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Home > Operation Research > Simplex method example
2. Simplex method example ( Enter your problem )
( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm (using `Z`-row method)
  3. Maximization example-1 (using `Z`-row method)
  4. Maximization example-2 (using `Z`-row method)
  5. Maximization example-3 (using `Z`-row method)
  6. BigM method Algorithm (using `Z`-row method)
  7. Minimization example-1 (using `Z`-row method)
  8. Minimization example-2 (using `Z`-row method)
  9. Minimization example-3 (using `Z`-row method)
  10. Degeneracy example-1 (Tie for leaving basic variable) (using `Z`-row method)
  11. Degeneracy example-2 (Tie first Artificial variable removed) (using `Z`-row method)
  12. Unrestricted variable example (using `Z`-row method)
  13. Multiple optimal solution example (using `Z`-row method)
  14. Infeasible solution example (using `Z`-row method)
  15. Unbounded solution example (using `Z`-row method)
  16. Algorithm (using `Z_j-C_j` method)
  17. Maximization example-1 (using `Z_j-C_j` method)
  18. Maximization example-2 (using `Z_j-C_j` method)
  19. Maximization example-3 (using `Z_j-C_j` method)
  20. BigM method Algorithm (using `Z_j-C_j` method)
  21. Minimization example-1 (using `Z_j-C_j` method)
  22. Minimization example-2 (using `Z_j-C_j` method)
  23. Minimization example-3 (using `Z_j-C_j` method)
  24. Degeneracy example-1 (Tie for leaving basic variable) (using `Z_j-C_j` method)
  25. Degeneracy example-2 (Tie first Artificial variable removed) (using `Z_j-C_j` method)
  26. Unrestricted variable example (using `Z_j-C_j` method)
  27. Multiple optimal solution example (using `Z_j-C_j` method)
  28. Infeasible solution example (using `Z_j-C_j` method)
  29. Unbounded solution example (using `Z_j-C_j` method)
  30. Algorithm (using `C_j-Z_j`method)
  31. Maximization example-1 (using `C_j-Z_j`method)
  32. Maximization example-2 (using `C_j-Z_j`method)
  33. Maximization example-3 (using `C_j-Z_j`method)
  34. BigM method Algorithm (using `C_j-Z_j`method)
  35. Minimization example-1 (using `C_j-Z_j`method)
  36. Minimization example-2 (using `C_j-Z_j`method)
  37. Minimization example-3 (using `C_j-Z_j`method)
  38. Degeneracy example-1 (Tie for leaving basic variable) (using `C_j-Z_j`method)
  39. Degeneracy example-2 (Tie first Artificial variable removed) (using `C_j-Z_j`method)
  40. Unrestricted variable example (using `C_j-Z_j`method)
  41. Multiple optimal solution example (using `C_j-Z_j`method)
  42. Infeasible solution example (using `C_j-Z_j`method)
  43. Unbounded solution example (using `C_j-Z_j`method)
 		Other related methods
  1. Formulate linear programming model
  2. Graphical method
  3. Simplex method (BigM method)
  4. Two-Phase method
  5. Primal to dual conversion
  6. Dual simplex method
  7. Integer simplex method
  8. Branch and Bound method
  9. 0-1 Integer programming problem
  10. Revised Simplex method
12. Unrestricted variable example (using `Z`-row method)
(Previous example)		14. Infeasible solution example (using `Z`-row method)
(Next example)

13. Multiple optimal solution example (using `Z`-row method)


Multiple optimal solution
In the final simplex table when all `z_j` imply optimal solution (for maximization all `z_j >= 0` and for minimization all `z_j <= 0`)
but if `z_j = 0` for some non-basic variable column, then this indicates that there are more than 1 optimal solution of the problem. Thus by entering this variable into the basis, we may obtain another alternative optimal solution.
Example
Find solution using Simplex method (BigM method)
MAX Z = 6x1 + 4x2
subject to
2x1 + 3x2 <= 30
3x1 + 2x2 <= 24
x1 + x2 >= 3
and x1,x2 >= 0;

Solution:
Problem is
Max `Z``=````6``x_1`` + ``4``x_2`
subject to
```2``x_1`` + ``3``x_2`≤`30`
```3``x_1`` + ``2``x_2`≤`24`
`````x_1`` + ````x_2`≥`3`
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`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_3` and add artificial variable `A_1`

After introducing slack,surplus,artificial variables
Max `Z``=````6``x_1`` + ``4``x_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
```2``x_1`` + ``3``x_2`` + ````S_1`=`30`
```3``x_1`` + ``2``x_2`` + ````S_2`=`24`
`````x_1`` + ````x_2`` - ````S_3`` + ````A_1`=`3`
and `x_1,x_2,S_1,S_2,S_3,A_1 >= 0`


`Z`` - ``6``x_1`` - ``4``x_2`` + ``M``A_1`=`0`
```2``x_1`` + ``3``x_2`` + ````S_1`=`30`
```3``x_1`` + ``2``x_2`` + ````S_2`=`24`
`````x_1`` + ````x_2`` - ````S_3`` + ````A_1`=`3`


Simplex tableau is
Tableau-0
`"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1` `Z``-6``-4``0``0``0``M``0`
`R_2` `S_1``2``3``1``0``0``0``30`
`R_3` `S_2``3``2``0``1``0``0``24`
`R_4` `A_1``1``1``0``0``-1``1``3`


Make the Z-row consistent with the rest of the table (set coefficient of basis variables to 0 in Z-row)
`R_1`(new)`= R_1`(old) - `M R_4`(old)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1`(old) = `-6``-4``0``0``0``M``0`
`R_4`(old) = `1``1``0``0``-1``1``3`
`M xx R_4`(new) = `-M``-M``0``0``M``-M``-3M`
`R_1`(new)`= R_1`(old) - `M R_4`(old)`-M-6``-M-4``0``0``M``0``-3M`


Tableau-1
`"Basis"``x_1``darr``x_2``S_1``S_2``S_3``A_1``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `Z``-M-6``-M-4``0``0``M``0``-3M`
`R_2` `S_1``2``3``1``0``0``0``30``(30)/(2)=15`
`R_3` `S_2``3``2``0``1``0``0``24``(24)/(3)=8`
`R_4` `A_1``(1)``1``0``0``-1``1``3``(3)/(1)=3``->`


Most Negative `Z` is `-M-6`. So, the entering variable is `x_1`.

Minimum ratio is `3`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `1`.

Entering `=x_1`, Departing `=A_1`, Key Element `=1`
`R_4`(new)`= R_4`(old)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_4`(old) = `1``1``0``0``-1``1``3`
`R_4`(new)`= R_4`(old)`1``1``0``0``-1``1``3`
`R_2`(new)`= R_2`(old) - `2 R_4`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_2`(old) = `2``3``1``0``0``0``30`
`R_4`(new) = `1``1``0``0``-1``1``3`
`2 xx R_4`(new) = `2``2``0``0``-2``2``6`
`R_2`(new)`= R_2`(old) - `2 R_4`(new)`0``1``1``0``2``-2``24`
`R_3`(new)`= R_3`(old) - `3 R_4`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_3`(old) = `3``2``0``1``0``0``24`
`R_4`(new) = `1``1``0``0``-1``1``3`
`3 xx R_4`(new) = `3``3``0``0``-3``3``9`
`R_3`(new)`= R_3`(old) - `3 R_4`(new)`0``-1``0``1``3``-3``15`
`R_1`(new)`= R_1`(old) - `(-M-6) R_4`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1`(old) = `-M-6``-M-4``0``0``M``0``-3M`
`R_4`(new) = `1``1``0``0``-1``1``3`
`(-M-6) xx R_4`(new) = `-M-6``-M-6``0``0``M+6``-M-6``-3M-18`
`R_1`(new)`= R_1`(old) - `(-M-6) R_4`(new)`0``2``0``0``-6``M+6``18`


Tableau-2
`"Basis"``x_1``x_2``S_1``S_2``S_3``darr``A_1``RHS``"Ratio"=(RHS)/(S_3)`
`R_1` `Z``0``2``0``0``-6``M+6``18`
`R_2` `S_1``0``1``1``0``2``-2``24``(24)/(2)=12`
`R_3` `S_2``0``-1``0``1``(3)``-3``15``(15)/(3)=5``->`
`R_4` `x_1``1``1``0``0``-1``1``3``(3)/(-1)` (ignore, denominator is -ve)


Most Negative `Z` is `-6`. So, the entering variable is `S_3`.

Minimum ratio is `5`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `3`.

Entering `=S_3`, Departing `=S_2`, Key Element `=3`
`R_3`(new)`= R_3`(old) `-: 3`
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_3`(old) = `0``-1``0``1``3``-3``15`
`R_3`(new)`= R_3`(old) `-: 3``0``-0.3333``0``0.3333``1``-1``5`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_2`(old) = `0``1``1``0``2``-2``24`
`R_3`(new) = `0``-0.3333``0``0.3333``1``-1``5`
`2 xx R_3`(new) = `0``-0.6667``0``0.6667``2``-2``10`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)`0``1.6667``1``-0.6667``0``0``14`
`R_4`(new)`= R_4`(old) + `R_3`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_4`(old) = `1``1``0``0``-1``1``3`
`R_3`(new) = `0``-0.3333``0``0.3333``1``-1``5`
`R_4`(new)`= R_4`(old) + `R_3`(new)`1``0.6667``0``0.3333``0``0``8`
`R_1`(new)`= R_1`(old) - `-6 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1`(old) = `0``2``0``0``-6``M+6``18`
`R_3`(new) = `0``-0.3333``0``0.3333``1``-1``5`
`-6 xx R_3`(new) = `0``2``0``-2``-6``6``-30`
`R_1`(new)`= R_1`(old) - `-6 R_3`(new)`0``0``0``2``0``M``48`


Tableau-3
`"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1` `Z``0``0``0``2``0``M``48`
`R_2` `S_1``0``1.6667``1``-0.6667``0``0``14`
`R_3` `S_3``0``-0.3333``0``0.3333``1``-1``5`
`R_4` `x_1``1``0.6667``0``0.3333``0``0``8`


Since all `Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=8,x_2=0`

Max `Z=48`


Here `Z_2=0` and `x_2` is not in the basis (i.e. `x_2=0`).

This indicates that there are more than 1 optimal solution of the problem.
Thus by entering `x_2` into the basis, we may obtain another alternative optimal solution.

Tableau-3
`"Basis"``x_1``x_2``darr``S_1``S_2``S_3``A_1``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `Z``0``0``0``2``0``M``48`
`R_2` `S_1``0``(1.6667)``1``-0.6667``0``0``14``(14)/(1.6667)=8.4``->`
`R_3` `S_3``0``-0.3333``0``0.3333``1``-1``5``(5)/(-0.3333)` (ignore, denominator is -ve)
`R_4` `x_1``1``0.6667``0``0.3333``0``0``8``(8)/(0.6667)=12`


So, the entering variable is `x_2`.

Minimum ratio is `8.4` and its row index is `1`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `1.6667`.

Entering `=x_2`, Departing `=S_1`, Key Element `=1.6667`
`R_2`(new)`= R_2`(old) `-: 1.6667`
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_2`(old) = `0``1.6667``1``-0.6667``0``0``14`
`R_2`(new)`= R_2`(old) `-: 1.6667``0``1``0.6``-0.4``0``0``8.4`
`R_3`(new)`= R_3`(old) + `0.3333 R_2`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_3`(old) = `0``-0.3333``0``0.3333``1``-1``5`
`R_2`(new) = `0``1``0.6``-0.4``0``0``8.4`
`0.3333 xx R_2`(new) = `0``0.3333``0.2``-0.1333``0``0``2.8`
`R_3`(new)`= R_3`(old) + `0.3333 R_2`(new)`0``0``0.2``0.2``1``-1``7.8`
`R_4`(new)`= R_4`(old) - `0.6667 R_2`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_4`(old) = `1``0.6667``0``0.3333``0``0``8`
`R_2`(new) = `0``1``0.6``-0.4``0``0``8.4`
`0.6667 xx R_2`(new) = `0``0.6667``0.4``-0.2667``0``0``5.6`
`R_4`(new)`= R_4`(old) - `0.6667 R_2`(new)`1``0``-0.4``0.6``0``0``2.4`
`R_1`(new)`= R_1`(old)
`x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1`(old) = `0``0``0``2``0``M``48`
`R_1`(new)`= R_1`(old)`0``0``0``2``0``M``48`


Tableau-4
`"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``RHS`
`R_1` `Z``0``0``0``2``0``M``48`
`R_2` `x_2``0``1``0.6``-0.4``0``0``8.4`
`R_3` `S_3``0``0``0.2``0.2``1``-1``7.8`
`R_4` `x_1``1``0``-0.4``0.6``0``0``2.4`


Since all `Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=2.4,x_2=8.4`

Max `Z=48`


This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then Submit Here


12. Unrestricted variable example (using `Z`-row method)
(Previous example)		14. Infeasible solution example (using `Z`-row method)
(Next example)


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