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

17. Maximization example-1 (using `Z_j-C_j` method)


Find solution using Simplex method (BigM method)
MAX Z = 3x1 + 5x2 + 4x3
subject to
2x1 + 3x2 <= 8
2x2 + 5x3 <= 10
3x1 + 2x2 + 4x3 <= 15
and x1,x2,x3 >= 0;

Solution:
Problem is
Max `Z``=````3``x_1`` + ``5``x_2`` + ``4``x_3`
subject to
```2``x_1`` + ``3``x_2`≤`8`
```2``x_2`` + ``5``x_3`≤`10`
```3``x_1`` + ``2``x_2`` + ``4``x_3`≤`15`
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``=````3``x_1`` + ``5``x_2`` + ``4``x_3`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`
subject to
```2``x_1`` + ``3``x_2`` + ````S_1`=`8`
```2``x_2`` + ``5``x_3`` + ````S_2`=`10`
```3``x_1`` + ``2``x_2`` + ``4``x_3`` + ````S_3`=`15`
and `x_1,x_2,x_3,S_1,S_2,S_3 >= 0`


Tableau-1`C_j``3``5``4``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` `0``S_1``2``(3)``0``1``0``0``8``(8)/(3)=2.6667``->`
`R_2` `0``S_2``0``2``5``0``1``0``10``(10)/(2)=5`
`R_3` `0``S_3``3``2``4``0``0``1``15``(15)/(2)=7.5`
`Z_j``0``0``0``0``0``0``Z=0`
`Z_j-C_j``-3``-5``uarr``-4``0``0``0`


Most Negative `Z_j-C_j` is `-5`. So, the entering variable is `x_2`.

Minimum ratio is `2.6667`. So, the leaving basis variable is `S_1`.

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

Entering `=x_2`, Departing `=S_1`, Key Element `=3`
`R_1`(new)`= R_1`(old) `-: 3`
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_1`(old) = `2``3``0``1``0``0``8`
`R_1`(new)`= R_1`(old) `-: 3``0.6667``1``0``0.3333``0``0``2.6667`
`R_2`(new)`= R_2`(old) - `2 R_1`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_2`(old) = `0``2``5``0``1``0``10`
`R_1`(new) = `0.6667``1``0``0.3333``0``0``2.6667`
`2 xx R_1`(new) = `1.3333``2``0``0.6667``0``0``5.3333`
`R_2`(new)`= R_2`(old) - `2 R_1`(new)`-1.3333``0``5``-0.6667``1``0``4.6667`
`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) = `3``2``4``0``0``1``15`
`R_1`(new) = `0.6667``1``0``0.3333``0``0``2.6667`
`2 xx R_1`(new) = `1.3333``2``0``0.6667``0``0``5.3333`
`R_3`(new)`= R_3`(old) - `2 R_1`(new)`1.6667``0``4``-0.6667``0``1``9.6667`


Tableau-2`C_j``3``5``4``0``0``0`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``RHS``"Ratio"=(RHS)/(x_3)`
`R_1` `5``x_2``0.6667``1``0``0.3333``0``0``2.6667``(2.6667)/(0)` (ignore, denominator is 0)
`R_2` `0``S_2``-1.3333``0``(5)``-0.6667``1``0``4.6667``(4.6667)/(5)=0.9333``->`
`R_3` `0``S_3``1.6667``0``4``-0.6667``0``1``9.6667``(9.6667)/(4)=2.4167`
`Z_j``3.3333``5``0``1.6667``0``0``Z=13.3333`
`Z_j-C_j``0.3333``0``-4``uarr``1.6667``0``0`


Most Negative `Z_j-C_j` is `-4`. So, the entering variable is `x_3`.

Minimum ratio is `0.9333`. So, the leaving basis variable is `S_2`.

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

Entering `=x_3`, Departing `=S_2`, Key Element `=5`
`R_2`(new)`= R_2`(old) `-: 5`
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_2`(old) = `-1.3333``0``5``-0.6667``1``0``4.6667`
`R_2`(new)`= R_2`(old) `-: 5``-0.2667``0``1``-0.1333``0.2``0``0.9333`
`R_1`(new)`= R_1`(old)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_1`(old) = `0.6667``1``0``0.3333``0``0``2.6667`
`R_1`(new)`= R_1`(old)`0.6667``1``0``0.3333``0``0``2.6667`
`R_3`(new)`= R_3`(old) - `4 R_2`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_3`(old) = `1.6667``0``4``-0.6667``0``1``9.6667`
`R_2`(new) = `-0.2667``0``1``-0.1333``0.2``0``0.9333`
`4 xx R_2`(new) = `-1.0667``0``4``-0.5333``0.8``0``3.7333`
`R_3`(new)`= R_3`(old) - `4 R_2`(new)`2.7333``0``0``-0.1333``-0.8``1``5.9333`


Tableau-3`C_j``3``5``4``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` `5``x_2``0.6667``1``0``0.3333``0``0``2.6667``(2.6667)/(0.6667)=4`
`R_2` `4``x_3``-0.2667``0``1``-0.1333``0.2``0``0.9333``(0.9333)/(-0.2667)` (ignore, denominator is -ve)
`R_3` `0``S_3``(2.7333)``0``0``-0.1333``-0.8``1``5.9333``(5.9333)/(2.7333)=2.1707``->`
`Z_j``2.2667``5``4``1.1333``0.8``0``Z=17.0667`
`Z_j-C_j``-0.7333``uarr``0``0``1.1333``0.8``0`


Most Negative `Z_j-C_j` is `-0.7333`. So, the entering variable is `x_1`.

Minimum ratio is `2.1707`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `2.7333`.

Entering `=x_1`, Departing `=S_3`, Key Element `=2.7333`
`R_3`(new)`= R_3`(old) `-: 2.7333`
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_3`(old) = `2.7333``0``0``-0.1333``-0.8``1``5.9333`
`R_3`(new)`= R_3`(old) `-: 2.7333``1``0``0``-0.0488``-0.2927``0.3659``2.1707`
`R_1`(new)`= R_1`(old) - `0.6667 R_3`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_1`(old) = `0.6667``1``0``0.3333``0``0``2.6667`
`R_3`(new) = `1``0``0``-0.0488``-0.2927``0.3659``2.1707`
`0.6667 xx R_3`(new) = `0.6667``0``0``-0.0325``-0.1951``0.2439``1.4472`
`R_1`(new)`= R_1`(old) - `0.6667 R_3`(new)`0``1``0``0.3659``0.1951``-0.2439``1.2195`
`R_2`(new)`= R_2`(old) + `0.2667 R_3`(new)
`x_1``x_2``x_3``S_1``S_2``S_3``RHS`
`R_2`(old) = `-0.2667``0``1``-0.1333``0.2``0``0.9333`
`R_3`(new) = `1``0``0``-0.0488``-0.2927``0.3659``2.1707`
`0.2667 xx R_3`(new) = `0.2667``0``0``-0.013``-0.078``0.0976``0.5789`
`R_2`(new)`= R_2`(old) + `0.2667 R_3`(new)`0``0``1``-0.1463``0.122``0.0976``1.5122`


Tableau-4`C_j``3``5``4``0``0``0`
`C_B``"Basis"``x_1``x_2``x_3``S_1``S_2``S_3``RHS``"Ratio"`
`R_1` `5``x_2``0``1``0``0.3659``0.1951``-0.2439``1.2195`
`R_2` `4``x_3``0``0``1``-0.1463``0.122``0.0976``1.5122`
`R_3` `3``x_1``1``0``0``-0.0488``-0.2927``0.3659``2.1707`
`Z_j``3``5``4``1.0976``0.5854``0.2683``Z=18.6585`
`Z_j-C_j``0``0``0``1.0976``0.5854``0.2683`


Since all `Z_j-C_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=2.1707,x_2=1.2195,x_3=1.5122`

Max `Z=18.6585`


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


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


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