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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
35. Minimization example-1 (using `C_j-Z_j`method)
(Previous example)		37. Minimization example-3 (using `C_j-Z_j`method)
(Next example)

36. Minimization example-2 (using `C_j-Z_j`method)


Find solution using Simplex method (BigM method)
MIN Z = 5x1 + 3x2
subject to
2x1 + 4x2 <= 12
2x1 + 2x2 = 10
5x1 + 2x2 >= 10
and x1,x2 >= 0;

Solution:
Problem is
Min `Z``=````5``x_1`` + ``3``x_2`
subject to
```2``x_1`` + ``4``x_2`≤`12`
```2``x_1`` + ``2``x_2`=`10`
```5``x_1`` + ``2``x_2`≥`10`
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 artificial variable `A_1`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

After introducing slack,surplus,artificial variables
Min `Z``=````5``x_1`` + ``3``x_2`` + ``0``S_1`` + ``0``S_2`` + ``M``A_1`` + ``M``A_2`
subject to
```2``x_1`` + ``4``x_2`` + ````S_1`=`12`
```2``x_1`` + ``2``x_2`` + ````A_1`=`10`
```5``x_1`` + ``2``x_2`` - ````S_2`` + ````A_2`=`10`
and `x_1,x_2,S_1,S_2,A_1,A_2 >= 0`


Tableau-1`C_j``5``3``0``0``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `0``S_1``2``4``1``0``0``0``12``(12)/(2)=6`
`R_2` `M``A_1``2``2``0``0``1``0``10``(10)/(2)=5`
`R_3` `M``A_2``(5)``2``0``-1``0``1``10``(10)/(5)=2``->`
`Z_j``7M``4M``0``-M``M``M``Z=20M`
`C_j-Z_j``-7M+5``uarr``-4M+3``0``M``0``0`


Most Negative `C_j-Z_j` is `-7M+5`. So, the entering variable is `x_1`.

Minimum ratio is `2`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `5`.

Entering `=x_1`, Departing `=A_2`, Key Element `=5`
`R_3`(new)`= R_3`(old) `-: 5`
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_3`(old) = `5``2``0``-1``0``1``10`
`R_3`(new)`= R_3`(old) `-: 5``1``0.4``0``-0.2``0``0.2``2`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `2``4``1``0``0``0``12`
`R_3`(new) = `1``0.4``0``-0.2``0``0.2``2`
`2 xx R_3`(new) = `2``0.8``0``-0.4``0``0.4``4`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)`0``3.2``1``0.4``0``-0.4``8`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_2`(old) = `2``2``0``0``1``0``10`
`R_3`(new) = `1``0.4``0``-0.2``0``0.2``2`
`2 xx R_3`(new) = `2``0.8``0``-0.4``0``0.4``4`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)`0``1.2``0``0.4``1``-0.4``6`


Tableau-2`C_j``5``3``0``0``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `0``S_1``0``(3.2)``1``0.4``0``-0.4``8``(8)/(3.2)=2.5``->`
`R_2` `M``A_1``0``1.2``0``0.4``1``-0.4``6``(6)/(1.2)=5`
`R_3` `5``x_1``1``0.4``0``-0.2``0``0.2``2``(2)/(0.4)=5`
`Z_j``5``1.2M+2``0``0.4M-1``M``-0.4M+1``Z=6M+10`
`C_j-Z_j``0``-1.2M+1``uarr``0``-0.4M+1``0``1.4M-1`


Most Negative `C_j-Z_j` is `-1.2M+1`. So, the entering variable is `x_2`.

Minimum ratio is `2.5`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `3.2`.

Entering `=x_2`, Departing `=S_1`, Key Element `=3.2`
`R_1`(new)`= R_1`(old) `-: 3.2`
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `0``3.2``1``0.4``0``-0.4``8`
`R_1`(new)`= R_1`(old) `-: 3.2``0``1``0.3125``0.125``0``-0.125``2.5`
`R_2`(new)`= R_2`(old) - `1.2 R_1`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_2`(old) = `0``1.2``0``0.4``1``-0.4``6`
`R_1`(new) = `0``1``0.3125``0.125``0``-0.125``2.5`
`1.2 xx R_1`(new) = `0``1.2``0.375``0.15``0``-0.15``3`
`R_2`(new)`= R_2`(old) - `1.2 R_1`(new)`0``0``-0.375``0.25``1``-0.25``3`
`R_3`(new)`= R_3`(old) - `0.4 R_1`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_3`(old) = `1``0.4``0``-0.2``0``0.2``2`
`R_1`(new) = `0``1``0.3125``0.125``0``-0.125``2.5`
`0.4 xx R_1`(new) = `0``0.4``0.125``0.05``0``-0.05``1`
`R_3`(new)`= R_3`(old) - `0.4 R_1`(new)`1``0``-0.125``-0.25``0``0.25``1`


Tableau-3`C_j``5``3``0``0``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS``"Ratio"=(RHS)/(S_2)`
`R_1` `3``x_2``0``1``0.3125``0.125``0``-0.125``2.5``(2.5)/(0.125)=20`
`R_2` `M``A_1``0``0``-0.375``(0.25)``1``-0.25``3``(3)/(0.25)=12``->`
`R_3` `5``x_1``1``0``-0.125``-0.25``0``0.25``1``(1)/(-0.25)` (ignore, denominator is -ve)
`Z_j``5``3``-0.375M+0.3125``0.25M-0.875``M``-0.25M+0.875``Z=3M+12.5`
`C_j-Z_j``0``0``0.375M-0.3125``-0.25M+0.875``uarr``0``1.25M-0.875`


Most Negative `C_j-Z_j` is `-0.25M+0.875`. So, the entering variable is `S_2`.

Minimum ratio is `12`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `0.25`.

Entering `=S_2`, Departing `=A_1`, Key Element `=0.25`
`R_2`(new)`= R_2`(old) `-: 0.25`
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_2`(old) = `0``0``-0.375``0.25``1``-0.25``3`
`R_2`(new)`= R_2`(old) `-: 0.25``0``0``-1.5``1``4``-1``12`
`R_1`(new)`= R_1`(old) - `0.125 R_2`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_1`(old) = `0``1``0.3125``0.125``0``-0.125``2.5`
`R_2`(new) = `0``0``-1.5``1``4``-1``12`
`0.125 xx R_2`(new) = `0``0``-0.1875``0.125``0.5``-0.125``1.5`
`R_1`(new)`= R_1`(old) - `0.125 R_2`(new)`0``1``0.5``0``-0.5``0``1`
`R_3`(new)`= R_3`(old) + `0.25 R_2`(new)
`x_1``x_2``S_1``S_2``A_1``A_2``RHS`
`R_3`(old) = `1``0``-0.125``-0.25``0``0.25``1`
`R_2`(new) = `0``0``-1.5``1``4``-1``12`
`0.25 xx R_2`(new) = `0``0``-0.375``0.25``1``-0.25``3`
`R_3`(new)`= R_3`(old) + `0.25 R_2`(new)`1``0``-0.5``0``1``0``4`


Tableau-4`C_j``5``3``0``0``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``A_1``A_2``RHS``"Ratio"`
`R_1` `3``x_2``0``1``0.5``0``-0.5``0``1`
`R_2` `0``S_2``0``0``-1.5``1``4``-1``12`
`R_3` `5``x_1``1``0``-0.5``0``1``0``4`
`Z_j``5``3``-1``0``3.5``0``Z=23`
`C_j-Z_j``0``0``1``0``M-3.5``M`


Since all `C_j-Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=4,x_2=1`

Min `Z=23`


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


35. Minimization example-1 (using `C_j-Z_j`method)
(Previous example)		37. Minimization example-3 (using `C_j-Z_j`method)
(Next example)


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