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Problem

Let the payoff matrix of a 2 x 2 game be characterized by the matrix

$\mathbf{A} = \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}$.

All entries are positive real numbers.

Part 1: Use the simplex method to show that the value of the matrix game is $v = \frac{\det{(A)}}{S}$, where $S = (a+d) - (b+c)$ and $\det{(A)} = ad -bc$.

Part 2: Determine the strategy of the row player and the column player.

Solution

Part 1

Construct the initial tableau.

Tableau 1

x y s1 s2 Z
s1 $a$ $b$ $1$ $0$ $0$ $1$
s2 $c$ $d$ $0$ $1$ $0$ $1$
Z $-1$ $-1$ $0$ $0$ $1$ $0$

Leaving variable is $x$.

Entering variable is $s_1$.

Divide row 1 by $a$: $\frac{1}{a} R_1$.

x y s1 s2 Z
s1 $1$ $b/a$ $1/a$ $0$ $0$ $1/a$
s2 $c$ $d$ $0$ $1$ $0$ $1$
Z $-1$ $-1$ $0$ $0$ $1$ $0$

Perform the following row operations to get tableau 2: $R_3 + R_1$ and $R_2 - c R_1$.

Tableau 2

x y s1 s2 Z
x $1$ $\frac{b}{a}$ $\frac{1}{a}$ $0$ $0$ $\frac{1}{a}$
s2 $0$ $\frac{ad-bc}{a}$ $\frac{-c}{a}$ $1$ $0$ $1 - \frac{c}{a}$
Z $0$ $\frac{b}{a} - 1$ $\frac{1}{a}$ $0$ $1$ $\frac{1}{a}$

Leaving variable is $y$.

Entering variable is $s_2$.

Divide row 1 by $\frac{ad-bc}{a}$: $\left(\frac{a}{ad - bc} \right) R_1$.

x y s1 s2 Z
x $1$ $\frac{b}{a}$ $\frac{1}{a}$ $0$ $0$ $\frac{1}{a}$
s2 $0$ $1$ $\frac{-c}{ad-bc}$ $\frac{a}{ad-bc}$ $0$ $\frac{a-c}{ad-bc}$
Z $0$ $\frac{b}{a} - 1$ $\frac{1}{a}$ $0$ $1$ $\frac{1}{a}$

Perform the following row operations to get tableau 3: $R_1 - \frac{b}{a} R_2$ and $R_3 - \left(\frac{b}{a} - 1 \right)R2$.

x y s1 s2 Z
x $1$ $0$ $\frac{ad}{ad-bc}$ $\frac{-b}{ad-bc}$ $0$ $\frac{d-b}{ad-bc}$
y $0$ $1$ $\frac{-c}{ad-bc}$ $\frac{a}{ad-bc}$ $0$ $\frac{a-c}{ad-bc}$
Z $0$ $0$ $\frac{d-c}{ad-bc}$ $\frac{a-b}{ad-bc}$ $1$ $\frac{(a+d)-(b+c)}{ad-bc}$

The value of the game is the reciprocal of $Z$ (the entry in the lower right corner). Therefore,

$v = \frac{ad - bc}{(a+d) - (b+c)}$
$v = \frac{\det{(A)}}{S}$.

Part 2

To obtain the strategy for the row player, take the entries of the bottom row under $s1$ and $s2$ and divide each by the value of $Z$.

$R = \begin{bmatrix} r_1 & r_2 \\ \end{bmatrix}$

where $r_1 = \frac{d-c}{S}$ and $r_2 = \frac{a-b}{S}$.

To obtain the strategy for the column player, take the entries of the right-most column on row $x$ and $y$ and divide each by the value of $Z$.

$C = \begin{bmatrix} c_1 \\ c_2 \\ \end{bmatrix}$

where $c_1 = \frac{d-b}{S}$ and $c_2 = \frac{a-c}{S}$.

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