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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
22. Minimization example-2 (using `Z_j-C_j` method)
(Previous example)		24. Degeneracy example-1 (Tie for leaving basic variable) (using `Z_j-C_j` method)
(Next example)

23. Minimization example-3 (using `Z_j-C_j` method)


Find solution using Simplex method (BigM method)
MIN Z = 2000x1 + 1500x2
subject to
6x1 + 2x2 >= 8
2x1 + 4x2 >= 12
4x1 + 12x2 >= 24
and x1,x2 >= 0;

Solution:
Problem is
Min `Z``=````2000``x_1`` + ``1500``x_2`
subject to
```6``x_1`` + ``2``x_2`≥`8`
```2``x_1`` + ``4``x_2`≥`12`
```4``x_1`` + ``12``x_2`≥`24`
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 subtract surplus variable `S_1` and add artificial variable `A_1`

2. As the constraint-2 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_3` and add artificial variable `A_3`

After introducing surplus,artificial variables
Min `Z``=````2000``x_1`` + ``1500``x_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` + ``M``A_1`` + ``M``A_2`` + ``M``A_3`
subject to
```6``x_1`` + ``2``x_2`` - ````S_1`` + ````A_1`=`8`
```2``x_1`` + ``4``x_2`` - ````S_2`` + ````A_2`=`12`
```4``x_1`` + ``12``x_2`` - ````S_3`` + ````A_3`=`24`
and `x_1,x_2,S_1,S_2,S_3,A_1,A_2,A_3 >= 0`


Tableau-1`C_j``2000``1500``0``0``0``M``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS``"Ratio"=(RHS)/(x_2)`
`R_1` `M``A_1``6``2``-1``0``0``1``0``0``8``(8)/(2)=4`
`R_2` `M``A_2``2``4``0``-1``0``0``1``0``12``(12)/(4)=3`
`R_3` `M``A_3``4``(12)``0``0``-1``0``0``1``24``(24)/(12)=2``->`
`Z_j``12M``18M``-M``-M``-M``M``M``M``Z=44M`
`Z_j-C_j``12M-2000``18M-1500``uarr``-M``-M``-M``0``0``0`


Most Positive `Z_j-C_j` is `18M-1500`. So, the entering variable is `x_2`.

Minimum ratio is `2`. So, the leaving basis variable is `A_3`.

`:.` The pivot element is `12`.

Entering `=x_2`, Departing `=A_3`, Key Element `=12`
`R_3`(new)`= R_3`(old) `-: 12`
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_3`(old) = `4``12``0``0``-1``0``0``1``24`
`R_3`(new)`= R_3`(old) `-: 12``0.3333``1``0``0``-0.0833``0``0``0.0833``2`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_1`(old) = `6``2``-1``0``0``1``0``0``8`
`R_3`(new) = `0.3333``1``0``0``-0.0833``0``0``0.0833``2`
`2 xx R_3`(new) = `0.6667``2``0``0``-0.1667``0``0``0.1667``4`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)`5.3333``0``-1``0``0.1667``1``0``-0.1667``4`
`R_2`(new)`= R_2`(old) - `4 R_3`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_2`(old) = `2``4``0``-1``0``0``1``0``12`
`R_3`(new) = `0.3333``1``0``0``-0.0833``0``0``0.0833``2`
`4 xx R_3`(new) = `1.3333``4``0``0``-0.3333``0``0``0.3333``8`
`R_2`(new)`= R_2`(old) - `4 R_3`(new)`0.6667``0``0``-1``0.3333``0``1``-0.3333``4`


Tableau-2`C_j``2000``1500``0``0``0``M``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS``"Ratio"=(RHS)/(x_1)`
`R_1` `M``A_1``(5.3333)``0``-1``0``0.1667``1``0``-0.1667``4``(4)/(5.3333)=0.75``->`
`R_2` `M``A_2``0.6667``0``0``-1``0.3333``0``1``-0.3333``4``(4)/(0.6667)=6`
`R_3` `1500``x_2``0.3333``1``0``0``-0.0833``0``0``0.0833``2``(2)/(0.3333)=6`
`Z_j``6M+500``1500``-M``-M``0.5M-125``M``M``-0.5M+125``Z=8M+3000`
`Z_j-C_j``6M-1500``uarr``0``-M``-M``0.5M-125``0``0``-1.5M+125`


Most Positive `Z_j-C_j` is `6M-1500`. So, the entering variable is `x_1`.

Minimum ratio is `0.75`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `5.3333`.

Entering `=x_1`, Departing `=A_1`, Key Element `=5.3333`
`R_1`(new)`= R_1`(old) `-: 5.3333`
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_1`(old) = `5.3333``0``-1``0``0.1667``1``0``-0.1667``4`
`R_1`(new)`= R_1`(old) `-: 5.3333``1``0``-0.1875``0``0.0312``0.1875``0``-0.0312``0.75`
`R_2`(new)`= R_2`(old) - `0.6667 R_1`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_2`(old) = `0.6667``0``0``-1``0.3333``0``1``-0.3333``4`
`R_1`(new) = `1``0``-0.1875``0``0.0312``0.1875``0``-0.0312``0.75`
`0.6667 xx R_1`(new) = `0.6667``0``-0.125``0``0.0208``0.125``0``-0.0208``0.5`
`R_2`(new)`= R_2`(old) - `0.6667 R_1`(new)`0``0``0.125``-1``0.3125``-0.125``1``-0.3125``3.5`
`R_3`(new)`= R_3`(old) - `0.3333 R_1`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_3`(old) = `0.3333``1``0``0``-0.0833``0``0``0.0833``2`
`R_1`(new) = `1``0``-0.1875``0``0.0312``0.1875``0``-0.0312``0.75`
`0.3333 xx R_1`(new) = `0.3333``0``-0.0625``0``0.0104``0.0625``0``-0.0104``0.25`
`R_3`(new)`= R_3`(old) - `0.3333 R_1`(new)`0``1``0.0625``0``-0.0938``-0.0625``0``0.0938``1.75`


Tableau-3`C_j``2000``1500``0``0``0``M``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS``"Ratio"=(RHS)/(S_3)`
`R_1` `2000``x_1``1``0``-0.1875``0``0.0312``0.1875``0``-0.0312``0.75``(0.75)/(0.0312)=24`
`R_2` `M``A_2``0``0``0.125``-1``(0.3125)``-0.125``1``-0.3125``3.5``(3.5)/(0.3125)=11.2``->`
`R_3` `1500``x_2``0``1``0.0625``0``-0.0938``-0.0625``0``0.0938``1.75``(1.75)/(-0.0938)` (ignore, denominator is -ve)
`Z_j``2000``1500``0.125M-281.25``-M``0.3125M-78.125``-0.125M+281.25``M``-0.3125M+78.125``Z=3.5M+4125`
`Z_j-C_j``0``0``0.125M-281.25``-M``0.3125M-78.125``uarr``-1.125M+281.25``0``-1.3125M+78.125`


Most Positive `Z_j-C_j` is `0.3125M-78.125`. So, the entering variable is `S_3`.

Minimum ratio is `11.2`. So, the leaving basis variable is `A_2`.

`:.` The pivot element is `0.3125`.

Entering `=S_3`, Departing `=A_2`, Key Element `=0.3125`
`R_2`(new)`= R_2`(old) `-: 0.3125`
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_2`(old) = `0``0``0.125``-1``0.3125``-0.125``1``-0.3125``3.5`
`R_2`(new)`= R_2`(old) `-: 0.3125``0``0``0.4``-3.2``1``-0.4``3.2``-1``11.2`
`R_1`(new)`= R_1`(old) - `0.0312 R_2`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_1`(old) = `1``0``-0.1875``0``0.0312``0.1875``0``-0.0312``0.75`
`R_2`(new) = `0``0``0.4``-3.2``1``-0.4``3.2``-1``11.2`
`0.0312 xx R_2`(new) = `0``0``0.0125``-0.1``0.0312``-0.0125``0.1``-0.0312``0.35`
`R_1`(new)`= R_1`(old) - `0.0312 R_2`(new)`1``0``-0.2``0.1``0``0.2``-0.1``0``0.4`
`R_3`(new)`= R_3`(old) + `0.0938 R_2`(new)
`x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS`
`R_3`(old) = `0``1``0.0625``0``-0.0938``-0.0625``0``0.0938``1.75`
`R_2`(new) = `0``0``0.4``-3.2``1``-0.4``3.2``-1``11.2`
`0.0938 xx R_2`(new) = `0``0``0.0375``-0.3``0.0938``-0.0375``0.3``-0.0938``1.05`
`R_3`(new)`= R_3`(old) + `0.0938 R_2`(new)`0``1``0.1``-0.3``0``-0.1``0.3``0``2.8`


Tableau-4`C_j``2000``1500``0``0``0``M``M``M`
`C_B``"Basis"``x_1``x_2``S_1``S_2``S_3``A_1``A_2``A_3``RHS``"Ratio"`
`R_1` `2000``x_1``1``0``-0.2``0.1``0``0.2``-0.1``0``0.4`
`R_2` `0``S_3``0``0``0.4``-3.2``1``-0.4``3.2``-1``11.2`
`R_3` `1500``x_2``0``1``0.1``-0.3``0``-0.1``0.3``0``2.8`
`Z_j``2000``1500``-250``-250``0``250``250``0``Z=5000`
`Z_j-C_j``0``0``-250``-250``0``-M+250``-M+250``-M`


Since all `Z_j-C_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=0.4,x_2=2.8`

Min `Z=5000`


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


22. Minimization example-2 (using `Z_j-C_j` method)
(Previous example)		24. Degeneracy example-1 (Tie for leaving basic variable) (using `Z_j-C_j` method)
(Next example)


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