Infeasible solution
If there is no feasible area
(there is no any point that satisfy all constraints of the problem),
then this solution is called infeasible solution.
Example
Find solution using graphical method
MAX Z = 6X1 - 4X2
subject to
2X1 + 4X2 <= 4
4X1 + 8X2 >= 16
and X1,X2 >= 0
Solution:
Problem is
| MAX `Z_x` | `=` | `` | `6` | `X_1` | ` - ` | `4` | `X_2` |
|
| subject to |
| `` | `2` | `X_1` | ` + ` | `4` | `X_2` | ≤ | `4` | | `` | `4` | `X_1` | ` + ` | `8` | `X_2` | ≥ | `16` |
|
| and `X_1,X_2 >= 0; ` |
Hint to draw constraints
1. To draw constraint `color{red}{2X_1+4X_2<=4 ->(1)}`
Treat it as `color{red}{2X_1+4X_2=4}`
When `X_1=0` then `X_2=?`
`=>2(0)+4X_2=4`
`=>4X_2=4`
`=>X_2=(4)/(4)=1`
When `X_2=0` then `X_1=?`
`=>2X_1+4(0)=4`
`=>2X_1=4`
`=>X_1=(4)/(2)=2`
2. To draw constraint `color{green}{4X_1+8X_2>=16 ->(2)}`
Treat it as `color{green}{4X_1+8X_2=16}`
When `X_1=0` then `X_2=?`
`=>4(0)+8X_2=16`
`=>8X_2=16`
`=>X_2=(16)/(8)=2`
When `X_2=0` then `X_1=?`
`=>4X_1+8(0)=16`
`=>4X_1=16`
`=>X_1=(16)/(4)=4`
Problem has an infeasible solution.
This material is intended as a summary. Use your textbook for detail explanation.
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