1. Chapters 19 – 20: Dealing With Uncertainties TPS Example: Uncertain outcome Decision rules for complete uncertainty Utility functions Expected value of perfect information Statistical decision theory The value-of-information procedure

2. TPS Decision Problem 7 OS Development Options: Option B-Conservative (BC) –Sure to work (cost=$600K) –Performance = 4000 tr/sec Option B-Bold (BB) –If successful,Cost=$600K Perf.= 4667 tr/sec –If not,Cost=$100K Perf.= 4000 tr/sec Which option should we choose?

3. Operating System Development Options Option BB (Bold) SuccessfulNot Successful Option BC (Conservative) Option A Performance (tr/sec)46674000 2400 Value ($0.3K per tr/sec)14001200 720 Basic cost600 170 Total cost600700600170 Net value NV800500600550 NV relative to Option A250-50500

4. Payoff Matrix for OS Options BB, BC State of Nature AlternativeFavorableUnfavorable BB (Bold)250-50 BC (Conservative)50

5. Decision Rules for Complete Uncertainty Maximin Rule. Determine minimum payoff for each option. Choose option maximizing the minimum payoff. Maximax Rule. Determine max-payoff for each option. Choose option maximizing max-payoff. Laplace Rule. Assume all states of nature equally likely. Choose option with maximum expected value. BB = 0.5(250) + 0.5(-50) = 100 BC = 0.5(250) + 0.5(50) = 50

6. Difficulty with Laplace Rule FavorableU1U1 U2U2 Expected Value BB250-50 50 BC50 U 1 : performance failure U 2 : reliability failure

7. Breakeven Analysis Treat uncertainty as a parameter P= Prob (Nature is favorable) Compute expected values as f(P) EV (BB) = P(250) + (1-P)(-50) = -50 + 300P EV (BC) = P(50) + (1-P)(50) = 50 Determine breakeven point P such that EV (BB) = EV (BC): -50 + 300P = 50, P=0.333 If we feel that actually P>0.333, Choose BB P<0.333, Choose BC

8. Utility Functions Is a Guaranteed $50K The same as A 1/3 chance for $250K and A 2/3 chance for -$50K?

9. Utility Function for Two Groups of Managers Utility 12345-2-3 10 20 -10 -20 Payoff ($M) Group 1 Group 2

10. Possible TPS Utility Function 10 20 -10 -20 -100 100200300 (-80,-25) (-50,-20) (50,5) (100,10) (170,15) (250,20)

11. Utility Functions in S/W Engineering Managers prefer loss-aversion –EV-approach unrealistic with losses Managers’ U.F.’S linear for positive payoffs –Can use EV-approach then People’s U.F.’S aren’t –Identical –Easy to predict –Constant

12. TPS Decision Problem 7 Available decision rules inadequate Need better information Info. has economic value State of Nature AlternativeFavorableUnfavorable BB (Bold)250-50 BC (Conservative)50

13. Expected Value of Perfect Information (EVPI) Build a Prototype for $10K –If prototype succeeds, choose BB Payoff: $250K – 10K = $240K –If prototype fails, choose BC Payoff: $50K – 10K = $40K If equally likely, Ev = 0.5 ($240K) + 0.5 ($40K) = $140K Could invest up to $50K and do better than before –thus, EVPI = $50K

14. However, Prototype Will Give Imperfect Information That is, P(IB|SF) = 0.0 Investigation (Prototype) says choose bold state of nature: Bold will fail P(IB|SS) = 1.0 Investigation (prototype) says choose bold state of nature: bold will succeed

15. Suppose we assess the prototype’s imperfections as P(IB|SF) = 0.20,P(IB|SS) = 0.90 And Suppose the states of nature are equally likely P(SF) = 0.50P(SS) = 0.50 We would like to compute the expected value of using the prototype EV(IB,IC) = P(IB) (Payoff if use bold) +P(IC) (Payoff if use conservative) = P(IB) [ P(SS|IB) ($250K) + P(SE|IB) (-$50K) ] + P(IC) ($50K) But these aren’t the probabilities we know

16. How to get the probabilities we need P(IB) = P(IB|SS) P(SS) + P(IB|SF) P(SF) P(IC) = 1 – P(IB) P(SS|IB) = P(IB|SS) P(SS)(Bayes’ formula) P(IB) P(SF|IB) = 1 – P (SS|IB) P(SS|IB) = Prob (we will choose Bold in a state of nature where it will succeed) Prob (we will choose Bold)

17. Net Expected Value of Prototype PROTO COST, $K P (PS|SF)P (PS|SS)EV, $KNET EV, $K 0600 50.300.8069.34.3 100.200.9078.28.2 200.100.9586.86.8 300.001.00900 8 4 0 102030 NET EV, $K PROTO COST, $K

18. Conditions for Successful Prototyping (or Other Info-Buying) 1.There exist alternatives whose payoffs vary greatly depending on some states of nature. 2.The critical states of nature have an appreciable probability of occurring. 3.The prototype has a high probability of accurately identifying the critical states of nature. 4.The required cost and schedule of the prototype does not overly curtail its net value. 5.There exist significant side benefits derived from building the prototype.

19. Pitfalls Associated by Success Conditions 1.Always build a prototype or simulation –May not satisfy conditions 3,4 2.Always build the software twice –May not satisfy conditions 1,2 3.Build the software purely top-down –May not satisfy conditions 1,2 4.Prove every piece of code correct –May not satisfy conditions 1,2,4 5.Nominal-case testing is sufficient –May need off-nominal testing to satisfy conditions 1,2,3

20. Statistical Decision Theory: Other S/W Engineering Applications How much should we invest in: –User information gathering –Make-or-buy information –Simulation –Testing –Program verification How much should our customers invest in: –MIS, query systems, traffic models, CAD, automatic test equipment, …

21. The Software Engineering Field Exists Because Processed Information Has Value