Degeneracy example-1 (Tie for leaving basic variable)
During solving LP problem, a situation may arise in which there is a tie between, 2 or more basic variables for leaving the basis. (means minimum ratios are same).
It is called degeneracy and to resolve this we can select any of them arbitrarily.
But if artificial variable is present then it must be removed first.
Example
Find solution using Simplex(BigM) method
MAX Z = 3x1 + 9x2
subject to
x1 + 4x2 <= 8
x1 + 2x2 <= 4
and x1,x2 >= 0
Solution:
Problem is
| Max `Z` | `=` | `` | `3` | `x_1` | ` + ` | `9` | `x_2` |
|
| subject to |
| `` | `` | `x_1` | ` + ` | `4` | `x_2` | ≤ | `8` | | `` | `` | `x_1` | ` + ` | `2` | `x_2` | ≤ | `4` |
|
| and `x_1,x_2 >= 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`
After introducing slack variables
| Max `Z` | `=` | `` | `3` | `x_1` | ` + ` | `9` | `x_2` | ` + ` | `0` | `S_1` | ` + ` | `0` | `S_2` |
|
| subject to |
| `` | `` | `x_1` | ` + ` | `4` | `x_2` | ` + ` | `` | `S_1` | | | | = | `8` | | `` | `` | `x_1` | ` + ` | `2` | `x_2` | | | | ` + ` | `` | `S_2` | = | `4` |
|
| and `x_1,x_2,S_1,S_2 >= 0` |
| Iteration-1 | | `C_j` | `3` | `9` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `x_1` | `x_2` Entering variable | `S_1` | `S_2` | MinRatio
`(X_B)/(x_2)` |
| `S_1` | `0` | `8` | `1` | `4` | `1` | `0` | `(8)/(4)=2` |
| `S_2` Leaving variable | `0` | `4` | `1` | `(2)` (pivot element) | `0` | `1` | `(4)/(2)=2``->` |
`Z=0` `0=`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `0` `0=0xx1+0xx1`
`Z_j=sum C_B x_1` | `0` `0=0xx4+0xx2`
`Z_j=sum C_B x_2` | `0` `0=0xx1+0xx0`
`Z_j=sum C_B S_1` | `0` `0=0xx0+0xx1`
`Z_j=sum C_B S_2` | |
| | | `C_j-Z_j` | `3` `3=3-0` | `9` `9=9-0``uarr` | `0` `0=0-0` | `0` `0=0-0` | |
Positive maximum `C_j-Z_j` is `9` and its column index is `2`. So, the entering variable is `x_2`.
Minimum ratio is `2` and its row index is `2`. So, the leaving basis variable is `S_2`.
`:.` The pivot element is `2`.
Entering `=x_2`, Departing `=S_2`, Key Element `=2`
`R_2`(new)`= R_2`(old)`-: 2`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new)
| Iteration-2 | | `C_j` | `3` | `9` | `0` | `0` | |
| `B` | `C_B` | `X_B` | `x_1` | `x_2` | `S_1` | `S_2` | MinRatio |
| `S_1` | `0` | `0` `0=8-4xx2`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new) | `-1` `-1=1-4xx1/2`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new) | `0` `0=4-4xx1`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new) | `1` `1=1-4xx0`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new) | `-2` `-2=0-4xx1/2`
`R_1`(new)`= R_1`(old)`- 4 R_2`(new) | |
| `x_2` | `9` | `2` `2=4-:2`
`R_2`(new)`= R_2`(old)`-: 2` | `1/2` `1/2=1-:2`
`R_2`(new)`= R_2`(old)`-: 2` | `1` `1=2-:2`
`R_2`(new)`= R_2`(old)`-: 2` | `0` `0=0-:2`
`R_2`(new)`= R_2`(old)`-: 2` | `1/2` `1/2=1-:2`
`R_2`(new)`= R_2`(old)`-: 2` | |
`Z=18` `18=9xx2`
`Z_j=sum C_B X_B` | | `Z_j` `Z_j=sum C_B x_j` | `9/2` `9/2=0xx(-1)+9xx1/2`
`Z_j=sum C_B x_1` | `9` `9=0xx0+9xx1`
`Z_j=sum C_B x_2` | `0` `0=0xx1+9xx0`
`Z_j=sum C_B S_1` | `9/2` `9/2=0xx(-2)+9xx1/2`
`Z_j=sum C_B S_2` | |
| | | `C_j-Z_j` | `-3/2` `-3/2=3-(9/2)` | `0` `0=9-9` | `0` `0=0-0` | `-9/2` `-9/2=0-(9/2)` | |
Since all `C_j-Z_j <= 0`
Hence, optimal solution is arrived with value of variables as :
`x_1=0,x_2=2`
Max `Z = 18`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
