Find Solution of game theory problem using linear programming method
| Player A\Player B | B1 | B2 | B3 |
| A1 | 1 | -1 | 3 |
| A2 | 3 | 5 | -3 |
| A3 | 6 | 2 | -2 |
Solution:1. Saddle point testing
Players
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | |
| Player `A` | `A_1` | | 1 | -1 | 3 | |
| `A_2` | | 3 | 5 | -3 | |
| `A_3` | | 6 | 2 | -2 | |
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 | [-1] | (3) | | `[-1]` |
| `A_2` | | 3 | 5 | -3 | | `-3` |
| `A_3` | | 6 | 2 | -2 | | `-2` |
| Column
Maximum | | `6` | `5` | `(3)` | | |
Select minimum from the maximum of columns
Column MiniMax = (3)
Select maximum from the minimum of rows
Row MaxiMin = [-1]
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 | -1 | 3 | |
| `A_2` | | 3 | 5 | -3 | |
| `A_3` | | 6 | 2 | -2 | |
Also, no course of action dominates the other
So the value of the game lies between -1 and 3
It is possible that the value of game may be negative or zero.
Thus, a constant k is added to all the elements of pay-off matrix.
Let k = 3, then the given pay-off matrix becomes:
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | |
| Player `A` | `A_1` | | 4 | 2 | 6 | |
| `A_2` | | 6 | 8 | 0 | |
| `A_3` | | 9 | 5 | 1 | |
Let V = value of the game
`p_1,p_2,p_3` = probabilities of selecting strategies `A_1,A_2,A_3` respectively.
`q_1,q_2,q_3` = probabilities of selecting strategies `B_1,B_2,B_3` respectively.
| | | Player `B` | | |
| | | `B_1` | `B_2` | `B_3` | | Probability |
| Player `A` | `A_1` | | 4 | 2 | 6 | | `p_1` |
| `A_2` | | 6 | 8 | 0 | | `p_2` |
| `A_3` | | 9 | 5 | 1 | | `p_3` |
| Probability | | `q_1` | `q_2` | `q_3` | | |
player A's objective is to maximize the expected gains, which can be achieved by maximizing V, i.e., it might gain more than V if company B adopts a poor strategy.
The expected gain for player A will be as follows
`4p_1+6p_2+9p_3>=V`
`2p_1+8p_2+5p_3>=V`
`6p_1+0p_2+p_3>=V`
Dividing the above constraints by V, we get
`4(p_1/V)+6(p_2/V)+9(p_3/V)>=1`
`2(p_1/V)+8(p_2/V)+5(p_3/V)>=1`
`6(p_1/V)+0(p_2/V)+(p_3/V)>=1`
To simplify the problem, we put
`p_1/V=x_1,p_2/V=x_2,p_3/V=x_3`
In order to maximize V, player A can
Minimize `Z_p=1/V = x_1+x_2+x_3`
subject to
`4x_1+6x_2+9x_3>=1`
`2x_1+8x_2+5x_3>=1`
`6x_1+0x_2+x_3>=1`
and `x_1,x_2,x_3>=0`
player B's objective is to minimize its expected losses, which can be reduced by minimizing V, i.e., player A adopts a poor strategy.
The expected loss for player B will be as follows
`4q_1+2q_2+6q_3<=V`
`6q_1+8q_2+0q_3<=V`
`9q_1+5q_2+q_3<=V`
Dividing the above constraints by V, we get
`4(q_1/V)+2(q_2/V)+6(q_3/V)<=1`
`6(q_1/V)+8(q_2/V)+0(q_3/V)<=1`
`9(q_1/V)+5(q_2/V)+(q_3/V)<=1`
To simplify the problem, we put
`q_1/V=y_1,q_2/V=y_2,q_3/V=y_3`
In order to minimize V, player B can
Maximize `Z_q=1/V = y_1+y_2+y_3`
subject to
`4y_1+2y_2+6y_3<=1`
`6y_1+8y_2+0y_3<=1`
`9y_1+5y_2+y_3<=1`
and `y_1,y_2,y_3>=0`
Now, solve this problem using simplex method.
Problem is | Max `Z_q` | `=` | `` | `` | `y_1` | ` + ` | `` | `y_2` | ` + ` | `` | `y_3` |
|
| subject to |
| `` | `4` | `y_1` | ` + ` | `2` | `y_2` | ` + ` | `6` | `y_3` | ≤ | `1` | | `` | `6` | `y_1` | ` + ` | `8` | `y_2` | | | | ≤ | `1` | | `` | `9` | `y_1` | ` + ` | `5` | `y_2` | ` + ` | `` | `y_3` | ≤ | `1` |
|
| and `y_1,y_2,y_3 >= 0; ` |
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate
1. As the constraint-1 is of type '`<=`' we should add slack variable `S_1`
2. As the constraint-2 is of type '`<=`' we should add slack variable `S_2`
3. As the constraint-3 is of type '`<=`' we should add slack variable `S_3`
After introducing slack variables| Max `Z_q` | `=` | `` | `` | `y_1` | ` + ` | `` | `y_2` | ` + ` | `` | `y_3` | ` + ` | `0` | `S_1` | ` + ` | `0` | `S_2` | ` + ` | `0` | `S_3` |
|
| subject to |
| `` | `4` | `y_1` | ` + ` | `2` | `y_2` | ` + ` | `6` | `y_3` | ` + ` | `` | `S_1` | | | | | | | = | `1` | | `` | `6` | `y_1` | ` + ` | `8` | `y_2` | | | | | | | ` + ` | `` | `S_2` | | | | = | `1` | | `` | `9` | `y_1` | ` + ` | `5` | `y_2` | ` + ` | `` | `y_3` | | | | | | | ` + ` | `` | `S_3` | = | `1` |
|
| and `y_1,y_2,y_3,S_1,S_2,S_3 >= 0` |
| Iteration-1 | | `C_j` | `1` | `1` | `1` | `0` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `y_1` Entering variable | `y_2` | `y_3` | `S_1` | `S_2` | `S_3` | MinRatio
`(X_B)/(y_1)` |
| `S_1` | `0` | `1` | `4` | `2` | `6` | `1` | `0` | `0` | `(1)/(4)=0.25` |
| `S_2` | `0` | `1` | `6` | `8` | `0` | `0` | `1` | `0` | `(1)/(6)=0.1667` |
| `S_3` Leaving variable | `0` | `1` | `(9)` (pivot element) | `5` | `1` | `0` | `0` | `1` | `(1)/(9)=0.1111``->` |
`Z_q=0` `0=`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `0` `0=0xx4+0xx6+0xx9`
`Z_j=sum C_B y_1` | `0` `0=0xx2+0xx8+0xx5`
`Z_j=sum C_B y_2` | `0` `0=0xx6+0xx0+0xx1`
`Z_j=sum C_B y_3` | `0` `0=0xx1+0xx0+0xx0`
`Z_j=sum C_B S_1` | `0` `0=0xx0+0xx1+0xx0`
`Z_j=sum C_B S_2` | `0` `0=0xx0+0xx0+0xx1`
`Z_j=sum C_B S_3` | |
| | | `C_j-Z_j` | `1` `1=1-0``uarr` | `1` `1=1-0` | `1` `1=1-0` | `0` `0=0-0` | `0` `0=0-0` | `0` `0=0-0` | |
Positive maximum `C_j-Z_j` is `1` and its column index is `1`. So,
the entering variable is `y_1`.
Minimum ratio is `0.1111` and its row index is `3`. So,
the leaving basis variable is `S_3`.
`:.`
The pivot element is `9`.
Entering `=y_1`, Departing `=S_3`, Key Element `=9`
`R_3`(new)`= R_3`(old)`-: 9`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new)
`R_2`(new)`= R_2`(old)`- 6 R_3`(new)
| Iteration-2 | | `C_j` | `1` | `1` | `1` | `0` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `y_1` | `y_2` | `y_3` Entering variable | `S_1` | `S_2` | `S_3` | MinRatio
`(X_B)/(y_3)` |
| `S_1` Leaving variable | `0` | `5/9` `5/9=1-4xx1/9`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `0` `0=4-4xx1`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `-2/9` `-2/9=2-4xx5/9`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `(50/9)` `50/9=6-4xx1/9` (pivot element)
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `1` `1=1-4xx0`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `0` `0=0-4xx0`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `-4/9` `-4/9=0-4xx1/9`
`R_1`(new)`= R_1`(old)`- 4 R_3`(new) | `(5/9)/(50/9)=0.1``->` |
| `S_2` | `0` | `1/3` `1/3=1-6xx1/9`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `0` `0=6-6xx1`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `14/3` `14/3=8-6xx5/9`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `-2/3` `-2/3=0-6xx1/9`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `0` `0=0-6xx0`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `1` `1=1-6xx0`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | `-2/3` `-2/3=0-6xx1/9`
`R_2`(new)`= R_2`(old)`- 6 R_3`(new) | --- |
| `y_1` | `1` | `1/9` `1/9=1-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `1` `1=9-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `5/9` `5/9=5-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `1/9` `1/9=1-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `0` `0=0-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `0` `0=0-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `1/9` `1/9=1-:9`
`R_3`(new)`= R_3`(old)`-: 9` | `(1/9)/(1/9)=1` |
`Z_q=1/9` `1/9=1xx1/9`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `1` `1=0xx0+0xx0+1xx1`
`Z_j=sum C_B y_1` | `5/9` `5/9=0xx(-2/9)+0xx14/3+1xx5/9`
`Z_j=sum C_B y_2` | `1/9` `1/9=0xx50/9+0xx(-2/3)+1xx1/9`
`Z_j=sum C_B y_3` | `0` `0=0xx1+0xx0+1xx0`
`Z_j=sum C_B S_1` | `0` `0=0xx0+0xx1+1xx0`
`Z_j=sum C_B S_2` | `1/9` `1/9=0xx(-4/9)+0xx(-2/3)+1xx1/9`
`Z_j=sum C_B S_3` | |
| | | `C_j-Z_j` | `0` `0=1-1` | `4/9` `4/9=1-(5/9)` | `8/9` `8/9=1-(1/9)``uarr` | `0` `0=0-0` | `0` `0=0-0` | `-1/9` `-1/9=0-(1/9)` | |
Positive maximum `C_j-Z_j` is `8/9` and its column index is `3`. So,
the entering variable is `y_3`.
Minimum ratio is `0.1` and its row index is `1`. So,
the leaving basis variable is `S_1`.
`:.`
The pivot element is `50/9`.
Entering `=y_3`, Departing `=S_1`, Key Element `=50/9`
`R_1`(new)`= R_1`(old)`xx9/50`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new)
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new)
| Iteration-3 | | `C_j` | `1` | `1` | `1` | `0` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `y_1` | `y_2` Entering variable | `y_3` | `S_1` | `S_2` | `S_3` | MinRatio
`(X_B)/(y_2)` |
| `y_3` | `1` | `1/10` `1/10=5/9xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `0` `0=0xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `-1/25` `-1/25=(-2/9)xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `1` `1=50/9xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `9/50` `9/50=1xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `0` `0=0xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | `-2/25` `-2/25=(-4/9)xx9/50`
`R_1`(new)`= R_1`(old)`xx9/50` | --- |
| `S_2` Leaving variable | `0` | `2/5` `2/5=1/3+2/3xx1/10`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `0` `0=0+2/3xx0`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `(116/25)` `116/25=14/3+2/3xx(-1/25)` (pivot element)
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `0` `0=(-2/3)+2/3xx1`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `3/25` `3/25=0+2/3xx9/50`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `1` `1=1+2/3xx0`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `-18/25` `-18/25=(-2/3)+2/3xx(-2/25)`
`R_2`(new)`= R_2`(old)`+ 2/3 R_1`(new) | `(2/5)/(116/25)=0.0862``->` |
| `y_1` | `1` | `1/10` `1/10=1/9-1/9xx1/10`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `1` `1=1-1/9xx0`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `14/25` `14/25=5/9-1/9xx(-1/25)`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `0` `0=1/9-1/9xx1`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `-1/50` `-1/50=0-1/9xx9/50`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `0` `0=0-1/9xx0`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `3/25` `3/25=1/9-1/9xx(-2/25)`
`R_3`(new)`= R_3`(old)`- 1/9 R_1`(new) | `(1/10)/(14/25)=0.1786` |
`Z_q=1/5` `1/5=1xx1/10+1xx1/10`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `1` `1=1xx0+0xx0+1xx1`
`Z_j=sum C_B y_1` | `13/25` `13/25=1xx(-1/25)+0xx116/25+1xx14/25`
`Z_j=sum C_B y_2` | `1` `1=1xx1+0xx0+1xx0`
`Z_j=sum C_B y_3` | `4/25` `4/25=1xx9/50+0xx3/25+1xx(-1/50)`
`Z_j=sum C_B S_1` | `0` `0=1xx0+0xx1+1xx0`
`Z_j=sum C_B S_2` | `1/25` `1/25=1xx(-2/25)+0xx(-18/25)+1xx3/25`
`Z_j=sum C_B S_3` | |
| | | `C_j-Z_j` | `0` `0=1-1` | `12/25` `12/25=1-(13/25)``uarr` | `0` `0=1-1` | `-4/25` `-4/25=0-(4/25)` | `0` `0=0-0` | `-1/25` `-1/25=0-(1/25)` | |
Positive maximum `C_j-Z_j` is `12/25` and its column index is `2`. So,
the entering variable is `y_2`.
Minimum ratio is `0.0862` and its row index is `2`. So,
the leaving basis variable is `S_2`.
`:.`
The pivot element is `116/25`.
Entering `=y_2`, Departing `=S_2`, Key Element `=116/25`
`R_2`(new)`= R_2`(old)`xx25/116`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new)
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new)
| Iteration-4 | | `C_j` | `1` | `1` | `1` | `0` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `y_1` | `y_2` | `y_3` | `S_1` | `S_2` | `S_3` Entering variable | MinRatio
`(X_B)/(S_3)` |
| `y_3` | `1` | `3/29` `3/29=1/10+1/25xx5/58`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `0` `0=0+1/25xx0`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `0` `0=(-1/25)+1/25xx1`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `1` `1=1+1/25xx0`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `21/116` `21/116=9/50+1/25xx3/116`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `1/116` `1/116=0+1/25xx25/116`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | `-5/58` `-5/58=(-2/25)+1/25xx(-9/58)`
`R_1`(new)`= R_1`(old)`+ 1/25 R_2`(new) | --- |
| `y_2` | `1` | `5/58` `5/58=2/5xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `0` `0=0xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `1` `1=116/25xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `0` `0=0xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `3/116` `3/116=3/25xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `25/116` `25/116=1xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | `-9/58` `-9/58=(-18/25)xx25/116`
`R_2`(new)`= R_2`(old)`xx25/116` | --- |
| `y_1` Leaving variable | `1` | `3/58` `3/58=1/10-14/25xx5/58`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `1` `1=1-14/25xx0`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `0` `0=14/25-14/25xx1`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `0` `0=0-14/25xx0`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `-1/29` `-1/29=(-1/50)-14/25xx3/116`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `-7/58` `-7/58=0-14/25xx25/116`
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `(6/29)` `6/29=3/25-14/25xx(-9/58)` (pivot element)
`R_3`(new)`= R_3`(old)`- 14/25 R_2`(new) | `(3/58)/(6/29)=0.25``->` |
`Z_q=7/29` `7/29=1xx3/29+1xx5/58+1xx3/58`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `1` `1=1xx0+1xx0+1xx1`
`Z_j=sum C_B y_1` | `1` `1=1xx0+1xx1+1xx0`
`Z_j=sum C_B y_2` | `1` `1=1xx1+1xx0+1xx0`
`Z_j=sum C_B y_3` | `5/29` `5/29=1xx21/116+1xx3/116+1xx(-1/29)`
`Z_j=sum C_B S_1` | `3/29` `3/29=1xx1/116+1xx25/116+1xx(-7/58)`
`Z_j=sum C_B S_2` | `-1/29` `-1/29=1xx(-5/58)+1xx(-9/58)+1xx6/29`
`Z_j=sum C_B S_3` | |
| | | `C_j-Z_j` | `0` `0=1-1` | `0` `0=1-1` | `0` `0=1-1` | `-5/29` `-5/29=0-(5/29)` | `-3/29` `-3/29=0-(3/29)` | `1/29` `1/29=0-(-1/29)``uarr` | |
Positive maximum `C_j-Z_j` is `1/29` and its column index is `6`. So,
the entering variable is `S_3`.
Minimum ratio is `0.25` and its row index is `3`. So,
the leaving basis variable is `y_1`.
`:.`
The pivot element is `6/29`.
Entering `=S_3`, Departing `=y_1`, Key Element `=6/29`
`R_3`(new)`= R_3`(old)`xx29/6`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new)
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new)
| Iteration-5 | | `C_j` | `1` | `1` | `1` | `0` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `y_1` | `y_2` | `y_3` | `S_1` | `S_2` | `S_3` | MinRatio |
| `y_3` | `1` | `1/8` `1/8=3/29+5/58xx1/4`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `5/12` `5/12=0+5/58xx29/6`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `0` `0=0+5/58xx0`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `1` `1=1+5/58xx0`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `1/6` `1/6=21/116+5/58xx(-1/6)`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `-1/24` `-1/24=1/116+5/58xx(-7/12)`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | `0` `0=(-5/58)+5/58xx1`
`R_1`(new)`= R_1`(old)`+ 5/58 R_3`(new) | |
| `y_2` | `1` | `1/8` `1/8=5/58+9/58xx1/4`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `3/4` `3/4=0+9/58xx29/6`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `1` `1=1+9/58xx0`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `0` `0=0+9/58xx0`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `0` `0=3/116+9/58xx(-1/6)`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `1/8` `1/8=25/116+9/58xx(-7/12)`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | `0` `0=(-9/58)+9/58xx1`
`R_2`(new)`= R_2`(old)`+ 9/58 R_3`(new) | |
| `S_3` | `0` | `1/4` `1/4=3/58xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `29/6` `29/6=1xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `0` `0=0xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `0` `0=0xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `-1/6` `-1/6=(-1/29)xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `-7/12` `-7/12=(-7/58)xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | `1` `1=6/29xx29/6`
`R_3`(new)`= R_3`(old)`xx29/6` | |
`Z_q=1/4` `1/4=1xx1/8+1xx1/8`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `7/6` `7/6=1xx5/12+1xx3/4+0xx29/6`
`Z_j=sum C_B y_1` | `1` `1=1xx0+1xx1+0xx0`
`Z_j=sum C_B y_2` | `1` `1=1xx1+1xx0+0xx0`
`Z_j=sum C_B y_3` | `1/6` `1/6=1xx1/6+1xx0+0xx(-1/6)`
`Z_j=sum C_B S_1` | `1/12` `1/12=1xx(-1/24)+1xx1/8+0xx(-7/12)`
`Z_j=sum C_B S_2` | `0` `0=1xx0+1xx0+0xx1`
`Z_j=sum C_B S_3` | |
| | | `C_j-Z_j` | `-1/6` `-1/6=1-(7/6)` | `0` `0=1-1` | `0` `0=1-1` | `-1/6` `-1/6=0-(1/6)` | `-1/12` `-1/12=0-(1/12)` | `0` `0=0-0` | |
Since all `C_j-Z_j <= 0`
Hence, optimal solution is arrived with value of variables as :
`y_1=0,y_2=1/8,y_3=1/8`
Max `Z_q = 1/4`
`:. Z_q=1/V=1/4`
`:. V=4`
player B's optimal strategy
`q_1=V xx y_1=4 xx 0=0`
`q_2=V xx y_2=4 xx 1/8=1/2`
`q_3=V xx y_3=4 xx 1/8=1/2`
Hence, player B's `(B_1,B_2,B_3)` optimal strategy is `(0,1/2,1/2)`.
player A's optimal strategy
The values for `x_1,x_2,x_3` can be obtained from the `z_j-c_j` row of final simplex table
`x_1=1/6,x_2=1/12,x_3=0`
`p_1=V xx x_1=4 xx 1/6=2/3`
`p_2=V xx x_2=4 xx 1/12=1/3`
`p_3=V xx x_3=4 xx 0=0`
Hence, player A's `(A_1,A_2,A_3)` optimal strategy is `(2/3,1/3,0)`.
So, finally player B's `(B_1,B_2,B_3)` optimal strategy is `(0,1/2,1/2)`.
and player A's `(A_1,A_2,A_3)` optimal strategy is `(2/3,1/3,0)`.
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