Solutions Manual for Engineering Economy, 9th Edition by Leland Blank, Anthony Tarquin

All19 ChaptersCovered
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SOLUTIONS

,TableofContents
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1. FoundationsofEngineering Economy k k k




2. Factors:HowTimeandInterestAffectMoney
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3. CombiningFactors and SpreadsheetFunctions k k k k




4. Nominal and Effective Interest Rates
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5. AnalysisUsing PresentWorthandFuture Worth Values
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6. AnnualWorthAnalysis
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7. Rateof Return Analysis: One Project
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8. Rateof Return Analysis: Multiple Alternatives
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9. Benefit/CostAnalysisand Public Sector Economics k k k k k




10. Project Financing and Noneconomic Attributes
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11. ReplacementandRetentionDecisions k k k




12. Independent Projectswith BudgetLimitation k k k k




13. BreakevenandPaybackAnalysis k k k




14. EffectsofInflation
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15. CostEstimationandIndirect CostAllocation
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16. Depreciation andDepletion Methods k k k




17. After-TaxEconomic Analysis k k




18. SensitivityAnalysis and Staged Decisions
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19. Decision Making under Risk
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, Chapter 1 k




Foundations of Engineering Economy k k k




Basic Concepts
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1.1 Financial units for economically best. k k k k




1.2 Morale,goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
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1.3 Measure of worth is a criterion used to select the economically best alternative.
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Some measuresarepresentworth,rateofreturn,paybackperiod,benefit/cost
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ratio.
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1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
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1.5 Sustainability: Intangible; installation cost: tangible; transportation cost: k k k k k k




tangible; simplicity: intangible; taxes: tangible; resale value: tangible;
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morale:intangible;
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rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others:
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kintangible; ethics: intangible. k k




1.6 Examplesare: house purchase; car purchase, credit card(whichones to use); personal loans
(and their rate of interest and repayment schedule); investment decisions of all types; whento
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sellahouseorcar.
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Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
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(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and
deceptive acts are clearly present.
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(b) The Code for Engineer’s is only partially useful to the owners in determining sound
basessince thecontractoris not an engineer. Much ofthelanguage of the Code is
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orientedtowardrepresentation,qualifications,etc.,notspecificacts ofdeceitand
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fraudulent behavior.Code sections may be somewhatdifficultto interpret in
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constructionofahouse.
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(c) Probably a better source would be a Code for Contractor’s or consulting with a real
estate attorney.
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1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
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1.9 Example actions are: k k




• Try to talk them out of doing it now, explaining itis stealing
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• Try to get them to pay for their drinks
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, • Pay for all the drinks himself
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• Walk away and not associate with them again
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1.10 This is structured to be a discussion question; many responses are acceptable. Responses can
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vary from the ethical (stating the truth and accepting the consequences) to unethical (continuing to
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deceive himself and the instructor and devise some on-the-spot excuse).
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Lessons can be learned from the experience. A few of them are:
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• Think before he cheats again. k k k k




• Think about the longer-term consequences of unethical decisions. k k k k k k k




• 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 k k k k k k k k k k k




practices.
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Interest Rate and Rate of Return k k k k k




1.11 Extra amount received = 2865 - 25.80*100 = $285
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Rate of return = 285/2580 k k k k k




= 0.110 (11%) k




Total invested + fee 2865 + 50 = $2915 k k k k k k k k




k Amount required for 11% return = 2915*1.11 k k k k k k




= $3235.65 k




1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
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(b) Interest = total amount paid – principal
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= 1,936,000- 1,600,000 k k




= $336,000 k




1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
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1.14 Interest rate = interest paid/principal k k k k




= (312,000/2,600,000) k




= 0.12 (12%) k




1.15 i = (1125/12,500)*100 = 9% i
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= (6160/56,000)*100 = 11% i
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= (7600/95,000)*100 = 8%
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The $56,000 investment has the highest rate of return
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1.16 Interest on loan = 45,800(0.10) = $4,580 k k k k k k




Default insurance = $900 k k k




Set-upfee=45,800(0.01) =458 k k k k k k




Total amount paid = 4,580 + 900 + 458 = $5938 k k k k k k k k k k




Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of
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McGraw-Hill Education.
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