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  • Typewritten TextSOLUTIONMANUAL

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    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. (b) Sensitivity analysis is a procedure that involves changing various estimates to see if/how they affect the economic decision. 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; morale: non-economic; dependability: non-economic; inflation: economic; profit: economic; acceptance: non-economic; ethics: non-economic; interest rate: economic. 1.5 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b. 1.6 Example actions are:

    Try to talk them out of doing it now, explaining it is stealing Try to get them to pay for their drinks Pay for all the drinks himself Walk away and not associate with them again

    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 can feel justified by the legal fault and punishment he receives; he can get angry because it WAS an accident; he can become tormented over time due to the stress caused by accidently causing a childs death.

    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).

    Lessons can be learned from the experience. A few of them are:

    Think before he cheats again. Think about the longer-term consequences of unethical decisions. Face ethical-dilemma situations honestly and make better decisions in real time.

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    Alternatively, Claude may learn nothing from the experience and continue his unethical practices.

    1.9 i = [(3,885,000 - 3,500,000)/3,500,000]*100% = 11% per year 1.10 (a) Amount paid first four years = 900,000(0.12) = $108,000 (b) Final payment = 900,000 + 900,000(0.12) = $1,008,000 1.11 i = (1125/12,500)*100 = 9% i = (6160/56,000)*100 = 11% i = (7600/95,000)*100 = 8% The $56,000 investment has the highest rate of return. 1.12 Interest on loan = 23,800(0.10) = $2,380 Default insurance = 23,800(0.05) = $1190 Set-up fee = $300 Total amount paid = 2380 + 1190 + 300 = $3870 Effective interest rate = (3870/23,800)*100 = 16.3% 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 1.14 PW = present worth; PV = present value; NPV = net present value; DCF = discounted cash flow; and CC = capitalized cost 1.15 P = $150,000; F = ?; i = 11%; n = 7 1.16 P = ?; F = $100,000; i = 12%; n = 2 1.17 P = $3.4 million; A = ?; i = 10%; n = 8 1.18 F = ?; A = $100,000 + $125,000?; i = 15%; n = 3 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: Inflow; water and sodas: outflow; advertising: outflow; landing fees: outflow; seat preference fees: inflow.

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    1.21 End-of-period amount for June = 50 + 70 + 120 + 20 = $260 End-of-period amount for Dec = 150 + 90 + 40 + 110 = $390 1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $1000

    Jan 500 300 +200 Feb 800 500 +300 Mar 200 400 -200

    Apr 120 400 -280 May 600 500 +100

    June 900 600 +300 July 800 300 +500 Aug 700 300 +400 Sept 900 500 +400 Oct 500 400 +100 Nov 400 400 0 Dec 1800 700 +1100

    Net Cash flow = $2,920 ($2,920,000) 1.23

    1.24

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    1.25

    1.26 Amount now = F = 100,000 + 100,000(0.15) = $115,000 1.27 Equivalent present amount = 1,000,000/(1 + 0.15) = $869,565 Discount = 790,000 869,565 = $79,565 1.28 5000(40 )(1 + i) = 225,000 1 + i = 1.125 i = 0.125 = 12.5% per year 1.29 Total bonus next year = 8,000 + 8,000(1.08) = $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 F2 = 1,100,000 + 1,100,000(0.10) = $1,210,000 1.32 90,000 = 60,000 + 60,000(5)(i) 300,000 i = 30,000 i = 0.10 (10% per year) 1.33 (a) F = 1,800,000(1 + 0.10) (1 + 0.10) = $2,178,000 (b) Interest = 2,178,000 1,800,000 = $378,000

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    1.34 F = 6,000,000(1 + 0.09) (1 + 0.09) (1 + 0.09) = $7,770,174 1.35 4,600,000 = P(1 + 0.10)(1 + 0.10) P = $3,801,653 1.36 86,400 = 50,000(1 + 0.20)n log (86,400/50,000) = n(log 1.20) 0.23754 = 0.07918n n = 3 years 1.37 Simple: F = 10,000 + 10,000(3)(0.10) = $13,000 Compound: 13,000 = 10,000(1 + i) (1 + i) (1 + i) (1 + i)3 = 1.3000 3log(1 + i) = log 1.3 3log (1 + i) = 0.1139 log(1 + i) = 0.03798 1 + i = 1.091 i = 9.1% per year 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 term loan debt; capital advance from friend debt; cash on hand equity; credit card debt; home equity loan - debt. 1.40 WACC = 0.30(8%) + 0.70(13%) = 11.5% 1.41 WACC = 10%(0.09) + 90%(0.16) = 15.3% The company should undertake the inventory, technology, and warehouse projects. 1.42 (a) PV(i%,n,A,F) finds the present value P (b) FV(i%,n,A,P) finds the future value F (c) RATE(n,A,P,F) finds the compound interest rate i (d) IRR(first_cell:last_cell) finds the compound interest rate i

    (e) PMT(i%,n,P,F) finds the equal periodic payment A (f) NPER(i%,A,P,F) finds the number of periods n

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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 = ? (c) RATE(10,1000,-12000,2000): n = 10; A = $1000; P = $-12,000; F = $2000; i = ? (d) PMT(11%,20,,14000): i = 11%; n = 20; F = $14,000; A = ?

    (e) PV(8%,15,-1000,800): i = 8%; n = 15; A = $-1000; F = $800; P = ? 1.44 (a) PMT is A (b) FV is F (c) NPER is n (d) PV is P (e) IRR is i 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 function to solve a particular problem, it can be left blank (omitted) because it is an end parameter. (b) When the parameter involved is an interior one (like P in the PMT function), a comma must be put in its position. 1.46 Spreadsheet shows relations only in cell reference format. Cell E10 will indicate $64 more than cell C10.

    1.47 Answer is (b) 1.48 Answer is (d) 1.49 Answer is (a) 1.50 Answer is (d) 1.51 Upper limit = (12,300 10,700)/10,700 = 15% Lower limit = (10,700 8,900)/10,700 = 16.8% Answer is (c) 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) 1.54 2P = P + P(n)(0.04) 1 = 0.04n n = 25 Answer is (b) 1.55 Answer is (a) 1.56 WACC = 0.70(16%) + 0.30(12%) = 14,8% Answer is (c)

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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. X(0.5847) 0.1027 = 0.1022 + ---------------------- (5.052 B)(0.5847) X = $2.526 million

    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