Q 5-2 a.

1. Q 5-2 a. E = Efficiency score wi = Weight applied to i ’s input and output resources by the composite hospital

2. Q 5-2 a. cont’d

3. Q 5-2 b. OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables) Variable Value Cost VAR 1(E) 0.9235 1.0000 VAR 2(wa) 0.0745 0.0000 VAR 4(wc) 0.4362 0.0000 VAR 6(we) 0.4894 0.0000 Slack Variables CONSTR 2 0.8376 0.0000 CONSTR 5 33.9691 0.0000 CONSTR 6 34.5612 0.0000 CONSTR 7 150.549 0.0000 Objective Function Value = 0.923523 E* = 0.924, wa* = 0.075, wc* = 0.436, we* = 0.489 wb*= wd*= wf*=wg* =0

4. Q 5-2 c. d. e (c) Hospital D is relatively inefficient (7.6%). Efficiency score (Objective) is 92.4 %. (d) Patient-days (65 or older) 55.31(0.0745) + 32.91(0.4362) + 32.48(0.4894) = 34.37 Patient-days (under 65) 49.52(0.0745) + 25.77(0.4362) + 55.3(0.4894) = 41.99 (e) Hospitals: A, C, and E

5. Q 5-3 a. E = Efficiency Score wi = Weight applied to i ’s input and output resources by the composite hospital

6. Q 5-3 a. cont’d 206.4 E

7. Q 5-3 b. OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables) Variable Value Cost VAR 1(E) 1.0000 1.0000 VAR 6(we) 1.0000 0.0000 Slack Variables CONSTR 6 53.6000 0.0000 Objective Function Value = 1 E* = 1, we* = 1; All the other weights* = 0

8. Q 5-3 c. d. (c) Hospital E is efficient. (d) Hospital E is the only hospital in the composite. If a hospital is relatively efficient, the hospital will make up the composite hospital with weight equal to 1.

9. Q 5-4 a. Let, E = Efficiency Score wb = Weight applied to Bardstown’s input and output resources by the composite restaurant wc = Weight applied to Clarksville’s input and output resources by the composite restaurant wj = Weight applied to Jeffersonville’s input and output resources by the composite restaurant wn = Wight applied to New Albany’s input and output resources by the composite restaurant ws = weight applied to St. Matthews’s input and output resources by the composite restaurant

10. Q 5-4 a. cont’d

11. Q 5-4 b. OPTIMAL SOLUTION - SUMMARY REPORT (Nonzero Variables) Variable Value Cost VAR 1(E) 0.960 1.0000 VAR 2(wb) 0.175 0.0000 VAR 4(wj) 0.575 0.0000 VAR 5(wn) 0.250 0.0000 Constraint Slack/Surplus Dual Price CONSTR 1 0.000 0.200 CONSTR 2 220.000 0.000 CONSTR 3 0.000 -0.004 CONSTR 4 0.000 -0.123 CONATR 5 0.000 0.009 CONSTR 6 1.710 0.000 CONSTR 7 129.614 0.000 Objective Function Value = 0.960 E* = 0.960, wb* = 0.175, wj* = 0.575, wn* = 0.250, wc*, ws* =0

12. Q 5-4 c. E = 0.96 indicates the 96% level of efficiency and 4% inefficiency.

13. Q 5-4 d. More Output ($220: more profit per week) is needed. Less Input Hours of Operation: 110(0.96) = 105.6 FTE Staff: 22(0.96) – 1.71 = 19.41 Supply Expense: 1400(0.96)– 129.614 = 1214.39

14. Q 5-4 e. wb = 0.175, wj = 0.575, and wn = 0.250. So, from Bardstown, Jeffersonville, and New Albany restaurants.

15. Game Theory

16. Game theory is a mathematical theory that deals with the general features of competitive situations. The final outcome depends primarily upon the combination of strategies selected by the adversaries.

17. Two key Assumptions: (a) Both players are rational (b) Both players choose their strategies solely to increase their own welfare.

18. Payoff Table Player 2 Strategy 1 2 3 1 2 3 1 2 4 1 0 5 0 1 -1 Player 1 Each entry in the payoff table for player 1 represents the utility to player 1 (or the negative utility to player 2) of the outcome resulting from the corresponding strategies used by the two players.

19. A strategy is dominated by a second strategy if the second strategy is always at least as good regardless of what the opponent does. A dominated strategy can be eliminated immediately from further consideration. Player 2 Strategy 1 2 3 1 2 3 1 2 4 1 0 5 0 1 -1 Player 1 For player 1, strategy 3 can be eliminated. ( 1 > 0, 2 > 1, 4 > -1)

20. 1 2 3 1 2 1 2 4 1 0 5 For player 2, strategy 3 can be eliminated. ( 1 < 4, 1 < 5 ) 1 2 1 2 1 2 1 0 For player 1, strategy 2 can be eliminated. ( 1 = 1, 2 < 0 )

21. 1 2 1 1 2 For player 2, strategy 2 can be eliminated. ( 1 < 2 ) Consequently, both players should select their strategy 1. A game that has a value of 0 is said to be a fair game.

22. Minimax criterion: To minimize his maximum losses whenever resulting choice of strategy cannot be exploited by the opponent to then improve his position. Player 2 Strategy Minimum 1 2 3 -3 0 -4 1 2 3 -3 -2 6 2 0 2 5 -2 -4 Player 1 5 0 6 Maximum: Minimax value Maximin value

23. The value of the game is 0, so this is fair game Saddle Point: A Saddle point is an entry that is both the maximin and minimax. Player 2 Strategy Minimum 1 2 3 -3 0 -4 1 2 3 -3 -2 6 2 0 2 5 -2 -4 Player 1 5 0 6 Maximum: Saddle point

24. There is no saddle point. An unstable solution Player 2 Strategy Minimum 1 2 3 -2 -3 -4 1 2 3 0 -2 2 5 4 -3 2 3 -4 Player 1 5 4 2 Maximum:

25. Mixed Strategies = probability that player 1 will use strategy i ( i = 1,2,…,m), = probability that player 2 will use strategy j ( j = 1,2,…,n), Expected payoff for player 1 =

26. Minimax theorem: If mixed strategies are allowed, the pair of mixed strategies that is optimal according to the minimax criterion provides a stable solution with (the value of the game), so that neither player can do better by unilaterally changing her or his strategy. = maximin value = minimax value

27. Solving by Linear Programming Expected payoff for player 1 = The strategy is optimal if

28. For each of the strategies where one and the rest equal 0. Substituting these values into the inequality yields Because the are probabilities,

29. The two remaining difficulties are (1) is unknown (2) the linear programming problem has no objective function. Replacing the unknown constant by the variable and then maximizing , so that automatically will equal at the optimal solution for the LP problem.

32. Player 2 Example Probability Pure Strategy Probability 1 2 3 1 2 0 -2 2 5 4 -3 Player 1

33. Player 2 The dual Probability Pure Strategy Probability 1 2 3 1 2 0 -2 2 5 4 -3 Player 1

34. Home Work • Problem 5-13 • Problem 5-15 • Due Date: September 30

35. Question 1 (Optional: Not Home Work) Consider the game having the following payoff table. (a) Formulate the problem of finding optimal mixed strategies according to the minimax criterion as a linear programming problem. (b) Use the simplex method to find these optimal mixed strategies.