Advanced Engineering Economics
Contemporary Engineering Economics, 5th edition, © 2010
Chapter
Opening Story
 General Motors Finances
Ethanol Maker Coskata$400 million in cellulosic
ethanol plant to produce
100 million gallons a year.
 At Issue: What would be
the GM’s financial risk in
investing an Ethanol
project? How should GM
factor the future fluctuation
and uncertainty of gasoline
prices into the analysis?
Contemporary Engineering Economics, 5th edition, © 2010
Methods of Describing Project Risk
 Sensitivity Analysis: a procedure of identifying the
project variables which, when varied, have the
greatest effect on project acceptability.
 Break-Even Analysis: a procedure of identifying the
value of a particular project variable that causes the
project to exactly break even.
 Scenario Analysis: a procedure of comparing a “base
case” to one or more additional scenarios, such as
best and worst cases, to identify the extreme and
most likely project outcomes.
Contemporary Engineering Economics, 5th edition, © 2010
Example 12.1
Transmission-Housing
Project by Boston Metal
Company
Financial Facts:
Known Facts
Required investment =
$125,000
Project Life = 5 years
Income tax rate = 40%
MARR = 15%
Unknown but Predictable (Most
Likely Values)
Unit variable cost = $15 per
unit
Number of units = 2,000 units
Unit Price = $50 per unit
Salvage value = $40,000
Fixed cost = $10,000/Yr
Required: Determine the acceptability of
the investment
Contemporary Engineering Economics, 5th edition, © 2010
Sensitivity Analysis for Five Key Input
Variables
Deviation
-20% -15%
-10%
-5%
0%
Unit price
5%
10%
15%
20%
$57 $9,999
$20,055
$30,111
$40,169
Demand
12,010 19,049
26,088
33,130
40,169
47,208
54,247
61,286
68,325
Variable
cost
52,236 49,219
46,202
43,186
40,169
37,152
34,135
31,118
28,101
Fixed cost
44,191 43,185
42,179
41,175
40,169
39,163
38,157
37,151
36,145
Salvage
value
37,782 38,378
38,974
39,573
40,169
40,765
41,361
41,957
42,553
$50,225 $60,281 $70,337 $80,393
 Variable most sensitive to NPW – Unit price
 Variable least sensitive to NPW – Salvage value
Contemporary Engineering Economics, 5th edition, © 2010
Break-Even Analysis
Breakeven analysis is a tool used to determine when a
business will be able to cover all its expenses and begin
to make a profit from a project.
Excel using a Goal Seek function
Analytical Approach
Contemporary Engineering Economics, 5th edition, © 2010
Using a Goal Seek
Function in Excel
 PW of Inflow: 100.5650X + $44,490
 PW of Outflow: 30.1694X + $145,113
 NPW = 70.3956X - $100,623
Contemporary Engineering Economics, 5th edition, © 2010
Analytical
Approach
The NPW:
PW (15%) =
100.5650X + $44,490 (30.1694X + $145,113)
=70.3956X - $100,623.
 Breakeven volume:

PW (15%)= 70.3956X $100,623 = 0
Xb =1,430 units.
 PW of cash inflows
PW(15%)Inflow= (PW of after-tax net revenue)
+ (PW of net salvage value)
+ (PW of tax savings from depreciation
= 30X(P/A, 15%, 5) + $37,389(P/F, 15%, 5)
+ $7,145(P/F, 15%,1) + $12,245(P/F, 15%, 2)
+ $8,745(P/F, 15%, 3) + $6,245(P/F, 15%, 4)
+ $2,230(P/F, 15%,5)
= 30X(P/A, 15%, 5) + $44,490
= 100.5650X + $44,490
 PW of cash outflows:
PW(15%)Outflow = (PW of capital expenditure
+ (PW) of after-tax expenses
= $125,000 + (9X+$6,000)(P/A, 15%, 5)
= 30.1694X + $145,113
Contemporary Engineering Economics, 5th edition, © 2010
Scenario Analysis
 Scenario analysis is a process of analyzing possible
future outcomes by considering alternative possible
events (scenarios). The analysis is designed to allow
improved decision-making by allowing more complete
consideration of outcomes and their implications.
Source: Wikipedia
Contemporary Engineering Economics, 5th edition, © 2010
Example 12.3 Scenario Analysis
Variable
Considered
WorstCase
Scenario
Most-Likely-Case
Scenario
Best-Case
Scenario
Unit demand
1,600
2,000
2,400
Unit price ($)
48
50
53
Variable cost ($)
17
15
12
Fixed Cost ($)
11,000
10,000
8,000
Salvage value ($)
30,000
40,000
50,000
PW (15%)
-$5,856
$40,169
$104,295
Contemporary Engineering Economics, 5th edition, © 2010
Constructing a
Decision Tree
A Company is considering
marketing a new product. Once
the product is introduced, there
is a 70% chance of encountering
a competitive product. Two
options are available each
situation.
Option 1 (with competitive
product): Raise your price and
see how your competitor
responds. If the competitor
raises price, your profit will be
$60. If they lower the price,
you will lose $20.
Option 2 (without
competitive product): You still
two options: raise your price
or lower your price.
The conditional profits
associated with each event
along with the likelihood of
each event is shown in the
decision tree.
Conditional
Profit
Competitor’s price
Decision Points
(0.5)
Events
(
$60
High
Our Price
(0.5)
-$20
Low
High
) Probability
Competitive
Product (0.7)
Market
$40
High
(0.2)
Low
No Competitive
Product (0.3)
Do not market
Low
(0.8)
$10
High
$100
Low
$0
First Decision Point
Contemporary Engineering Economics, 5th edition, © 2010
$30
Second Decision Point
Rollback Procedure
 To analyze a decision tree, we begin at the end of the
tree and work backward.
 For each chance node, we calculate the expected
monetary value (EMV), and place it in the node to
indicate that it is the expected value calculated over all
branches emanating from that node.
 For each decision node, we select the one with the
highest EMV (or minimum cost). Then those decision
alternatives not selected are eliminated from further
consideration.
Contemporary Engineering Economics, 5th edition, © 2010
Making Sequential Investment Decisions
$60
High
(0.5)
$20
Set High Price
$44
Market
$44
(0.5)
-$20
Low
Competitive $20
Product (0.7)
$40
High
(0.2)
Low
$16
No Competitive
Product (0.3)
Do not market
Low
(0.8)
$10
Set High Price
$100
$100
Low
$0
Contemporary Engineering Economics, 5th edition, © 2010
$30
Decision Rules
Market the new product.
Whether or not you encounter a competitive
product, raise your price.
The expected monetary value associated
with marketing the new product is $44.
Contemporary Engineering Economics, 5th edition, © 2010
Practice Problem
 A company is considering the purchase of a new labor-
saving machine.
 The machine’s cost will turn out to be $55 per day. Each
hour of labor that is saved reduces costs by $5.
However, there is some uncertainty over the number of
hours that actually will be saved.
 It is judged that the hours of labor saved per day will be
10, 11, or 12, with probabilities of 0.10, 0.60, 0.30,
respectively.
 Let us define “profit” as the excess of labor-cost savings
over the machine cost.
Contemporary Engineering Economics, 5th edition, © 2010
Construct a Decision Tree
-$5
0.10
$2.5
$2.5
Invest
10
11
0.60
0
12
Do not invest
0.30
$10
0
EMV = $2.5
Decision: Purchase the equipment
Contemporary Engineering Economics, 5th edition, © 2010
Expected Value of Perfect Information (EVPI)
 What is EVPI? This is equivalent to asking yourself how much you can
improve your decision if you had perfect information.
 Mathematical Relationship:
EVPI = EPPI – EMV = EOL
where EPPI (Expected profit with perfect information) is the expected
profit you could obtain if you had perfect information, and EMV
(Expected monetary value) is the expected profit you could obtain
based on your own judgment. This is equivalent to expected
opportunity loss (EOL).
Contemporary Engineering Economics, 5th edition, © 2010
Expected Value of Perfect Information (EVPI)
State of
Nature
10
11
12
Best
Strategy
Don’t Buy
Indifferent
Buy
Maximum
Payoff
Probability
the State of
Nature
Occurs
Expected
Payoff or
each State
0
0
10
0.10
0.60
0.30
0
0
3
Expected Profit with Perfect Information (EPPI):
(0.10)(0) + (0.60)(0) + (0.30(10) = $3
 Expected Value of Perfect Information (EVPI) = EPPI – EMV
$3 - $2.5 = $0.5

Contemporary Engineering Economics, 5th edition, © 2010
Bill’s Decision
Problem –
$50,000 to Invest
 Decision Problem:
Buying a highly
speculative stock (d1) with
three potential levels of
return – High (50%),
Medium (9%), and Low (30%).
 Buying a very safe U.S.
Treasury bond (d2) with a
guaranteed 7.5% return.
 Seek advice from an expert?
 Seek professional advice
before making the decision
 Do not seek professional
advice – do on his own.
Contemporary Engineering Economics, 5th edition, © 2010
Decision Tree for Bill’s Investment Problem
-
Contemporary Engineering Economics, 5th edition, © 2010
Expected Value of Perfect Information
Decision Option
Potential
Return Level
Probability
Option1:
Invest in
Stock
(Prior
Optimal)
Option 2:
Invest in
Bonds
Optimal
Choice with
Perfect
Information
Opportunity
Loss
Associated
with Investing
in Bonds
High (A)
0.25
$16,510
$898
Stock
$15,612
Medium (B)
0.40
890
898
Bond
0
Low(C)
0.35
-13,967
898
Bond
0
-$405
$898
EMV
EPPI = (0.25)($16,510) + (0.40)($898)
+ (0.35)($898) = $4,801
EVPI = EPPI – EV
= $4,801 - $898
= $3,903
Contemporary Engineering Economics, 5th edition, © 2010
$3,903
EOL = (0.25)($15,612)
+ (0.40)(0) + (0.35)(0)
= $3,903
Bill’s Investment Problem with an Option of Getting
Professional Advice
Updating Conditional Profit (or Loss) after
Paying a Fee to the Expert (Fee = $200)
Revised Decision Tree
Contemporary Engineering Economics, 5th edition, © 2010
Conditional Probabilities of the Expert’s Prediction,
Given a Potential Return on the Stock
Given Level of Stock Performance
What the Report
Will Say
High
(A)
Medium
(B)
Low
(C)
Favorable (F)
0.80
0.65
0.20
Unfavorable (UF)
0.20
0.35
0.80
Contemporary Engineering Economics, 5th edition, © 2010
Nature’s Tree: Conditional Probabilities and
Joint Probabilities
Nature’s Tree
Joint & Marginal Probabilities
 P(A,F) = P(F|A)P(A) = (0.80)(0.25) = 0.20
 P(A,UF|A)P(A) = (0.20)(0.25) = 0.05
 P(B,F) = P(F|B)P(B) = (0.65)(0.40) = 0.26
 P(B,UF) = P(UF|B)P(B) = (0.35)(0.40) = 0.14
 P(F) = 0.20 + 0.26 + 0.07 = 0.53
 P(UF) = 1 – P(F) = 1 – 0.53 =
0.47
Contemporary Engineering Economics, 5th edition, © 2010
Joint and Marginal Probabilities
What the Report Will Say
Joint Probabilities
When Potential Level
of Return is Given
Favorable (F)
Unfavorable (UF)
Marginal
Probabilities of
Return Level
High (A)
0.20
0.05
0.25
Medium (B)
0.26
0.14
0.40
Low (C)
0.07
0.28
0.35
Marginal
Probabilities
0.53
0.47
1.00
Contemporary Engineering Economics, 5th edition, © 2010
Determining Revised Probabilities
P(A|F) = P(A,F)/P(F) = 0.20/0.53 = 0.38
P(B|F) = P(B,F)/P(F) = 0.26/0.53 = 0.49
P(C|F) = P(C,F)/P(F) = 0.07/0.53 = 0.13
P(A|UF) = P(A,UF)/P(UF) = 0.05/0.47 = 0.30
P(B|UF) + P(B,UF)/P(UF) = 0.14/0.47 = 0.30
P(C|UF) = P(C,UF)/P(UF) = 0.28/0.47 = 0.59
Contemporary Engineering Economics, 5th edition, © 2010
Decision Making after Having Imperfect Information
Contemporary Engineering Economics, 5th edition, © 2010
Contemporary Engineering Economics, 5th
edition, © 2010