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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
18. Maximization example-2 (using `Z_j-C_j` method)
(Previous example)		20. BigM method Algorithm (using `Z_j-C_j` method)
(Next example)

19. Maximization example-3 (using `Z_j-C_j` method)


Find solution using Simplex method (BigM method)
MAX Z = 22x1 + 6x2 + 2x3
subject to
10x1 + 2x2 + x3 <= 100
7x1 + 3x2 + 2x3 <= 72
2x1 + 4x2 + x3 <= 80
and x1,x2,x3 >= 0;

Solution:
Problem is
Max `Z``=````22``x_1`` + ``6``x_2`` + ``2``x_3`
subject to
```10``x_1`` + ``2``x_2`` + ````x_3`≤`100`
```7``x_1`` + ``3``x_2`` + ``2``x_3`≤`72`
```2``x_1`` + ``4``x_2`` + ````x_3`≤`80`
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 add slack variable `S_3`

After introducing slack variables
Max `Z``=````22``x_1`` + ``6``x_2`` + ``2``x_3`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`
subject to
```10``x_1`` + ``2``x_2`` + ````x_3`` + ````S_1`=`100`
```7``x_1`` + ``3``x_2`` + ``2``x_3`` + ````S_2`=`72`
```2``x_1`` + ``4``x_2`` + ````x_3`` + ````S_3`=`80`
and `x_1,x_2,x_3,S_1,S_2,S_3 >= 0`


Tableau-1`C_j``22``6``2``0``0``0`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `0``S_1``(10)``2``1``1``0``0``100``(100)/(10)=10``->`
`R_2` `0``S_2``7``3``2``0``1``0``72``(72)/(7)=10.2857`
`R_3` `0``S_3``2``4``1``0``0``1``80``(80)/(2)=40`
`Z_j``0``0``0``0``0``0``Z=0`
`Z_j-C_j``-22``uarr``-6``-2``0``0``0`


Most Negative `Z_j-C_j` is `-22`. So, the entering variable is `x_1`.

Minimum ratio is `10`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `10`.

Entering `=x_1`, Departing `=S_1`, Key Element `=10`
`R_1`(new)`= R_1`(old) `-: 10`
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_1`(old) = `10``2``1``1``0``0``100`
`R_1`(new)`= R_1`(old) `-: 10``1``0.2``0.1``0.1``0``0``10`
`R_2`(new)`= R_2`(old) - `7 R_1`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_2`(old) = `7``3``2``0``1``0``72`
`R_1`(new) = `1``0.2``0.1``0.1``0``0``10`
`7 xx R_1`(new) = `7``1.4``0.7``0.7``0``0``70`
`R_2`(new)`= R_2`(old) - `7 R_1`(new)`0``1.6``1.3``-0.7``1``0``2`
`R_3`(new)`= R_3`(old) - `2 R_1`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_3`(old) = `2``4``1``0``0``1``80`
`R_1`(new) = `1``0.2``0.1``0.1``0``0``10`
`2 xx R_1`(new) = `2``0.4``0.2``0.2``0``0``20`
`R_3`(new)`= R_3`(old) - `2 R_1`(new)`0``3.6``0.8``-0.2``0``1``60`


Tableau-2`C_j``22``6``2``0``0``0`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `22``x_1``1``0.2``0.1``0.1``0``0``10``(10)/(0.2)=50`
`R_2` `0``S_2``0``(1.6)``1.3``-0.7``1``0``2``(2)/(1.6)=1.25``->`
`R_3` `0``S_3``0``3.6``0.8``-0.2``0``1``60``(60)/(3.6)=16.6667`
`Z_j``22``4.4``2.2``2.2``0``0``Z=220`
`Z_j-C_j``0``-1.6``uarr``0.2``2.2``0``0`


Most Negative `Z_j-C_j` is `-1.6`. So, the entering variable is `x_2`.

Minimum ratio is `1.25`. So, the leaving basis variable is `S_2`.

`:.` The pivot element is `1.6`.

Entering `=x_2`, Departing `=S_2`, Key Element `=1.6`
`R_2`(new)`= R_2`(old) `-: 1.6`
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_2`(old) = `0``1.6``1.3``-0.7``1``0``2`
`R_2`(new)`= R_2`(old) `-: 1.6``0``1``0.8125``-0.4375``0.625``0``1.25`
`R_1`(new)`= R_1`(old) - `0.2 R_2`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_1`(old) = `1``0.2``0.1``0.1``0``0``10`
`R_2`(new) = `0``1``0.8125``-0.4375``0.625``0``1.25`
`0.2 xx R_2`(new) = `0``0.2``0.1625``-0.0875``0.125``0``0.25`
`R_1`(new)`= R_1`(old) - `0.2 R_2`(new)`1``0``-0.0625``0.1875``-0.125``0``9.75`
`R_3`(new)`= R_3`(old) - `3.6 R_2`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_3`(old) = `0``3.6``0.8``-0.2``0``1``60`
`R_2`(new) = `0``1``0.8125``-0.4375``0.625``0``1.25`
`3.6 xx R_2`(new) = `0``3.6``2.925``-1.575``2.25``0``4.5`
`R_3`(new)`= R_3`(old) - `3.6 R_2`(new)`0``0``-2.125``1.375``-2.25``1``55.5`


Tableau-3`C_j``22``6``2``0``0``0`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``RHS``"Ratio"`
`R_1` `22``x_1``1``0``-0.0625``0.1875``-0.125``0``9.75`
`R_2` `6``x_2``0``1``0.8125``-0.4375``0.625``0``1.25`
`R_3` `0``S_3``0``0``-2.125``1.375``-2.25``1``55.5`
`Z_j``22``6``3.5``1.5``1``0``Z=222`
`Z_j-C_j``0``0``1.5``1.5``1``0`


Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=9.75,x_2=1.25,x_3=0`

Max `Z=222`


This material is intended as a summary. Use your textbook for detail explanation.
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18. Maximization example-2 (using `Z_j-C_j` method)
(Previous example)		20. BigM method Algorithm (using `Z_j-C_j` method)
(Next example)


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