1. 1 CHAPTER 6 (handout) Decision Trees

2. 2 6.1. Introduction Sequential decision making w sequence of chance-dependent decisions w presentation of analysis can be complex Decision Trees w Pictorial device to represent problem & calculations w Useful for problems with small no. of sequential decisions

3. 3 6.3. Another Decision Tree Ex. 2 boxes, externally identical Must decide which box w a 1 : box 1: 6 black balls, 4 white balls w a 2 : box 2: 8 black balls, 2 white balls w Correct guessReceive $100 w Wrong guessReceive $0 Prior Probability w P(  1 ) = 0.5 w P(  2 ) = 0.5

4. 4 Decision Tree w A connected set of nodes & arcs w Nodes:join arcs w Arcs:have direction (L to R) w Branch:arc & all elements that follow it w 2 branches from same initial node cannot have elements in common w 2 nodes cannot be joined by > 1 arc

5. 5 Example of a Decision Tree

6. 6 A diagram which is not a tree

7. 7 Types of nodes w Decision point Choosing next action (branch) w Chance node Uncontrollable probabilistic event w Terminal node Specifies final payoff

8. 8 Example of Sequential Decision Problem Car Exchange Problem A person must decide whether to keep or exchange his car in a showroom. There are 2 decisions: a 1 : keep cost = 1400 SR a 2 : exchange, has 2 possibilities: good buyP(G) = 0.6cost = 1200 SR bad buyP(B) = 0.4cost = 1600 SR Good or bad buy can be identified only after buying and using the car. What he should do to minimize his cost?

9. 9 Car Exchange Problem (no information) Payoff (Cost) Matrix  P(  )a 1 : keep a 2 : exchange  1 : Good0.614001200  2 : Bad0.414001600 EV14001360

10. 10 Car exchange decision tree Keep Exchange G: 0.6 B: 0.4 $1400 G: 0.6 B: 0.4 $1200 $1600

11. 11 Car exchange decision tree Keep Exchange G: 0.6 B: 0.4 $1400 G: 0.6 B: 0.4 $1200 $1600 $1400 $1360

12. 12 6.2. A Sequential Test Problem Car Exchange Problem Assume the person has 5 options for deciding whether to keep or exchange his car. (i)Decide without extra information (ii)Decide on basis of free road (driving) test (iii)Decide after oil consumption test costing $25 (iv)Decide after combined road/oil test costing $10 (v)Decide sequentially: road test then possibly oil test costing $10 In (iv), both tests must be taken In (v), oil test is optional, depending on road test

13. 13 Car Exchange Problem (with information) The decision tree is complicated Cannot fit in 1 slide 5 branches: 5 options Probabilities after extra information are conditional (posterior) To illustrate, we choose the branch of option (v) Road test then, depending on result, possible oil test costing $10

14. 14 Car Exchange Problem (with information) Result of road test: y 1 : fair p(y 1 ) = 0.5 y 2 : poorp(y 2 ) = 0.5 Result of oil consumption test: Z 1 : highp(Z 1 |y) Z 2 : mediump(Z 2 |y) Z 3 : lowp(Z 3 |y)

15. 15 Car exchange decision tree (with information) y 1 : 0.5 y 2 : 0.5 No test Oil test Z3Z3 Z1Z1 Z2Z2 No test Oil test Z3Z3 Z1Z1 Z2Z2 Road test

16. 16 Car exchange decision tree with information (y 1 branch) y 1 : 0.5 No test Z 1 : 0.28 Oil test a2a2 a1a1 0.6 0.4 1400 0.6 0.4 1200 1600 a2a2 a1a1 0.43 0.57 1410 0.43 0.57 1210 1610 a2a2 a1a1 0.5 1410 0.5 1210 1610 a2a2 a1a1 0.75 0.25 1410 0.75 0.25 1210 1610 Z 2 : 0.24 Z 3 : 0.48

17. 17 Car exchange decision tree with information (y 1 branch) y 1 : 0.5 No test Z 1 : 0.28 Oil test a2a2 a1a1 0.6 0.4 1400 0.6 0.4 1200 1600 a2a2 a1a1 0.43 0.57 1410 0.43 0.57 1210 1610 a2a2 a1a1 0.5 1410 0.5 1210 1610 a2a2 a1a1 0.75 0.25 1410 0.75 0.25 1210 1610 Z 2 : 0.24 Z 3 : 0.48 1400 1360 1410 1439 1410 1310 1360 1410 1310 1362

18. 18 Car exchange decision tree with information (y 2 branch) y 2 : 0.5 No test Z 1 : 0.32 Oil test a2a2 a1a1 0.6 0.4 1400 0.4 0.6 1200 1600 a2a2 a1a1 0.25 0. 75 1410 0.25 0.75 1210 1610 a2a2 a1a1 0.31 0.69 1410 0.31 0.69 1210 1610 a2a2 a1a1 0.57 0.43 1410 0.57 0.43 1210 1610 Z 2 : 0.26 Z 3 : 0.42

19. 19 Car exchange decision tree with information (y 2 branch) y 2 : 0.5 No test Z 1 : 0.32 Oil test a2a2 a1a1 0.6 0.4 1400 0.4 0.6 1200 1600 a2a2 a1a1 0.25 0. 75 1410 0.25 0.75 1210 1610 a2a2 a1a1 0.31 0.69 1410 0.31 0.69 1210 1610 a2a2 a1a1 0.57 0.43 1410 0.57 0.43 1210 1610 Z 2 : 0.26 Z 3 : 0.42 1400 1440 1410 1510 1410 1487 1410 1381 1400 1410 1381 1398

20. 20 Decision Tree Calculations Tree is developed from left to right Calculations are made from right to left Many calculation are redundant For inferior solutions Not needed in final solution Probabilities after extra information (road or oil tests) are conditional (posterior) Calculated by Bayes’ theorem

21. 21 Initial Payoff Data (no information) Payoff (Reward) Matrix  P(  )a 1 : Box 1 a 2 : Box 2  1 : Box 10.51000  2 : Box 20.50100 EV5050

22. 22 Initial Probability Data (no information) Prior Probability Matrix  P(  )B: Black W: White  1 : Box 10.50.60.4  2 : Box 20.50.80.2

23. 23 Decision tree without information Box 1 Box 2  1 : 0.5  2 : 0.5 $100 $0  1 : 0.5  2 : 0.5 $0 $100 $50

24. 24 Decision Tree Example with information w Samples from box can be taken w Ball is returned to the box w Up to 2 samples are allowed w Cost = $3 per sample w What is the optimal plan?

25. 25 Posterior probabilities for sample 1 Probability Calculations  P(  )P(B)P(W)JointPosterior  1 : 0.50.60.40.30.20.430.67  2 : 0.50.80.20.40.10.570.33  1.00.70.31.001.00

26. 26 Decision tree with information No sample Sample 1 B: 0.7 $50 $ W: 0.3 $ $ a 1 or a 2 $ Sample 2 No sample No information

27. 27 Posterior probabilities for sample 2 when sample 1 is Black Probability Calculations  P(  )P(B)P(W)JointPosterior  1 : 0.430.60.40.260.170.360.61  2 : 0.570.80.20.460.110.640.39  1.00.720.281.001.00

28. 28 Sample 1 Black, No Sample 2 No 2 nd sample Sample 2 1: 0.43 2: 0.57 $97 $-3 B: 0.72 W: 0.28 $ $ 1: 0.43 2: 0.57 $-3 $97 a1a1 a2a2 Black sample 1 40 54

29. 29 Samples 1 & 2 Both Black Black sample 2 Sample 2 1: 0.36 2: 0.64 $94 $-6 B: 0.72 W: 0.28 $ 1: 0.36 2: 0.64 $-6 $94 a1a1 a2a2 Black sample 1 30 58 No Sample $54 58

30. 30 Sample 1 Black, Sample 2 White White sample 2 Sample 2 1: 0.61 2: 0.39 $94 $-6 W: 0.28 B: 0.72 $58 1: 0.61 2: 0.39 $-6 $94 a1a1 a2a2 Black sample 1 55 33 No Sample $54 55 57.16

31. 31 Posterior probabilities for sample 2 when sample 1 is White Probability Calculations  P(  )P(B)P(W)JointPosterior  1 : 0.670.60.40.400.270.610.79  2 : 0.330.80.20.260.070.390.21  1.00.660.341.001.00

32. 32 Sample 1 White, No Sample 2 No 2 nd sample Sample 2 1: 0.67 2: 0.33 $97 $-3 B: 0.66 W: 0.34 $ $ 1: 0.67 2: 0.33 $-3 $97 a1a1 a2a2 White sample 1 64 30 64

33. 33 Sample 1 White, Sample 2 Black Black sample 2 Sample 2 1: 0.61 2: 0.39 $94 $-6 B: 0.66 W: 0.34 $ 1: 0.61 2: 0.39 $-6 $94 a1a1 a2a2 White sample 1 55 33 No Sample $64 55

34. 34 Samples 1 & 2 Both White White sample 2 Sample 2 1: 0.79 2: 0.21 $94 $-6 W: 0.34 B: 0.66 $55 1: 0.79 2: 0.21 $-6 $94 a1a1 a2a2 White sample 1 73 15 No Sample $64 73 61.12

35. 35 Decision tree summary of results No samples Sample 1 B: 0.7 $50 $55 W: 0.3 a 1 or a 2 $54 Sample 2 No 2 nd sample No information $64 Sample 2 No 2 nd sample $58 $55 W, 0.28: a 1 B, 0.72: a 2 a2a2 B, 0.66: a 1 W, 0.34: a 1 a1a1 $73 57.2 61.1 64 57.2 59.2 a 1 : 6B, 4W a 2 : 8B, 2W

36. 36 Decision Tree with Fixed Costs w Example of fixed cost: sampling cost = 3/sample in previous example w If objective is to maximize expected payoff, w Constant costs can be deducted either from: Terminal node payoffs Expected values

37. 37 Example: Including fixed costs Sample 1 Black, cost = $3  1 : 0.43  2 : 0.57 $100 $0 43 – 3 a1a1 Sample 1 Black, cost = $3  1 : 0.43  2 : 0.57 $97 $– 3 40 a1a1 Recall Slide 9

38. 38 Fixed Costs & Utilities w Utilities can be used instead of payoffs w If objective is to maximize expected utility Constant costs must be deducted from terminal node payoffs Net payoffs are converted to net utilities Expected values are taken of utilities of net payoffs

39. 39 Including fixed costs Sample 1 Black, cost = $3  1 : 0.43  2 : 0.57 U(100) U(0) EU–U(3) a1a1 Sample 1 Black, cost = $3  1 : 0.43  2 : 0.57 U(97) U(– 3) EU a1a1 Incorrect Correct

40. 40 Allowing an optional 3 rd sample w Suppose now a 3 rd sample is allowed w Sample cost = $3 w Assume the decision whether or not to take sample 3 depends on results of samples 1 and 2 w What is the optimal plan?

41. 41 Posterior probabilities for sample 3 After 2 blacks (slide 8)slide 8  P(  )P(B)P(W)JointPosterior  1 : 0.360.60.40.220.140. 30.52  2 : 0.640.80.20.510.130. 70.48  1.00.730.271.001.00

42. 42 Decision tree with optional sample 3 Sample 1B: 0.7 $50 W: 0.3 $54 Sample 2 No 2 nd sample No sample $ $57.2 Sample 3 No 3 rd sample $64 Sample 2 No 2 nd sample $ $61.1 Sample 3 No 3 rd sample

43. 43 Fixing the number of samples w Suppose now a 3 rd sample is allowed w Sample cost = $3 w Assume we must decide the number of samples in advance: 0, 1, 2, or 3 w What is the optimal plan?

44. 44 Zero samples a 1 : Box 1  1 : 0.5  2 : 0.5 $100 $0  1 : 0.5  2 : 0.5 $0 $100 $50 a 2 : Box 2 50 No samples

45. 45 One Sample B: 0.7 W: 0.3 1: 0.43 2: 0.57 $97 $-3 1: 0.43 2: 0.57 $-3 $97 a1a1 a2a2 Sample once 40 54 1: 0.67 2: 0.33 $97 $-3 1: 0.67 2: 0.33 $-3 $97 a1a1 a2a2 64 30 64 57

46. 46 Posterior probabilities for 2 samples Examples: P(BB|  1 ) = P(BB) = 0.6(0.6)= 0.36 P(BW|  1 ) = P(BW) + P(WB) = 0.6*0.4 + 0.4*0.6= 0.48 P(WW|  1 ) = P(WW) = 0.4(0.4)= 0.16  P(  )BBBWWWJoint  1 : 0.50.360.480.160.180.240.08  2 : 0.50.640.320.040.320.160.02  0.500.400.10 Post  1 : 0.360.600.80  2 : 0.640.400.20

47. 47 Two Samples BB: 0.5 WW: 0.1 1: 0.36 2: 0.64 $94 $-6 a1a1 a2a2 Sample twice 30 58 1: 0.36 2: 0.64 $-6 $94 58 1: 0.6 2: 0.4 $94 $-6 a1a1 a2a2 54 1: 0.6 2: 0.4 $-6 $94 34 1: 0.8 2: 0.2 $94 $-6 a1a1 a2a2 74 1: 0.8 2: 0.2 $-6 $94 14 BW: 0.4

48. 48 Posterior probabilities for 3 samples P(BBB|  1 )= 0.6(0.6)(0.6)= 0.216 P(BBW|  1 )= P(BBW) + P(BWB) + P(WBB) = 3*0.6*0.6*0.4 = 0.432 P(BWW|  1 )= P(BWW) + P(WBW) + P(WWB) = 3*0.6*0.4*0.4 = 0.288 P(WWW|  1 )= 0.4(0.4)(0.4) = 0.064  P BBBBBWBWWWWWJoint  1 :0.5 0.2160.4320.2880.0640.1080.2160.1440.032  2 :0.5 0.5120.3840.0960.0080. 2560.1920.0480.004  0.3640.4080.1920.036 Post  1 : 0.300.530.750.89  2 : 0.700.470.250.11

49. 49 Three Samples BBB: 0.36 WWW: 0.04 1: 0.3 2: 0.7 $91 $-9 a1a1 a2a2 Sample 3 times 21 61 55.7 1: 0.3 2: 0.7 $-9 $91 61 BBW: 0.41 1: 0.53 2: 0.47 $91 $-9 a1a1 a2a2 44 1: 0.53 2: 0.47 $-9 $91 38 1: 0.75 2: 0.25 $91 $-9 a1a1 a2a2 66 1: 0.75 2: 0.25 $-9 $91 16 1: 0.89 2: 0.11 $91 $-9 a1a1 a2a2 80 1: 0.89 2: 0.11 $-9 $91 2 BWW: 0.19

50. 50 Summary of results with fixed number of samples $50 $57 1 sample 0 samples $55.7 $58 2 samples 3 Samples

51. 51 Value of Sample (new information w Results of previous example With sequential samples (slide 23)slide 23 With fixed no. of samples (slide 31)slide 31 w 3 rd Sample is never needed w Questions: How many samples should be taken? Is it better to decide immediately or after more information?

52. 52 Expected Value of Information w AssumeP(  1 ) = p,P(  2 ) = 1 – p w Then P(  )P(B)P(W)Joint  1 :p0.60.40.6p0.4p  2 :1–p0.80.20.8(1-p)0.2(1-p)  1.0(4-p)/5(1+p)/5 Posterior 3p/(4-p)2p/(1+p) 4(1-p)/(4-p)(1-p)/(1+p)

53. 53 Expected payoff w Best payoff if Black = 100[ max{3p/(4-p), 4(1-p)/(4-p)} ] w Best payoff if White = 100[ max{2p/(1+p), (1-p)/(1+p)} ] w Expected outcome w F(p) = 100 (4-p)/5 [ max{3p/(4-p), 4(1-p)/(4-p)} ] + 100 (1+p)/5[ max{2p/(1+p), (1-p)/(1+p)} ] w F(p) = 100[ max{0.6p, 0.8(1-p)} + max{0.4p, 0.2(1-p)} ] w F(p) = max{60p, 80(1-p)} + max{40p, 20(1-p)} w F(p) = max{a, b} + max{c, d}

54. 54 Graph of expected payoff p 100 1 80 4/71/3

55. 55 Maximum Expected Payoff w To maximize F(p) on 0 < p < 1, w Graphical solution gives 0 < p < 1/3F(p) = 100(1 – p)b + d 1/3 < p < 4/7F(p) = 80 – 40pb + c 4/7 < p < 1F(p) = 100pa + c w For 1 st and 3 rd ranges, solution is same as expected payoff given only P(  1 ) = p, P(  2 ) = 1 – p. w Only 2 nd range has improvement in expected payoff w Sample should be taken only if: 1/3 < p < 4/7

56. 56 Expected Value of Sample Information w Value of sample information = Expected improvement in payoff = 80 – 40p – (100 – 100p),0 < p < 0.5 = 80 – 40p – (100p), 0.5 < p < 1 Or = 60p – 20, 0 < p < 0.5 = 80 – 140p,0.5 < p < 1

57. 57 Range of p for sample cost = 3 w For sample cost = 3 w Sample should be taken only improvement is > 3 60p – 20 > 3 p > 0.383 80 – 140p > 3 p < 0.55 w Thus, 0.383 < p < 0.55

58. 58 For fixed no. of samples Posteriors after 2 samples (slide 27)slide 27 BBBWWW P(  1 ) = p 0.360.600.80 Since all probabilities are outside the range ( 0.383 < p < 0.55 ) A 3 rd sample should not be taken

59. 59 How many samples? w So far, analysis is for the value of 1 sample w We can estimate value of several samples w Max. no. of samples Expected payoff with no information= 50 Payoff with perfect information= 100 Max. no. of samples = (100 – 50)/3= 16