www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net Contemporary Engineering Economics Fourth Edition www.elsolucionario.net Instructor’s Manual Chan S Park Auburn University Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Preface • • • • • Contemporary Engineering Economics contains nearly 570 problems Many of them come with multiple-part questions, bringing the total number of questions to nearly 900 The Park’s online OneKey Course Management Site provides another 280 selftest questions This OneKey site is designed primarily to help students develop a working knowledge of the concepts and principles of engineering economics Most questions in this guide are structured in multiple-choice format as these types of exam questions are tested on the Fundamentals of Engineering (FE) exam and, increasingly, in introductory engineering economics courses The problems listed under ``Short Case Studies" require more involved assumptions, solution steps, and calculations Please note that your independent solutions to the text problems may yield answers slightly different from mine due to rounding differences In some cases, the interpretations and assumptions students bring to the problems may differ from my own in creating them, again resulting in different answers A comprehensive set of lecture notes in PowerPoint format is also available to instructors who adopt the text Having made these disclaimers, I wish to emphasize that it is my and the publisher's intention to provide the most accurate solutions possible Thus, if you discover typo errors, or disagree strongly with the interpretation and assumptions required of a particular solution, please not hesitate to report your concerns to me at the e-mail address below so they may be corrected in subsequent printings I will also plan on posting any errors on the book's web site under "Instructor Resources." Finally, I want to thank Major Hyun Jin Han for his assistance in preparing this manual www.elsolucionario.net This Instructor's Manual to Contemporary Engineering Economics contains detailed solutions to all the end-of-chapter problems The problem solutions follow topical headings to indicate the generic content of each problem The headings are provided to aid you in selecting your preferred mix of problem types for homework assignments Chan S Park Department of Industrial & Systems Engineering Auburn University, Auburn, AL 36849 Voice mail: (334) 844-1428 E-mail: park@eng.auburn.edu Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Basics of Financial Decisions www.elsolucionario.net Part One Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Chapter Engineering Economic Decisions There are no end-of-chapter questions in this introductory chapter However, the following questions could be added as a part of instruction: Work in small groups and brainstorm ideas about how a common appliance, device or tool could be redesigned to improve it in some way Identify the steps involved and the economic factors, which you would need to consider prior to making a decision to manufacture the redesigned product A detailed design and actual cost estimates are not required Some items, which could be considered for this redesign exercise, are: a shopping cart, telephone, can opener, screwdriver, etc Many oil price forecasts in the early 2000 indicated that the price of oil in the year 2005 would be in excess of $50 per barrel What is the price today? Why are these prices so difficult to predict? Imagine what the consequences would be if you used these pessimistic estimates in your economic analysis What would be some practical ways to handle project uncertainty? www.elsolucionario.net Ask students to review the contents of The Wall Street Journal for the past months Then, identify and categorize the types of investment decisions appeared in the journal according to the types of strategic economics decisions discussed in Section1.5 Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Chapter Understanding Financial Statements Financial Statement 2.1 (a) • • • Current assets = $150,000 + $200,000 + $150,000 + $50,000 + $30,000 = $580,000 Current liabilities = $50,000 + $100,000 + $80,000 = $230,000 Working capital = $580,000 - $230,000 = $350,000 Shareholder’s equity = $100,000 + $150,000 + $150,000 + $70,000 = $470,000 (b) EPS = $500,000/10,000 = $50 per share (c) Par value = $15; capital surplus = $150,000/10,000 = $15; Market price = $15 + $15 = $30 per share 2.2 (a) Working capital = Current assets – Current liabilities; Working capital requirements = Changes in current assets – Changes in current liabilities WC req = (+$100,000 - $20,000) – (+$30,000 - $40,000) = $90,000 (b) Taxable income = $1,500,000 - $650,000 - $150,000 - $20,000 = $680,000 (c) Net income = $680,000 - $272,000 = $408,000 (d) Net cash flow: A Operating activities = net income + depreciation – WC = $408,000 + $200,000 - $90,000 = $518,000 B Investing activities = equipment purchase = ($400,000) C Financing activities = borrowed fund = $200,000 D Net cash flow = $518,000 - $400,000 + $200,000 = $318,000 www.elsolucionario.net • 2.3 (a) Debt ratio = $55,663/$513,378 = 10.84% (b) Time-interest-earned ratio: not defined (c) Current ratio = $377,833/$55,663 = 6.79 Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net (d) Quick ratio = ($377,833 – 0)/$55,633 = 6.79 (e) Inventory turn over ratio: not defined (f) DSO = ($24,582 - $632)/($102,606/360) = 84.03 days (g) Total assets turnover ratio = $102,606/$513,378 = 0.20 times (h) Profit margin on sales = -$9,034/$102,606 = -8.8% (i) Return on total assets = (-$9,034 + 0)/(($513,378 + $36,671)/2) = -3.28% (k) Price-earning ratio = $56.67/-0.10 = -566.70 (l) Book value per share = $457,715,000/90,340,000 = $5.07 (Note: The total number of outstanding shares in year 2005 was 90,340,000.) 2.4 (a) Debt ratio = ($922,653 + $113,186)/$4,834,690 = 42.10% (b) Time-interest-earned ratio = ($432,342 + $36,479)/$36,479 = 12.85 times (c) Current ratio = $3,994,084/$1,113,186 = 3.59 times (d) Quick ratio = ($3,994,084 - $1,080,083)/$1,113,186 = 2.62 times (e) Inventory turn over ratio = $8,391,409/(($1,080,083 + $788,519)/2) = 8.98 times (f) DSO = ($1,123,901 - $5,580)/($8,391,409/360) = 47.98 days www.elsolucionario.net (j) Return on common equity = -$9,034/(($457,713 + $17,064)/2) = -3.81% (g) Total assets turnover ratio = $8,391,409/$4,834,696 = 1.74 times (h) Profit margin on sales = $293,935/$8,391,409 = 3.5% (i) Return on total assets = ($239,935 + $36,479(1 – 0.3201))/(($4,834,696 + $2,410,568)/2) = 8.80% (j) Return on common equity = $293,935/(($2,793,091 + $1,181,326)/2) = 14.79% Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net (k) Price-Earning ratio = $65/$1.13 = $57.52 Note: Assumed a share price of $65 The stock prices were fluctuating between $78.93 and $52.25 during the fourth quarter (l) Book value per share = $2,793,091,000/247,004,200 = $11.30 (Note: The total outstanding number of shares in year 2005 was 247,004,200.) • • • • Given Olson’s EPS = $8 per share; Cash dividend = $4 per share; Book value per share = $80; Changes in the retained earnings = $24 million; Total debt = $240 million; Find debt ratio = total debt/total assets Net Income EPS = = $8 X Where X = the number of outstanding shares Total shareholders' equity = $80 X Retained earnings = Net income – Cash dividend; Net income = 8X from EPS relationship and the total cash dividend = 4X, so we rewrite 8X – 4X = $24 million, or X = million shares Book value = • From book value per share, we know that total shareholders’ equity = 80X, or $480 million; Total assets = Total liabilities + Total shareholders’ equity = $240 million + $480 million = $720 million • Debt ratio = $240 million/$720 million = 0.33 2.6 (b) 2.7 (b) 2.8 (d) 2.9 (b) www.elsolucionario.net 2.5 Short Case Studies ST2.1 & ST2.2 – Not given Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Chapter 3: Interest Rate and Economic Equivalence Types of Interest 3.1 $10,000 = $5,000(1 + 0.08 N ) • Simple interest: (1 + 0.08 N ) = = 12.5 years N= 0.08 • Compound interest: (1 + 0.07) N = N = 10.2 years • Simple interest: I = iPN = (0.08)($1,000)(5) = $400 • Compound interest: I = P[(1 + i ) N − 1] = $1,000(1.4693 − 1) = $469 • Option 1: Compound interest with 8%: F = $3,000(1 + 0.08) = $3,000(1.4693) = $4,408 • Option 2: Simple interest with 9% $3,000(1 + 0.09 × 5) = $3,000(1.45) = $4,350 www.elsolucionario.net $10,000 = $5,000(1 + 0.07) N 3.2 3.3 • Option is better 3.4 End of Year Principal Repayment Interest payment $1,671 $1,821 $1,985 $2,164 $2,359 $900 $750 $586 $407 $212 Remaining Balance $10,000 $8,329 $6,508 $4,523 $2,359 $0 Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Equivalence Concept 3.5 3.6 P = $12,000( P / F , 5%, 5) = $12,000(0.7835) = $9,402 F = $20,000( F / P, 8%, 2) = $20,000(1.1664) = $23,328 3.7 (a) F = $5,000( F / P, 5%, 8) = $7,388 (b) F = $2,250( F / P, 3%,12) = $3,208 (c) F = $8,000( F / P, 7%, 31) = $65,161 (d) F = $25,000( F / P, 9%, 7) = $45,700 3.8 (a) P = $5,500( P / F ,10%, 6) = $3,105 (b) P = $8,000( P / F , 6%,15) = $3,338 (c) P = $30,000( P / F , 8%, 5) = $20,418 (d) P = $15,000( P / F ,12%, 8) = $3,851 www.elsolucionario.net Single Payments (Use of F/P or P/F Factors) 3.9 (a) P = $10,000( P / F ,13%, 5) = $5,428 (b) F = $25,000( F / P,13%, 4) = $40,763 3.10 F = 3P = P (1 + 0.12) N log = N log(1.12) N = 9.69 years Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net (b) The floatation costs and the number of common stocks to raise $8,500,000: Flotation cost = $8,500, 000 − $8,500, 000 = $749,184 − 0.081 Number of shares = $8,500, 000 = 205,537 shares (1 − 0.081)($45) (c) The floatation costs and the number of $1,000 bonds to raise $10.5 million: $10,500, 000 − $10,500, 000 = $203,364 − 0.019 Number of bonds = $10,500, 000 = 11,893units (1 − 0.019)($900) www.elsolucionario.net Flotation cost = Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be 18 obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net (a) The net cash flow from the cogeneration project with bond financing - 11 12 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $6,120,000 $480,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $500,000 $1,000,000 $6,400 $1,280,000 $500,000 $950,000 $855,000 $770,000 $693,000 $623,000 $590,500 $100,000 $160,000 $96,000 $57,600 $57,600 $28,800 $1,070,370 $1,070,370 $1,070,370 $1,070,370 $1,070,370 $1,070,370 $1,070,370 $295,000 $1,070,370 Taxable Income Income Taxes (36%) $2,143,230 $1,633,230 $1,792,230 $1,915,630 $1,992,630 $2,091,430 $2,152,730 $771,563 $587,963 $645,203 $689,627 $717,347 $752,915 $774,983 $2,448,230 $881,363 Net Income $1,371,667 $1,045,267 $1,147,027 $1,226,003 $1,275,283 $1,338,515 $1,377,747 $1,566,867 $1,371,667 $1,045,267 $1,147,027 $1,226,003 $1,275,283 $1,338,515 $1,377,747 $1,566,867 Cash Flow Statement Cash from operation Net Income Depreciation Unit Inter Equipment Investment/Salvage Unit Inter Equipment Gains Tax Loan Repayment Net cash flow PW(27%) = $500,000 $100,000 $950,000 $160,000 $855,000 $96,000 $770,000 $57,600 $693,000 $57,600 $623,000 $28,800 $590,500 ($10,000,000) ($500,000) $10,500,000 $0 $1,971,667 $2,155,267 $2,098,027 $2,053,603 $2,025,883 $1,990,315 $1,968,247 $295,000 $1,000,000 $490,140 ($11,893,000) ($8,540,993) $6,516,321 Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 19 www.elsolucionario.net ST 15.5 Income Statement Revenue Electricity Bill Excess power Expenses O&M Misc Standby power Overhead Depreciation Unit Inter Equipment Interest (9%) www.elsolucionario.net (b) The maximum annual lease amount that ACC is willing to pay is $933,358 (By Excel solver.) - 11 12 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $6,120,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $480,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $500,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $1,000,000 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $6,400 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $1,280,000 $933,358 $933,358 $933,358 $933,358 $933,358 $933,358 $933,358 $933,358 Taxable Income $2,880,242 $2,880,242 $2,880,242 $2,880,242 $2,880,242 $2,880,242 $2,880,242 $2,880,242 Income Taxes (36%) $1,036,887 $1,036,887 $1,036,887 $1,036,887 $1,036,887 $1,036,887 $1,036,887 $1,036,887 Net Income $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 Cash Flow Statement Cash from operation Net Income $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 Net cash flow $0 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 $1,843,355 PW(27%) = $6,516,321 Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 20 www.elsolucionario.net Income Statement Revenue Electricity Bill Excess power Expenses O&M Misc Standby power Overhead Lease www.elsolucionario.net Chapter 16 Economic Analysis in the Service Sector Cost-Effectiveness Analysis 16.1 Cost effectiveness of the alternatives Type of Treatment Antibiotic A Antibiotic B Antibiotic C Cost Effectiveness 12,000/75 =160 168.75 180.49 16.2 • The summary of three mutually exclusive alternatives CER: Strategy Cost Nothing Simple Complex $0 $5,000 $50,000 Effectiveness Cost Effectiveness years years 5.5 years 1,000 9,091 Incremental CER 1,000 90,000 • Since there is no clear dominance we can draw a cost-effective diagram Life-Year www.elsolucionario.net ∴ The best treatment option is Antibiotic A $0 $10,000 $20,000 $30,000 $40,000 $50,000 $60,000 Investment cost Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Budget Available ($) Less than $5,000 $5,000 $5,000 - $50,000 $50,000 or larger Treatment Option To Be Implemented As much of Simple treatment as budget allows 100% of Simple treatment Simple treatment and as much of Complex treatment as budget allows 100% of Complex treatment 16.3 (a) • User’s benefits: - Prevention (or retardation) of highway corrosion: resulting in lower highway maintenance cost This lower maintenance cost implies lower users’ taxes on gasoline, and so forth - Prevention of rust on vehicles: resulting in lower repair and maintenance costs and higher resale value of vehicles - Prevention of corrosion to utility lines and damages to water supplies: resulting in lower utility rates - Prevention of damages to vegetation and soil surrounding areas: increasing land values and agriculture yields • Users’ costs: - Paying higher taxes - Unknown environmental damages due to using CMA (b) The state of Michigan may declare certain sections of highway for experimental purpose CMA may be used exclusively for a designated area and road salts for another area for an extended period time Then, it investigates the impact of CMA on vegetation yields, which can be compared with those of areas from road salt use The difference in vegetation yields may be quantified in terms of market value, and so forth www.elsolucionario.net Valuation of Benefits and Costs 16.4 This is an open-end type question (No solution is given.) 16.5 This is an open-end type question (No solution is given.) Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net Benefit-Cost Analysis 16.6 (a) BC (i ) analysis: • Design A: I = $400, 000 C ' = $50, 000( P / A,8%,15) = $427,974 B = $85, 000 I = $300, 000 C ' = $80, 000( P / A,8%,15) = $684, 758 B = $85, 000 • Incremental analysis: Fee collections in the amount of $85,000 will be the same for both alternatives Therefore, we will not be able to compute the BC (i ) ratio If this happens, we may select the best alternative based on either the lease cost ( I + C ' ) criterion or the incremental B 'C (i ) criterion Using the B 'C (i ) criterion, ∆B C (8%) A− B ' ∆B − ∆C ' − ($427,974 − $684, 758) = = = 2.56 > ∆I $100, 000 ∴Select design A (b) Incremental analysis (A – C): ∆B 'C (8%) A−C = ∆B − ∆C ' − ($427,974 − $556,366) = = 2.57 > ∆I $50, 000 ∴Select design A www.elsolucionario.net • Design B: 16.7 • Building X: BX = $1,960, 000( P / A,10%, 20) = $16, 686, 656 C X = $8, 000, 000 + $240, 000( P / A,10%, 20) − $4,800, 000( P / F ,10%, 20) = $9,329,984 $16, 686, 656 BC (10%) X = = 1.79 > $9,329,984 Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net • Building Y: BY = $1,320, 000( P / A,10%, 20) = $11, 237,952 CY = $12, 000, 000 + $180, 000( P / A,10%, 20) − $7, 200, 000( P / F ,10%, 20) = $12, 462,528 $11, 237,904 BC (10%)Y = = 0.90 < $12, 462,528 16.8 Incremental BC (i ) analysis: Present Worth I B C BC (i ) A1 $100 $400 $100 Proposals A2 $300 $700 $200 1.4 A3 $200 $500 $150 1.43 Incremental A3-A1 A2-A1 $100 $200 $100 $300 $50 $100 0.67 C $5,656 $1,600 $2,922 1.25 Incremental C-B A-B -$1,414 $754 $720 $1,560 -$472 $471 -5.7 0.37 ∴ Select either A1 or A2 16.9 Incremental BC (i ) analysis Present Worth I B C’ BC (i ) A $7,284 $2,440 $3,865 1.24 Design B $7,070 $880 $3,394 1.65 www.elsolucionario.net Since Building Y is not desirable at the outset, we don’t need to conduct an incremental analysis Building X becomes the better choice ∴ Select Design B Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 16.10 (a) The benefit-cost ratio for each alternative: • Alternative A: B = ($1, 000, 000 + $250, 000 + $350, 000 + $100, 000)( P / A,10%,50) = $16,855,185 C = $8, 000, 000 + $200, 000( P / A,10%,50) = $9,982,963 BC (10%) A = $16,855,185 = 1.69 > $9,982,963 B = ($1, 200, 000 + $350, 000 + $450, 000 + $200, 000)( P / A,10%,50) = $21,812,592 C = $10, 000, 000 + $250, 000( P / A,10%,50) = $12, 478, 704 BC (10%) B = $21,812,592 = 1.75 > $12, 478, 704 • Alternative C: B = ($1,800, 000 + $500, 000 + $600, 000 + $350, 000)( P / A,10%,50) = $32, 223,147 C = $15, 000, 000 + $350, 000( P / A,10%,50) = $18, 470,185 BC (10%) B = $32, 223,147 = 1.74 > $18, 470,185 www.elsolucionario.net • Alternative B: (b) Select the best alternative based on BC (i ) : $21,812,592 − $16,855,185 $12, 478, 704 − $9,982,963 = 1.99 > ( Select B ) BC (10%) B − A = $32, 223,147 − $21,812,592 $18, 470,185 − $12, 478, 704 = 1.74 > ( Select C ) BC (10%)C − B = Comments: You could select the best alternative based on B 'C (i ) : Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net A $8,000,000 $1,982,963 1.86 I C’ ' B C (i ) Alternative B $10,000,000 $2,478,704 1.93 C $15,000,000 $3,470,185 1.92 ($21,812,592 − $16,855,185) − ($2, 478, 704 − $1,982,963) $10, 000, 000 − $8, 000, 000 = 2.23 > ( Select B ) B 'C (10%) B − A = ($32, 223,147 − $21,812,592) − ($3, 470,185 − $2, 478, 704) $15, 000, 000 − $10, 000, 000 = 1.88 > ( Select C ) 16.11 • Option – The “long” route: user's annual cost = 22 miles × $0.25 per mile × 400,000 cars = $2, 200, 000 sponsor's annual cost = $21, 000, 000( A / P,10%, 40) + $140, 000 = $2, 287, 448 • Option – Shortcut: user's annual cost = 10 miles × $0.25 per mile × 400,000 cars = $1, 000, 000 sponsor's annual cost = $45, 000, 000( A / P,10%, 40) + $165, 000 = $4, 766, 674 www.elsolucionario.net B 'C (10%)C − B = • Incremental analysis (Option - Option 1): Incremental user's benefit = $2, 200, 000 − $1, 000, 000 = $1, 200, 000 $1, 200, 000 BC (10%) −1 = $4, 766, 674 − $2, 287, 448 = 0.48 < ∴ Assuming that there is no do-nothing alternative, select option Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 16.12 • Multiple alternatives: Projects PW of Benefits A1 $40 A2 $150 A3 $70 A4 $120 PW of Costs $85 $110 $25 $73 Net PW -$45 $40 $45 $47 B/C ratio 0.47 1.36 2.80 1.64 Since the BC ratio for project A1 is less than 1, we delete it from our comparison A3 vs A4: $120 − $70 $73 − $25 = 1.04 > BC (10%) A 4− A3 = Select A4 A2 vs A4: $150 − $120 $110 − $73 = 0.81 < BC (10%) A2− A4 = Select A4 Short Case Studies ST 16.1 Capital allocation decision, assuming that the government will be able to raise the required funds at 10% interest: District I II III Project 27th Street Holden Avenue Forest City Road Fairbanks Avenue Oak Ridge Road University Blvd Hiawassee Road Lake Avenue Apopka-Ocoee Road 10 Kaley Avenue PW(10%) $1,606,431 $3,438,531 $2,682,758 $2,652,473 $1,672,473 $5,258,050 $4,130,824 $2,958,052 $552,475 $4,459,032 www.elsolucionario.net • Incremental analysis Investment $980,000 $3,500,000 $2,800,000 $1,400,000 $2,380,000 $5,040,000 $2,520,000 $4,900,000 $1,365,000 $2,100,000 Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net 11.Apopka-Vineland Road 12 Washington Street 13 Mercy Drive 14 Apopka Road 15 Old Dixie Highway 16 Old Apopka Road IV $1,166,557 $1,788,245 $5,066,566 $2,338,635 $1,213,846 $1,899,946 $1,170,000 $1,120,000 $2,800,000 $1,690,000 $975,000 $1,462,500 District I II III IV Projects 1,2,4 5,7 9,10,11,12 13,14,16 NPW $7,697,488 $5,803,297 $5,966,309 $9,305,147 (b) $15 million to districts I & II and $9 million to districts III & IV: District I & II III & IV Projects 2,4,5,6,7 10,12,13,14,15 NPW $17,152,400 $12,866,320 Investment $14,840,000 $8,685,000 (c) If $24 million were allocated based on project merit alone, the optimal solution would be: Total investment = $23,587,500 Total net present value = $31,852,543 Projects selected = 1, 2, 4, 6, 7, 10, 12, 13, 14, 15, 16 ST 16.2 Given i = 8% , g = 10% , garbage amount/day = 300 tons (a) The operating cost of the current system in terms of $/ton of solid waste: www.elsolucionario.net (a) $6 million to each district: • Annual garbage collection required (assuming 365 days): Total amount of garbage = 300 tons × 365 days = 109,500 tons • Equivalent annual operating and maintenance cost: PW (8%) = $905, 400( P / A1 ,10%,8%, 20) = $20, 071,500 AEC (8%) = $20, 071,500( A / P,8%, 20) = $2, 044,300 Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net • Operating cost per ton: cost per ton= $2,044,300 =$18.67/ton 109,500 (b) The economics of each solid-waste disposal alternative in terms of $/ton: • Site 1: = $653, 000 $653,000 cost per ton= =$5.96/ton 109,500 • Site 2: AEC (8%)2 = $4,384, 000( A / P,8%, 20) + $480, 000 − ($14, 700 + $99,300) = $812,520 $812,520 cost per ton= =$7.42/ton 109,500 • Site 3: AEC (8%)3 = $4, 764, 000( A / P,8%, 20) + $414, 000 − ($15,300 + $103,500) = $780, 424 $780,424 cost per ton= =$7.13/ton 109,500 • Site 4: AEC (8%)2 = $5, 454, 000( A / P,8%, 20) + $408, 000 − ($17,100 + $119, 400) www.elsolucionario.net AEC (8%)1 = $4, 053, 000( A / P,8%, 20) + $342, 000 − ($13, 200 + $87, 600) = $827, 000 $827,000 cost per ton= =$7.55/ton 109,500 ∴ Site is the most economical choice Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net ST 16.3 (a) Let’s define the following variables to compute the equivalent annual cost Ala = initial land cost Aeq = initial equipment cost Ast = initial structure cost Apu = initial pumping equipment cost Aen = initial annual energy cost in today's dollars Alb = initial annual labor cost in today's dollars • Land: PW (10%)land = Ala − Ala (1.03 /1.1)120 = $0.99963 Ala • Equipment: Let’s define the following additional variables I15 n = replacement cost in year 15n S15 n = salvage value in year 15n C15 n = net replacement cost in year 15n where n = 1, 2, 3, 4, 5, 6, and The total replacement cost over the analysis period is calculated as follows: I15 = Aeq (1.05)15 = 2.07893 Aeq S15 = 0.5 Aeq www.elsolucionario.net Aen = initial annual repair cost in today's dollars C15 = (2.07893 − 0.5) Aeq = 1.57893 Aeq C15 n = (1.57893 Aeq )(1.05)15( n −1) S15 n = 0.5 Aeq (1.05)15 n PW (10%)equipment = Aeq + ∑ C15 n − S120 n =1 (1.57893 Aeq )(1.05)15( n −1) n =1 (1.1)15 n = Aeq + ∑ − 0.5 Aeq (1.05)120 (1.1)120 = 1.745 Aeq Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be 10 obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net • Structure: ⎡ 1 ⎤ 0.6 Ast PW (10%) structure = Ast + (0.40) Ast ⎢ + − 40 (1.1)80 ⎥⎦ (1.1)120 ⎣ (1.1) = 1.00902 Ast • Pumping: PW (10%) pumping = 1.745 Apu • Energy: 120 PW (10%)energy = ∑ Aen (1.05 /1.1) j = 20.92302 Aen • Labor: 120 PW (10%)labor = ∑ Alb (1.04 /1.1) j = 17.3113 Alb j =1 • Repair: 120 PW (10%)repair = ∑ Are (1.02 /1.1) j = 12.748 Are j =1 • Present worth of the life-cycle cost: PW (10%) = 0.99963 Ala + 1.745 Aeq + 1.00902 Ast + 1.745 Apu + 20.92302 Aen + 17.3113 Alb + 12.748 Are Parameters Ala Aeq Ast Apu Aen Alb Are PW(10%) AEC(10%) Option $2,400,000 $49,000 $49,000 $400,000 $500,000 $500,000 $400,000 $175,000 $700,000 $2,100,000 $2,463,000 $1,750,000 $100,000 0 $100,000 $200,000 $95,000 $30,000 $10,364,300 $1,036,440 $125,000 $65,000 $20,000 $7,036,290 $703,637 $100,000 $53,000 $15,000 $6,433,460 $643,353 $50,000 $37,000 $5,000 $4,395,790 $439,584 www.elsolucionario.net j =1 ∴ Option is the least cost alternative Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be 11 obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 www.elsolucionario.net (b) Cost / gallon = $439,584 / 2,000,000(365) = $0.0006 per gallon Monthly charge = (0.0006)(400)(30) = $7.23 per month • Users’ benefits (1) Reduced travel time (2) Reduced fuel consumption (3) Reduced air pollution (4) Reduced number of accidents • Users’ disbenefits: Increased automobile purchase and maintenance costs (b) Sponsor’s cost: • Development costs associated with computerized dashboard navigational systems, roadside sensors, and automated steering and speed controls • Implementation and maintenance costs • Public promotional and educational costs (c) On a national level, the sponsor’s costs are estimated to be as follows: • R&D costs = $2.5 billion • Implementation costs = $18 billion • Maintenance costs = $4 billion per year Comments: However, the users’ benefits are sketchy, except the level of reduction possible in the area of travel time, fuel consumption, and air pollution Ask the students to quantify these in dollar terms by consulting various government publications on public transportation Once these figures are estimated, the benefit-cost ratio can be easily derived www.elsolucionario.net ST 16.4 (a) Users benefits and disbenefits: Contemporary Engineering Economics, Fourth Edition, by Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected by Copyright and written permission should be 12 obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by means, electronic, mechanical, photocopying, recording, or likewise For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458 ... problems listed under ``Short Case Studies" require more involved assumptions, solution steps, and calculations Please note that your independent solutions to the text problems may yield answers slightly... another 280 selftest questions This OneKey site is designed primarily to help students develop a working knowledge of the concepts and principles of engineering economics Most questions in this guide... decisions discussed in Section1.5 Contemporary Engineering Economics, Fourth Edition, By Chan S Park ISBN 0-13-187628-7 © 2007 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This