978-0073523439 Chapter 1

subject Type Homework Help
subject Pages 9
subject Words 1425
subject Authors Anthony Tarquin, Leland Blank

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Solutions to end-of-chapter problems
Engineering Economy, 8th edition
Leland Blank and Anthony Tarquin
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost: tangible;
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client
and deceptive acts are clearly present.
1.9 Example actions are:
Try to talk them out of doing it now, explaining it is stealing
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Pay for all the drinks himself
Walk away and not associate with them again
1.10 This is structured to be a discussion question; many responses are acceptable. Responses
can vary from the ethical (stating the truth and accepting the consequences) to unethical
(continuing to deceive himself and the instructor and devise some on-the-spot excuse).
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $285
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
= $336,000
1.14 Interest rate = interest paid/principal
1.15 i = (1125/12,500)*100 = 9%
1.16 Interest on loan = 45,800(0.10) = $4,580
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Effective interest rate = (5,938/45,800)*100 = 12.97%
Terms and Symbols
1.17 P = ?; F = 8*240,000 = $1,920,000; n = 2; i = 0.10
Cash Flows
1.22 Well drilling: outflow; maintenance: outflow; water sales: inflow; accounting: outflow;
1.23 Let Rev = Revenues; Exp = Expenses
Year 1 2 3 4 5 Total
1.24 Month Receipts, $1000 Disbursements, $1000 NCF, $1000
Jan 300 500 -200
Feb 950 500 +450
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+3,170
Net cash flow = $3,170 ($3,170,000)
1.25 End-of-period amount for March: 50 + 70 = $120; Interest = 120*0.03 = $3.60
1.26
1.27
Equivalence
1.28 (a) i = (5000-4275)/4275 = 0.17 (17%)
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(d) Price one year earlier = 28,000/1.04 = $26,923
(e) Jackson: Interest rate = (2750/20,000)*100
= 13.75%
Henri: Interest rate = (2295/15,000)*100
= 15.30%
(f) 81,000 = 75,000 + 75,000(i)
i = 6,000/75,000
= 0.08 (8%)
1.29 (a) Profit = 8,000,000*0.28
1.30 P + P(0.10) = 1,600,000
P = $1,454,545
1.31 Equivalent present amount = 1,000,000/(1 + 0.10)
1.33 (a) Early-bird: 20,000 – 20,000(0.10) = $18,000
Simple and Compound Interest
1.34 (a) F = P + Pni
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(b) P(1+i)(1+i)(1+i) = 1,000,000
P = 1,000,000/[(1+0.20)(1+0.20)(1+0.20)]
= $578,704
1.35 F = P + Pni
1.36 F = 240,000(1+ 0.12)3
1.37 (a) F = P + Pni
1.38 (a) Total due; compound interest = 150,000(1.05)(1.05)(1.05)
1.39 90,000 = 60,000 + 60,000(5)(i)
1.40 Simple: F = 10,000 + 10,000(3)(0.10)
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3log (1 + i) = 0.1139
log(1 + i) = 0.03798
1 + i = 1.091
i = 9.1% per year
1.41 Follow plan 4, Example 1.16 as a model
1.42 (a) Simple: F = P + Pni
(b) Compound: F = P(1 + i) (1 + i) (1 + i) (1 + i)
MARR and Opportunity Cost
1.43 Bonds - debt; stock sales – equity; retained earnings – equity; venture capital – debt; short
1.44 WACC = 0.40(10%) + 0.60(16%) = 13.60%
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0.13 = x(0.28) + (1-x)(0.06)
0.22x = 0.07
x = 31.8%
Recommendation: debt-equity mix should be 31.8% debt and 68.2% equity financing
1.47 Money: The opportunity cost is the loss of the use of the $5000 plus the $100 interest.
Exercises for Spreadsheets
1.48 (a) PV is P; (b) PMT is A; (c) NPER is n; (d) IRR is i; (e) FV is F; (f) RATE is i
1.49 (a) PV(i%,n,A,F) finds the present value P
1.50 (a) (1) F = ?; i = 8%; n = 10; A = $3000; P = $8000
(b) (1) negative
1.51 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more
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Additional Problems and FE Review Questions
1.52 Answer is (d)
1.56 F = P + Pni
1.58 Move both cash flows to year 0 and solve for i
1.59 F in year 2 at 20% compound interest = P(1.20)(1.20) = 1.44P
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i = 22%
Answer is (c)
1.60 WACC = 0.70(16%) + 0.30(12%)
1.61 Amount available = total principal in year 0 + interest for 2 years + principal added year 1
Solution to Case Study, Chapter 1
There is no definitive answer to case study exercises. The following is only an example.
Renewable Energy Sources for Electricity Generation
3. LCOE approximation uses 1/(1.05)11 = 0.5847 and LCOE last year = 0.1022.

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