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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
14. Infeasible solution example
(Previous example)		4. Two-Phase method
(Next method)

15. Unbounded solution example


Unbounded solution
In simplex table, if a variable should enter into the basis, but all the coefficients in that column are negative or zero. So this variable can not be entered into the basis, because for minimum ratio, negative value in denominator can not be considered and zero value in denominator would result `oo`.
Hence, the solution to the given problem is unbounded.
Example
Find solution using Simplex(BigM) method
MAX Z = 3x1 + 5x2
subject to
x1 - 2x2 <= 6
x1 <= 10
x2 >= 1
and x1,x2 >= 0

Solution:
Problem is
Max `Z``=````3``x_1`` + ``5``x_2`
subject to
`````x_1`` - ``2``x_2`≤`6`
`````x_1`≤`10`
`````x_2`≥`1`
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``=````3``x_1`` + ``5``x_2`` + ``0``S_1`` + ``0``S_2`` + ``0``S_3`` - ``M``A_1`
subject to
`````x_1`` - ``2``x_2`` + ````S_1`=`6`
`````x_1`` + ````S_2`=`10`
`````x_2`` - ````S_3`` + ````A_1`=`1`
and `x_1,x_2,S_1,S_2,S_3,A_1 >= 0`


Iteration-1 `C_j``3``5``0``0``0``-M`
`B``C_B``X_B``x_1` `x_2` Entering variable`S_1``S_2``S_3``A_1`MinRatio
`(X_B)/(x_2)`
`S_1``0``6``1``-2``1``0``0``0`---
`S_2``0``10``1``0``0``1``0``0`---
 `A_1` Leaving variable`-M``1``0` `(1)`  (pivot element)`0``0``-1``1``(1)/(1)=1``->`
 `Z=0` `0=`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `0` `0=0xx1+0xx1+(-M)xx0`
`Z_j=sum C_B x_1`		 `-M` `-M=0xx(-2)+0xx0+(-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` `3` `3=3-0` `M+5` `M+5=5-(-M)``uarr` `0` `0=0-0` `0` `0=0-0` `-M` `-M=0-M` `0` `0=(-M)-(-M)`


Positive maximum `C_j-Z_j` is `M+5` and its column index is `2`. So, the entering variable is `x_2`.

Minimum ratio is `1` and its row index is `3`. So, the leaving basis variable is `A_1`.

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

Entering `=x_2`, 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)

Iteration-2 `C_j``3``5``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` `8` `8=6+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=(-2)+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)` (pivot element)
`R_1`(new)`= R_1`(old)`+ 2 R_3`(new)		---
`S_2``0` `10` `10=10`
`R_2`(new)`= R_2`(old)		 `1` `1=1`
`R_2`(new)`= R_2`(old)		 `0` `0=0`
`R_2`(new)`= R_2`(old)		 `0` `0=0`
`R_2`(new)`= R_2`(old)		 `1` `1=1`
`R_2`(new)`= R_2`(old)		 `0` `0=0`
`R_2`(new)`= R_2`(old)		---
`x_2``5` `1` `1=1`
`R_3`(new)`= R_3`(old)		 `0` `0=0`
`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=5` `5=5xx1`
`Z_j=sum C_B X_B`		  `Z_j` `Z_j=sum C_B x_j` `0` `0=0xx1+0xx1+5xx0`
`Z_j=sum C_B x_1`		 `5` `5=0xx0+0xx0+5xx1`
`Z_j=sum C_B x_2`		 `0` `0=0xx1+0xx0+5xx0`
`Z_j=sum C_B S_1`		 `0` `0=0xx0+0xx1+5xx0`
`Z_j=sum C_B S_2`		 `-5` `-5=0xx(-2)+0xx0+5xx(-1)`
`Z_j=sum C_B S_3`
`C_j-Z_j` `3` `3=3-0` `0` `0=5-5` `0` `0=0-0` `0` `0=0-0` `5` `5=0-(-5)``uarr`


Variable `S_3` should enter into the basis, but all the coefficients in the `S_3` column are negative or zero. So `S_3` can not be entered into the basis.

Hence, the solution to the given problem is unbounded.


This material is intended as a summary. Use your textbook for detail explanation.
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14. Infeasible solution example
(Previous example)		4. Two-Phase method
(Next method)


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