purchase $320,000 $8500 $80,000 7

Lease $40,000 $7000 1

Which alternative should be recommended based on the present worth method? Use a MARR of 9%.

Answer: PW(Purchase) = -$319,020.50 PW(Lease) = -$254,669.80 Therefore, the facility should be leased.

Explanation: PW(Purchase) = -320,000 - 8500(P/A, 9%, 7) + 80,000(P/F, 9%, 7) = -320,000 - 8500(5.033) + 80,000(0.547) = -319,020.50 A lease payment is made at the beginning of the year, whereas annual operating expense is paid at the end of the year.

PW(Lease) = (-40,000)(P/A, 9%, 7)(F/P, 9%, 1) - (7000)(P/A, 9%, 7) = -40,000(5.033)(1.09) - (7000)(5.033) = -254,669.80 PW (Lease) > PW(Purchase); therefore, the facility should be leased.

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27) A manufacturer of Neighborhood Electric Vehicles (NEVs) plans to upgrade its parts inventory tracking system. Two alternatives are under consideration. The estimated costs of each alternative are provided below. At the end of 9 years, alternative E will need to be upgraded with additional costs of $19,300 over the initial cost of the system. The upgrade will reduce the annual maintenance costs by $2000 over the next 9 years. Which alternative should be selected on the basis of their future worth at an interest rate of 17% per year and a study period of 18 years? Assume negligible salvage value.

Alternative First costs Annual maintenance costs Life, years

E $386,000

F $401,000

$60,000 9

$58,500 18

Answer: FW(E) = -$13,748,192.81 FW(F) = -$12,232,706.60 The alternative with the least negative future worth should be selected.

Explanation: FW(E) = -386,000(F/P, 17%, 18) - 60,000(F/A, 17%, 9)(F/P, 17%, 9) - 405,300(F/P, 17%, 9) - 58,000(F/A, 17%, 9) = -386,000(16.879) - 60,000(18.2847)(4.1084) - 405,300(4.1084) - 58,000( 18.2847) = -13,748,192.81

FW(F) = -401,000(F/P, 17%, 18) - 58,500(F/A, 17%, 18) = -401,000(16.879) - 58,500(93.4056) = -12,232,706.60 Select the alternative with the least negative future worth.

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28) In response to the new federal regulation to provide the public with safe and wholesome seafood, an Alaska seafood processor is considering two new sanitation control systems to help monitor its seafood processing operations. System X has a useful life of 20 years and requires an installed cost of $33,000 and annual maintenance cost of $7000. Some of the equipment can be sold at $2200 at the end its useful life. System Y has a useful life of 10 years. The system will cost $16,500 to install and will involve an annual maintenance fee of $8500. At the end of year 10, system Y can be upgraded for $23,100 to have the same capability and will last another 10 years. However, the upgraded system will require an annual maintenance fee of $8700. The salvage value for the system Y is negligible. Which system should be selected based on the present worth method? Assume the company uses a MARR of 5% per year.

28)

Answer: PW(A) = -$119,406.22 PW(B) = -$137,556.60 Therefore, system X should be selected.

Explanation: PW(X) = -33,000 - 7000(P/A, 5%, 20) + 2200(P/F, 5%, 20) = -33,000 - 7000(12.4622) + 2200(0.3769) = -119,406.22

PW (Y) = -16,500 - 8500(P/A, 5%, 10) - 23,100(P/F, 5%, 10) - 8700(P/A, 5%, 10)(P/F, 5%, 10) = -16,500 - 8500(7.7217) - 23,100(0.6139) - 8700(7.7217)(0.6139) = -137,556.60 PW(X) > PW(Y); therefore, system X should be selected.

29) A manufacturing firm is considering two models of lathes. Model A will have an initial cost of $29,000, an operating cost of $2750, and a salvage value of $6500 after 6 years.

Model B will have an initial cost of $36,500, an operating cost of $2200, and a $7750 resale value after 12 years. At an interest rate of 10% per year, which model should the consulting firm buy?

Answer: AW(A) = -$8566.00 AW(B) = -$7195.50 The alternative with the least negative annual worth should be selected.

Explanation: Use the annual worth method.

AW(A) = -29,000(A/P, 10%, 6) - 2750 + 6500(A/F, 10%, 6) = -29,000(0.2296) - 2750 + 6500(0.1296) = -8566.00 AW(B) = -36,500(A/P, 10%, 12) - 2200 + 7750(A/F, 10%, 12) = -36,500(0.1468)- 2200 + 7750(0.0468) = -7195.50 Select the alternative with the least negative annual worth.

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30) Compare the alternatives shown below on the basis of their future worth, using an interest rate of 14% per year, compounded quarterly. Which alternative should be selected?

Alternative Initial costs Quarterly revenues Quarterly expenses Disposal costs Life, years

A $421,000 $46,000 $17,000 $6750 12

B $1,171,000 $71,000 $18,500 $6250 24

Answer: FW(A) = $8,012,971.00 FW(B) = $7,436,476.25 The alternative with the largest positive future worth should be selected.

Explanation: Effective interest rate = 0.14/4 = 0.0350 or 3.5% per quarter FW(A) = -421,000(F/P, 3.5%, 96) + [(46,000) - (17,000)](F/A, 3.5%, 96) - [ 421,000 + 6750](F/P, 3.5%, 48) - 6750 = -421,000(27.1815) + 29,000(748.0431) - (427,750)(5.2136) - 6750 = 8,012,971.00 FW(B) = -1,171,000(F/P, 3.5%, 96) + [(71,000)-(18,500)](F/A, 3.5%, 96) - 6250 = -1,171,000(27.1815) + (52,500)(748.0431) - 6250 = 7,436,476.25 Select the alternative with the largest positive future worth.

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Answer Key Testname: C6

1) AW(X) = $41,438.00 AW(Y) = $84,840.00 AW(Z) = $105,926.00 Select the alternative with the largest positive annual worth.

2) Incremental IRR (M-D) = 54.22% > 9%; Discard Dynasty Incremental IRR (Z-M) = 44.32% > 9%; Discard Mighty-E Zenn should be selected.

3) AW (Purchase) = -$101,744.50 AW (Lease) = -$54,188.40 Select the alternative with the least negative annual worth.

4) AW(M) = $79,645.00 AW(N) = $99,640.00 AW(P) = $85,506.00

The alternative with the largest positive annual worth should be selected.

5) MV at the end of year 8 = $16,792.91 AW(X) = $4997.36 AW(Y) = $6647.76 Select machine Y. 6) AW (X) = -$18,254.90 AW(Y) = -$17,466.50 AW (Z) = -$17,984.57 Select the alternative with the least negative annual worth.

7) CW (Fund) = -$487,500,000.00 CW (Building) = -$81,564,500.00 Select the alternative with the least negative capitalized worth.

8) AW (R-Q) = 0 = -12,000(A/P, i%, 5) + 1900 + 350 (A/G, i%, 5) IRR (R-Q) = 2.50% < 4%; therefore, discard material R.

AW(S-Q) = 0 = -15,000(A/P, i%, 5) + 3450 - 100 (A/G, i%, 5) IRR (S-Q) = 2.79% < 4%; therefore, discard material S. Select material Q. 9) AW(A) = -$10,104.55 AW(B) = -$10,275.85 Therefore machine A should be recommended.

10) FW(A) = -$5,053,855.29 FW(B) = -$5,682,668.31 The alternative with the least negative future worth should be selected.

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Answer Key Testname: C6

11) FW(quarterly payment) = $231,621.33 FW(annual payment) = $220,021.10 Continuing to make quarterly payment offers more money at the time of retirement.

12) FW(L) = $277,411.15 FW(M) = $2,724,278.40 FW(N) = -$543,414.65 The alternative with the largest positive future worth should be selected.

13) IRR (V) = 2.66% < 6%; Discard vacuum forming.

Incremental IRR (F-D) = 19.43% > 6%; Discard drape forming.

Incremental IRR (Z-M) = 8.30% > 6%; Discard free blowing.

Pressure forming should be selected.

14) PW(ST) = $15,955,736.00 PW(LT) = $16,414,129.10 PW(GT) = $16,639,503.70 Therefore, Gentle Touch should be selected.

15) PW(A) = $40,849.49 PW(B) = $41,381.95 Bond B should be purchased.

16) FW(Electric) = -$176,217.48 FW(Gas) = -$91,517.46 FW(Oil) = -$153,104.10

The alternative with the least negative future worth should be selected.

17) PW(CM/GC) = -$10,208.40 PW(DB) = $23,411.35 Therefore, the Design-Build method should be selected.

18) PW(A) = -$102,140.80 PW(B) = -$83,550.73 PW(C) = -$83,349.30

Therefore, plan C should be selected.

19) PW(Delta-T) = $6370.50 PW(Alpha-B) = $10,487.50 Therefore Alpha-B should be recommended.

20) CW (A) = -$57,715.00 CW (B) = -$2,885,750.00 Design B should be selected.

21) PW(A) = $9799.40 PW(B) = $9355.42 Therefore, alternative A should be selected.

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Engineering Economy 16th Edition Sullivan Test Bank

Answer Key Testname: C6

22) Incremental IRR (X-Y) = 14.83% > 4%; discard machine Y Machine X should be selected.

23) ERR (M) = 4.11% > 4% ERR (N-M) = 5.55% > 4% Select project N. 24) AW(Separate Lease) = -53,550 AW(Combined Lease) = -63,000 AW(Purchase) = -$38,076.00

The alternative with the least negative annual worth should be selected.

25) PW(TX) = -$4,491,402.00 PW(UT) = -$4,987,058.00 PW(MN) = -$5,071,780.40 Therefore, the Texas plant should be selected.

26) PW(Purchase) = -$319,020.50 PW(Lease) = -$254,669.80 Therefore, the facility should be leased.

27) FW(E) = -$13,748,192.81 FW(F) = -$12,232,706.60 The alternative with the least negative future worth should be selected.

28) PW(A) = -$119,406.22 PW(B) = -$137,556.60 Therefore, system X should be selected.

29) AW(A) = -$8566.00 AW(B) = -$7195.50 The alternative with the least negative annual worth should be selected.

30) FW(A) = $8,012,971.00 FW(B) = $7,436,476.25 The alternative with the largest positive future worth should be selected.

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