FANDOM


Problem

Payoff matrix

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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