Answer to Question #203620 in Operations Research for Agar

Question

Solve the (4x3) game with pay off matrix.

A=

8 5 8

8 6 5

7 4 5

6 5 6

 At each stage, clearly explain the steps involved.

Answer 1

Question

Solve game with pay off matrix is 4 x 3.

$A=\begin{bmatrix} 8& 5& 8\\ 8 &6& 5\\ 7 &4& 5\\ 6 &5 &6\\ \end{bmatrix}$

Solution

 Dominance rule to reduce the size of the payoff matrix;

Using dominance property;

$player B$

$B_1$ $B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ A_3 \\ A_4\\ \end{matrix}\space\begin{bmatrix} 8& 5& 8\\ 8 &6& 5\\ 7 &4& 5\\ 6 &5 &6\\ \end{bmatrix}$

$row-4 ≤ row-1, so\space remove \space row-4, (A4≤A1:6≤8,5≤5,6≤8)$

$player B$

$B_1$ $B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ A_3 \\ \end{matrix}\space\begin{bmatrix} 8& 5& 8\\ 8 &6& 5\\ 7 &4& 5\\ \end{bmatrix}$

$row-3 ≤ row-2, so\space remove \space row-3, (A3≤A2:7≤8,4≤6,5≤5)$

$player B$

$B_1$ $B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 8& 5& 8\\ 8 &6& 5\\ \end{bmatrix}$

$column-1 ≥ column-3, so \space remove\space column-1. (B1≥B3:8≥8,8≥5)$

$player B$

$B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 5& 8\\ 6& 5\\ \end{bmatrix}$

now we solve last matrix

for the solution, we write difference of elements of row to front of other row

similarly we write difference of elements of column to below of other column

such that

$player B$

$B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 5& 8\\ 6& 5\\ \end{bmatrix}\begin{matrix} |6-5| \\ |5-8|\\ \end{matrix}$

|8-5| |5-6|

$player B$

$B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 5& 8\\ 6& 5\\ \end{bmatrix}\begin{matrix} 1\\ 3\\ \end{matrix}$

|3| |1|

$player B$

$B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 5& 8\\ 6& 5\\ \end{bmatrix}\begin{matrix} 1 & \frac{1}{1+3}\\ 3& \frac{3}{1+3}\\ \end{matrix}$

|3| |1|

$\frac{3}{3+1}$ $\frac{1}{3+1}$

$player B$

$B_2$ $B_3$

$player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 5& 8\\ 6& 5\\ \end{bmatrix}\begin{matrix} 1 & \frac{1}{4}\\ 3& \frac{3}{4}\\ \end{matrix}$

|3| |1|

$\frac{3}{4}$ $\frac{1}{4}$

Hence

strategy of A=$\{\frac{1}{4},\frac{3}{4},0,0\}$

strategy of B=$\{0,\frac{3}{4},\frac{1}{4}\}$

value of game =$(5\times \frac{1}{4})+(6\times \frac{3}{4})$

$=\frac{5}{4}+\frac{18}{4} =\frac{23}{4}$