SUPPLEMENT TO
CHAPTER 5
Decision Analysis
Supplement Outline
Introduction and Decision Tables, 2
Decision Trees and Influence
Diagrams, 4
Value of Information, 7
Sensitivity Analysis, 9
OM in Action: Oglethorpe Power, 11
Summary, 12
Key Terms, 12
Solved Problems, 12
Discussion and Review
Questions, 14
Internet Exercises, 15
Problems, 15
Mini-Case: Southern Company, 21
Mini-Case: ICI Canada, 22
Learning Objectives
After completing this supplement,
you should be able to:
Model a single-stage
decision problem as a
decision table and use the
expected value approach to
solve it.
LO 2 Construct a decision tree and
use it to analyze a multistage
problem; describe an
influence diagram.
LO 3 Calculate the expected value
of perfect and imperfect
(sample) information.
LO 4 Conduct sensitivity analysis
on a simple decision
problem.
LO 1
2
PART THREE System Design
Introduction and Decision Tables
LO 1
Decision analysis is a scientific and practical method for making important decisions. It
was introduced in the 1960s (see http://decision.stanford.edu/library/ronald-a.-howard/­
Decision%20Analysis-%20Applied%20Decision%20Theory.pdf). Decision analysis
involves identification, clear representation, and formal assessment of important aspects
of a decision and then determination of the best decision by applying the maximum expected
value criterion.
Decision analysis is suitable for a wide range of operations management decisions where
uncertainty is present. Among them are capacity planning, product design, equipment
selection, and location planning. Decisions that lend themselves to a decision analysis
approach tend to be characterized by the following elements:
1. There is at least one random variable with a set of possible values (or future conditions
or states of nature) that will have a bearing on the result of the decision.
2. There is a decision with a list of alternatives for the manager/decision maker to choose
from.
3. There is a known payoff (or value) for each alternative under each possible future
condition.
To use this approach, a decision maker would employ the following process:
1. Determine the goal or objective, e.g., maximize expected profit or net present value,
or minimize expect cost.
2. Develop a list of possible alternatives for the decision in order to achieve the goal.
3. Identify possible future conditions or states of nature for each random variable (e.g.,
demand will be low, medium, or high; the equipment will or will not fail; the competitor will or will not introduce a new product) that will affect the goal.
4. Determine or estimate the payoff (or value) associated with each alternative for every
possible future condition.
5. Estimate the likelihood of each possible future condition for each random variable.
6. Evaluate the alternatives according to the goal or decision criterion, and select the
best alternative.
payoff table Table showing
the payoffs (values) for each
alternative in every possible
state of nature of a onerandom-variable problem.
The data for a one random-variable problem can be summarized in a payoff table,
which shows the payoffs (value) for each alternative under the various possible states of
nature of the variable. These tables are helpful in choosing among alternatives because they
facilitate comparison of alternatives. Consider the following payoff table, which illustrates
a capacity planning problem.
Possible Future Demand
Alternatives
Low
Moderate
High
Small facility
$10*
$10
$10
Medium facility
7
12
12
Large facility
(4)
2
16
*Net present value in $ millions.
The payoffs are shown in the body of the table. In this instance, the payoffs are in terms
of net present values, which represent equivalent current dollar values of estimated future
income less costs. This is a convenient measure because it places all alternatives on a
comparable basis. If a small facility is built, the payoff will be $10 million for all three
possible states of nature. For a medium facility, low demand will have net present value
of $7 million, whereas both moderate and high demand will have net present values of
SUPPLEMENT TO CHAPTER 5 Decision Analysis
3
$12 million. A large facility will have loss of $4 million if demand is low, net present
value of $2 million if demand is moderate, and net present value of $16 million if demand
is high.
The problem for the decision maker is to select one of the alternatives, taking the net
present values into account. In order to do so, one needs to determine the probability
of occurrence for each state of nature (unless one of the alternatives dominates the rest,
i.e., would have the highest NPV under each state of nature. Because the states of nature
for a random variable should be mutually exclusive and collectively exhaustive, these
probabilities must add upto 1.00. These probabilities are usually subjective opinions of
experts. The decision analyst interviewing an expert should be careful not to ask leading questions and should be aware of psychological biases such as overconfidence or
intentional underestimation by marketing/sales managers (to make meeting these goals
easier).
Given the probabilities for each state of nature, a widely used approach is the expected
value method. The expected value is calculated for each alternative, and the one with the
highest expected value is selected (when more is better such as for NPV). The expected
value for an alternative is the sum over the states of nature of the payoff multiplied by
probability of each state of nature.
Using the expected value method, identify the best alternative for the following payoff table
given these probabilities for demand: low = .30, moderate = .50, and high = .20.
Example S-1
Possible Future Demand
Alternatives
Low
Moderate
High
Small facility
$10*
$10
$10
Medium facility
7
12
12
Large facility
(4)
2
16
*Net present value in $ millions.
Find the expected value of each alternative by multiplying the probability of occurrence
for each state of nature by the payoff of the alternative for that state of nature and summing them:
EVsmall = .30($10) + .50($10) + .20($10) = $10
EVmedium = .30($7) + .50($12) + .20($12) = $10.5
EVlarge = .30(−$4) + .50($2) + .20($16) = $3
Hence, choose the medium facility because it has the highest expected net present
value.
The expected value method is most appropriate when a decision maker is neither risk
averse nor risk seeking, but is risk neutral. Typically, well-established organizations facing decisions of this nature tend to use the expected-value method because it provides the
long-run average payoff. That is, the expected-value amount (e.g., $10.5 million in Example
S-1) is not an actual payoff but an expected or average amount that would be achieved if
a large number of identical decisions were to be made. Hence, if a decision maker applies
this method to a large number of similar decisions, the expected payoff for the total will
approximate the sum of the individual expected payoffs. If the decision maker is not risk
neural, then he tends to consider the spread or risk of payoffs, and even individual payoffs.
For a risk-averse decision maker, the minimum payoff, if the medium facility is built, $7
million, may still be acceptable. If it is not, he may look for a new alternative (e.g., seek
Solution
PART THREE System Design
4
FIGURE 5S-1
of
State
Format of a decision tree
1
eA
State
s
oo
Ch
�
e A1
of na
ture 2
Ch
oo
se
Choos
A
2
State
Decision point
ure 1
f nat
o
State
of na
ture 2
Payoff 2
2
Possible
second
decisions
Initial
decision
1
Payoff 1
e1
natur
Choos
e A2�
Payoff 3
�
e A3
Payoff 4
Choos
2
Choos
e A4�
Payoff 5
Payoff 6
Chance event
more information) or choose to build a small facility that has a higher minimum payoff
($10 million). An approach to deal with risk-averse decision makers is called the utility
theory.
More complex decision problems will have more than one decision variable (i.e., a
multistage decision problem) and/or more than one random variable. In this case, the
payoffs cannot be represented by a table. Instead, decision trees are used to graphically
model the decision problem.
Decision Trees and Influence Diagrams
LO 2
decision tree A graphical
representation of the decision
variables, random variables,
probabilities, and payoffs.
Example S-2
A decision tree is a graphical representation of the decision variables, random variables
and their probabilities, and the payoffs. The term gets its name from the treelike appearance of the diagram (see Figure 5S-1). Decision trees are particularly useful for analyzing
situations that involve sequential or multistage decisions. For instance, a manager may
initially decide to build a small facility but she has to allow for the possibility that demand
may be higher than anticipated. In this case, the manager may plan to make a subsequent
decision on whether to expand or build an additional facility.
A decision tree is composed of a number of nodes that have branches emanating from
them (see Figure 5S-1). Square nodes denote decision points, and circular nodes denote
chance events. Read the tree from left to right. Branches leaving square nodes represent
alternatives; branches leaving circular nodes represent the states of nature.
After a decision tree has been drawn and necessary data are determined, it is analyzed
from right to left; that is, starting with the last decision that might be made it is “rolled”
back. For each decision, choose the alternative that will yield the greatest return (or the
lowest cost). For each chance node, calculate the expected value of the payoffs of its
states of nature. If chance events follow a decision, choose the alternative that has the
highest expected value (or lowest expected cost).
A manager must decide on the size of a video arcade to construct. The manager has narrowed the choices to two: large or small. Information has been collected on payoffs, and
the following decision tree has been constructed. Analyze the decision tree and determine
SUPPLEMENT TO CHAPTER 5 Decision Analysis
5
which initial alternative (build small or build large) should be chosen in order to maximize
expected value.
Low
all
High
m
ds
uil
B
nd
a
dem
(.4)
dem
and
$40
D
(.6)
$40
g
hin
ot
on
Exp
and
ing
oth
n
Do
Bu
ild
lar
ge
and
em
wd
Lo
High
dem
and
(.4)
Red
uce
(.6)
pric
es
$55
($10)
$50
$70
The dollar amounts at the branch ends indicate the estimated payoffs if the sequence of
decisions and chance events occurs. For example, if the initial decision is to build a small
facility and it turns out that demand is low, the payoff will be $40 (thousand). Similarly, if
a small facility is built, and demand turns out high, and a later decision is made to expand,
the payoff will be $55. The figures in parentheses on branches leaving the chance nodes
indicate the probabilities of those states of nature. Hence, the probability of low demand
is .4, and the probability of high demand is .6. Payoffs in parentheses indicate losses.
Analyze the decisions from right to left:
1. Determine which alternative would be selected for each possible second decision.
For a small facility with high demand, there are two choices: do nothing, or expand.
Because expand has higher payoff, you would choose it. Indicate this by placing a
double slash through do nothing alternative. Similarly, for a large facility with low
demand, there are two choices: do nothing or reduce prices. You would choose reduce
prices because it has the higher expected value, so a double slash is placed on the
other branch.
2. Determine the product of the chance probabilities and their respective payoffs for the
remaining branches:
Build small
Low demand .4($40) = $16
High demand .6($55) = 33
Build large
Low demand .4($50) = 20
High demand .6($70) = 42
3. Determine the expected value of each initial alternative:
Build small $16 + $33 = $49
Build large $20 + $42 = $62
Hence, the choice should be to build a large facility because it has higher expected
value than the small facility.
Solution
PART THREE System Design
6
The above problem was entered in the decision analysis software DPL, from Syncopation Software (www.syncopation.com). The solution is shown below:
Small
Demand
[49.000]
Low
[40.000]
40%
40.000
High
React to high Demand
[55.000]
60%
Build
[62.000]
Large
Demand
[62.000]
React to low demand
[50.000]
Low
Do nothing
[40.000]
Expand
[55.000]
40.000
Do nothing
Reduce prices
40%
55.000
[–10.000]
–10.000
[50.000]
50.000
High
[70.000]
60%
70.000
How to use DPL software In the DPL’s window, the top pane is for the influence
diagram (described next), and the bottom pane is for the decision tree (see the snapshot
below). The decision tree was drawn by first adding decision nodes (by clicking on the
square icon at the top and moving it into the top pane) and discrete chance nodes (by
clicking on the green circle icon at the top and moving it into the top pane), and then
moving them to the bottom pane in order to draw the decision tree. Each node is then
double-clicked and its properties defined (name, alternatives/outcomes, probabilities,
nature of node’s branches: symmetric vs. asymmetric). Next, each payoff is specified by
double-clicking each branch. Finally, the expected value is computed by clicking on Run
(at the top) and then Decision Analysis.
React to
Demand
Build
Demand
React to
low
demand
Demand
Low
40
Build
React to
Demand
Do nothing
Small
40
High
Expand
55
Large
Demand
Do nothing
–10
Low
Reduce prices
50
High
70
SUPPLEMENT TO CHAPTER 5 Decision Analysis
Influence Diagrams
Influence diagrams (see example below) can graphically represent complex decision problems that have many random variables (chance events) and one or more decision variables.
Influence diagrams are more concise than decision trees because they do not show the
alternatives branches coming out of the decision nodes and the states of nature branches
coming out of the chance nodes. Constructing and validating an influence diagram improves
communication and consensus building at the beginning of the decision modelling process.
The following is an example of the influence diagram representing the decision of whether
or not to introduce a new product. The green circles show the random variables (chance
events) and the rounded yellow squares show the payoff or part of it. This influence diagram
for a new product decision also involves a pricing decision. The uncertainties (i.e., random
variables) are units sold, which are affected by the pricing decision, fixed cost, and variable
cost. Profit is the ultimate payoff, which is influenced by the total cost and revenue.
Decision analysis software such as DPL are able to measure the effect of variability of
each random variable on the payoff (e.g., profit) and identify those crucial random variables
that have the greatest influence on profit. The result is plotted in a vertical bar chart called
a Tornado diagram, named as such because the more important random variables, those
that affect the payoff most and hence have a longer bar, are drawn on the top, making the
chart look like a tornado (see e.g., http://en.wikipedia.org/wiki/Tornado_diagram).
Fixed
Cost
Units
Sold
Price?
Variable
Cost
Total
Cost
Revenue
Introduce
Product?
Profit
Source: K. Chelst, “Can’t See the Forest Because of the Decision Trees: A Critique of Decision
Analysis in Survey Texts,” Interfaces (28)2, March–April 1998, pp. 80–98.
Value of Information
LO 3
In certain situations, it is possible to know with more certainty which state of nature of a
random variable will actually occur in the future or be able to prepare better for various states
of nature. For instance, the choice of location for a restaurant may weigh heavily on whether
a new highway will be constructed nearby. A decision maker may have probabilities for this
random variable; however, it may be possible to delay a decision until it is more clear which
state of nature will occur. This might involve taking an option to buy the land. If the state
of nature is favourable, the option can be exercised; if it is unfavourable, the option can be
allowed to expire. The question to consider is whether the cost of the option will be less than
the expected gain due to delaying the decision. Possible ways of obtaining information about
a random variable depend on the nature of the random variable. Information about consumer
preferences might come from market research, and information about potential of a product
could come from product testing. If the information is perfect (i.e., it is 100 percent accurate
and will pinpoint the state of nature of the random variable), then the expected gain in payoff
is called the expected value of perfect information, or EVPI.
7
8
PART THREE System Design
expected value of perfect
information (EVPI) The
difference between the
expected payoff with perfect
information and the expected
payoff without the information.
Expected value of perfect information (EVPI) is the difference between the expected
payoff with perfect information (under certainty) and the expected payoff without the
information (under risk).
Example S-3
Solution
Expected value of
Expected payoff
Expected payoff
=
−
perfect information under certainty
under risk
(5S-1)
Using the information from Example S-1, determine the expected value of perfect information using Formula 5S-1.
First, calculate the expected payoff under certainty. To do this, identify the best payoff under
each state of nature. Then combine these by weighting each payoff by the probability of
that state of nature and adding the amounts. Thus, the best payoff under low demand is $10,
the best payoff under moderate demand is $12, and the best payoff under high demand is
$16. The expected payoff under certainty is, then:
.30($10) + .50($12) + .20($16) = $12.2
Possible Future Demand
Alternatives
Low
Moderate
High
Small facility
$10*
$10
$10
Medium facility
7
12
12
Large facility
(4)
2
16
*Net present value in $ millions.
The expected payoff under risk, as calculated in Example S-1, is $10.5. The EVPI is
the difference between these:
EVPI = $12.2 − $10.5 = $1.7
expected value of sample
information (EVSI) The
difference between the
expected payoff with sample
(imperfect) information and
the expected payoff without
sample information.
EVPI indicates the upper limit on the amount the decision maker should be willing to
spend to obtain information. Thus, if the cost exceeds EVPI, the decision maker would be
better off not spending additional money and simply going with the alternative that has
the highest expected payoff.
In most cases, the information that can be obtained about a random variable is useful but
not perfect. In these cases, a statistical result called Bayes Rule can be used to update the
(prior) probability distribution of the states of nature of the random variable. The difference
between the expected payoff with imperfect (sample) information and the expected payoff
without sample information is called the expected value of sample information (EVSI).
Bayes Rule: Let A be an “information” event, and let B1 and B2 be mutually exclusive and
exhaustive states of nature. Let P(B1) and P(B2) be the prior probabilities of the states of
nature, and P(A | B1) and P(A | B2) be the likelihood of observing A given B1 and B2, respectively (i.e., the conditional probabilities). Then, the posterior probabilities P(B1 | A) and
P(B2 | A) are:
(
) (P (A | B ) P (B ) + P (A | B ) P (B ))
P (B 1 | A ) = P (A | B1 ) P (B1 )
P (B 2 | A ) = 1 − P (B 1 | A )
1
1
Note: P(A) = P(A | B1) P(B1) + P(A | B2) P(B2 ).
2
2
SUPPLEMENT TO CHAPTER 5 Decision Analysis
A manager is considering random drug testing of his employees. However, the test is not
100 percent accurate. The conditional probabilities are as follows: If a person uses drugs,
then the test will be positive 93 percent of the time, whereas if he is not, the test will be
negative 97 percent of the time. Suppose that 5 percent of the employees are drug users.
9
Example S-4
a. What are the posterior probabilities?
b. If the cost of not identifying (and dismissing) a drug user is $1,000, cost of falsely
accusing a non-user is $200, and other costs are zero, what is the maximum the manager
should be willing to pay for a drug test, i.e., EVSI?
a. Consider a random employee. Let B1 = the event that he is a drug user, B2 = the event
that he is not a drug user. We have P(B1) = .05 and hence P(B2) = 1 – .05 = .95. Let
A + = the event that test result is positive. We have P(A + | B1) = .93 and P(A + | B2) =
1 – .97 = .03. Then,
(
) ( .93 ( .05 ) + .03 ( .95 )) = .62
P ( B 1 | A + ) = .93 ( .05 )
Solution
and
P ( B2 | A + ) = 1 − .62 = .38.
Also, P(A +) = P(A + | B1) P(B1) + P(A + | B2) P(B2) = .93(.05) + .03(.95) = .075,
And P(A−) = 1 – P(A+) = 1 – .075 = .925
ote the relatively low (62 percent) probability that a drug user will be identified and a
N
relatively high (38 percent) probability that a non-drug user will be implicated.
Similarly (A – | B1) = 1 – .93 = .07 and P(A – | B2) = .97, and
(
) ( .07 ( .05 ) + .97 ( .95 )) = .0038
P ( B 1 | A − ) = .07 ( .05 )
and
P ( B2 | A − ) = 1 − .0038 = .9962.
b. EVSI can be determined from the following decision tree. Note that negative payoffs
are costs.
Positive
Yes
Test result
[–9.215]
Test
[–9.215]
Employee 1
[–76.000]
7.50%
User
62.00%
Non-user
38.00%
Negative
Employee 2
[–3.800]
User
0.38%
Non-user
92.50%
99.62%
No
Employee 3
[–50.000]
User
5.00%
Non-user
95.00%
[–1000.000]
–1000.000
[0.000]
0.000
The maximum that the manager should pay for a drug test (i.e., EVSI) is
­$50 - $9.215 = $40.785.
Sensitivity Analysis
LO 4
Generally speaking, both the payoffs and the probabilities in a decision problem are estimated values. Consequently, it can be useful for the decision maker to have some indication
[0.000]
0.000
[–200.000]
–200.000
[–1000.000]
–1000.000
[0.000]
0.000
10
sensitivity analysis Determining the range of
probability for which an
alternative has the best
expected payoff.
Example S-5
PART THREE System Design
of how sensitive the choice of an alternative is to changes in one or more of these values.
Unfortunately, it is impossible to consider all possible combinations of every change in
a typical problem. Nevertheless, the sensitivity of the chosen alternative to probability
estimates can be determined for a simple problem.
Sensitivity analysis provides the range of probability over which an alternative has the
best expected payoff. The approach illustrated here is useful when there are two states of
nature. It involves constructing a graph and then using algebra to determine the range of
probabilities for which a given alternative is best. In effect, the graph provides a visual
indication of the range of probability over which various alternatives are optimal, and the
algebra provides exact values of the endpoints of the ranges. Example S-5 illustrates the
procedure.
Given the following (revenue) payoff table, determine the range of probability for state
of nature #2, that is, P2, for which each alternative is optimal using the expected value
method.
State of Nature
Alternative
Solution
#1
#2
A
4
12
B
16
2
C
12
8
First, plot the expected payoff of each alternative relative to P2. To do this, plot the #1 payoff
on the left side of the graph and the #2 payoff on the right side. For instance, for alternative
A, plot 4 on the left side of the graph and 12 on the right side. Then, connect these two points
with a straight line. The three alternatives are plotted on the graph as shown below.
The graph shows the range of values of P2 over which each alternative is optimal. Thus,
for low values of P2 (and thus high values of P1, since P1 + P2 = 1.0), alternative B will
have the highest expected value; for intermediate values of P2 alternative C is best; and
for higher values of P2 alternative A is best.
To find exact values of the ranges, determine where the upper parts of the lines intersect. Note that at the intersections, the two alternatives represented by the lines would be
equivalent in terms of expected value; hence, the decision maker would be indifferent
between the two at that point. To determine the intersections, you must obtain the equation
of each line. Because these are straight lines, they have the form y = a + bx, where a is
the y-intercept value at the left axis, b is the slope of the line, and x is P2. Slope is defined
as the change in y for a one-unit change in x. In this case, the distance between the two
vertical axes is 1.0. Consequently, the slope of each line is equal to the right-hand value
minus the left-hand value. The slopes and equations are:
16
16
B
14
14
12
10
#1
Payoff
A
C
12
10
8
#2
8 Payoff
6
6
4
4
2
B best
0
.2
C best
.4
.6
P (2)
2
A best
.8
1.0
SUPPLEMENT TO CHAPTER 5 Decision Analysis
#1
#2
Slope
A
4
12
B
16
2
2 − 16 = −14
16 − 14 P 2
C
12
8
8 − 12 = −4
12 − 4 P 2
12 − 4 = +8
11
Equation
4 + 8 P 2
From the graph, we can see that alternative B is best from the point P2 = 0 to the point
where the alternative B line intersects the alternative C line. To find that point, solve for
the value of P2 at their intersection. This requires setting the two equations equal to each
other and solving for P2. Thus,
16 − 14 P2 = 12 − 4 P2
Rearranging terms yields
4 = 10 P2
Solving yields P2 = .40. Thus, alternative B is best from P2 = 0 up to P2 = .40. Alternatives B and C are equivalent at P2 = .40.
Alternative C is best from that point until its line intersects alternative A’s line. To find
that intersection, set those two equations equal and solve for P2. Thus,
4 + 8 P2 = 12 − 4 P2
Rearranging terms results in
12 P2 = 8
Solving yields P2 = .67. Thus, alternative C is best from P2 > .40 up to P2 = .67, where
alternatives A and C are equivalent. For values of P2 greater than .67 up to P2 = 1.0, alternative A is best.
Note: If a problem calls for ranges with respect to P1, you could find the P2 ranges as
above, and then subtract each P2 from 1.00 (e.g., .40 becomes .60, and .67 becomes .33).
OM in Action Oglethorpe Power
O
glethorpe Power is an electric generation and distribution cooperative supplying 20 percent of the electricity
used in Georgia. Most of the remaining demand for electricity
in Georgia is supplied by Georgia Power Co. Oglethorpe and
Georgia Power jointly own most of the power transmission lines
in Georgia, and supply their excess electricity to Florida.
Some years ago, Florida Power and Light (FPL) indicated
to Oglethorpe that they were interested in the construction of
another major transmission line between Florida and Georgia.
Oglethorpe invited the consulting company Applied Decision
Analysis Inc. to assist it in making a decision.
A team was formed to investigate the decision variables,
random variables, and their payoffs. Brainstorming resulted in
an influence diagram and identification of the three decision
variables (each with alternatives): line (joint with Georgia Power,
alone, or no line), nature of control (Oglethorpe or FPL), and
whether to upgrade the associated facilities (joint with Georgia Power, alone, or no upgrade). Five random variables (each
with states of nature) were identified: construction cost (low,
medium, or high), competitive situation in Florida (good, fair,
or bad), Florida demand (low, medium, or high), Oglethorpe’s
share of demand (very low, low, medium, high, or very high),
and future spot price for electricity (low, medium, or high). The
payoffs were measured in terms of net present value.
The team estimated the probability of the states of nature
for each random variable and the payoff for each combination
of decision alternatives and states of nature. Using the DPL
software, the alternative (in brackets) for each decision variable
with highest expected value was: line (alone), control (Oglethorpe), and upgrade (no). Then, the risk profile (the distribution
of NPV) of this solution was determined by DPL. Because this
showed considerable possible negative NPV, the team identified
the random variable that most affected the downside risk: the
competitive situation in Florida. The next step was to collect
further information on this random variable in order to reduce the
range of values for the payoff. The team reported all its findings
to top management, who started negotiating with FPL.
Applied Decision Analysis, Inc. is now part of PricewaterhouseCoopers.
Source: A. Borison, “Oglethorpe Power Corporation Decides about
Investing in a Major Transmission System,” Interfaces (25)2, March–
April 1995, pp. 25–36.
PART THREE System Design
12
Summary
Key Terms
Solved Problems
Decision analysis is a formal approach to decision making that is useful in operations management.
Decision analysis provides a framework for the analysis of decisions. It involves identifying the decision alternatives, chance events (random variables), and payoffs. A simple decision problem can be
represented as a decision table. The common alternative selection method is the expected value method.
Two visual tools useful for representing decision problems are decision trees and influence diagrams.
Information can be used to reduce the variability of the important random variables. Two tools that
can be used in deciding whether or not to buy information are expected value of perfect information
and expected value of sample information. Graphical sensitivity analysis can be used to determine the
effect of variability of probabilities assigned to states of nature on the best alternative.
payoff table, 2
sensitivity analysis, 10
decision tree, 4
expected value of perfect information
(EVPI), 8
expected value of sample information
(EVSI), 8
«Xtags error: Style name
ambiguous: tag @...:»Solved
The following solved problems 1–4 refer to this payoff (profits) table:
Alternative capacity
for new store
New Bridge Built
No New Bridge
A
1
14
B
2
10
C
4
6
where A = small, B = medium, and C = large.
Problem 1
Using graphical sensitivity analysis, determine the probability for “No New Bridge” state of
nature for which each alternative would be optimal.
Solution
Plot a straight line for each alternative. Do this by plotting the payoff for “new bridge built” on
the left axis and the payoff for “no new bridge” on the right axis, and then connecting the two
points with a straight line. Each line represents the expected profit for an alternative for the entire
range of probability of “no new bridge.” Because the lines represent expected profit, the line that
is highest for a given value of P(no new bridge) is optimal. Thus, from the graph, you can see
that for low values of this probability, alternative C is best, and for higher values, alternative A is
best (alternative B is never the highest line, so it is never optimal, i.e., it is dominated).
Payoff if
new bridge
14
A
B
Payoff if
no new
bridge
10
C
6
4
2
1
C best
0
A best
1.0
.27
P (no new bridge)
0
SUPPLEMENT TO CHAPTER 5 Decision Analysis
13
The dividing line between the ranges where alternatives C and A are optimal occurs where
the two lines intersect. To find that probability, first formulate the equation for each line. To do
this, let the intersection with the left axis be the y intercept; the slope equals the right-side payoff
minus the left-side payoff. Thus, for C you have 4 + (6 − 4)P, which is 4 + 2P. For A, 1 + (14 − 1)
P, which is 1 + 13P. Setting these two equal to each other, you can solve for P:
4 + 2P = 1 + 13P
Solving, P = 3/11 = .27. Therefore, the ranges for P(no new bridge) for each alternative to be best are:
A: .27 ≤ P ≤ 1.00
B: never optimal
C: 0 ≤ P ≤ .27
Using the probabilities of .60 for a new bridge and .40 for no new bridge, calculate the expected
value of each alternative, and identify the alternative that would be selected using the expected value
method.
A: .60(1) + .40(14) = 6.20 [best]
Problem 2
Solution
B: .60(2) + .40(10) = 5.20
C: .60(4) + .40(6) = 4.80
Calculate the EVPI using the data from the previous problem.
Problem 3
Using Formula 5S-1, the EVPI is the expected payoff under certainty minus the expected payoff
under risk (i.e., the maximum ­expected value). The expected payoff under certainty involves multiplying the best payoff in each column by the column probability and then summing those amounts.
The best payoff in the first column is 4, and the best in the second column is 14. Thus,
Solution
Expected payoff under certainty = .60(4) + .40(14) = 8.00
The maximum expected value is 6.20 from the previous problem. Therefore,
EVPI = 8.00 − 6.20 = 1.80
Suppose that the values in the payoff table represent costs instead of profits.
Problem 4
a. Using sensitivity analysis, determine the range of P(no new bridge) for which each alternative
would be optimal.
b. If P(new bridge) = .60 and P(no new bridge) = .40, find the alternative to minimize expected
cost.
a. The graph is identical to that shown in Solved Problem 1. However, the lines now represent
expected costs, so the best alternative for a given value of P(no new bridge) is the lowest line.
Hence, for very low values of P(no new bridge), A is best; for intermediate values, B is best;
and for high values, C is best. You can set the equations of A and B, and B and C, equal to each
other in order to determine the values of P(no new bridge) at their intersections.
Thus,
A = B: 1 + 13P = 2 + 8P; solving, P = 1/5 = .20
B = C: 2 + 8P = 4 + 2P; solving, P = 2/6 = .33
Hence, the ranges are:
A best: 0 ≤ P < .20
B best: .20 < P ≤ .33
C best: .33 < P ≤ 1.00
Solution
PART THREE System Design
14
b. Expected value calculations are the same whether the values represent costs or profits. Hence, the
expected payoffs for costs are the same as the expected payoffs for profits that were calculated
in Solved Problem 1. However, now you want the alternative that has the lowest expected payoff
rather than the one with the highest payoff. Consequently, alternative C is the best because its
expected payoff (4.8) is the lowest of the three (6.2, 5.2, 4.8).
Cost with
new bridge
14
A
Cost with
no new
bridge
10
B
C
6
4
2
1
A
best
0
B
best
.20
C best
.33
P (no new bridge)
1.0
0
Problem 5
A diagnostic test of a certain disease has 95 percent sensitivity (i.e., probability that the test will
be positive for a patient with the disease) and 96 percent specificity (i.e., probability that the test
will be negative for a healthy person). Only 1 percent of the population has the disease in question.
If the diagnostic test reports that a person chosen at random tests positive, what is the conditional
probability that the person does, in fact, have the disease?
Solution
P(disease | + test)
= ( P ( + test | disease ) P ( disease )) / ( P ( + test | disease ) P ( disease ) + P ( + test | no disease ) P ( no disease )
=
.95 (.01 )
= 0.1935
.95 (.01 ) + ( 1 − .96 )( 1 − .01 )
or 19.35 percent, not a large probability.
Discussion and
Review Questions
1. List the steps in the decision analysis process. (LO1)
2. What information is contained in a payoff table? (LO1)
3. Under what circumstances is expected value method appropriate? When isn’t it appropriate?
(LO1)
4. What information does a decision maker need in order to perform an expected value analysis
of a problem? What options are available to the decision maker if the probabilities of the
states of nature are unknown? Can you think of a way you might use sensitivity analysis in
such a case? (LO1 & 4)
5. Suppose a manager is using the expected-value method as a basis for making a capacity decision and obtains a result in which there is a virtual tie between two of the seven alternatives.
How is the manager to make the decision? (LO1)
6. What is the difference between a decision tree and an influence diagram? When should each
be used? (LO2)
7. Will expected value of perfect information be always greater than or equal to expected value
of sample information? Briefly explain. (LO3)
8. What is sensitivity analysis, and how can it be useful to a decision maker? (LO4)
SUPPLEMENT TO CHAPTER 5 Decision Analysis
1. View the video http://www.youtube.com/chevron#p/u/12/JRCxZA6ay3M and describe how
decision analysis has helped Chevron. (LO1–4)
15
Internet Exercises
2. Visit http://gunston.gmu.edu/healthscience/730/IntroductionToDecisionAnalysis.
asp?E=0#Organization_of_the_Book, browse through the “Introduction to Decision Analysis” section, find out why some decisions are hard to make, and list them. (LO1)
3. Visit http://www.treeage.com/resources/caseStudies.html, choose a case study, and summarize
how decision analysis is being used. Visit http://www.palisade.com/cases/bucknell.asp,
choose a case study, and summarize how decision analysis is being used. (LO1 & 2)
4. Visit http://www.syncopation.com/­casestudies.html, choose a case study, and summarize
how decision analysis is being used. (LO1 & 2)
5. Visit http://www.lionhrtpub.com/orms/surveys/das/das.html and find out the educational price
of DPL. (LO2)
1. A small building contractor has recently experienced two successive years in which work
opportunities exceeded the firm’s capacity. The contractor must now make a decision on
capacity for next year. Estimated profit for each alternative under each of the two possible
states of nature for next year’s demand is shown below. (LO1–3)
Next Year’s Demand
Alternative
Low
High
Do nothing
$50*
$60
Expand
20
80
Subcontract
40
70
*Profit in $ thousands.
Suppose after a certain amount of discussion, the contractor is able to subjectively assess
the probabilities of low and high demand: P(low) = .3 and P(high) = .7.
a. Determine the expected profit of each alternative. Which alternative is best?
b. Analyze the problem using a decision tree. Show the expected profit of each alternative
on the tree.
c. Calculate the expected value of perfect information. How could the contractor use this
knowledge?
2. Refer to Problem 1. Construct a graph that will enable you to perform sensitivity analysis on
the problem. Over what range of P(high) would each of the alternatives be best? (LO4)
3. A company that plans to expand its product line must decide whether to build a small or a
large facility to produce the new products. If it builds a small facility and demand is low,
the net present value after deducting for building costs will be $400,000; If demand is high,
the company can either maintain the small facility or expand it. Expansion would have a net
present value of $450,000, and maintaining the small facility would have a net present value
of $350,000. (LO2–4)
If a large facility is built and demand is high, the estimated net present value is $800,000;
If demand turns out to be low, the net present value will be −$10,000.
The probability that demand will be high is estimated to be .60, and the probability of low
demand is estimated to be .40.
a. Analyze this decision problem using a decision tree.
b. Calculate the EVPI. How could this information be used?
c. Determine the range of P(demand low) over which each alternative would be better.
Problems
PART THREE System Design
4. Determine the course of action that has the highest expected payoff for the decision tree
below. (LO2)
$1.0*
)
d
an
l
al
Sm
(.4
M
Larg
e
5)
nd (.
ema
md
ediu
Expa
nd
and
g
(.1)
Do n
Expand
Build
n
ema
g
all d
Medium demand (.5)
Sm
Larg
ed
ema
nd
(.1)
Other use #1
Othe
r use
#2
g
othin
Do n
Subcontract
ild
Bu
all d
Sm
Med
ium
dem
and
(.5)
La
rg
ed
em
an
d
(.1
)
$0.7
$1.5
$1.0
g
Other use #1
Othe
r use
Othe
($0.9)
$1.4
$1.0
g
$1.0
Other use #1
r use
$1.5
#2
othin
Do n
$1.6
$1.7
othin
Do n
(.4)
$1.6
$1.6
Build
nd
ema
$1.5
$1.8
othin
Do n
$1.3
$1.3
othin
dem
4)
d (.
Expand
g
othin
Do n
m
de
Su
bc
on
tra
ct
16
#2
$1.1
$0.9
$2.4
* Net present value in millions
5. The lease of an amusement park is about to expire. Management must decide whether to
renew the lease for another 10 years or to relocate near the site of a proposed resort. The town
planning board is currently debating the merits of granting approval to the resort. A consultant
has estimated the net present value of amusement park’s two alternatives under each state of
nature as shown below. (LO1–3)
Options
Renew (current location)
Relocate near resort
Resort Approved
Resort Rejected
$3,000,000
$4,000,000
5,000,000
2,000,000
Suppose that the management of the amusement park has decided that there is a 75% probability that the resort’s application will be approved.
a. If the management uses the expected value method, which alternative should it
choose?
SUPPLEMENT TO CHAPTER 5 Decision Analysis
b. Represent this problem in the form of a decision tree and solve it.
c. If the management has been offered the option of a temporary lease while the town planning board considers the resort’s application, would you advise the management to sign
the lease? The temporary lease will cost $200,000.
6. Construct a graph that can be used for sensitivity analysis of the preceding problem.
(LO1 & 4)
a. How sensitive is the solution to the problem in terms of the probability estimate
of .75?
b. Suppose that, after consulting with a member of the town planning board, management
decides that the estimate of resort approval should be .60. How sensitive is the solution to
this revised estimate? Explain.
c. Suppose that management is confident of all the estimated payoffs except for $5 million.
If the probability of resort approval is .60, for what range of payoff for relocate/approved
will the relocate alternative remain optimal?
7. A company must decide whether to construct a small, medium, or large plant. A consultant’s
report indicates a .20 probability that demand will be low and a .80 probability that demand
will be high.
If the company builds a small facility and demand turns out to be low, the net present value
will be $42 million; if demand turns out to be high, the company can either subcontract some
production and realize the net present value of $43 million or expand for a net present value
of $48 million.
The company could build a medium-size facility: if demand turns out to be low, its net
present value is estimated at $22 million; if demand turns out to be high, the company could
do nothing and realize a net present value of $46 million, or it could expand and realize a net
present value of $50 million.
If the company builds a large facility and demand is low, the net present value will be −$20
million, whereas high demand will result in a net present value of $72 million. (LO2–4)
a. Analyze this problem using a decision tree. What is the best alternative?
b. Calculate the EVPI and interpret it.
c. Perform sensitivity analysis on P(high).
8. A manager must decide how many machines of a certain type to buy. The machines will
be used to manufacture a new gear for which there is increased demand. The manager has
narrowed the decision to two alternatives: buy one machine or buy two machines. If only
one machine is purchased and demand is more than it can handle, a second machine can be
purchased at a later time. However, the cost per machine would be lower if the two machines
were purchased at the same time.
The estimated probability of low demand is .30, and the estimated probability of high
demand is .70.
The net present value associated with the purchase of two machines initially is $75,000 if
demand is low and $130,000 if demand is high.
The net present value for one machine and low demand is $90,000; if demand is high,
there are three options: one option is to do nothing, which would have a net present value of
$90,000; a second option is to subcontract some production, which would have a net present
value of $110,000; the third option is to purchase a second machine, which would have a net
present value of $100,000.
How many machines should the manager purchase initially? Use a decision tree to analyze
this problem. (LO2)
9. Determine the course of action that has the highest expected value for the following decision
tree. (LO2)
17
18
PART THREE System Design
1/3
1/3
1/3
.30
0
60
90
40
Alte
tiv
rna
.50
eA
44
.20
60
1/3
Al
1/3
ter
na
1/3
e
tiv
B
.30
(45)
45
99
40
.50
50
30
.20
1/2
1/
2
40
50
10. The director of social services of a province has learned that a new act has mandated additional
information requirements. This will place additional burden on the agency. The director has
identified three alternatives to handle the increased workload. One alternative is to reassign
present staff members, the second is to hire and train two new workers, and the third is to
redesign current practice so that workers can readily collect the information with little additional effort. An unknown factor is the caseload for the coming year when the new data will
be collected on a trial basis. The estimated costs for various options and caseloads are shown
in the following table: (LO1–3)
Caseload
Moderate
High
Very High
$50*
60
85
New staff
60
60
60
Redesign data collection
40
50
90
Reassign staff
*Cost in $ thousands.
he director has decided that reasonable caseload probabilities are .10 for moderate, .30 for
T
high, and .60 for very high.
a. Which alternative will yield minimum expected cost?
b. Construct a decision tree for this problem and solve it.
c. Determine the expected value of perfect information.
Suppose the director has the option of hiring an additional worker if one worker is hired initially and the caseload turns out to be high or very high. Under that plan, the cost if demand is
moderate will be 40, the cost if demand is high will be 70, and the cost if demand is very high
will also be 70. Construct a decision tree that shows this additional alternative and determine
which alternative will minimize expected cost.
SUPPLEMENT TO CHAPTER 5 Decision Analysis
11. A manager has compiled estimated profits for various capacity alternatives but is reluctant to
assign probabilities to the states of nature. The payoff table is: (LO4)
State of Nature
Alternative
#1
#2
A
$20*
140
B
120
80
C
100
40
*In $ thousands.
a. Plot the expected value lines on a graph.
b. Is there any alternative that would never be appropriate in terms of maximizing expected
profit? Explain on the basis of your graph.
c. For what range of P(#2) would alternative A be the best choice if the goal is to maximize
expected profit?
d. For what range of P(#1) would alternative A be the best choice if the goal is to maximize
expected profit?
12. Repeat all parts of Problem 11, assuming the values in the payoff table are estimated costs
and the goal is to minimize expected cost. (LO4)
13. Consider the following payoff table of estimated NPV for four alternatives: (LO4)
State of Nature 1
State of Nature 2
$10*
−2
#1
Alternative
#2
8
3
#3
5
5
#4
0
7
*In $ thousands.
Relative to the probability of state of nature 2, determine the range of probability for which
each of the alternatives would maximize the expected NPV.
14. Given the following payoff table: (LO4)
State of Nature
#1
A
Alternative
$120*
#2
20
B
60
40
C
10
110
D
90
90
*In $ thousands.
a. Determine the range of P(#2) for which each alternative would be best, treating the payoffs
as profits.
b. Answer part a, treating the payoffs as costs.
15. A law firm is representing a company that is being sued by a customer. The decision is whether
to go through litigation or settle the case out of court. The cost of losing the case is estimated
to be $1 million, whereas the cost of settling the case is $200,000. Winning the case in court
would result in no loss. What is the minimum chance of winning for which the company
should contest the case? (LO1 & 4)
16. An 18-year-old youth has just arrived at a hospital complaining of abdominal pain. The medical
findings are consistent with appendicitis but not totally typical of appendicitis. The lab and
X-ray test results are not clear. The surgeon is wondering whether to operate or wait 12 hours
19
20
PART THREE System Design
(in case it is not appendicitis). In this case, if the pain does not recede, then it must be appendicitis and the surgeon will operate. The probability that it is appendicitis is 56 percent using
past experience with similar cases. Also, from similar cases, the probability that the appendicitis
will perforate during 12 hours of wait is 6 percent. The payoffs are measured in terms of death
rate. The death rate of operating when appendicitis is present is 0.09 percent; it is 0.04 percent
when appendicitis is not present. The death rate of operating a perforated appendicitis is 0.64
percent. Draw the decision tree and determine the best course of action in this case.1 (LO2)
17. An oil company has some land that may contain oil.2 The oil reserve and its probability (based
on experience in the area) are (LO1):
Oil reserve (in 1000 barrels)
500
200
50
0
Probability
.1
.15
.25
.50
The company has to decide whether to drill for oil or lease the land. Cost of drilling will
be $500,000 if it is a dry well or $750,000 if it is oil producing. The profit per barrel will be
$45. The revenue for leasing the land will be $3,000,000. Determine the best decision for
the oil company.
18. The board of governors of Santa Clara University is contemplating mandatory testing of their
student athletes.3 The test is not 100 percent accurate. The conditional probabilities are as
follows: If an athlete uses drugs, then the test will be positive 94 percent of the time, whereas
if he does not (i.e., he is a non-user), the test will be negative 98 percent of the time. Suppose
the board suspects that 4 percent of the athletes use drugs. (LO2 & 3)
a. Calculate the posterior probabilities.
b. If a test costs $50, the cost of not identifying (and barring) a drug user is $1,000, the cost
of falsely accusing a non-user is $100, and other costs are zero, should the university test
any athlete?
19. An electric utility is trying to decide whether to replace the PCB transformer in a generating
station with a new and safer one.4 To evaluate this decision, the utility needs information about
the likelihood of an incident, such as a fire, the cost of such an incident, and the cost of replacing the transformer. Suppose that the total cost of replacement is $75,000. If the transformer is
replaced, there is virtually no chance of a fire. However, if the current transformer is retained,
the probability of a fire is assessed to be 0.0025. If a fire occurs, then the clean-up cost could
be high ($80 million) or low ($5 million). The probability of a high clean-up cost, given that
a fire occurs, is assessed to be 0.2. (LO2 & 4)
a. Should the company replace the transformer?
b. Perform sensitivity analysis on the probability of a fire. Does the optimal decision from
part (a) remain optimal for a wide range of values for this probability?
20. A 59–year-old executive has just completed his annual physical exam.5 (The exam included
a test on a sample of his stool. The purpose of this test is to screen for cancer of the large
intestine (colon cancer). The test looks for small amounts of blood in the stool. The executive
is informed that the test showed presence of blood in his stool. The executive is quite anxious.
The likelihood of any 59–year-old man having colon cancer is 1 in 1,000. The test detects 85
percent of cases of colon cancer, but 2 percent of patients without cancer have a small amount
of blood in their stool. What is the probability that the executive has colon cancer? (LO3)
21. An offshore underwater structure has tested positive for oil by two exploratory wells drilled
near the crest of the structure.6 Management is now trying to determine whether it is wise
1
. Clarke, “The Application of Decision Analysis to Clinical Medicine,” Interfaces 17(2), March–April 1987,
R
pp. 27–34.
2
F. S. Hillier and G. J. Lieberman, Operations Research, 2nd ed, 1974, San Francisco: Holden-Day.
3
C. D. Feinstein, “Deciding Whether to Test Student Athletes for Drug Use,” Interfaces 20(3), 1990, pp. 80–87.
4
W. E. Balson et al, “Using Decision Analysis and Risk Analysis to Manage Utility Environmental Risk,”
Interfaces 22(6), 1992, pp. 126–139.
5
P. H. Hill et al, Making Decisions, Massachusetts: Addison-Wesley, 1980, p. 170.
6
P. D. Newendorp, Decision Analysis for Petroleum Exploration, Tulsa, Oklahoma: PennWell Publishing, 1975,
p. 524.
SUPPLEMENT TO CHAPTER 5 Decision Analysis
21
to make a definite commitment to begin building a platform, and, if so, what size, or to
defer the decision to allow for the drilling of an additional exploratory well. The principal
uncertainty is the size of the field. If the structure is nearly filled with oil, the area of the
field will be about 15 square miles and it would require installation of a large platform
costing $50 million. If the structure is only partially filled with oil, the area will be much
smaller and the field could be depleted with fewer wells and a smaller platform costing
$30 million. Based on the subjective judgment of company’s geologists, the likelihood of
the structure being completely filled is 0.4 (i.e., field is large). If a small platform is built
and the field turns out to be large, a second small platform will have to be built at the
same cost. If another exploratory well is drilled along the flanks of the structure, there is
a 10 percent chance that the result will be erroneous. However, management will base its
platform size decision on the result of the exploratory well. The exploratory well will cost
$2 million. (LO2 & 3)
a. Calculate the posterior probabilities.
b. Which decision has the lowest expected cost?
22. The management of a printed circuit boards (PCB) manufacturer is concerned about the significant number of bad (defective) PCBs that are sent to the company’s customers.7 The quality
control procedure for the PCBs is as follows: each PCB is sent to a testing station where it
is tested using an electronic instrument; those that pass are sent to the customers; those that
fail are retested; those that pass the second test are sent to the customers; those that fail the
second time are retested again; those that pass the third test are sent to customers, and those
that fail the third time are discarded. A consultant is brought in to examine the situation. She
studied 1,136 PCBs. Of these, 1,087 passed the quality test at the first attempt. The 49 that
failed were retested, and 21 passed. The remaining 28 PCBs all failed the third time and were
discarded. The consultant asked for retesting of those that passed the first time. A sample of
200 was retested and 7 failed. (LO3)
a. Calculate the probability of a good (non-defective) PCB passing the test the first time and the
probability of a bad (defective) PCB failing the test the first time. Also, calculate the proportion of PCBs that are good (non-defective) and the proportion that are bad (defective).
b. Propose a testing procedure that improves the quality of PCBs sent to the customers.
Mini-Case
Southern Company
F
leet managers have a large pool of cars and trucks to maintain.8 One approach to the vehicle maintenance is to use
oil analysis, where the oil from the engine and transmission
are subjected periodically to a test. These tests can sometimes
signal an impending failure (for example, too much iron in
the oil), and preventive maintenance is then performed (at
a relatively low cost), eliminating the risk of failure (failure
would result in a relatively high cost). However, oil analysis
costs money, and it is not perfect—it can indicate that a unit
is defective when in fact it is not, and it can indicate that a
unit is non-defective when in fact it is. As a possible substitute for oil analysis, the company could simply change the oil
periodically, thereby reducing the probability of failure. The
fleet manager for the Southern Company, an electrical utility
7
based in Atlanta, has four alternatives: (1) do nothing, (2) use
oil analysis only, (3) replace oil only, or (4) replace oil and do
oil analysis. For option (1) the probability of failure is 0.1, and
the cost of failure is $1,200 (includes cost of an oil change).
For option (2), the probability of failure remains at 0.1. If the
unit is about to fail, the oil analysis will indicate this with probability 0.7; if the unit is not about to fail, the oil analysis will
indicate this with probability 0.8. The oil analysis itself costs
$20, and if it indicates that failure is about to occur, the oil will
be changed at the cost of $14.80 and preventive maintenance
will be performed. The cost of preventative maintenance to
restore a unit that is about to fail is $500, whereas the cost of
maintenance for a unit that is not about to fail is $250. The
only difference between options (3) and (4) is that probability
of failure decreases from 0.1 to 0.04 if oil is changed periodically. Analyze this decision problem.
. C. Bell, Management Science/Operations Research: A Strategic Perspective, Cincinnati: South-Western
P
College Publishing, 1999, p. 90.
8
J. M. Mellichamp, “The Southern Company Uses a Probability Model for Cost Justification of Oil Sample
Analysis,” Interfaces 23(3), 1993, pp. 118–124.
PART THREE System Design
22
Mini-Case
ICI Canada
Significant market and technical feasibility (P1)
Board sanctions plant (P2)
0.8 ± 0.2
CI’s Canadian subsidiary (now part of AkzoNobel) discovered
a new, but unpatentable, application for a chemical agent
to reduce pulp-mill water pollution.9 However, everything was
quite uncertain, and the management was trying to decide
whether to go ahead with its R&D or abandon the product. The
following questions indicate the primary risks:
Commercial success (P3)
0.8 ± 0.2
I
• Would market tests confirm that there is a significant market for the product and could the company develop a new
process for making this product—that is, is it technically
feasible?
• After a production process is developed, would the company’s
board sanction production on a commercial scale?
• Would the venture turn out to be commercially successful?
The management team assumed that each of these questions
had a yes or no answer. The probabilities of yes answers are
shown below. The plus-or-minus value indicates management’s
uncertainty about the true probabilities.
9
0.36 ± 0.09
The primary economic factors and their expense/gain (in
million dollars) were the following:
• The marketing development cost to determine whether
there is a significant market and research expenses to
identify a new production process for the product (C1):
$1 ± 25%
• The process development costs, including pre-sanction engineering & commercial development (C2): $3.5 ± 25%
• The commercial development costs after the board’s sanction
(C3): $1.0 ± 25%
• The venture value (net present value) if successful (R):
$25 ± 50%
Again, the plus-or-minus values indicate management’s
c­ onsiderable uncertainty about the values. Should management
go-ahead with R&D for this product?
. W. Hess, “Swinging on the Branch of a Tree: Project Selection Applications,” Interfaces 23(6), 1993,
S
pp. 5–12).