1. Steps to Good Decisions
Define problem and influencing factors
Establish decision criteria
Select decision-making tool (model)
Identify and evaluate alternatives using decision-making tool (model)
Select best alternative
Implement decision
Evaluate the outcome
A major point to make here is that this process should be considered iterative. Once we have actually built and exercised the model, we may realize we did not understand the problem in the first place!

2. The Decision-Making Process
Problem
Decision
Quantitative Analysis
Logic
Historical Data
Marketing Research
Scientific Analysis
Modeling
Qualitative Analysis
Emotions
Intuition
Personal Experience
and Motivation
Rumors

3. Ways of Displaying a Decision Problem
Alternatives
States of Nature
Out- comes
Decision trees
Decision tables
It can be useful here to explain what decision trees and decision tables are, and then ask students to suggest some problems that lend themselves to the use of one or the other of these models.

4. Decision Theory
Is a general approach to decision making when outcomes associated with alternatives are often in doubt.
It helps op. manager with decisions on process, capacity, locations & inventory because such decisions are about an uncertain future.

5. The procedure is as follows:
List the feasible alternatives
( one of them do nothing )
List the events ( chance events )
Calculate the payoff for each alternative in each event. ( payoff is the total cost or total profit )
Esitimate the likelihood of each event
Select a decision rule to evaluate the alternatives

6. Fundamentals of Decision Theory
The three types of decision models:
Decision making under uncertainty
Decision making under risk
Decision making under certainty
It is usually worthwhile developing the difference between risk and uncertainty.

7. Fundamentals of Decision Theory - continued
Terms:
Alternative: course of action or choice
State of nature: an occurrence over which the decision maker has no control
Symbols used in decision tree:
A decision node from which one of several alternatives may be selected
A state of nature node out of which one state of nature will occur

8. Decision Table States of Nature Alternatives State 1 State 2
Outcome 1
Outcome 2
Alternative 2
Outcome 3
Outcome 4

9. Decision Making Under Certainty
Decision rule here is pick the alternative with the best payoff for the known event.
Costs the lowest
Profits the highest

10. Example
A manager is deciding whether to build a small or a large facility.
The manager knows with certainty the payoffs that will result under each alternative shown in the following payoff table
The payoffs in $ are the present value of the future revenues – costs for each alternative in each event.

11. What is the best choice if the future demand be low? Solution :
Alternatives
Possible future demand
Low
High
Small
200
270
Large
160
800
Do nothing
What is the best choice if the future demand be low?
Solution :
the highest value in the low future demand is
200,000 which is the small facility.

12. Decision Making Under Uncertainty
Maximax - Choose the alternative that maximizes the maximum outcome for every alternative (Optimistic criterion)
Maximin - Choose the alternative that maximizes the minimum outcome for every alternative (Pessimistic criterion)
Equally likely ( laplace) - chose the alternative with the highest average outcome.
Minimax regret – the best “worst regret”
Calculate table of regrets ( opportunity losses) in which the rows represent the alternative & the columns represent the events
Regret is the difference between a given payoff & the best payoff in the same column.

13. Example Reconsider the payoff table of the previous example
what is the best alternative for each decision rule?
( maximin, maximax, laplace, minimax regret)

14. Maximin The best of the worst The worst payoffs are
The best of the worst is $200,000 small
Alternative
Worst payoffs
Small
200
large
160

15. Maximax Best of the best The best is 800,000 large Alternative
best payoffs
Small
270
large
800

16. Laplace ( Equally Likely )
We have here two events so we assign each a probability of ½
But if we have 3 events we assign each a probability of 1/3
The best is 480,000 , Large
Alternative
Weighted payoff
Small
½ X ½ X 270 = 235
Large
½ X ½ X 800 = 480

17. Minimax regret The worst regret appears in the maximum regret column
Alternative
Regret
Maximum regret
Low
High
Small =0
=530
530
large
=40 =0 40
The worst regret appears in the maximum regret column
To minimize the maximum regret we choose 40 ( large )

18. Example - Decision Making Under Uncertainty
Maximax
Maximin
Equally likely

19. Decision Making Under Risk
Probabilistic decision situation
States of nature have probabilities of occurrence
Select alternative with largest expected monetary value (EMV)
EMV = Average return for alternative if decision were repeated many times

20. Expected Monetary Value Equation
Probability of payoff
EMV A V P i N ( ) =  * + 1 2
Number of states of nature
Value of Payoff
Alternative i
...

21. Example Reconsider the previous example
If the probability of the low demand is estimated to be 0.4 & the probability of the high demand is estimated to be 0.6 .
Choose the Large Facility ( the highest EMV )
Alternatives
Expected Monetary Value ( EMV )
Small
Emv= 0.4 X X 270 = 242
Large
Emv= 0.4 X X 800 = 544

22. Example - Decision Making Under Risk
Best choice

23. Decision Trees Graphical display of decision process
Used for solving problems
With 1 set of alternatives and states of nature, decision tables can be used also
With several sets of alternatives and states of nature (sequential decisions), decision tables cannot be used
EMV is criterion most often used

24. Analyzing Problems with Decision Trees
Define the problem
Structure or draw the decision tree
Assign probabilities to the states of nature
Estimate payoffs for each possible combination of alternatives and states of nature
Solve the problem by computing expected monetary values for each state-of-nature node

25. Decision trees
Introduction
In many problems chance (or probability) plays an important role. Decision analysis is the general name that is given to techniques for analyze problems containing
risk/uncertainty/probabilities.
Decision trees are one specific decision analysis technique and we will illustrate the technique by use of an example.

26. Decision Tree 1 2 State 1 E1( P(E1)) State 2 E2(P(E2)) State 1 State 2
Alternative 1
Alternative 2
Decision Node
Outcome 1
Outcome 2
Outcome 3
Outcome 4
State of Nature Node

27. After drawing the decision tree, we solve it by working from right to left, calculating the expected payoff for each node as follows:
For event node , we multiply the payoff of each event branch by event’s probability. We add these products to get the expected payoff.
For a decision node , we pick the alternative that has the best expected payoff. If an alternative leads to an event node , its payoff is equal to that node’s expected payoff( already calculated).

28. Example
Retailer must decide whether to build a small or large facility at a new location.
Demand can be either low or high, with probability estimated to be 0.4 and 0.6 respe.
If a small facility is built and demands prooves to be high the manager may choose not to expand payoff 223 or to expand payoff 270
If a small facility is built and demand is low, there is no reason to expand and the payoff is $200,000.
If large facility is built and demand proves to be low, the choice is do nothing $40,000 or to stimulate demand through local advertising.

29. Draw and analyze the decision tree.
Example continue
The response to advertising may be either modest or sizable, with their probability estimated to be 0.3 and 0.7,rspect.
If it is modest, the payoff is estimated to be only $20,000; the payoff grows to $220,000 if the response is sizable.
Finally , if a large facility is built and demand turns out to be high, the payoff is $800,000.
Draw and analyze the decision tree.

30. Format of a Decision Tree
Decision Point
Low demand [0.4] B
$200
High demand[0.6]
$ 223
$ 270 2
Don’t Expand
Expand
$ 800
High Demand [0.6] 3
$ 40
Do nothing
Advertise
Low Demand [0.4]
Small facility
Large facility 1
Chance Event
$ 242
$ 270
$ 544
$ 20
Modest [0.3]
$ 160
Sizable [0.7]
$ 160
$ 544
$ 220

31. Analyzing Solution
For the event node dealing with advertising the expected payoff is 160 or the sum of each event’s payoff weighted by its probability [ 0.3 ( 20 ) ( 220) ]
The expected payoff for decision node 3 is 160 because 160 is better than do nothing 40, Prune the do nothing Alternative.
And so on
The best alternative is to built a large facility.

32. Example
A company faces a decision with respect to a product (codenamed M997) developed by one of its research laboratories. It has to decide whether to proceed to test market M997 or whether to drop it completely. It is estimated that test marketing will cost $100. Past experience indicates that only 30% of products are successful in test market.
If M997 is successful at the test market stage then the company faces a further decision relating to the size of plant to set up to produce M997. A small plant will cost $150 to build and produce 2000 units a year whilst a large plant will cost $250 to build but produce 4000 units a year.
The marketing department have estimated that there is a 40% chance that the competition will respond with a similar product and that the price per unit sold (in $) will be as follows (assuming all production sold):
Large plant Small plant
Competition respond
Competition do not respond
Assuming that the life of the market for M997 is estimated to be 7 years and that the yearly plant running costs are $50 (both sizes of plant - to make the numbers easier!) should the company go ahead and test market M997?

34. In that figure we have three types of node represented:
Decision nodes;
Chance nodes; and
Terminal nodes.
Decision nodes represent points at which the company has to make a choice of one alternative from a number of possible alternatives e.g. at the first decision node the company has to choose one of the two alternatives "drop M997" or "test market M997".
Chance nodes represent points at which chance, or probability, plays a dominant role and reflect alternatives over which the company has (effectively) no control.
Terminal nodes represent the ends of paths from left to right through the decision tree.

35. path to terminal node 2 - we drop M997
Step 1
In this step we, for each path through the decision tree from the initial node to a terminal node of a branch, work out the profit (in $) involved in that path. Essentially in this step we work from the left-hand side of the diagram to the right-hand side.
path to terminal node 2 - we drop M997
Total revenue = 0 Total cost = 0 Total profit = 0
Note that we ignore here (and below) any money already spent on developing M997 (that being a sunk cost, i.e. a cost that cannot be altered no matter what our future decisions are, so logically has no part to play in deciding future decisions).

36. path to terminal node 4 - we test market M997 (cost $100) but then find it is not successful so we drop it
Total revenue = 0 Total cost = 100 Total profit = -100 (all figures in $)
path to terminal node 7 - we test market M997 (cost $100), find it is successful, build a small plant (cost $150) and find we are without competition (revenue for 7 years at 2000 units a year at
$65 per unit = $910)
Total revenue = 910 Total cost = x50 (running cost)
Total profit = 310

37. path to terminal node 8 - we test market M997 (cost $100), find it is successful, build a small plant (cost $150) and find we have competition (revenue for 7 years at 2000 units a year at $35 per unit = $490)
Total revenue = 490 Total cost = x50 Total profit = -110
path to terminal node 10 - we test market M997 (cost $100), find it is successful, build a large plant (cost $250) and find we are without competition (revenue for 7 years at 4000 units a year at $50 per unit = $1400)
Total revenue = Total cost = x50 Total profit = 700

38. Total revenue = 560 Total cost = 350 + 7x50 Total profit = -140
path to terminal node 11 - we test market M997 (cost $100), find it is successful, build a large plant (cost $250) and find we have competition (revenue for 7 years at 4000 units a year at $20 per unit = $560)
Total revenue = 560 Total cost = x50 Total profit = -140
path to terminal node 12 - we test market M997 (cost $100), find it is successful, but decide not to build a plant
Total revenue = 0 Total cost = 100 Total profit = -100
Note that, as mentioned previously, we include this option because, even if the product is successful in test market, we may not be able to make sufficient revenue from it to cover any plant construction and running costs.

39. Hence we can form the table below indicating, for each branch, the total profit involved in that branch from the initial node to the terminal node.
Terminal node Total profit (£K)
2 0
So far we have not made use of the probabilities in the problem - this we do in the second step where we work from the right-hand side of the diagram back to the left-hand side.

40. Step 2
Consider chance node 6 with branches to terminal nodes 7 and 8 emanating from it.
The branch to terminal node 7 occurs with probability 0.6 and total profit $310 whilst the branch to terminal node 8 occurs with probability 0.4 and total profit $-110.
Hence the expected monetary value (EMV) of this chance node is given by
0.6 x (310) x (-110) = 142 ($)
Essentially this figure represents the expected (or average) profit from this chance node (60% of the time we get $310 and 40% of the time we get $ -110 so on average we get
(0.6 x (310) x (-110)) = 142 ($)).

41. The EMV for any chance node is defined by
"sum over all branches, the probability of the branch multiplied by the monetary ($) value of the branch".
Hence the EMV for chance node 9 with branches to terminal nodes 10 and 11 emanating from it is given by
0.6 x (700) x (-140) = 364 ($) node 10 node 11

42. We can now picture the decision node relating to the size of plant to build as below where the chance nodes have been replaced by their corresponding EMV's.

43. Hence at the plant decision node we have the three alternatives:
Alternative 3: build small plant EMV = 142K
Alternative 4: build large plant EMV = 364K
Alternative 5: build no plant EMV = -100K

44. It is clear that, in $ terms, alternative number 4 is the most attractive alternative and so we can discard the other two alternatives, giving the revised decision tree shown below.

45. We can now repeat the process we carried out above.
The EMV for chance node 3 representing whether M997 is a success in test market or not is given by
0.3 x (364) x (-100) = 39.2 ($) plant decision node node 4
Hence at the decision node representing whether to test market M997 or not we have the two alternatives:
Alternative 1: drop M997 EMV = 0 Alternative 2: test market M997 EMV = 39.2$
It is clear that, in $ terms, alternative number 2 is preferable and so we should decide to test market M997.

46. Summary
Let us be clear then about what we have decided as a result of the above process:
We should test market M997 and this decision has an expected monetary value (EMV) of $39.2
If M997 is successful in test market then we anticipate, at this stage, building a large plant (recall the alternative we chose at the decision node relating to the size of plant to build). However it is plain that in real life we will review this once test marketing has been completed

47. Note here that the EMV of our decision (39
Note here that the EMV of our decision (39.2 in this case) DOES NOT reflect what will actually happen - it is merely an average or expected value if we were to have the tree many times - but if fact we have the tree once only.
If we follow the path suggested above of test marketing M997 then the actual monetary outcome will be one of
[-100, 310, -110, 700, -140, -100] corresponding to terminal
nodes 4,7,8,10,11 and 12 depending upon future decisions and chance events.