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Home > Operation Research calculators > Simplex method example
2. Simplex method example ( Enter your problem )
( Enter your problem )
  1. Structure of Linear programming problem
  2. Algorithm
  3. Maximization example-1
  4. Maximization example-2
  5. Maximization example-3
  6. BigM method Algorithm
  7. Minimization example-1
  8. Minimization example-2
  9. Minimization example-3
  10. Degeneracy example-1 (Tie for leaving basic variable)
  11. Degeneracy example-2 (Tie first Artificial variable removed)
  12. Unrestricted variable example
  13. Multiple optimal solution example
  14. Infeasible solution example
  15. Unbounded solution example
 		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
12. Unrestricted variable example
(Previous example)		14. Infeasible solution example
(Next example)

13. Multiple optimal solution example


Multiple optimal solution
In the final simplex table when all `c_j - z_j` imply optimal solution (for maximization all `c_j - z_j <= 0` and for minimization all `c_j - z_j >= 0`)
but if `c_j - z_j = 0` for some non-basic variable column, then this indicates that there are more than 1 optimal solution of the problem. Thus by entering this variable into the basis, we may obtain another alternative optimal solution.
Example
Find solution using Simplex(BigM) method
MAX Z = 6x1 + 4x2
subject to
2x1 + 3x2 <= 30
3x1 + 2x2 <= 24
x1 + x2 >= 3
and x1,x2 >= 0

Solution:
Problem is
Max `Z``=````6``x_1`` + ``4``x_2`
subject to
```2``x_1`` + ``3``x_2`≤`30`
```3``x_1`` + ``2``x_2`≤`24`
`````x_1`` + ````x_2`≥`3`
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 subtract surplus variable `S_3` and add artificial variable `A_1`

After introducing slack,surplus,artificial variables
Max `Z``=````6``x_1`` + ``4``x_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
```2``x_1`` + ``3``x_2`` + ````S_1`=`30`
```3``x_1`` + ``2``x_2`` + ````S_2`=`24`
`````x_1`` + ````x_2`` - ````S_3`` + ````A_1`=`3`
and `x_1,x_2,S_1,S_2,S_3,A_1 >= 0`


Iteration-1 `C_j``6``4``0``0``0``-M`
`B``C_B``X_B` `x_1` Entering variable`x_2``S_1``S_2``S_3``A_1`MinRatio
`(X_B)/(x_1)`
`S_1``0``30``2``3``1``0``0``0``(30)/(2)=15`
`S_2``0``24``3``2``0``1``0``0``(24)/(3)=8`
 `A_1` Leaving variable`-M``3` `(1)`  (pivot element)`1``0``0``-1``1``(3)/(1)=3``->`
 `Z=0` `0=`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `-M` `-M=0xx2+0xx3+(-M)xx1`
`Z_j=sum C_B x_1`		 `-M` `-M=0xx3+0xx2+(-M)xx1`
`Z_j=sum C_B x_2`		 `0` `0=0xx1+0xx0+(-M)xx0`
`Z_j=sum C_B S_1`		 `0` `0=0xx0+0xx1+(-M)xx0`
`Z_j=sum C_B S_2`		 `M` `M=0xx0+0xx0+(-M)xx(-1)`
`Z_j=sum C_B S_3`		 `-M` `-M=0xx0+0xx0+(-M)xx1`
`Z_j=sum C_B A_1`
`C_j-Z_j` `M+6` `M+6=6-(-M)``uarr` `M+4` `M+4=4-(-M)` `0` `0=0-0` `0` `0=0-0` `-M` `-M=0-M` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `M+6` and its column index is `1`. So, the entering variable is `x_1`.

Minimum ratio is `3` and its row index is `3`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `1`.

Entering `=x_1`, Departing `=A_1`, Key Element `=1`

`R_3`(new)`= R_3`(old)

`R_1`(new)`= R_1`(old)`- 2 R_3`(new)

`R_2`(new)`= R_2`(old)`- 3 R_3`(new)

Iteration-2 `C_j``6``4``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2` `S_3` Entering variableMinRatio
`(X_B)/(S_3)`
`S_1``0` `24` `24=30-2xx3`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		 `0` `0=2-2xx1`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		 `1` `1=3-2xx1`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		 `1` `1=1-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		 `2` `2=0-2xx(-1)`
`R_1`(new)`= R_1`(old)`- 2 R_3`(new)		`(24)/(2)=12`
 `S_2` Leaving variable`0` `15` `15=24-3xx3`
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		 `0` `0=3-3xx1`
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		 `-1` `-1=2-3xx1`
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		 `0` `0=0-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		 `1` `1=1-3xx0`
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		 `(3)` `3=0-3xx(-1)` (pivot element)
`R_2`(new)`= R_2`(old)`- 3 R_3`(new)		`(15)/(3)=5``->`
`x_1``6` `3` `3=3`
`R_3`(new)`= R_3`(old)		 `1` `1=1`
`R_3`(new)`= R_3`(old)		 `1` `1=1`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`R_3`(new)`= R_3`(old)		 `-1` `-1=-1`
`R_3`(new)`= R_3`(old)		---
 `Z=18` `18=6xx3`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `6` `6=0xx0+0xx0+6xx1`
`Z_j=sum C_B x_1`		 `6` `6=0xx1+0xx(-1)+6xx1`
`Z_j=sum C_B x_2`		 `0` `0=0xx1+0xx0+6xx0`
`Z_j=sum C_B S_1`		 `0` `0=0xx0+0xx1+6xx0`
`Z_j=sum C_B S_2`		 `-6` `-6=0xx2+0xx3+6xx(-1)`
`Z_j=sum C_B S_3`
`C_j-Z_j` `0` `0=6-6` `-2` `-2=4-6` `0` `0=0-0` `0` `0=0-0` `6` `6=0-(-6)``uarr`


Positive maximum `C_j-Z_j` is `6` and its column index is `5`. So, the entering variable is `S_3`.

Minimum ratio is `5` and its row index is `2`. So, the leaving basis variable is `S_2`.

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

Entering `=S_3`, Departing `=S_2`, Key Element `=3`

`R_2`(new)`= R_2`(old)`-: 3`

`R_1`(new)`= R_1`(old)`- 2 R_2`(new)

`R_3`(new)`= R_3`(old)`+ R_2`(new)

Iteration-3 `C_j``6``4``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2``S_3`MinRatio
`S_1``0` `14` `14=24-2xx5`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `5/3` `5/3=1-2xx(-1/3)`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `1` `1=1-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `-2/3` `-2/3=0-2xx1/3`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `0` `0=2-2xx1`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)
`S_3``0` `5` `5=15-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `0` `0=0-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `-1/3` `-1/3=(-1)-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `0` `0=0-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `1/3` `1/3=1-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `1` `1=3-:3`
`R_2`(new)`= R_2`(old)`-: 3`
`x_1``6` `8` `8=3+5`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `1` `1=1+0`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `2/3` `2/3=1+(-1/3)`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `0` `0=0+0`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `1/3` `1/3=0+1/3`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `0` `0=(-1)+1`
`R_3`(new)`= R_3`(old)`+ R_2`(new)
 `Z=48` `48=6xx8`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `6` `6=0xx0+0xx0+6xx1`
`Z_j=sum C_B x_1`		 `4` `4=0xx5/3+0xx(-1/3)+6xx2/3`
`Z_j=sum C_B x_2`		 `0` `0=0xx1+0xx0+6xx0`
`Z_j=sum C_B S_1`		 `2` `2=0xx(-2/3)+0xx1/3+6xx1/3`
`Z_j=sum C_B S_2`		 `0` `0=0xx0+0xx1+6xx0`
`Z_j=sum C_B S_3`
`C_j-Z_j` `0` `0=6-6` `0` `0=4-4` `0` `0=0-0` `-2` `-2=0-2` `0` `0=0-0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=8,x_2=0`

Max `Z = 48`


Here `C_2-Z_2=0` and `x_2` is not in the basis (i.e. `x_2=0`).

This indicates that there are more than 1 optimal solution of the problem.
Thus by entering `x_2` into the basis, we may obtain another alternative optimal solution.

Iteration-3 `C_j``6``4``0``0``0`
`B``C_B``X_B``x_1` `x_2` Entering variable`S_1``S_2``S_3`MinRatio
`(X_B)/(x_2)`
 `S_1` Leaving variable`0` `14` `14=24-2xx5`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `0` `0=0-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `(5/3)` `5/3=1-2xx(-1/3)` (pivot element)
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `1` `1=1-2xx0`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `-2/3` `-2/3=0-2xx1/3`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		 `0` `0=2-2xx1`
`R_1`(new)`= R_1`(old)`- 2 R_2`(new)		`(14)/(5/3)=8.4``->`
`S_3``0` `5` `5=15-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `0` `0=0-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `-1/3` `-1/3=(-1)-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `0` `0=0-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `1/3` `1/3=1-:3`
`R_2`(new)`= R_2`(old)`-: 3`		 `1` `1=3-:3`
`R_2`(new)`= R_2`(old)`-: 3`		---
`x_1``6` `8` `8=3+5`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `1` `1=1+0`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `2/3` `2/3=1+(-1/3)`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `0` `0=0+0`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `1/3` `1/3=0+1/3`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		 `0` `0=(-1)+1`
`R_3`(new)`= R_3`(old)`+ R_2`(new)		`(8)/(2/3)=12`
 `Z=48` `48=6xx8`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `6` `6=0xx0+0xx0+6xx1`
`Z_j=sum C_B x_1`		 `4` `4=0xx5/3+0xx(-1/3)+6xx2/3`
`Z_j=sum C_B x_2`		 `0` `0=0xx1+0xx0+6xx0`
`Z_j=sum C_B S_1`		 `2` `2=0xx(-2/3)+0xx1/3+6xx1/3`
`Z_j=sum C_B S_2`		 `0` `0=0xx0+0xx1+6xx0`
`Z_j=sum C_B S_3`
`C_j-Z_j` `0` `0=6-6` `0` `0=4-4``uarr` `0` `0=0-0` `-2` `-2=0-2` `0` `0=0-0`


So, the entering variable is `x_2`.

Minimum ratio is `8.4` and its row index is `1`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `5/3`.

Entering `=x_2`, Departing `=S_1`, Key Element `=5/3`

`R_1`(new)`= R_1`(old)`xx3/5`

`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)

`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)

Iteration-4 `C_j``6``4``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2``S_3`MinRatio
`x_2``4` `42/5` `42/5=14xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`		 `0` `0=0xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`		 `1` `1=5/3xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`		 `3/5` `3/5=1xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`		 `-2/5` `-2/5=(-2/3)xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`		 `0` `0=0xx3/5`
`R_1`(new)`= R_1`(old)`xx3/5`
`S_3``0` `39/5` `39/5=5+1/3xx42/5`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)		 `0` `0=0+1/3xx0`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)		 `0` `0=(-1/3)+1/3xx1`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)		 `1/5` `1/5=0+1/3xx3/5`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)		 `1/5` `1/5=1/3+1/3xx(-2/5)`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)		 `1` `1=1+1/3xx0`
`R_2`(new)`= R_2`(old)`+ 1/3 R_1`(new)
`x_1``6` `12/5` `12/5=8-2/3xx42/5`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)		 `1` `1=1-2/3xx0`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)		 `0` `0=2/3-2/3xx1`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)		 `-2/5` `-2/5=0-2/3xx3/5`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)		 `3/5` `3/5=1/3-2/3xx(-2/5)`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)		 `0` `0=0-2/3xx0`
`R_3`(new)`= R_3`(old)`- 2/3 R_1`(new)
 `Z=48` `48=4xx42/5+6xx12/5`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `6` `6=4xx0+0xx0+6xx1`
`Z_j=sum C_B x_1`		 `4` `4=4xx1+0xx0+6xx0`
`Z_j=sum C_B x_2`		 `0` `0=4xx3/5+0xx1/5+6xx(-2/5)`
`Z_j=sum C_B S_1`		 `2` `2=4xx(-2/5)+0xx1/5+6xx3/5`
`Z_j=sum C_B S_2`		 `0` `0=4xx0+0xx1+6xx0`
`Z_j=sum C_B S_3`
`C_j-Z_j` `0` `0=6-6` `0` `0=4-4` `0` `0=0-0` `-2` `-2=0-2` `0` `0=0-0`


Since all `C_j-Z_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=12/5,x_2=42/5`

Max `Z = 48`



This material is intended as a summary. Use your textbook for detail explanation.
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12. Unrestricted variable example
(Previous example)		14. Infeasible solution example
(Next example)


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