1. 1-1 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

2. 1-2 The Decision-Making Process ProblemDecision Quantitative Analysis Logic Historical Data Marketing Research Scientific Analysis Modeling Qualitative Analysis Emotions Intuition Personal Experience and Motivation Rumors

3. 1-3 Decision Problem Alternatives States of Nature Out- comes  Decision trees  Decision tables Ways of Displaying a Decision Problem

4. 1-4 Preference Matrix  Used in evaluating an alternative  Preference matrix is a table that allows the manager to rate an alternative according to several performance creteria. Example : The Following table shows the performance criteria, weights and scores for a new product:

5. 1-5 A thermal storage air Conditioner  If Mgt. wants to introduce just one new product & the highest total score of any of other product ideas is 800  Should firm pursue making the air conditioner.

6. 1-6 Performance Criteria WeightScoreWeighted score Market potential308240 Unit profit margin2010200 Op. Compatibility206120 Competitive adv.1510150 Investment Req.10220 Project risk5420 Totals100From 1-10750 So Mgt. would not pursue the thermal Storage air Cond. 750 < 800.

7. 1-7 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.

8. 1-8 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

9. 1-9 Fundamentals of Decision Theory The three types of decision models:  Decision making under uncertainty  Decision making under risk  Decision making under certainty

10. 1-10 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

11. 1-11 Decision Table States of Nature AlternativesState 1State 2 Alternative 1Outcome 1Outcome 2 Alternative 2Outcome 3Outcome 4

12. 1-12 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

13. 1-13 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.

14. 1-14  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. AlternativesPossible future demand LowHigh Small200270 Large160800 Do nothing00

15. 1-15 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.

16. 1-16 Example  Reconsider the payoff table of the previous example  what is the best alternative for each decision rule? ( maximin, maximax, laplace, minimax regret)

17. 1-17 Maximin  The best of the worst  The worst payoffs are The best of the worst is $200,000 small AlternativeWorst payoffs Small200 large160

18. 1-18 Maximax  Best of the best  The best is 800,000 large Alternativebest payoffs Small270 large800

19. 1-19 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 AlternativeWeighted payoff Small½ X 200 + ½ X 270 = 235 Large½ X 160 + ½ X 800 = 480

20. 1-20 Minimax regret AlternativeRegretMaximum regret LowHigh Small200-200=0800-270=530530 large200-160=40800-800=040 The worst regret appears in the maximum regret column To minimize the maximum regret we choose 40 ( large )

21. 1-21 Example - Decision Making Under Uncertainty MaximaxMaximin Equally likely

22. 1-22  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 Decision Making Under Risk

23. 1-23 Expected Monetary Value Equation Probability of payoff EMVAVPV VPVVPVVPV ii i i NN (() ()()() )= N =  * = * + * ++ * 1 1122 Number of states of nature Value of Payoff Alternative i...

24. 1-24 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 ) SmallEmv= 0.4 X 200 + 0.6 X 270 = 242 LargeEmv= 0.4 X 160 + 0.6 X 800 = 544

25. 1-25 Example - Decision Making Under Risk Best choice

26. 1-26  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 Decision Trees

27. 1-27 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

28. 1-28 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.

29. 1-29 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

30. 1-30  After drawing the decision tree, we solve it by working from right to left, calculating the expected payoff for each node as follows: 1.For event node, we multiply the payoff of each event branch by event’s probability. We add these products to get the expected payoff. 2.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).

31. 1-31 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.

32. 1-32 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.

33. 1-33 Format of a Decision Tree 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 Decision Point Chance Event $ 270 Modest [0.3] Sizable [0.7] $ 20 $ 220 $ 160 $ 544 $ 242 $ 544

34. 1-34 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 ) + 0.7 ( 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.

35. 1-35 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 20 35 Competition do not respond 50 65 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?

36. 1-36

37. 1-37 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.

38. 1-38 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).

39. 1-39 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 = 250 + 7x50 (running cost) Total profit = 310

40. 1-40 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 = 250 + 7x50 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 = 1400 Total cost = 350 + 7x50 Total profit = 700

41. 1-41 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 = 350 + 7x50 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.

42. 1-42 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 4 -100 7 310 8 -110 10 700 11 -140 12 -100 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.

43. 1-43 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 $-110. Hence the expected monetary value (EMV) of this chance node is given by 0.6 x (310) + 0.4 x (-110) = 142 ($) 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) + 0.4 x (-110)) = 142 ($)).

44. 1-44 The EMV for any chance node is defined by " sum over all branches, the probability of the branch multiplied by the monetary ($) value of the branch ". Hence the EMV for chance node 9 with branches to terminal nodes 10 and 11 emanating from it is given by 0.6 x (700) + 0.4 x (-140) = 364 ($) node 10 node 11

45. 1-45 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.

46. 1-46 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

47. 1-47 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.

48. 1-48 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) + 0.7 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.

49. 1-49 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

50. 1-50 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.

51. 1-51