CHAPTER 9 DECISION ANALYSIS
9.1 Introduction to Decision Analysis
Elements:
(1) Decision maker (“we”),
(2) Choices (decisions, strategies),
(3) Payoffs
Selection problems. Define a binary variable yj for each decision with yj = 1, if we make decision j, and 0 otherwise. A selection problem with n possible decisions will then feature the constraint y1 + y2 + …+ yn = 1.
Multiple criteria (MCDM = multiple criteria decision making vs multiple states of nature – decision analysis = games against nature.
Example: Consider a problem with three decisions d1, d2, and d3, and four states of nature s1, s2, s3, and s4. The payoffs are shown below.
s1 / s2 / s3 / s4d1 / 3 / 2 / 4 / 6
d2 / 2 / 0 / 4 / 1
d3 / 5 / 2 / 0 / 3
This game is
(1) asymmetric: the decision maker is rational (looks at the payoffs), while nature is a random player
(2) a simultaneous game (we do not know in advance what state of nature will be chosen).
Consider the continuum between
certainty ― … ― risk ― …― uncertainty
Certainty: we know exactly which strategy nature will play.
Risk: we know the probability distribution nature uses to play her strategies (e.g., by way of past observations)
Uncertainty: we do not know even the probability distribution of nature’s strategies (e.g., new untried product).
Certainty is trivial: Since we know what column nature plays, all we have to do is choose the row that leads to the highest payoff.
9.2 Visualizations of Decision Problems
Macro view: Influence diagrams.
Micro view: Decision trees.
Influence diagrams: Decision nodes, random nodes, consequence nodes. Shows general interrelations between decisions, chance events, & results/outcomes.
Example:
Decision / Random event / ConsequenceAdd electronics department / General economic conditions / Profit
Relocate department into a separate building / Local acceptance of services
The broken arcs: possible influences (local acceptance of an electronics department or store may be influenced by the existence of an electronics department in our department store and our competitors’ reaction to our introduction of the department).
Decision tree for the same problem:
9.3 Decision Rules Under Uncertainty and Risk
Start with uncertainty. (Little input on our part, only crude information will come out).
Example:
s1 / s2 / s3d1 / 2 / 2 / 5
d2 / 0 / 1 / 7
d3 / 2 / 1 / 1
d4 / 2 / 3 / 4
Before commencing, check for dominances. One decision (row) dominates another, if its payoffs are all greater or equal than those of a single other row.
Here, d1 dominates d4. Dominated decisions can be deleted. Column dominances do not exist.
Checking for dominances requires a total of ½m(m1) pairs of comparisons. General procedure for all decision rules: Determine anticipated outcomes for all decisions.
Decision rules under uncertainty:
(1) Wald’s rule (pessimist’s rule, maximin rule): for each decision, the anticipated outcome is the worst-case outcome. Among the anticipated outcomes, choose the best. The decision that leads to this outcome is chosen.
In our example, the anticipated outcomes are 2, 1, 1, & 3 (in case we did not eliminate decision d4), the maximum is 1, which belongs to d3. This is the chosen decision.
(2) Optimist’s rule. For each decision, the anticipated outcome is the state of nature with the highest payoff. Among those, we choose the decision with the highest anticipated payoff → a maximax rule.
In this example, the anticipated payoffs are 5, 7, 2, & 4. Choose d2 that―in our opinion―leads to a payoff of 7.
(3) Savage – Niehans rule (minimaxregret criterion).
First generate regrets, by comparing each payoff with the highest payoff given that state of nature (i.e., in that column). Then apply Wald’s rule to the regret matrix.
In this example, the regret matrix is
R =
Applying Wald’s rule means choosing the worst possible outcomes for each decision (i.e., the highest regrets) & then choose the best (i.e., the lowest) among them. This makes Wald’s rule applied to this regret (= cost) matrix a minimax rule.
Decision making under risk.
Given probabilities for all states of nature, e.g., p = [.5, .3, .2].
(4) Bayes’s rule: Choose the maximum of the expected values (a weighted maxisum rule).
In the example, the expected payoffs (or expected monetary valuesEMV):
EMV(d1) = 2(.5) 2(.3) + 5(.2) = 1.4,
EMV(d2) = 0(.5) 1(.3) + 7(.2) = 1.1,
EMV(d3) = 2(.5) + 1(.3) + 1(.2) = 1.5,
EMV(d4) = 2(.5) 3(.3) + 4(.2) = .9.
The anticipated payoffs are 1.4, 1.1, 1.5, and 0.9 → choose d3.
Newspaper Boy Problem. Decide how man papers to purchase, given that the demand is uncertain.
Example:The demand is 10, 20, 30, or 40 newspaper on any given day. Purchase a newspaper for 20¢ and sell for 90¢. The salvage value of an unsold newspaper is 5¢, while the opportunity cost for newspapers has been estimated to be 15¢ for each newspaper that could have been sold but was not due to the lack of supply. Suppose that the purchasing strategies are 10, 20, 30, & 40.
Payoff matrix
s1 s2 s3 s4
A =
Given probabilities of p = [.6, .2, .1, .1], the decisions have expected payoffs of $5.95, $8.45, $8.95, and $8.45,
→buy 30 newspapers expect a daily payoff of $8.95.
(5) Using target valuesT. Idea: Plot payoff values against the probability that the value can be achieved.
The original payoff matrix was
s1 / s2 / s3d1 / 2 / 2 / 5
d2 / 0 / 1 / 7
d3 / 2 / 1 / 1
d4 / 2 / 3 / 4
p = [.5, .3, .2].
d1: solid line
d2: broken line
d3: dotted line
Decision rules (by way of upper envelope):
If T2, any decision will achieve the target.
If T [2, 1], d2 and d3 are best. Both will reach
the target with a probability of 1.
If T [1, 1], d3 is best. It reaches the target with
a probability of 1.
If T [1, 2], d1 is best. It reaches the target with
a probability of .7.
If T [2, 5], d1 and d2 are best. Both achieve the
target with a probability of 0.2.
If T [5, 7], d2 is best. It achieves the target with
a probability of .2.
If T > 7, none of the decisions will be able to reach
the target.
9.4 Sensitivity Analyses
“What if” individual payoffs aij change?
s1 / s2 / s3d1 / 2 / 2 / 5
d2 / 0 / 1 / 7
d3 / 2 / 1 / 1
d4 / 2 / 3 / 4
p / .5 / .3 / .2
Suppose that we are uncertain about a23. Rewrite the payoff as a23 = 7 + with an unknown [2, 3], meaning that we expect the payoff to be between 5 & 10.
Expected payoffs:
EMV = .
Clearly, d1d4are dominated. The payoffs for the remaining strategies are shown in the figure below.
This leads to
If > 2 (i.e., a23 > 9), then decision d2 is best,
if 2 (i.e., a23 9), decision d3 is best.
Another source of uncertainty relates to the magnitude of the probabilities.
Suppose that we are unsure about p1. Similar to the above, we can use p1 + with some unknown . However: The sum of probabilities must equal 1, so if p1 increases, the other probabilities must decrease by . Assume that the other two probabilities decrease by the same amounts, i.e.
p = [.5 + , .3 ½, .2 ½].
Given the same payoff matrix
s1 / s2 / s3d1 / 2 / 2 / 5
d2 / 0 / 1 / 7
d3 / 2 / 1 / 1
d4 / 2 / 3 / 4
p / .5 / .3 / .2
we can compute the expected values as
EMV() = .
Suppose we estimate that p1will be between .3.6. (Alternatively, starting with p1 =.5, the change [.2, +.1].
Within this range, decision d3 dominates d1d4, which can be deleted.
The expected monetary values for d2d3 areshown in the figure below:
Decision rule:
If .1 (i.e., p1 .4), then decision d2 is best,
if .1 (i.e., p1 > .4), then decision d3 is best.
Different example: Same payoff matrix, but as p1, p2⅔p3⅓. The expected payoffs are then
.
Decision rule:
If .15 (i.e., p1 < .35), then decision d2 is optimal,
if .15 (i.e., p1 .35), then decision d3 is optimal.
9.5 Decision Trees and the Value of Information
Value of information. Extreme case first:
Expected value of perfect information (EVPI): Difference between the expected payoff with perfect information minus the expected payoff without information (beyond the prior probabilities p).
Previous example:
s1 / s2 / s3d1 / 2* / 2 / 5
d2 / 0 / 1 / 7*
d3 / 2* / 1* / 1
d4 / 2* / 3 / 4
p / .5 / .3 / .2
p = [.5, .3, .2].
The best responses to nature’s strategies (in this sequential game, we assume that while we cannot influence nature’s choice, we know about it before we make our move) are shown with asterisks. The expected payoff with perfect informationEPPI is then
EPPI = 2(.5) + 1(.3) + 7(.2) = 2.7.
As the expected payoff without perfect information (i.e., the expected monetary value of the best strategy EMV*) was 1.5 (achieved by using d3), the expected value of perfect information is then
EVPI = EPPIEMV* = 2.7 1.5 = 1.2.
Now imperfect information.
Image a forecasting institute that uses indicators I1, I2, … to forecast the states of nature s1, s2, … . Clearly, the indicators & the states of nature should be related. (typical examples are demand & wholesaler’s orders, the changes of the supply of money & the rate of inflation, etc.).
The relations between indicators & states of nature may be given in the form of conditional probabilities.
s1 / s2 / s3P(I|s): / I1 / .6 / .9 / .2
I2 / .4 / .1 / .8
Notice: The state s1 is linked only very weakly to the indicators.
Decision tree:
From left to right, the tree depicts the sequence of events. Decision modes (we decide) are squares, event nodes (nature makes a random choice) are circles, & terminal nodes (there are no further muves) as triangles.
Note: the lower part of the tree equals what we have already done in the matrix when we determined the EMV* strategy.
Numbers that are needed:
(1) Payoffs (at the terminal nodes)
(2) Indicator probabilitiesP(I) at nature’s choice of indicator
(3) Posterior probabilities P(s|I) at nature’s choice of the state of nature.
While the payoffs are known, the probabilities in (2) & (3) are not.
However, it is possible to use the known probabilities P(I|s) to compute the indicator probabilities P(I) & the posterior probabilities P(s|I). For that purpose, we use
Bayes’s theorem: P(si|I1) = P(I1|si)P(si)/P(I1) & similar for all other indicators.
Here:
For I1, we compute P(I1) and P(s|I1)
s / P(s) / P(I1|s) / P(I1|s)P(s) / P(s|I1)s1 / .5 / .6 / .30 / .4918
s2 / .3 / .9 / .27 / .4426
s3 / .2 / .2 / .04 / .0656
P(I1) = .61
For I2, we compute P(I2) and P(s|I2)
s / P(s) / P(I2|s) / P(I2|s)P(s) / P(s|I2)s1 / .5 / .4 / .20 / .5128
s2 / .3 / .1 / .03 / .0769
s3 / .2 / .8 / .16 / .4103
P(I2) =.39
The complete decision tree is then as follows:
The numbers next to the nodes are computed by backward recursion. The recursion starts at the triangular terminal nodes & works backwards to the root of the tree.
Two rules apply:
(1) Back into a decision node: choose the best strategy, i.e., the one with the highest (expected) payoff.
(2) Back into an event node: compute the expected value of all successor nodes.
Here, the result is the payoff with imperfect information EPII = 2.05.
Now compare: without information (beyond prior probabilities), we can get EMV* = 1.5. With additional information, we can get EPII = 2.05.
Hence the expected value of perfect information
EVSI = 2.05 1.5 = 0.55.
A standardized measure is efficiency E, which is
E = .55/1.2 = .4583.
The same example with different numbers:
s1 / s2 / s3P(I|s): / I1 / .9 / .6 / .2
I2 / .1 / .4 / .8
Note: This matrix is the same as before with some columns exchanges. However, now s1 is strongly linked to the indicators, & P(s1) = .5, so the result should be (at least somewhat) better.
Computation of P(I1) and P(s|I1):
s / P(s) / P(I1|s) / P(I1|s)P(s) / P(s|I1)s1 / .5 / .9 / .45 / .6716
s2 / .3 / .6 / .18 / .2687
s3 / .2 / .2 / .04 / .0597
P(I1) = .67
Computation of P(I2) and P(s|I2):
s / P(s) / P(I2|s) / P(I2|s)P(s) / P(s|I2)s1 / .5 / .1 / .05 / .1515
s2 / .3 / .4 / .12 / .3636
s3 / .2 / .8 / .16 / .4848
P(I2) =.33
We then obtain EPII = 2.12, so that
EVSI = 2.12 1.5 = .62 E= .62/1.2 = .5167.
Extreme examples:Random indicators result in the prior probabilities equaling the posterior probabilities & EVSI = 0.
On the other hand, while a forecast that is always correct will have EVSI = EVPIE = 1, a forecast that is always wrong has the same features → it is the consistency of the indicators that is important, not their actual meaning.
9.6. Utility Theory
Expected values are meaningful, only if decision makers are risk neutral. This means, they should be indifferent to either
(1) receiving $100,000 in cash, no questions asked, or
(2) playing the lottery with a 50% of winning $200,000 & a 50% chance of winning nothing.
Most people would prefer (1) i.e., they are not risk neutral.
Once the utilities have been determined, they can be used instead of payoffs. All procedures remain unchanged.