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6. Matrix method example ( Enter your problem )
  1. Method & Example-1
  2. Example-2
Other related methods
  1. Saddle point
  2. Dominance method
  3. Algebraic method
  4. Calculus method
  5. Arithmetic method
  6. Matrix method
  7. 2Xn Games
  8. Graphical method
  9. Linear programming method
  10. Bimatrix method

1. Method & Example-1



Method
Step-1: Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[p_1,p_2]]`
Step-2: Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]]) = [[q_1,q_2]]`
Step-3: Value of the game = (Player A's optimal strategies) `xx` (Payoff matrix P) `xx` (Player B's optimal strategies)
`V = [[p_1,p_2]] xx P xx [[q_1],[q_2]]`

Example-1
1. Find Solution of game theory problem using matrix method
Player A\Player BB1B2B3
A1172
A2627
A3516


Solution:
1. Saddle point testing
Players
Player `B`
`B_1``B_2``B_3`
Player `A``A_1` 1  7  2 
`A_2` 6  2  7 
`A_3` 5  1  6 


We apply the maximin (minimax) principle to analyze the game.

Player `B`
`B_1``B_2``B_3`Row
Minimum
Player `A``A_1` 1  7  2 `1`
`A_2` (6)  [2]  7 `[2]`
`A_3` 5  1  6 `1`
Column
Maximum
`(6)``7``7`


Select minimum from the maximum of columns
Column MiniMax = (6)

Select maximum from the minimum of rows
Row MaxiMin = [2]

Here, Column MiniMax `!=` Row MaxiMin

`:.` This game has no saddle point.



2. Dominance rule to reduce the size of the payoff matrix
Using dominance property
Player `B`
`B_1``B_2``B_3`
Player `A``A_1` 1  7  2 
`A_2` 6  2  7 
`A_3` 5  1  6 


row-3 `<=` row-2, so remove row-3

Player `B`
`B_1``B_2``B_3`
Player `A``A_1` 1  7  2 
`A_2` 6  2  7 


column-3 `>=` column-1, so remove column-3

Player `B`
`B_1``B_2`
Player `A``A_1` 1  7 
`A_2` 6  2 




reduced matrix
Player `B`
`B_1``B_2`
Player `A``A_1` 1  7 
`A_2` 6  2 


For this reduced matrix, calculate `P_(Adj)` and `P_(Cof)`

`P_(Adj) = [[2,-7],[-6,1]]`

and `P_(Cof) = [[2,-6],[-7,1]]`

Player A's optimal strategies `=([[1,1]] xx P_(Adj))/([[1,1]] xx P_(Adj) xx [[1],[1]])`

`=([[1,1]][[2,-7],[-6,1]])/([[1,1]][[2,-7],[-6,1]][[1],[1]])`

`=([[-4,-6]])/(-10)`

`=[[2/5,3/5]]`

`p_1=2/5` and `p_2=3/5`, where `p_1` and `p_2` represent the probabilities of player A's, using his strategies `A_1` and `A_2` respectively.


Similarly,
Player B's optimal strategies `=([[1,1]] xx P_(Cof))/([[1,1]] xx P_(Adj) xx [[1],[1]])`

`=([[1,1]][[2,-6],[-7,1]])/([[1,1]][[2,-7],[-6,1]][[1],[1]])`

`=([[-5,-5]])/(-10)`

`=[[1/2,1/2]]`

`q_1=1/2` and `q_2=1/2`, where `q_1` and `q_2` represent the probabilities of player B's, using his strategies `B_1` and `B_2` respectively.


Hence, Value of the game `V` = (Player A's optimal strategies) `xx` (Payoff matrix `P_(ij)`) `xx` (Player B's optimal strategies)

`V=[[2/5,3/5]][[1,7],[6,2]][[1/2],[1/2]]=4`


This material is intended as a summary. Use your textbook for detail explanation.
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