978-0073523439 Chapter 1

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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

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.