Unbounded solution
In maximization problem, if shaded area is open-ended. This means that the maximization is not possible and the LPP has no finite solution. Hence the solution of the given problem is unbounded.
Example
Find solution using graphical method
MAX Z = 5X1 + 4X2
subject to
X1 - 2X2 <= 1
X1 + 2X2 >= 3
and X1,X2 >= 0
Solution:
Problem is
| MAX `Z_x` | `=` | `` | `5` | `X_1` | ` + ` | `4` | `X_2` |
|
| subject to |
| `` | `` | `X_1` | ` - ` | `2` | `X_2` | ≤ | `1` | | `` | `` | `X_1` | ` + ` | `2` | `X_2` | ≥ | `3` |
|
| and `X_1,X_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{X_1-2X_2<=1 ->(1)}`
Treat it as `color{red}{X_1-2X_2=1}`
When `X_1=0` then `X_2=?`
`=>(0)-2X_2=1`
`=>-2X_2=1`
`=>X_2=(1)/(-2)=-0.5`
When `X_2=0` then `X_1=?`
`=>X_1-2(0)=1`
`=>X_1=1`
2. To draw constraint `color{green}{X_1+2X_2>=3 ->(2)}`
Treat it as `color{green}{X_1+2X_2=3}`
When `X_1=0` then `X_2=?`
`=>(0)+2X_2=3`
`=>2X_2=3`
`=>X_2=(3)/(2)=1.5`
When `X_2=0` then `X_1=?`
`=>X_1+2(0)=3`
`=>X_1=3`
The value of the objective function at each of these extreme points is as follows:
Extreme Point
Coordinates
(`X_1`,`X_2`) | Lines through Extreme Point | Objective function value
`Z=5X_1 + 4X_2` |
| `color{red}{A(0,1.5)}` | `color{green}{2->X_1+2X_2>=3}`
`color{black}{3->X_1>=0}` | `5(0)+4(1.5)=6` |
| `color{green}{B(2,0.5)}` | `color{red}{1->X_1-2X_2<=1}`
`color{green}{2->X_1+2X_2>=3}` | `5(2)+4(0.5)=12` |
Problem has an unbounded solution.
This material is intended as a summary. Use your textbook for detail explanation.
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