Solutions to end–of–chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 1
Foundations of Engineering Economy
1.1 The four elements are cash flows, time of occurrence of cash flows, interest rates, and
measure of economic worth.
1.2 (a) Capital funds are money used to finance projects. It is usually limited in the amount
of money available.
1.3 Any of the following are measures of worth: present worth, future worth, annual worth, rate
of return, benefit/cost ratio, capitalized cost, payback period, economic value added.
1.4 First cost: economic; leadership: non-economic; taxes: economic; salvage value: economic;
1.6 Example actions are:
• Try to talk them out of doing it now, explaining it is stealing
1.7 This is structured to be a discussion question; many responses are acceptable. It is an
ethical question, but also a guilt-related situation. He can justify the result as an accident; he
1.8 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).
1.9 i = [(3,885,000 – 3,500,000)/3,500,000]*100% = 11% per year
1.11 i = (1125/12,500)*100 = 9%
1.12 Interest on loan = 23,800(0.10) = $2,380
Default insurance = 23,800(0.05) = $1190
1.13 The market interest rate is usually 3 – 4 % above the expected inflation rate. Therefore,
Market rate is in the range 3 + 8 to 4 + 8 = 11 to 12% per year
1.15 P = $150,000; F = ?; i = 11%; n = 7
1.16 P = ?; F = $100,000; i = 12%; n = 2
1.19 End-of-period convention means that all cash flows are assumed to take place at the end of
the interest period in which they occur.
1.20 fuel cost: outflow; pension plan contributions: outflow; passenger fares: inflow;
maintenance: outflow; freight revenue: inflow; cargo revenue: inflow; extra bag charges:
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1.21 End-of-period amount for June = 50 + 70 + 120 + 20 = $260
1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $1000
Jan 500 300 +200
Feb 800 500 +300
Mar 200 400 -200
1.23
1.24
1.25
1.26 Amount now = F = 100,000 + 100,000(0.15) = $115,000
1.28 5000(40 )(1 + i) = 225,000
1 + i = 1.125
= $16,640
1.30 (a) Early-bird payment = 10,000 – 10,000(0.10) = $9000
(b) Equivalent future amount = 9000(1 + 0.10) = $9900
Savings = 10,000 – 9900 = $100
1.31 F1 = 1,000,000 + 1,000,000(0.10)
= 1,100,000
1.32 90,000 = 60,000 + 60,000(5)(i)
1.33 (a) F = 1,800,000(1 + 0.10) (1 + 0.10) = $2,178,000
1.35 4,600,000 = P(1 + 0.10)(1 + 0.10)
P = $3,801,653
1.37 Simple: F = 10,000 + 10,000(3)(0.10)
= $13,000
1.38 Minimum attractive rate of return is also referred to as hurdle rate, cutoff rate, benchmark
rate, and minimum acceptable rate of return.
1.39 bonds – debt; stock sales – equity; retained earnings – equity; venture capital – debt; short
1.40 WACC = 0.30(8%) + 0.70(13%) = 11.5%
1.42 (a) PV(i%,n,A,F) finds the present value P
(b) FV(i%,n,A,P) finds the future value F
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1.43 (a) NPER(8%,-1500,8000,2000): i = 8%; A = $-1500; P = $8000; F = $2000; n = ?
(b) FV(6%,10,2000,-9000): i = 6%; n = 10; A = $2000; P = $-9000; F = ?
1.45 (a) For built–in functions, a parameter that does not apply can be left blank when
it is not an interior one. For example, if there is no F involved when using the PMT
1.46 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more
than cell C10.
1.48 Answer is (d)
1.50 Answer is (d)
1.52 Amount one year ago = 10,000/(1 + 0.10) = $9090.90
Answer is (b)
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1.53 Answer is (c)
1.55 Answer is (a)
1.56 WACC = 0.70(16%) + 0.30(12%)
= 14,8%
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Solution to Case Studies, Chapter 1
There is no definitive answer to case study exercises. The following are examples only.
Renewable Energy Sources for Electricity Generation
3. LEC approximation uses (1.05)11 = 0.5847, X = P11 + A11 + C11 and LEC last year = 0.1022.
Refrigerator Shells
1. The first four steps are: Define objective, information collection, alternative definition and
estimates, and criteria for decision–making.
Objective: Select the most economic alternative that also meets requirements such as
production rate, quality specifications, manufacturability for design specifications, etc.
Information: Each alternative must have estimates for life (likely 10 years), AOC and other
2. Consider these and others like them:
Debt capital availability and cost
Competition and size of market share required
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3. With the addition of C, this is now a make/buy decision. Economic estimates needed are:
Cost of lease arrangement or unit cost, whatever is quoted.
Readiness of the company to enter the market now versus later