ISE 195
Introduction to Industrial
Engineering
Lecture 4
Decision Analysis
Decision Analysis
What is the hardest decision you have ever had to
make?
Since we all have to make decisions, we are all
Decision Makers of a sort and can benefit from
the study of decision making.
Have you ever had to make a decision and then
later have to explain or defend that decision?
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Decision Domains
Personal domain
 Where to live; college to attend; car to buy; etc
Business domain
 Introduce the new product; bid on a contract; hire
Government domain
 How to allocate money; where to get involved
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Decision Roles
Those who study decisions will be referred to as decision
analysts while those that make the decisions will be
referred to as the decision makers.
Why do you think we would want to separate the roles of
the decision analyst and the decision maker?
Proper decision making requires collaboration among the
decision makers and the decision analysts in order to
find the best solution based on insights versus position
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Why Decisions Are Hard
Decisions are hard for a number of structural,
emotional, and organizational reasons
 Structural – uncertainty, trade-offs, complexity
 Emotional – anxiety, multiple objectives, competition
 Organizational – lack of consensus, differing
perspectives
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Why Decisions Are Hard
Do you think your personal decisions are going to be
easier or harder than the decisions you might be faced
with in business (engineering)?
What might be some of the reasons, both obvious and
less obvious, for this difference in level of complexity
between decisions from the personal domain and
decisions from the business or government domain?
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Why Decisions Are Hard
There are other reasons decisions are hard
 Consequences
 Uncertainty
 Ambiguity
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Why Decisions Are Hard
Consequences
MEDIUM
HIGH
LOW
Uncertainty
Ambiguity
CAU Model, Skinner
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Why Decisions Are Hard
Consequences
Uncertainty
Ambiguity
CAU Model, Skinner
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What Makes A Good Decision
What is a good decision?
What is a good outcome?
Does a good decision always lead to a good
outcome?
 Name some examples. . .
A good decision emerges as the result of valid
decision making process (of which there are a few as
we will see)
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“When you come to a fork
in the road, take it”
- Yogi Berra
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History
Operational research, quantitative management, based
on repetitive actions
 Focused on optimizing objectives and meeting constraints
Failed to focus on needs of executive decision making
 In particular their more complex, strategic problems
Technique needed for logical guidance on complex,
uncertain situations
DA combines systems analysis and statistical
decision theory
13
History
Problems typical of DA application are:
 Unique
 Important
 Contain uncertainty
 Have long-run implications
 Contain complex preferences
DA arose in the late 60s, early 70s and balances the
following OR considerations:
 Mathematical modeling
 Computer implementation
 Quantitative analysis and decision making
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History
DA also incorporated the following aspects of
human decision making
 Management experience
 Management judgment
 Management preferences
The art of DA involves “capturing” the above
from the managers and decision makers
 The techniques used to capture the above are
sometimes controversial within the operational
research / systems engineering field
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Terminology
Decision
 A conscious irrevocable allocation of resources with the purpose of
achieving a desired objective
Uncertainty
 Something that is unknown or not perfectly known
Outcomes
 Depend on alternative chosen and the uncertainties impacting it
Value
 Something the decision maker wants and can tradeoff
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Terminology
Objective
 Something specific the decision maker wants to achieve
Decision Maker
 Anyone with the authority to allocate the necessary resources for
the decision being made
Subjective Probability
 Classical approach to probability called the “frequentist” approach
 Subjective approach, the Bayesian, allows that each of us can
provide valid probabilities
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Probabilistic Methods
(Pay attention is ISE 301!)
•These assume the possible outcomes (states of
nature) can be assigned probabilities that represent
their likelihood of occurrence.
-Also referred to as methods for decision making
under “risk”
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Expected Monetary Value

Selects alternative with the largest expected monetary
value (EMV)
EMVi   rij p j
j
rij  payoff for alternative i under the j th state of nature
p j  the probability of the jth state of nature

EMVi is the average payoff we would receive if we
faced the same decision problem numerous times and
always selected alternative i.
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Decision Trees
Graphical means for displaying a decision problem that
shows, in chronological order:
 the alternatives available to the decision maker;
 the futures that could be experienced; and
 the consequences of choosing between alternatives
Trees consist of:
 Branches — lines representing possible “decision paths”
 Decision Forks — “nodes” which represent choices to be made
by the decision maker; and
 Chance Forks — nodes which represent possible futures that
are modeled as selected by nature
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Decision Trees (continued)
To “evaluate” a tree one must:
 assign values of an appropriate evaluation measure
to each branch (often summarized at the end of the
branch); and
 choose branches appropriately at each decision
node, working from right to left
 When making decisions under risk, this entails:
– assigning probabilities to each branch emanating from a chance fork;
– computing expected values at each chance node; and
– finding the branch that maximizes the expected value from among all
branches emanating from a decision fork.
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Example – Electronics Firm
An electronics firm makes components that are sold and
shipped to an automobile manufacturer.
 Five percent of all components produced are defective due to
poor solder connections.
 Can’t tell if defective until after it is installed on a car.
– Auto maker will charge the electronics firm $800 per defective component to cover
the cost of repair.
A proposal: double-solder each component before before
it is shipped to the automobile maker.
 Will cost $50 per component to double-solder but is sure to
eliminate this cause of defective components
– i.e., no double-soldered components will be defective.
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Example – Electronics Firm (Continued)
Is the proposal worthwhile?
 Assume electronics firm seeks to minimize its
expected cost and consider using our structure:
 Actions:
 Outcomes:
1 -- double solder before shipping
2 -- do not double solder
1 -- component is defective
2 -- component is good
 Prior Probabilities:
P1 = 0.05; P2 = 0.95
– Note that these probabilities apply only if the component is not
double-soldered!
 Value Function:
E11 = -50; E12 = -50;
E21 = -800; E22 = 0
“Values”
are
“negative”
costs here
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Electronics Firm – Decision Tree
 Expected Values:
Double-Solder:
E(V)1 = -50
Do Not: E(V)2 = -800(0.05) + 0(0.95) = -40
--
» No, it would not be worthwhile to double-solder every component since the
maximum expected value (minimum expected cost) is obtained for action 2
(do not double-solder).
 Decision Tree:
-40
Double Solder
X
-40 Defective (0.05)
Do Not
Good (0.95)
-50
-800
0
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Example – Doing Better?
To this point, we have assumed that the firm is unable to
tell if a component is defective until after it is installed on a
car.
 Obviously, if the firm were to know in advance that a component was
defective, it would double-solder that component.
 A reasonable strategy, then, might be to attempt to determine
whether or not a component is defective before the decision to
double-solder or not is made.
 How much should the firm be willing to pay to for this sort of
information?
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Example – Paying for More Information
 Without any advance info about components, the
firm’s best strategy is to not double-solder any
components
– This has an expected cost of $40 per component.
 With advance info, however, the firm should:
 double-solder all defective components at a cost of $50 each,
and
 not double-solder the rest (the good components).
 Since 5% of all components are defective, the
expected cost of this strategy would be:
– $50(0.05) + 0(0.95) = $2.50 per component.
 Thus, the most the firm should be willing to pay for
this advance info is the difference between these, or
$40 - 2.50 = $37.50 per component.
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Getting Advance Information
“Advance information” can often be obtained by
performing some sort of test before making a decision
 If so, then the initial choice we must make is whether or not to do
the testing
The ideal situation would be one in which the testing
enables us to correctly predict the future
 In our example, this would mean that the test is 100%
accurate in classifying components as good or defective
– e.g., if the test classifies a component as “defective,” then that
component is indeed defective.
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Decision Tree w/ Perfect Testing
Do Not Test
Double Solder
-40 X
-40 Defective (0.05)
No Action
Good (0.95)
Perform
Test
(Get
Advance
Info)
Classify as
Double Solder
-50
Defective (0.05)
-800 Defective (1.00)
X No Action
-2.5
Good (0.00)
Classify as
Good (0.95)
0 X
Double Solder
0 Defective (0.00)
No Action
Good (1.00)
-50
-800
0
-50
-800
0
-50
-800
0
We assume here that “testing” is “perfect,” so that all components will be
correctly classified and, thus, 95% will be classified as “good” while 5% will
be classified as “defective”
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Bayesian Decision Making
A method for accounting for the effects of advance testing
in decision making
Based on Bayes’ Theorem which provides us a way to
“revise” our initial “prior” probabilities for the occurrence of
each possible future given the results of testing
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Example – Revisited
Suppose now that the firm can choose to test each
component, at a cost of $20 apiece, to see if the
component might be defective before the decision to
double-solder or not is made.
 The test is not perfect, but they have a track record
 Based on the results of testing known good and known defective
components, it is determined that:
– the test will incorrectly classify 15% of all defective components as good, and
– incorrectly classify 10% of all good components as defective.
 Is it worthwhile for the firm to perform this test?
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Decision Tree w/ Testing
Do Not Test
Double Solder
-40 X
-40 Defective (0.05)
No Action
Good (0.95)
-20
Perform
Test
Classify as
Defective (???)
Double Solder
Defective (???)
No Action
Classify as
Good (???)
Good (???)
Double Solder
Defective (???)
No Action
Good (???)
To evaluate this tree and decide what to do, we need to fill in
appropriate probabilities at all chance forks.
-50
-800
0
-50
-800
0
-50
-800
0
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Example – Description of Test
 Note that there are two possible results when a component is
subjected to the proposed test:
Result 1: The component is classified as defective
Result 2: The component is classified as good
 The particular result to be obtained will depend on both:
 the state of nature (the condition of the component being tested),
and,
 since the experiment is not perfect, also on chance.
 What we know about the accuracy of the test is captured by the
conditional probabilities of obtaining a particular result given a
particular state of nature . . .
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Description of Test – Continued
 That is, we “know” that our experiment will incorrectly classify
15% of all defective components as good and 10% of all good
components as defective.
 We denote this using the notation:
» P[component classified as defective | it is defective]
=
P[Result 1 | State of nature 1]  Q1|1 = 0.85
» P[component classified as good | it is defective]
=
P[Result 2 | State of nature 1]  Q2|1 = 0.15
» P[component classified as defective | it is good]
=
P[Result 1 | State of nature 2]  Q1|2 = 0.10
» P[component classified as good | it is good]
= P[Result 2 | State of nature 2]  Q2|2 = 0.90
– Unfortunately, these are not the probabilities we need to complete the decision
tree!
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Bayes’ Theorem
To compute the conditional probabilities of encountering each possible
future given the results of the test, we combine previous results to obtain:
 Bayes’ Theorem: If
are n mutually exclusive and exhaustive
events defined over a sample space and E is any other event with P[E] > 0, then
F1 , F2 ,
Bayes’
Rule
The Law of

“Total
Probability”


Pr Fj E 
, Fn
Pr E  Fj 
Pr E



Pr E F j Pr Fj 
Pr E
Bottom Line: the posterior probability of encountering state of nature j (j = 1, 2, . . ., m)
given
testEproduces
(k = E
1, 2,
., r) F
can 
be found
Pr that
E the
 Pr
F Pr result
F k Pr
F . .Pr
 Prvia:E F Pr F
 

1
  
Pj k 
1

2
  
2

n
  
n
Qk j Pj
Qk
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Example – Posterior Probabilities
 For the electronics firm, we can then compute:
P1|1
= P[component is defective | classified as defective]
= P[State of nature 1|Result 1]
= P[Result 1|State of nature 1]  P[State of nature 1]
 P[Result 1]
= Q1|1P1/Q1
= (0.85)(0.05)/(0.1375)  0.3091
 Likewise
P2|1 = P[component is good |classified as defective]
= P[State of nature 2|Result 1]
= Q1|2P2/Q1
= (0.10)(0.95)/(0.1375)  0.6909
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Example – Posterior Probabilities (Continued)
 Similarly
P1|2 = P[component is defective |classified as good]
= Q2|1P1/Q2
= (0.15)(0.05)/(0.8675)  0.0087
and
P2|2 = P[component is good |classified as good]
= Q2|2P2/Q2
= (0.90)(0.95)/(0.8675)  0.9913
 Terminology: the quantity Qk|jPj formed in the preceding computations is often
called the joint probability of encountering state of nature j and obtaining result
k from the experiment (since it is the probability of both the two events occurring).
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Example – Decision Tree (Revisited)
Do Not Test
X
Double Solder
-40 X
-50
-40 Defective (0.05)
No Action
Good (0.95)
-800
0
-50 Double Solder
-32.88
Classify as
-50
Defective
(0.1375)
-20
-247.47 Defective (0.3091)
Perform
-800
No
Action
-12.88
Test
X
Good (0.6909)
0
-6.96 Double Solder
Classify as
X
-50
Good (0.8625)
-6.96 Defective (0.0087)
-800
No Action
Good
(0.9913)
0
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Discrete Probability Assessment
Three methods for assessing discrete probability
 Direct question
 Assumes familiarity with probability
 Usually means decision maker used to providing probabilities;
based on similar experiences
 Betting method
 Most people bet in some fashion
 “odds” provide perception of likelihood of the outcome
 Use the odds to derive the probability
 Reference lottery
 Find probability yielding indifference point
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Experts and Assessments
Reliance on experts important in complex problems
Important to avoid bias in assessment and collection
Protocol for expert assessment







Background
Identify and recruit experts
Motivate the experts
Structure and decompose the problem
Probability assessment training
Probability elicitation and verification
Aggregation of distributions
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Theoretical Probability Models
Subjective probabilities may be difficult to get
Alternative is to use some theoretical distribution
 Actually making a subjective assessment via your choice
A variety of distributions apply in a variety of
applications




Binomial
Normal
Exponential
Triangular
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Other Decision Factors: Risk
Have not worried about “risk”
 Decision makers may actually have differing attitudes toward risk
 Would like a model to map outcomes into measures that
incorporate attitudes towards risk
 In decision analysis this is accomplished using utility functions
The corresponding outcomes, not measured in utilities,
may provide different alternative selections than those
not using utilities
 Changes due to incorporation of risk
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Other Decision Factors:
Multi-attribute Decisions
Have focused on a single attribute
Most decisions are multi-attribute in nature
 Trade-off between weight and redundancy
 Trade-off between reliability and maintenance
Some multi-attribute models assume independence




Assess each attribute
Develop a weighting scheme for each attribute
Use weighted sum of scores
Called an “additive model”
42
Multiple Attributes
Reality in most multi-attribute models requires
some form of interaction
 Attributes are not independent
 Need to derive a utility surface
 Techniques for determining the surface are extensions
of independent techniques
 Complications come during elicitation as the expert is
asked to specifically consider dependencies
43
ISE 195: Overview of Decision Analysis
Questions?