# rec6v - MASSACHUSETTS INSTITUTE OF TECHNOLOGY 15.053...

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M ASSACHUSETTS I NSTITUTE OF T ECHNOLOGY 15.053 – Optimization Methods in Management Science (Spring 2007) Virtual Recitation 6, 2007 Welcome to Recitation 6. As always we start with a quiz. You have 15 minutes to complete the quiz. OK. That was a joke. This virtual recitation is being given in place of holding an actual recitation on March 22 th , 2007. We have found that too many students take the day off before spring break or are more focused on gearing up for vacation. Hence, we offer this tutorial instead. It should be thought of as a regular recitation. All of the problems should be worked through, and their solutions understood before attempting Problem Set 6. However, if you struggle, do not worry. As in accordance with our new recitation policy, we will be creating a virtual cyber cast or movie of the recitation. Problem 1: Start With the Basics In this first problem, we start to look at the structure of games and develop strategies. This problem is considered very basic; when you are solving the game, make sure you understand what you are doing and not just the mechanics of solving the problem. It will be very important later on when we study more difficult topics. Find the value and optimal minimax and maximin strategies for the game in the table below. Use the graphical approach. 21 13 Solution: We let x be the probability that the row player chooses Row 1, and we let y be the probability that the column player chooses column 1. Then 1 – x = Probability that the row player chooses row 2 1 – y = Probability column player chooses column 2 We now compute the cases: Case 1: If the column player chooses column one the if the row player uses a strategy (x,1-x) then the expected payoff to the row player is: 2x+(1-x) = x+1.

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We can graph the reward as function of x. This is shown below, in red: x Payoff to Row Case 2: If the column player chooses column two, and if the row player uses a strategy (x, 1-x) then the expected payoff to the row player is: x + 3(1-x) = -2x + 3. We can graph the reward as function of x. This is shown below, in green: x Payoff to Row Player The row player can guarantee a payoff of the lower envelope of the above graph, regardless of what the column player chooses. This is shown below:
x Payoff to Row Player Thus since the row player wants to maximize his profits, he will choose the highest point on the blue line. This is the point where the red and green lines intersect, and is the point (2/3, 5/3). This shows the optimal strategy for the row player is to choose row one 2/3 of the time and row two 1/3 of the time. To find the optimal strategy for the column player we repeat the same procedure as above: Case 1: If the row player chooses row one and if the column player uses a strategy (y, 1-y) then the expected payoff to the row player is: 2y + (1-y) = y+1 We can graph the reward as function of y. This is shown below, in green: y Payoff to Row Player

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Case 2: If the row player chooses row two, and if the column player uses a strategy (y, 1-y) then the expected payoff to the column player is: y +3(1-y) = -2y + 3 We can graph the reward as function of y. This is shown below, in plum:
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rec6v - MASSACHUSETTS INSTITUTE OF TECHNOLOGY 15.053...

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