1. Method & Example-1
Method
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graphical method Steps (Rule)
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Step-1:
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This method can only be used in games with no saddle point, and having a pay-off matrix of type n`xx`2 or 2`xx`n.
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Step-2:
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The example is used to explain the procedure.
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Example-1
Find Solution of game theory problem using graphical method
| Player A\Player B | B1 | B2 | | A1 | 1 | -3 | | A2 | 3 | 5 | | A3 | -1 | 6 | | A4 | 4 | 1 | | A5 | 2 | 2 | | A6 | -5 | 0 |
Solution:
1. Saddle point testing
Players
| | | Player `B` | | | | | | `B_1` | `B_2` | | | | Player `A` | `A_1` | | 1 | -3 | | | `A_2` | | 3 | 5 | | | `A_3` | | -1 | 6 | | | `A_4` | | 4 | 1 | | | `A_5` | | 2 | 2 | | | `A_6` | | -5 | 0 | |
We apply the maximin (minimax) principle to analyze the game.
| | | Player `B` | | | | | | `B_1` | `B_2` | | Row
Minimum | | Player `A` | `A_1` | | 1 | -3 | | `-3` | | `A_2` | | [3] | 5 | | `[3]` | | `A_3` | | -1 | 6 | | `-1` | | `A_4` | | (4) | 1 | | `1` | | `A_5` | | 2 | 2 | | `2` | | `A_6` | | -5 | 0 | | `-5` | | Column
Maximum | | `(4)` | `6` | | |
Select minimum from the maximum of columns
Column MiniMax = (4)
Select maximum from the minimum of rows
Row MaxiMin = [3]
Here, Column MiniMax `!=` Row MaxiMin
`:.` This game has no saddle point.
Solution using graphical method
First, we draw two parallel lines 1 unit distance apart and mark a scale on each.
The two parallel lines represent strategies of player `B`.
If player `A` selects strategy `A_1`, player `B` can win 1 or -3 units depending on B's selection of strategies.
The value 1 is plotted along the vertical axis under strategy `B_1` and the value -3 is plotted along the vertical axis under strategy `B_2`.
A straight line joining the two points is then drawn.
Similarly, we can plot strategies `A_2,A_3,A_4,A_5,A_6` also. The problem is graphed in the following figure.
The lowest point `V` in the shaded region indicates the value of game. From the above figure, the value of the game is `3.4` units.
1. The point of optimal solution occurs at the intersection of two lines
`E_2 = 3p_1 +5p_2`
`E_4 = 4p_1 +p_2`
Comparing the above two equations, we have
`3p_1 +5p_2 = 4p_1 +p_2`
Substituting `p_2 = 1 - p_1`
`3p_1 +5(1 - p_1) = 4p_1 +(1 - p_1)`
Solving `p_1=4/5`
`p_2=1-p_1=1-4/5=1/5`
Substituting the values of `p_1` and `p_2` in equation `E_2`
`V = 3(4/5) +5(1/5)=17/5`
2. The point of optimal solution occurs at the intersection of two lines
`L_2 = 3q_1 +4q_2`
`L_4 = 5q_1 +q_2`
Comparing the above two equations, we have
`3q_1 +4q_2 = 5q_1 +q_2`
Substituting `q_2 = 1 - q_1`
`3q_1 +4(1 - q_1) = 5q_1 +(1 - q_1)`
Solving `q_1=3/5`
`q_2=1-q_1=1-3/5=2/5`
Substituting the values of `q_1` and `q_2` in equation `L_2`
`V = 3(3/5) +4(2/5)=17/5`
This material is intended as a summary. Use your textbook for detail explanation.
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