Answer to Question #173505 in Operations Research for ANJU JAYACHANDRAN

Question

8(b) Using graphical method, solve the game whose pay-off matrix is given as: (4) 

 

 Player B

 I II III IV 

 I 1 3 − 3 7 

Player A 

 II 2 5 4 − 6

Answer 1

Solutions

Graphical method is applicable to only those games in which one of the players has two

strategies only. The advantage of this method is that is that it solves the problem relatively

faster.

The graphical method consists of two graphs.

(i) the pay-off (gains) available to player ‘A’ against his strategies and options.

(ii) the pay-off (losses) faced by player ‘B’ against his strategies and options.

The pay-off matrix is given below.

The problem has no saddle point.

∴ There are no pure strategies and mixed strategies are to be adopted.

Player $‘A’$ adopts the probabilities $x_i$ and $x_{ii}$ for strategies 1 and 2. for player ‘B’ the

mixed strategies with probabilities are $y_i , y_{ii}, y_{iii}$ and $y_{iv}$ respectively. At the optimal level with

the value of the game, as $V$ the following relationship can be established.

$$x_i + x_{ii} = 1----------------------{1}\\ y_i + y_{ii} + y_{iii}+y_{iv} = 1------------------{2}\\ x_i + 2x_{ii} ≥ V----------------------{3}\\ 3x_i + x_{ii} ≥ V----------------------{4}\\ -3x_i + 4x_{ii} ≥ V---------------------{5}\\ 7x_i - 6x_{ii} ≥ V----------------------{6}\\ y_i + 3y_{ii} -3 y_{iii}+ 7y_{iv} \leq V-----------------{7}\\ 2y_i + 5y_{ii} + 4y_{iii} -6 y_{iv} \leq V-----------------{8}\\$$

The above equations can be written in terms of the player having two strategies.

i.e. in terms of player A.

$$∴\ x_{ii} = 1 – x_i$$

Substituting the value of $x_{ii}$ in equations (3), (4), (5) and (6). we get:

Equation (3) can be written as:

$x_i + 2x_{ii} ≥ V\\ \therefore x_i + 2(1 – x_i)≥ V\\ \implies V + x_i \leq 2-------------------{9}$

Equation (4) can be written as:

$3x_i + 5x_{ii} ≥ V\\ \therefore 3x_i + 5(1 – x_i)≥ V\\ \implies V + 2x_i \leq 5------------------{10}$

Equation (5) can be written as:

$-3x_i + 4(1 – x_i) ≥ V\\ \therefore 4 - 7x_i≥ V\\ \implies V + 7x_i \leq 4------------------{11}$

Equation (6) can be written as:

$7x_i - 6(1 – x_i) ≥ V\\ \therefore x_i - 6 ≥ V\\ \implies x_i - V \leq 6------------------{12}$

Player $A’s$ objective is to maximize the value of $‘V’$ and to find the combination of $x_i$ and $x_{ii}$ which gives the maximum value.

The graph of $x_i$ versus $V$ can be drawn with the relationships in equations (8), (9), (10) and

(11) by plotting $x_i$ on x-axis and $‘V’$ on y-axis. The range of $x_i$ is between 0 and 1, and so we

plot the graph within 0 and 1 of $x_i$ .

Equation (8) gives,

When

$x_i = 0 ;\ V= 2\\ x_i = 1 ;\ V= 1\\$

Equation (9) gives,

When

$x_i = 0 ;\ V= 5\\ x_i = 1 ;\ V= 3\\$

Equation (10) gives,

When

$x_i = 0 ;\ V= 4\\ x_i = 1 ;\ V= -3\\$

Equation (11) gives,

When

$x_i = 0 ;\ V= -6\\ x_i = 1 ;\ V= -5\\$

Plotting these equations on the graph, as shown below the feasible region is given by the area below AGD.

The maximum value of ‘V’ in this region is given at point G. At this point $x_i=\frac{1}{3}$ and $x_{ii}= \frac{2}{3}.$