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CHAPTER 1 Solutions Manual For

Basics of Engineering Economy, 1e Leland Blank, PhD, PE Texas A&M University and American University of Sharjah, UAE

Anthony Tarquin, PhD, PE University of Texas at El Paso

PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.

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Solutions to End-of-Chapter Problems Chapter 1 1.1 Engineering economy will help you pick the most economical alternative from the ones that have been identified. It will not help you identify what the alternatives are. 1.2 Revenues and costs are examples of cash flows. 1.3 The analysis techniques that are used in engineering economic analysis are only as good as the accuracy of the cash flow estimates. 1.4 Evaluation criterion refers to the measurement standard or procedure that is used to determine which alternative is “best”. 1.5 In engineering economy, the evaluation criterion is financial units (dollars, pesos, etc). 1.6 Intangible factors are non-economic considerations that may have to be taken into account in identifying the best alternative. 1.7 Examples of intangible factors are goodwill, morale, convenience, friendship, implementation know-how. 1.8 Interest is a manifestation of the time value of money. 1.9 The most important fundamental dimension in economic analysis is time, because of the time value of money. 1.10 The term that describes compensation for “renting” of money is time value of money, which manifests itself as interest. 1.11 When an interest rate does not include the time period, the time period is assumed to be one year. 1.12 The original amount of money in a loan transaction is known as principal. 1.13 When the yield on a U.S. Government Bond is 3% per year, investors are expecting the inflation rate to be near zero. 1.14 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000 (b) Interest = total amt paid – principal = 1,936,000- 1,600,000 = $336,000

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1.15 F = P(1 + i) 2,420,000 = 2,000,000(1 + i) i = 21% 1.16 F = P(1 + i) 82,000,000 = x(1 + 0.08) x = $75,925,926 1.17 Amount of earnings in year one = 200,000,000(0.30) = $60,000,000 1.18 Amt at end of year 1 = 280,000(1.15) = $322,000 Amt at end of year 2 = 322,000(1.15) = $370,300 Amt at end of year 3 = 370,300(1.15) = $425,845

< $425,000 < $425,000 > $425,000

Time = 3 years 1.19 (a) Amt owed, comp’d interest = 150,000(1.05)(1.05)(1.05) = $173,644 Amt owed, simple interest = P + Pni = 150,000 + 150,000(3)(0.055) = 150,000 + 24,750 = $174,750 Company should accept 5% compound interest rate. (b) Difference = 174,173,644 = $1106 1.20 F = P + Pni 850,000 = P + P(4)(0.10) 1.4P = 850,000 P = $607,143 1.21 Amt original certificate = 580,000 + 580,000(2)(0.09) = $684,400 Total amt after two more years = 684,400(1.09)(1.09) = $813,136 1.22 (a) F = P + Pni 2,800,000 = 2,000,000 + 2,000,000(4)(i) Interest = 10% per year (b) F = P(1 + i) (1 + i) (1 + i) (1 + i) 2,800,000 = 2,000,000(1 + i)4 (1 + i)4 = 1.4000

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log(1 + i)4 = log1.400 4log(1 + i) = 0.146 log(1 + i) = 0.0365 (1 + i) = 100.0365 (1 + i) = 1.0877 i = 8.77% 1.23 F = P + P(n)(i) 3P = P + P(n)(0.20) n = 10 years 1.24 (a) F = P + P(n)(i) 2P = P + P(4)(i) i = 25% (b) F = P(1 + i) (1 + i) (1 + i) (1 + i) 2P = P(1 + i)4 2 = (1 + i)4 log 2 = log(1 + i)4 0.301 = 4 log(1 + i) log(1 + i) = 0.0753 (1 + i) = 100.0753 i = 18.93% 1.25 (a) Loan A: interest/year = 500,000(0.10) = $50,000 Loan B: interest/year = 500,000(0.10) = $50,000 The same amount of interest was be paid on each loan (b) There was no difference paid between the two loans 1.26 All engineering economy problems will involve i and n 1.27 F= $200,000,000; n = 5; i = 20% per year; A = ? 1.28 Office supplies = outflow; GPS surveying equipment = outflow; Auctioning of used earth-moving equipment = inflow; Staff salaries = outflow; Fees for services rendered = inflow; Interest from bank deposits = inflow 1.29 A = $225,000; n = 3; i = 15% per year; F = ? 1.30 F = $400,000; n = 2; i = 20% per year; P = “?” 1.31 P = $225,000; n = 4; i = 15% per year; A =”?”

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1.32 P = $1,800,000; n = 6; i = 12% per year; A =”?” 1.33 P = $16,000,000; A = $3,800,000; i = 18% per year; n = ? 1.34 Period refers to interest period. 1.35 The difference between cash inflows and cash outflows is known as net cash flow. 1.36 Assuming down is negative: down arrow of $10,000 in year 0; up arrows in the amount of $3000 in years 1 thru 5; i = 10% per year; arrow in year 5 identified as “F = ?” 1.37 Assuming down is negative: down arrow in year 0 identified as “P=?”; i = 15% per year; up arrows in years 1 thru 5 in the amount of $75,000 - $30,000 = $45,000 1.38 (a) FV = F (b) PMT = A (c) Nper = n

(d) IRR = i

(e) PV = P

1.39 (a) F = ?; i = 8%; n = 10; A = $2000; P = $10,000 (b) A = ?; i = 12%; n = 30; P = $16,000; F = 0 (c) P = ?; i = 9%; n = 15; A = $1000; F = $700 1.40 (a) FV(i%,n,A,P) finds the future value, F (b) IRR(first_cell:last_cell) finds the compound interest rate, i (c) PMT(i%,n,P,F) finds the equal periodic payment, A (d) PV(i%,n,A,F) finds the present value, P Problems for Test Review and FE Exam Review 1.41 Answer is (c) 1.42 Answer is (c) 1.43 Answer is (b) 1.44 Move both cash flows to year 0 and solve for i. 1000(1 + i) = 1345.60/(1 + i) (1 + i)2 = 1345.60/1000 (1 + i) = 1.16 i = 16% Answer is (d) 1.45 F in year 2 at 20% compound interest = P(1.20)(1.20) = 1.44P For simple interest, F = P + Pni = P(1 + ni) P(1 + 2i) = 1.44P (1 + 2i) = 1.44P i = 22%

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Answer is (c) 1.46 Answer is (d)

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