Solutions Manual for Engineering Economy (9th Edition) by Leland Blank and Anthony Tarquin

All 19 Chapters Covered
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SOLUTIONS

,Table ofContents
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1. Foundations of Engineering Economy dr dr dr



2. Factors: How Time and Interest Affect Money
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3. Combining Factors and Spreadsheet Functions dr dr dr dr




4. Nominal and Effective Interest Rates
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5. Analysis Using Present Worth and Future Worth Values
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6. Annual Worth Analysis
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7. Rate of Return Analysis: One Project
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8. Rate of Return Analysis: Multiple Alternatives
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9. Benefit/Cost Analysis and Public Sector Economics dr dr dr dr dr




10. Project Financing and Noneconomic Attributes
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11. Replacement and Retention Decisions dr dr dr




12. Independent Projects with Budget Limitation dr dr dr dr




13. Breakeven and Payback Analysis dr dr dr




14. Effects of Inflation
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15. Cost Estimation and Indirect Cost Allocation
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16. Depreciation and Depletion Methods dr dr dr




17. After-Tax Economic Analysis dr dr




18. Sensitivity Analysis and Staged Decisions dr dr dr dr




19. Decision Making under Risk dr dr dr

, Chapter 1 dr



Foundations of Engineering Economy dr dr dr




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




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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dr Some measures are present worth, rate ofreturn, payback period, benefit/cost
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dr ratio.
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:dr dr dr dr dr dr


dr tangible; simplicity: intangible; taxes: tangible; resale value:
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dr tangible; morale: intangible; dr dr




rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others:
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intangible; ethics: intangible.
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1.6 Examples a re: house purchase; car purchase, credit card (which ones to use); personal
loans (and their r ate of interest and repayment schedule); investment decisions of all
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types; when to sell ah ouse or c ar.
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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 bases since the contractor is not an engineer. Much of the language of
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the Code is oriented toward representation, q ualifications, etc., not specific acts
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of deceit and fraudulent behavior. Code sections may be somewhat difficult to
interpret in construction of a house.
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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: dr dr


 Try to talk them out of doing it now, explaining it is 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.
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Responses can vary from the ethical (stating the truth and accepting the consequences) to
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unethical (continuing to deceive himself and the instructor and devise some on-the-spot
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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. dr dr dr dr


 Think about the longer-term consequences of unethical decisions.
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 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
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practices.
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Interest Rate and Rate of Returndr dr dr dr dr




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


= 0.110 (11%) dr




Total invested + fee 2865 + 50 = $2915
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Amount required for 11% return =
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2915*1.11
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= $3235.65 dr




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 dr dr


= $336,000 dr




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 dr dr dr dr


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


= 0.12 (12%) dr




1.15 i = (1125/12,500)*100 =
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9% i = (6160/56,000)*100
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= 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 dr dr dr dr dr dr


Default insurance = dr dr

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