Question #173505

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


1
Expert's answer
2021-04-13T14:28:13-0400

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‘A’ adopts the probabilities xix_i and xiix_{ii} for strategies 1 and 2. for player ‘B’ the

mixed strategies with probabilities are yi,yii,yiiiy_i , y_{ii}, y_{iii} and yivy_{iv} respectively. At the optimal level with

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


xi+xii=11yi+yii+yiii+yiv=12xi+2xiiV33xi+xiiV43xi+4xiiV57xi6xiiV6yi+3yii3yiii+7yivV72yi+5yii+4yiii6yivV8x_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.


 xii=1xi∴\ x_{ii} = 1 – x_i

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

Equation (3) can be written as:

xi+2xiiVxi+2(1xi)V    V+xi29x_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:

3xi+5xiiV3xi+5(1xi)V    V+2xi5103x_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:

3xi+4(1xi)V47xiV    V+7xi411-3x_i + 4(1 – x_i) ≥ V\\ \therefore 4 - 7x_i≥ V\\ \implies V + 7x_i \leq 4------------------{11}

Equation (6) can be written as:

7xi6(1xi)Vxi6V    xiV6127x_i - 6(1 – x_i) ≥ V\\ \therefore x_i - 6 ≥ V\\ \implies x_i - V \leq 6------------------{12}


Player AsA’s objective is to maximize the value of V‘V’ and to find the combination of xix_i and xiix_{ii} which gives the maximum value.

The graph of xix_i versus VV can be drawn with the relationships in equations (8), (9), (10) and

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

plot the graph within 0 and 1 of xix_i .

Equation (8) gives,

When

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

Equation (9) gives,

When

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

Equation (10) gives,

When

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

Equation (11) gives,

When

xi=0; V=6xi=1; V=5x_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 xi=13x_i=\frac{1}{3} and xii=23.x_{ii}= \frac{2}{3}.








Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS