All 19 Chapters Covered
SOLUTIONS
,
Table of Contents
1. Foundations of Engineering Economy
2. Factors: How Time and Interest Affect Money
3. Combining Factors and Spreadsheet Functions
4. Nominal and Effective Interest Rates
5. Analysis Using Present Worth and Future Worth Values
6. Annual Worth Analysis
7. Rate of Return Analysis: One Project
8. Rate of Return Analysis: Multiple Alternatives
9. Benefit/Cost Analysis and Public Sector Economics
10. Project Financing and Noneconomic Attributes
11. Replacement and Retention Decisions
12. Independent Projects with Budget Limitation
13. Breakeven and Payback Analysis
14. Effects of Inflation
15. Cost Estimation and Indirect Cost Allocation
16. Depreciation and Depletion Methods
17. After-Tax Economic Analysis
18. Sensitivity Analysis and Staged Decisions
19. Decision Making under Risk
,
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
1.3 Measure of worth is a criterion used to select the economically best alternative. Some
measures are present worth, rate of return, payback period, benefit/cost ratio.
1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost:
tangible; simplicity: intangible; taxes: tangible; resale value: tangible;
morale: intangible;
rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others: intangible;
ethics: intangible.
1.6 Examples are: house purchase; car purchase, credit card (which ones to use); personal loans
(and their rate of interest and repayment schedule); investment decisions of all types; when to
sell a house or car.
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and
deceptive acts are clearly present.
(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 the Code is
oriented toward representation, qualifications, etc., not specific acts of deceit and
fraudulent behavior. Code sections may be somewhat difficult to interpret in
construction of a house.
(c) Probably a better source would be a Code for Contractor’s or consulting with a real
estate attorney.
1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
1.9 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
1.10 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.
Alternatively, Claude may learn nothing from the experience and continue his unethical
practices.
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $285
Rate of return = 285/2580
= 0.110 (11%)
Total invested + fee 2865 + 50 = $2915
Amount required for 11% return = 2915*1.11
= $3235.65
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
(b) Interest = total amount paid – principal
= 1,936,000- 1,600,000
= $336,000
1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
1.14 Interest rate = interest paid/principal
= (312,000/2,600,000)
= 0.12 (12%)
1.15 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
1.16 Interest on loan = 45,800(0.10) = $4,580
Default insurance = $900
Set-up fee = 45,800(0.01) = 458
Total amount paid = 4,580 + 900 + 458 = $5938
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
SOLUTIONS
,
Table of Contents
1. Foundations of Engineering Economy
2. Factors: How Time and Interest Affect Money
3. Combining Factors and Spreadsheet Functions
4. Nominal and Effective Interest Rates
5. Analysis Using Present Worth and Future Worth Values
6. Annual Worth Analysis
7. Rate of Return Analysis: One Project
8. Rate of Return Analysis: Multiple Alternatives
9. Benefit/Cost Analysis and Public Sector Economics
10. Project Financing and Noneconomic Attributes
11. Replacement and Retention Decisions
12. Independent Projects with Budget Limitation
13. Breakeven and Payback Analysis
14. Effects of Inflation
15. Cost Estimation and Indirect Cost Allocation
16. Depreciation and Depletion Methods
17. After-Tax Economic Analysis
18. Sensitivity Analysis and Staged Decisions
19. Decision Making under Risk
,
Chapter 1
Foundations of Engineering Economy
Basic Concepts
1.1 Financial units for economically best.
1.2 Morale, goodwill, dependability, acceptance, friendship, convenience, aesthetics, etc.
1.3 Measure of worth is a criterion used to select the economically best alternative. Some
measures are present worth, rate of return, payback period, benefit/cost ratio.
1.4 The color I like, best fuel rating, roomiest, safest, most stylish, fastest, etc.
1.5 Sustainability: Intangible; installation cost: tangible; transportation cost:
tangible; simplicity: intangible; taxes: tangible; resale value: tangible;
morale: intangible;
rate of return: tangible; dependability: intangible; inflation: tangible; acceptance by others: intangible;
ethics: intangible.
1.6 Examples are: house purchase; car purchase, credit card (which ones to use); personal loans
(and their rate of interest and repayment schedule); investment decisions of all types; when to
sell a house or car.
Ethics
1.7 This problem can be used as a discussion topic for a team-based exercise in class.
(a) Most obvious are the violations of Canons number 4 and 5. Unfaithfulness to the client and
deceptive acts are clearly present.
(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 the Code is
oriented toward representation, qualifications, etc., not specific acts of deceit and
fraudulent behavior. Code sections may be somewhat difficult to interpret in
construction of a house.
(c) Probably a better source would be a Code for Contractor’s or consulting with a real
estate attorney.
1.8 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.
1.9 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
1.10 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.
Alternatively, Claude may learn nothing from the experience and continue his unethical
practices.
Interest Rate and Rate of Return
1.11 Extra amount received = 2865 - 25.80*100 = $285
Rate of return = 285/2580
= 0.110 (11%)
Total invested + fee 2865 + 50 = $2915
Amount required for 11% return = 2915*1.11
= $3235.65
1.12 (a) Payment = 1,600,000(1.10)(1.10) = $1,936,000
(b) Interest = total amount paid – principal
= 1,936,000- 1,600,000
= $336,000
1.13 i = [(5,184,000 – 4,800,000)/4,800,000]*100% = 8% per year
1.14 Interest rate = interest paid/principal
= (312,000/2,600,000)
= 0.12 (12%)
1.15 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
1.16 Interest on loan = 45,800(0.10) = $4,580
Default insurance = $900
Set-up fee = 45,800(0.01) = 458
Total amount paid = 4,580 + 900 + 458 = $5938
Copyright © 2018 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent
of McGraw-Hill Education.
@Seismicisolation
2@Seismicisolation
