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Solve a game by simplex method - Linear Programming Approach To Game Theory

   Posted On :  25.06.2018 02:00 am

Solve the following game by simplex method for LPP:

Solve a game by simplex method

Problem 1

 
Solve the following game by simplex method for LPP:




So, Maximum of {Row minima}   Minimum of {Column maxima}.
 
Therefore the given game has no saddle point.  It is a mixed game.
 
Let us convert the given game into a LPP.
 

Problem formulation

 
Let V denote the value of the game. Let the probability that the player B will use his first strategy be r and second strategy be s. Let V denote the value of the game.

When A follows his first strategy
 
The expected payoff to A (i.e., the expected loss to B) = 3 r + 6 s.
 
This pay-off cannot exceed V. So we have  =3 r + 6 s V        (1)
 
When A follows his second strategy
 
The expected pay-off to A (i.e., expected loss to B) = 5 r + 2 s.
 
This cannot exceed V. Hence we obtain the condition = 5 r + 2 s V        (2)
 
From (1) and (2) we have
 


Solution of LPP




Consider the negative elements in the objective function row. They are –1, -1. The absolute values are 1, 1. There is a tie between these coefficients. To resolve the tie, we select the variable x. We take the new basic variable as x. Consider the ratio of b-value to x-value. We have the following ratios:





Problem 2

 
Solve the following game



Solution

The game has no saddle point. It is a mixed game. Let the probability that B will use his first strategy be r. Let the probability that B will use his second strategy be s. Let V be the value of the game.

When A follows his first strategy
 
The expected payoff to A (i.e., the expected loss to B) = 2 r +5 s.
 
The pay-off to A cannot exceed V. So we have       = 2 r + 5 s V          (I)

 When A follows his second strategy
 
The expected pay-off to A (i.e., expected loss to B) = 4 r + s.
 
The pay-off to A cannot exceed V. Hence we obtain the condition = 4 r + s V (II)
 
From (I) and (II) we have




Consider the negative elements in the objective function row. They are –1, -1. The absolute value are 1, 1. There is a tie between these coefficients. To resolve the tie, we select the variable x. We take the new basic variable as x. Consider the ratio of b-value to x-value. We have the following ratios:




Problem 3

 
Solve the following game by simplex method for LPP



So, Maximum of {Row minima}   Minimum of {Column maxima}.
 
Therefore the given game has no saddle point.  It is a mixed game. Let us convert the given game into a LPP.
 

Problem formulation

 
Let V denote the value of the game. Let the probability that the player B will use his first strategy be r and second strategy be s. Let V denote the value of the game.
 

When A follows his first strategy

 

 

The expected payoff to A (i.e., the expected loss to B) =

- 48 r + 2 s.

This pay-off cannot exceed V. So we have =- 48 r + 2 s

V

(1)’


When A follows his second strategy
 
The expected pay-off to A (i.e., expected loss to B) = 6 r - 4 s. 

This cannot exceed V. Hence we obtain the condition = 6 r - 4 s V (2)’




Consider the negative elements in the objective function row. They are –1, -1. The absolute value are 1, 1. There is a tie between these coefficients. To resolve the tie, we select the variable x. We take the new basic variable as x. Consider the ratio of b-value to x-value. We have the following ratios:





Problem 4

 
Transform the following game into an LPP
 
s = 9/10

Solution
 
We have to determine the optimal strategy for player B. Using the entries of the given matrix, we obtain the inequalities


Tags : Operations Management - Game Theory, Goal Programming & Queuing Theory
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