Answer to Question #203620 in Operations Research for Agar

Question #203620

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.



1
Expert's answer
2021-06-08T10:20:18-0400

Question

Solve game with pay off matrix is 4 x 3.

A=[858865745656]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;

playerBplayer B

B1B_1 B2B_2 B3B_3

playerAA1A2A3A4 [858865745656]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}


row4row1,so remove row4,(A4A1:68,55,68)row-4 ≤ row-1, so\space remove \space row-4, (A4≤A1:6≤8,5≤5,6≤8)



playerBplayer B

B1B_1 B2B_2 B3B_3

playerAA1A2A3 [858865745]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}


row3row2,so remove row3,(A3A2:78,46,55)row-3 ≤ row-2, so\space remove \space row-3, (A3≤A2:7≤8,4≤6,5≤5)



playerBplayer B

B1B_1 B2B_2 B3B_3

playerAA1A2 [858865]player A \begin{matrix} A_1 \\ A_2\\ \end{matrix}\space\begin{bmatrix} 8& 5& 8\\ 8 &6& 5\\ \end{bmatrix}


column1column3,so remove column1.(B1B3:88,85)column-1 ≥ column-3, so \space remove\space column-1. (B1≥B3:8≥8,8≥5)




playerBplayer B

B2B_2 B3B_3

playerAA1A2 [5865]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



playerBplayer B

B2B_2 B3B_3

playerAA1A2 [5865]6558player 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|




playerBplayer B

B2B_2 B3B_3

playerAA1A2 [5865]13player 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|


playerBplayer B

B2B_2 B3B_3

playerAA1A2 [5865]111+3331+3player 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|

33+1\frac{3}{3+1} 13+1\frac{1}{3+1}


playerBplayer B

B2B_2 B3B_3

playerAA1A2 [5865]114334player 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|

34\frac{3}{4} 14\frac{1}{4}

Hence

strategy of A={14,34,0,0}\{\frac{1}{4},\frac{3}{4},0,0\}

strategy of B={0,34,14}\{0,\frac{3}{4},\frac{1}{4}\}


value of game =(5×14)+(6×34)(5\times \frac{1}{4})+(6\times \frac{3}{4})

=54+184=234=\frac{5}{4}+\frac{18}{4} =\frac{23}{4}


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