Method

graphical method Steps (Rule)
Step-1:	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.
Step-2:	The example is used to explain the procedure.

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Example-1
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.

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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.

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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`

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