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Engineering Economy (7th Edition, Leland Blank & Anthony Tarquin) – Complete Solutions Manual

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This document provides the full solutions manual for Engineering Economy (7th edition) by Leland Blank and Anthony Tarquin. It contains detailed, step-by-step solutions to all end-of-chapter problems across the textbook. Topics include time value of money, cost estimation, cash flow analysis, interest rates, depreciation methods, project evaluation, replacement analysis, and economic decision-making. This resource is structured chapter by chapter and is ideal for exam preparation, homework support, and mastering engineering economy problem-solving techniques.

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Engineering Economy 7th Edition
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Engineering Economy 7th Edition

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Subido en
29 de septiembre de 2025
Número de páginas
369
Escrito en
2025/2026
Tipo
Examen
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SOLUTION MANUAL
Engineering Economy,
7th edition Leland Blank and Anthony Tarquin




1

, Solutions to end-of-chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin

Chapter 1
Foundations of Engineering Economy
1.1 The four elements are cash flows, time of occurrence of cash flows, interest rates, and
measure of economic worth.

1.2 (a) Capital funds are money used to finance projects. It is usually limited in the amount
of money available.

(b) Sensitivity analysis is a procedure that involves changing various estimates to see if/how
they affect the economic decision.

1.3 Any of the following are measures of worth: present worth, future worth, annual worth, rate
of return, benefit/cost ratio, capitalized cost, payback period, economic value added.

1.4 First cost: economic; leadership: non-economic; taxes: economic; salvage value: economic;
morale: non-economic; dependability: non-economic; inflation: economic; profit: economic;
acceptance: non-economic; ethics: non-economic; interest rate: economic.

1.5 Many sections could be identified. Some are: I.b; II.2.a and b; III.9.a and b.

1.6 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

1.7 This is structured to be a discussion question; many responses are acceptable. It is an
ethical question, but also a guilt-related situation. He can justify the result as an accident; he
can feel justified by the legal fault and punishment he receives; he can get angry because it
WAS an accident; he can become tormented over time due to the stress caused by accidently
causing a child’s death.

1.8 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.


2

, Alternatively, Claude may learn nothing from the experience and continue his unethical
practices.

1.9 i = [(3,885,000 - 3,500,000)/3,500,000]*100% = 11% per year

1.10 (a) Amount paid first four years = 900,000(0.12) = $108,000

(b) Final payment = 900,000 + 900,000(0.12) = $1,008,000

1.11 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.

1.12 Interest on loan = 23,800(0.10) = $2,380
Default insurance = 23,800(0.05) = $1190
Set-up fee = $300

Total amount paid = 2380 + 1190 + 300 = $3870

Effective interest rate = (3870/23,800)*100 = 16.3%

1.13 The market interest rate is usually 3 – 4 % above the expected inflation rate. Therefore,

Market rate is in the range 3 + 8 to 4 + 8 = 11 to 12% per year

1.14 PW = present worth; PV = present value; NPV = net present value; DCF = discounted cash
flow; and CC = capitalized cost

1.15 P = $150,000; F = ?; i = 11%; n = 7

1.16 P = ?; F = $100,000; i = 12%; n = 2

1.17 P = $3.4 million; A = ?; i = 10%; n = 8

1.18 F = ?; A = $100,000 + $125,000?; i = 15%; n = 3

1.19 End-of-period convention means that all cash flows are assumed to take place at the end of
the interest period in which they occur.

1.20 fuel cost: outflow; pension plan contributions: outflow; passenger fares: inflow;
maintenance: outflow; freight revenue: inflow; cargo revenue: inflow; extra bag charges:
Inflow; water and sodas: outflow; advertising: outflow; landing fees: outflow; seat
preference fees: inflow.


3

, 1.21 End-of-period amount for June = 50 + 70 + 120 + 20 = $260
End-of-period amount for Dec = 150 + 90 + 40 + 110 = $390

1.22 Month Receipts, $1000 Disbursements, $1000 Net CF, $1000
Jan 500 300 +200
Feb 800 500 +300
Mar 200 400 -200
Apr 120 400 -280
May 600 500 +100
June 900 600 +300
July 800 300 +500
Aug 700 300 +400
Sept 900 500 +400
Oct 500 400 +100
Nov 400 400 0
Dec 1800 700 +1100

Net Cash flow = $2,920 ($2,920,000)

1.23




1.24




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