COMPUTATIONAL TECHNIQUES

1. COMPUTATIONAL TECHNIQUES

2. 1.6 Computational Techniques • Some details on the Simplex Method approach • 2x2 games • 2xn and mx2 games • Recall: First try pure strategies. If there are no saddle points use mixed strategies.

3. Linear Programming

4. 1.6.1 Example (page 32) • L = max {-3,-1,-1} = -1 • U = min {3,1,2} = 1 • L < U thus no saddle points, need mixed strategies. • The value of this game may not be positive, since L ≤ v ≤ U, we have –1 ≤ v ≤ 1.

5. Note that • Adding a constant, r, to every element of a payoff matrix does not change the optimal strategies, but the new game given by this new matrix will have r added to the value of the game. • Making one row strictly positive is sufficient (see problem sheet) to give a game with positive v. • Thus we add to every element of the matrix say r = 2 to make sure the value is strictly positive:

6. é ù 0 3 1 - ê ú V ' V 2 1 1 4 = + = ê ú ë û 5 2 1

7. LP Formulations Player I Player II • We prefer Player II’s formulation. Why?

11. v’ = 60/33; y’* = (1/60, 23/60, 9/60) y* = (60/33) (1/60, 23/60, 9/60) = (1/33, 23/33, 9/33) x’* = (9/60, 1/4, 9/60); x* = (60/33) (9/60, 1/4, 9/60) = (3/11, 5/11, 3/11). v = 60/33 – 2 = –2/11.

12. More graph ideas

13. 1.6.3 2xn and mx2 Games • If there are only two pure strategies to one of the players, the decision making problem is “one dimensional” (why?). • Thus, it is easy to describe and solve the problem graphically. • We shall assume a 2xn game. • There are many important problems of this kind: “Yes” vs “No”.

14. Analysis for 2xn • We look at player 1, as only two choices. • v1 = max {min {xV.j : j = 1,...,n}: x in S} • = max {min {(x1, x2)(v1j,v2j)t: j = 1,...,n} x1+x2 = 1, x1,x2 ≥ 0} • = max {min (x1,1– x1 )(v1j,v2j)t: j = 1,...,n}, 0 ≤ x1 ≤ 1} • = max {min{x1(v1j – v2j) + v2j}: j = 1,...,n} 0 ≤ x1 ≤ 1} How do we solve optimization problems of this kind?

15. v1 = max {min{x1(v1j – v2j) + v2j}: j = 1,...,n} 0 ≤ x1 ≤ 1} v1 = max { g(x1): 0 ≤ x1 ≤ 1} where g(x1):= min{x1(v1j – v2j) + v2j } : j = 1,...,n} Observation: • g(x1) is piecewise linear with x1, with vertices where two or more linear functions intersect. • The maximum of this function will therefore be attained at one of these vertices • or at x1= 0 or x1 = 1. • Identifying this vertex will yield a 2x2 subgame which we can then easily solve.

16. 1.6.4 Example (page 41) • Note that this is an mx2 case, so here we will look at player 2, thus will have min max. (compare with previous analysis) • There are no saddle points • No dominated pure strategies

17. v2 = min{max{Vi.y: i = 1,2,3}, y1+y2 = 1, y1,y2 ≥ 0} = min{max{8y2, 9y1+4y2, 8y1+6y2}, for y1+y2 = 1, y1,y2 ≥ 0} = min{max{8 – 8y1, 5y1 + 4, 2y1 + 6}, y1 ≥ 0} = min{g(y1): 0 ≤ y1 ≤ 1} g(y1): = max{8 – 8y1, 5y1 + 4, 2y1 + 6}: We can plot g(y1) and identify the linear segments that define the optimal solution.

18. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6}

19. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6}

20. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6}

21. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6}

22. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6}

23. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6} v2 = min {g(y1): 0 ≤ y1 ≤ 1} minmax

24. 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 g(y1):=max{8 – 8y1, 5y1+ 4, 2y1 + 6} v2 = min {g(y1): 0 ≤ y1 ≤ 1}

25. g(y1):=max{8 – 8y1, 5y1+ 4, 2y1+ 6} 8 7 6 5 4 3 2 1 y1 0 0 .1 .2 .3 .4 .5 .6 .7 .8 .9 1 v2 = min {g(y1): 0 ≤ y1 ≤ 1} • Optimal y1: 8 – 8y1 = 2y1+ 6 , y1 = 0.2, • y2 = 1– y1 = 0.8, • y* = (0.2, 0.8)

26. Player I • Since the optimal solution to Player II involves the 1st and 3rd rows of V, we conclude that the optimal solution to Player I can be obtained from the reduced payoff matrix consisting of these two rows. • We thus set x2 = 0.

27. By Theorem 1.6.2, (recipe for 2x2 games) x’ = (0.2, 0.8) v’ = 32/5 • Thus, the optimal strategy for the “full” game is x* = (0.2, 0, 0.8) y* = (0.2, 0.8) • Also see notes about evaluating all 2x2 subgames as another approach.

28. Summing up so far

29. Overview zero-sum two-person games • Assumption - play from pessimistic viewpoint. • L = Player I’s largest security level = lower value of game. • U = Player II’s lowest security level = upper value of game. • In general L ≤ U. • Iff L = Uwe have a saddle point, say (ai*, Aj*) which gives stable (equilib.) solution, and value v = L = U. • Solution - pure strategies: • X* = (0, 0, ...0, 1, 0 ..., 0), Y* = (0, ...0, 1, 0, ... 0) • Player I can choose row i* and II choose column j* and play these all the time and be happy.

30. Case L < U. Not possible to find a pair of pure strategies with the solution in equilibrium. • We determine an optimal strategy based on what a person will gain on average -thinking of playing game repeatedly many times. • Players mix up the strategies they use to maximize their expected payoff. • Expected payoff E(X, Y) = XVY. • v1= I’s largest security level =maxX in S minY in T XVY • v2 = II’s largest sec level = minY in T maxX in S XVY • v1 and v2compare to L and U in the case L = U.

31. There is always an X* & Y* pair such that v1 = v2= v = X*VY*, the value of the game and this pair is in equilibrium, • i.e. XVY* ≤ X*VY* ≤ X*VY for all X in S and all Y in T.

32. To solve a 2-person, zero-sum game Check for saddle point. If saddle, give saddle point and value of game. If no saddle Try to reduce size using dominance. If reduced game of size 2x2, use formulae. If 2xm or nx2 use graph + formulae or LP. Otherwise use linear programming methods.