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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
3. Maximization example-1 (using `Z`-row method)
(Previous example)		5. Maximization example-3 (using `Z`-row method)
(Next example)

4. Maximization example-2 (using `Z`-row method)


Find solution using Simplex method (BigM method)
MAX Z = 30x1 + 40x2
subject to
3x1 + 2x2 <= 600
3x1 + 5x2 <= 800
5x1 + 6x2 <= 1100
and x1,x2 >= 0;

Solution:
Problem is
Max `Z``=````30``x_1`` + ``40``x_2`
subject to
```3``x_1`` + ``2``x_2`≤`600`
```3``x_1`` + ``5``x_2`≤`800`
```5``x_1`` + ``6``x_2`≤`1100`
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 add slack variable `S_3`

After introducing slack variables
Max `Z``=````30``x_1`` + ``40``x_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`
subject to
```3``x_1`` + ``2``x_2`` + ````S_1`=`600`
```3``x_1`` + ``5``x_2`` + ````S_2`=`800`
```5``x_1`` + ``6``x_2`` + ````S_3`=`1100`
and `x_1,x_2,S_1,S_2,S_3 >= 0`


`Z`` - ``30``x_1`` - ``40``x_2`=`0`
```3``x_1`` + ``2``x_2`` + ````S_1`=`600`
```3``x_1`` + ``5``x_2`` + ````S_2`=`800`
```5``x_1`` + ``6``x_2`` + ````S_3`=`1100`


Simplex tableau is
Tableau-0
`"Basis"``x_1``x_2``S_1``S_2``S_3``RHS`
`R_1` `Z``-30``-40``0``0``0``0`
`R_2` `S_1``3``2``1``0``0``600`
`R_3` `S_2``3``5``0``1``0``800`
`R_4` `S_3``5``6``0``0``1``1100`


Tableau-1
`"Basis"``x_1``x_2``darr``S_1``S_2``S_3``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `Z``-30``-40``0``0``0``0`
`R_2` `S_1``3``2``1``0``0``600``(600)/(2)=300`
`R_3` `S_2``3``(5)``0``1``0``800``(800)/(5)=160``->`
`R_4` `S_3``5``6``0``0``1``1100``(1100)/(6)=183.3333`


Most Negative `Z` is `-40`. So, the entering variable is `x_2`.

Minimum ratio is `160`. So, the leaving basis variable is `S_2`.

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

Entering `=x_2`, Departing `=S_2`, Key Element `=5`
`R_3`(new)`= R_3`(old) `-: 5`
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_3`(old) = `3``5``0``1``0``800`
`R_3`(new)`= R_3`(old) `-: 5``0.6``1``0``0.2``0``160`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_2`(old) = `3``2``1``0``0``600`
`R_3`(new) = `0.6``1``0``0.2``0``160`
`2 xx R_3`(new) = `1.2``2``0``0.4``0``320`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)`1.8``0``1``-0.4``0``280`
`R_4`(new)`= R_4`(old) - `6 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_4`(old) = `5``6``0``0``1``1100`
`R_3`(new) = `0.6``1``0``0.2``0``160`
`6 xx R_3`(new) = `3.6``6``0``1.2``0``960`
`R_4`(new)`= R_4`(old) - `6 R_3`(new)`1.4``0``0``-1.2``1``140`
`R_1`(new)`= R_1`(old) - `-40 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_1`(old) = `-30``-40``0``0``0``0`
`R_3`(new) = `0.6``1``0``0.2``0``160`
`-40 xx R_3`(new) = `-24``-40``0``-8``0``-6400`
`R_1`(new)`= R_1`(old) - `-40 R_3`(new)`-6``0``0``8``0``6400`


Tableau-2
`"Basis"``x_1``darr``x_2``S_1``S_2``S_3``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `Z``-6``0``0``8``0``6400`
`R_2` `S_1``1.8``0``1``-0.4``0``280``(280)/(1.8)=155.5556`
`R_3` `x_2``0.6``1``0``0.2``0``160``(160)/(0.6)=266.6667`
`R_4` `S_3``(1.4)``0``0``-1.2``1``140``(140)/(1.4)=100``->`


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

Minimum ratio is `100`. So, the leaving basis variable is `S_3`.

`:.` The pivot element is `1.4`.

Entering `=x_1`, Departing `=S_3`, Key Element `=1.4`
`R_4`(new)`= R_4`(old) `-: 1.4`
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_4`(old) = `1.4``0``0``-1.2``1``140`
`R_4`(new)`= R_4`(old) `-: 1.4``1``0``0``-0.8571``0.7143``100`
`R_2`(new)`= R_2`(old) - `1.8 R_4`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_2`(old) = `1.8``0``1``-0.4``0``280`
`R_4`(new) = `1``0``0``-0.8571``0.7143``100`
`1.8 xx R_4`(new) = `1.8``0``0``-1.5429``1.2857``180`
`R_2`(new)`= R_2`(old) - `1.8 R_4`(new)`0``0``1``1.1429``-1.2857``100`
`R_3`(new)`= R_3`(old) - `0.6 R_4`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_3`(old) = `0.6``1``0``0.2``0``160`
`R_4`(new) = `1``0``0``-0.8571``0.7143``100`
`0.6 xx R_4`(new) = `0.6``0``0``-0.5143``0.4286``60`
`R_3`(new)`= R_3`(old) - `0.6 R_4`(new)`0``1``0``0.7143``-0.4286``100`
`R_1`(new)`= R_1`(old) - `-6 R_4`(new)
`x_1``x_2``S_1``S_2``S_3``RHS`
`R_1`(old) = `-6``0``0``8``0``6400`
`R_4`(new) = `1``0``0``-0.8571``0.7143``100`
`-6 xx R_4`(new) = `-6``0``0``5.1429``-4.2857``-600`
`R_1`(new)`= R_1`(old) - `-6 R_4`(new)`0``0``0``2.8571``4.2857``7000`


Tableau-3
`"Basis"``x_1``x_2``S_1``S_2``S_3``RHS`
`R_1` `Z``0``0``0``2.8571``4.2857``7000`
`R_2` `S_1``0``0``1``1.1429``-1.2857``100`
`R_3` `x_2``0``1``0``0.7143``-0.4286``100`
`R_4` `x_1``1``0``0``-0.8571``0.7143``100`


Since all `Z_j >= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=100,x_2=100`

Max `Z=7000`


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


3. Maximization example-1 (using `Z`-row method)
(Previous example)		5. Maximization example-3 (using `Z`-row method)
(Next example)


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