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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
38. Degeneracy example-1 (Tie for leaving basic variable) (using `C_j-Z_j`method)
(Previous example)		40. Unrestricted variable example (using `C_j-Z_j`method)
(Next example)

39. Degeneracy example-2 (Tie first Artificial variable removed) (using `C_j-Z_j`method)


Degeneracy example-2 (Tie - first Artificial variable removed)
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 method (BigM method)
MAX z = 750x1 + 900x2 - 450x3
subject to
x1 + 2x2 <= 70
2x1 + 3x2 - x3 <= 100
x1 >= 20
x2 >= 25
and x1,x2,x3 >= 0;

Solution:
Problem is
Max `z``=````750``x_1`` + ``900``x_2`` - ``450``x_3`
subject to
`````x_1`` + ``2``x_2`≤`70`
```2``x_1`` + ``3``x_2`` - ````x_3`≤`100`
`````x_1`≥`20`
`````x_2`≥`25`
and `x_1,x_2,x_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 subtract surplus variable `S_3` and add artificial variable `A_1`

4. As the constraint-4 is of type '`>=`' we should subtract surplus variable `S_4` and add artificial variable `A_2`

After introducing slack,surplus,artificial variables
Max `z``=````750``x_1`` + ``900``x_2`` - ``450``x_3`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``0``S_4`` - ``M``A_1`` - ``M``A_2`
subject to
`````x_1`` + ``2``x_2`` + ````S_1`=`70`
```2``x_1`` + ``3``x_2`` - ````x_3`` + ````S_2`=`100`
`````x_1`` - ````S_3`` + ````A_1`=`20`
`````x_2`` - ````S_4`` + ````A_2`=`25`
and `x_1,x_2,x_3,S_1,S_2,S_3,S_4,A_1,A_2 >= 0`


Tableau-1`C_j``750``900``-450``0``0``0``0``-M``-M`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `0``S_1``1``2``0``1``0``0``0``0``0``70``(70)/(2)=35`
`R_2` `0``S_2``2``3``-1``0``1``0``0``0``0``100``(100)/(3)=33.3333`
`R_3` `-M``A_1``1``0``0``0``0``-1``0``1``0``20``(20)/(0)` (ignore, denominator is 0)
`R_4` `-M``A_2``0``(1)``0``0``0``0``-1``0``1``25``(25)/(1)=25``->`
`Z_j``-M``-M``0``0``0``M``M``-M``-M``z=-45M`
`C_j-Z_j``M+750``M+900``uarr``-450``0``0``-M``-M``0``0`


Most Positive `C_j-Z_j` is `M+900`. So, the entering variable is `x_2`.

Minimum ratio is `25`. So, the leaving basis variable is `A_2`.

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

Entering `=x_2`, Departing `=A_2`, Key Element `=1`
`R_4`(new)`= R_4`(old)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_4`(old) = `0``1``0``0``0``0``-1``0``1``25`
`R_4`(new)`= R_4`(old)`0``1``0``0``0``0``-1``0``1``25`
`R_1`(new)`= R_1`(old) - `2 R_4`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_1`(old) = `1``2``0``1``0``0``0``0``0``70`
`R_4`(new) = `0``1``0``0``0``0``-1``0``1``25`
`2 xx R_4`(new) = `0``2``0``0``0``0``-2``0``2``50`
`R_1`(new)`= R_1`(old) - `2 R_4`(new)`1``0``0``1``0``0``2``0``-2``20`
`R_2`(new)`= R_2`(old) - `3 R_4`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_2`(old) = `2``3``-1``0``1``0``0``0``0``100`
`R_4`(new) = `0``1``0``0``0``0``-1``0``1``25`
`3 xx R_4`(new) = `0``3``0``0``0``0``-3``0``3``75`
`R_2`(new)`= R_2`(old) - `3 R_4`(new)`2``0``-1``0``1``0``3``0``-3``25`
`R_3`(new)`= R_3`(old)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_3`(old) = `1``0``0``0``0``-1``0``1``0``20`
`R_3`(new)`= R_3`(old)`1``0``0``0``0``-1``0``1``0``20`


Tableau-2`C_j``750``900``-450``0``0``0``0``-M``-M`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `0``S_1``1``0``0``1``0``0``2``0``-2``20``(20)/(1)=20`
`R_2` `0``S_2``(2)``0``-1``0``1``0``3``0``-3``25``(25)/(2)=12.5``->`
`R_3` `-M``A_1``1``0``0``0``0``-1``0``1``0``20``(20)/(1)=20`
`R_4` `900``x_2``0``1``0``0``0``0``-1``0``1``25``(25)/(0)` (ignore, denominator is 0)
`Z_j``-M``900``0``0``0``M``-900``-M``900``z=-20M+22500`
`C_j-Z_j``M+750``uarr``0``-450``0``0``-M``900``0``-M-900`


Most Positive `C_j-Z_j` is `M+750`. So, the entering variable is `x_1`.

Minimum ratio is `12.5`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `2`.

Entering `=x_1`, Departing `=S_2`, Key Element `=2`
`R_2`(new)`= R_2`(old) `-: 2`
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_2`(old) = `2``0``-1``0``1``0``3``0``-3``25`
`R_2`(new)`= R_2`(old) `-: 2``1``0``-0.5``0``0.5``0``1.5``0``-1.5``12.5`
`R_1`(new)`= R_1`(old) - `R_2`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_1`(old) = `1``0``0``1``0``0``2``0``-2``20`
`R_2`(new) = `1``0``-0.5``0``0.5``0``1.5``0``-1.5``12.5`
`R_1`(new)`= R_1`(old) - `R_2`(new)`0``0``0.5``1``-0.5``0``0.5``0``-0.5``7.5`
`R_3`(new)`= R_3`(old) - `R_2`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_3`(old) = `1``0``0``0``0``-1``0``1``0``20`
`R_2`(new) = `1``0``-0.5``0``0.5``0``1.5``0``-1.5``12.5`
`R_3`(new)`= R_3`(old) - `R_2`(new)`0``0``0.5``0``-0.5``-1``-1.5``1``1.5``7.5`
`R_4`(new)`= R_4`(old)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_4`(old) = `0``1``0``0``0``0``-1``0``1``25`
`R_4`(new)`= R_4`(old)`0``1``0``0``0``0``-1``0``1``25`


Tableau-3`C_j``750``900``-450``0``0``0``0``-M``-M`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS``"Ratio"=(RHS)/(x_3)`
`R_1` `0``S_1``0``0``0.5``1``-0.5``0``0.5``0``-0.5``7.5``(7.5)/(0.5)=15`
`R_2` `750``x_1``1``0``-0.5``0``0.5``0``1.5``0``-1.5``12.5``(12.5)/(-0.5)` (ignore, denominator is -ve)
`R_3` `-M``A_1``0``0``(0.5)``0``-0.5``-1``-1.5``1``1.5``7.5``(7.5)/(0.5)=15``->`
`R_4` `900``x_2``0``1``0``0``0``0``-1``0``1``25``(25)/(0)` (ignore, denominator is 0)
`Z_j``750``900``-0.5M-375``0``0.5M+375``M``1.5M+225``-M``-1.5M-225``z=-7.5M+31875`
`C_j-Z_j``0``0``0.5M-75``uarr``0``-0.5M-375``-M``-1.5M-225``0``0.5M+225`


Most Positive `C_j-Z_j` is `0.5M-75`. So, the entering variable is `x_3`.

Minimum ratio is `15`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `0.5`.

Entering `=x_3`, Departing `=A_1`, Key Element `=0.5`
`R_3`(new)`= R_3`(old) `-: 0.5`
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_3`(old) = `0``0``0.5``0``-0.5``-1``-1.5``1``1.5``7.5`
`R_3`(new)`= R_3`(old) `-: 0.5``0``0``1``0``-1``-2``-3``2``3``15`
`R_1`(new)`= R_1`(old) - `0.5 R_3`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_1`(old) = `0``0``0.5``1``-0.5``0``0.5``0``-0.5``7.5`
`R_3`(new) = `0``0``1``0``-1``-2``-3``2``3``15`
`0.5 xx R_3`(new) = `0``0``0.5``0``-0.5``-1``-1.5``1``1.5``7.5`
`R_1`(new)`= R_1`(old) - `0.5 R_3`(new)`0``0``0``1``0``1``2``-1``-2``0`
`R_2`(new)`= R_2`(old) + `0.5 R_3`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_2`(old) = `1``0``-0.5``0``0.5``0``1.5``0``-1.5``12.5`
`R_3`(new) = `0``0``1``0``-1``-2``-3``2``3``15`
`0.5 xx R_3`(new) = `0``0``0.5``0``-0.5``-1``-1.5``1``1.5``7.5`
`R_2`(new)`= R_2`(old) + `0.5 R_3`(new)`1``0``0``0``0``-1``0``1``0``20`
`R_4`(new)`= R_4`(old)
`x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS`
`R_4`(old) = `0``1``0``0``0``0``-1``0``1``25`
`R_4`(new)`= R_4`(old)`0``1``0``0``0``0``-1``0``1``25`


Tableau-4`C_j``750``900``-450``0``0``0``0``-M``-M`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``S_4``A_1``A_2``RHS``"Ratio"`
`R_1` `0``S_1``0``0``0``1``0``1``2``-1``-2``0`
`R_2` `750``x_1``1``0``0``0``0``-1``0``1``0``20`
`R_3` `-450``x_3``0``0``1``0``-1``-2``-3``2``3``15`
`R_4` `900``x_2``0``1``0``0``0``0``-1``0``1``25`
`Z_j``750``900``-450``0``450``150``450``-150``-450``z=30750`
`C_j-Z_j``0``0``0``0``-450``-150``-450``-M+150``-M+450`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=20,x_2=25,x_3=15`

Max `z=30750`


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


38. Degeneracy example-1 (Tie for leaving basic variable) (using `C_j-Z_j`method)
(Previous example)		40. Unrestricted variable example (using `C_j-Z_j`method)
(Next example)


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