final review 2011

7/31/2019 Final Review 2011

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ENGINEERING ECONOMY

Solved Problems

Text Book: Basics of Engineering Economy; Leland Blank and Anthony Tarquin; McGraw-Hill, 2008

Chapter 9:

9.23 AWX = -82,000(A/P,15%,2) 30,000 + 42,000(A/F,15%,2)

= -82,000(0.61512)30,000 + 42,000(0.46512)

= $-60,905

AWY = -97,000(A/P,15%,2) 27,000 + 51,000(A/F,15%,2)

= -97,000(0.61512)27,000 + 51,000(0.46512)

= $-62,946

Purchase robot X

9.5 AW1 = -10,000(A/P,10%,1) 1000 + 7000(A/F,10%,1) = $-5000

AW2 = -10,000(A/P,10%,2) 1000(P/F,10%,1)(A/P,10%,2)

+ (5000 1200)(A/F,10%,2) = $-4476

AW3 = -10,000(A/P,10%,3) [1000(P/F,10%,1) +1200(P/F,10% 2)](A/P,10%,3)

+ (4200 1500)(A/F,10%,3) = $-3970

AW4 = -10,000(A/P,10%,4) [1000(P/F,10%,1) +1200(P/F,10% 2)

+ 1500(P/F,10%,3)](A/P,10%,4) + (3000 2000)(A/F,10%,4) = $-3894

AW5 = -10,000(A/P,10%,5) [1000(P/F,10%,1) + 1200(P/F,10% 2)

+ 1500(P/F,10%,3) + 2000(P/F,10%,4)](A/P,10%,5)

+ (2000 3000)(A/F,10%,5) = $-3961

Therefore, ESL is 4 years with AW = $-3894


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Chapter 8:

Example: A company has two machine alternatives whose economic lives are 6 years. The

price and annual income of these machines are given in the following table. According to the

no return payback period, determine the alternative the company should invest.

Alternatives Cost (SR) Annual income (SR/yr)

Machine A 200 000 45 000

Machine B 300 000 60 000

Payback Period1= 200 000/45 000 = 4.4 years

Payback Period2= 300 000/60 000 = 5 years

According to the payback periods, first alternative should be preferred.

Example: A company wants to buy a production device for their new factory. They have

two alternatives, whose cash flows are given in the following table. According to these

cash flows,

determine the no return payback period of these alternatives.

Alternative A Alternative B

Cost 3 000 000 SR 3 500 000 SR

Annual income 1 200 000 SR first year,

decreasing by 300.000 SR

per year thereafter

100 000 SR for the first year,

increasing 300 000 SR per

year thereafter.

Useful life 4 years 8 years

Alternative A

Years 0 1 2 3 4

Cash Flow -3 000 000 1 200 000 900 000 600 000 300 000

Cumulative value -3 000 000 -1 800 000 -900 000 -300 000 0

PBA= 4 years


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Alternative B

Years 0 1 2 3 4 5

Cash Flow -3 500 000 100 000 400 000 700 000 1 000 000 1 300 000

Cumulative value -3 500 000 -3 400 000 -3 000 000 -2 300 000 -1 300 000 0

PBB= 5 years

According to the payback periods, alternative A should be preferred.

8.2 (a) TC = FC + vQ

= 2.58 million + 395Q

(b) QBE = 2.58 million/(550-395)

= 16,645

(c) 1.2Q = 1.2(16,645) = 19,974

Profit = R TC

= 19,974(550) 2,580,000 - 395(19,974)

= $515,970

8.44 (a) Yes; cash flows sum to $139,100, which exceeds the $75,000 first cost.

(b) Solve PW = 0 relation for i*.

PW = -75,000 -10,500(P/F,i,1) + + 105,000(P/F,i ,5) = 0i* = 13.96% (IRR function)

(c) Calculate PW at 7% by year to determine when PW turns positive. Start withN = 3 years.

N = 3: PW = -75,000 -10,500(P/F,7%,1) +18,600(P/F,7%,2) -2000(P/F,7%,3)= -75,000 -10,500(0.9346) +18,600(0.8734) -2000(0.8163)= $-70,201

N = 4: PW = -70,201 +28,000(P/F,7%,4)= $-48,840

N = 5: PW = -48,840 +105,000(P/F,7%,5)= $26,025

Investment is paid back plus 7% during year 5, in part due to large cash flow at

sale time.


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Chapter 7:

CONVENTION B/C=

..+,

MODIFIED B/C=+

.., where I.C.: costs or initial cost, Sv: savings and S: salvage value.

7.11 Convert all cash flows to AW

B = 3,800,000(A/P,8%,20)

= 3,800,000(0.10185)

= $387,030

D = $65,000

C = 1,200,000(A/P,8%,20) + 300,000

= 1,200,000(0.10185) + 300,000

= $422,220

B/C = (387,030 65,000)/ 422,220

= 0.76< 1.07.16 Use annual worth, since most of the cash flows are in annual dollars.

(a) Conventional B/C ratio

B = 300,000(0.06) + 100,000

= 18,000 + 100,000

= $118,000

D = $40,000

C = 1,500,000(0.06) + 200,000(P/F,6%,3)(0.06)

= 90,000 + 200,000(0.8396)(0.06)

= $100,075

Sv = 70,000

B/C = (118,000 40,000)/(100,075 70,000)

= 2.59

(b) Modified B/C ratio = (B D + Sv)/C

= (118,000 40,000 + 70,000)/100,075

= 1.48


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7.20 B = 1250(140) = $175,000 per year

C = 595,000(A/P,6%,5) + 56,000 + 19,000

= 595,000(0.23740) + 75,000

= $216,253

B/C = 0.81< 1.0 Therefore, the utility should send the samples out.

Final Exam: Chapters 1, 2, 4, 5, 7, 8, and 9. (50 Marks)

Chapter 9:

Read Sections 9.1-9.2, and 9.5.

Read and solve examples 9.4, 9.2.

Read and solve Problems: 9.5, 9.23.

HWK_CH. 9: PROPLEMS 9.6, 9.7, 9.8, 9.9, 9.22.

Chapter 8:

Read Sections 8.1, and 8.5.

Read and solve examples 8.1, 8.7.

Read and solve Problems: 8.2, 8.44.

HWK_CH. 8: PROPLEMS 8.12, 8.13, 8.45, 8.54 to 8.57.

Chapter 7:

Read Sections 7.1, and 7.2.

Read and solve example 7.2.

Read and solve Problems: 7.11, 7.16 AND 7.20.

HWK_CH. 7: PROPLEMS 7.10, 7.12, 7.14, 7.15, 7.18, 7.32 to 7.35.

Chapter 5:


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Read Sections 5.1, 5.2 and 5.3.

Read and solve examples 5.1, 5.2, 5.3, 5.4 and 5.6.

HWK_CH. 5: PROPLEMS 5.2, 5.4, 5.7, 5.12, 5.13, 5.17, 5.20, 5.21 and 5.25 to 5.27.

Chapter 4:

Read Sections 4.1, 4.2, 4.3 and 4.5.

Read and solve examples 4.1, 4.3 and 4.7.

HWK_CH. 4: PROPLEMS 4.9, 4.15, 4.23, 4.28, 4.36, 4.42 and 4.45 to 4.48.

Chapter 2:

Read Sections 2.1 to 2.4.

Read and solve examples 2.1 to 2.11.

HWK_CH. 2: PROPLEMS 2.12, 2.35, 2.36, 2.40, 2.54, Find (P0)in Fig. 2.16a-b (page 44), 2.64, 2.74, 2.80

and 2.85.

Chapter 1:

Read Sections 1.1 to 1.7.

Read and solve examples 1.1 to 1.13.

HWK_CH. 1: PROPLEMS 1.19, 1.22, 1.28, 1.37, 1.41 to 1.45, and ) In the plan 5, example 1.6 (page 12):

Show that equal end-of-year payment is $1252.28.

final review 2011

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