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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
7. Minimization example-1
(Previous example)		9. Minimization example-3
(Next example)

8. Minimization example-2


Find solution using Simplex method (BigM method)
MIN Z = 5x1 + 3x2
subject to
2x1 + 4x2 <= 12
2x1 + 2x2 = 10
5x1 + 2x2 >= 10
and x1,x2 >= 0

Solution:
Problem is
Min `Z``=````5``x_1`` + ``3``x_2`
subject to
```2``x_1`` + ``4``x_2`≤`12`
```2``x_1`` + ``2``x_2`=`10`
```5``x_1`` + ``2``x_2`≥`10`
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 artificial variable `A_1`

3. As the constraint-3 is of type '`>=`' we should subtract surplus variable `S_2` and add artificial variable `A_2`

After introducing slack,surplus,artificial variables
Min `Z``=````5``x_1`` + ``3``x_2`` + ``0``S_1`` + ``0``S_2`` + ``M``A_1`` + ``M``A_2`
subject to
```2``x_1`` + ``4``x_2`` + ````S_1`=`12`
```2``x_1`` + ``2``x_2`` + ````A_1`=`10`
```5``x_1`` + ``2``x_2`` - ````S_2`` + ````A_2`=`10`
and `x_1,x_2,S_1,S_2,A_1,A_2 >= 0`


Iteration-1 `C_j``5``3``0``0``M``M`
`B``C_B``X_B``x_1``x_2``S_1``S_2``A_1``A_2`MinRatio
`(X_B)/(x_1)`
`S_1``0``12``2``4``1``0``0``0``(12)/(2)=6`
`A_1``M``10``2``2``0``0``1``0``(10)/(2)=5`
`A_2``M``10``(5)``2``0``-1``0``1``(10)/(5)=2``->`
`Z=20M` `Z_j``7M``4M``0``-M``M``M`
`Z_j-C_j``7M-5``uarr``4M-3``0``-M``0``0`


Positive maximum `Z_j-C_j` is `7M-5` and its column index is `1`. So, the entering variable is `x_1`.

Minimum ratio is `2` and its row index is `3`. So, the leaving basis variable is `A_2`.

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

Entering `=x_1`, Departing `=A_2`, Key Element `=5`

`R_3`(new)`= R_3`(old) `-: 5`
`R_3`(old) = `10``5``2``0``-1``0`
`R_3`(new)`= R_3`(old) `-: 5``2``1``2/5``0``-1/5``0`


`R_1`(new)`= R_1`(old) - `2 R_3`(new)
`R_1`(old) = `12``2``4``1``0``0`
`R_3`(new) = `2``1``2/5``0``-1/5``0`
`2 xx R_3`(new) = `4``2``4/5``0``-2/5``0`
`R_1`(new)`= R_1`(old) - `2 R_3`(new)`8``0``16/5``1``2/5``0`


`R_2`(new)`= R_2`(old) - `2 R_3`(new)
`R_2`(old) = `10``2``2``0``0``1`
`R_3`(new) = `2``1``2/5``0``-1/5``0`
`2 xx R_3`(new) = `4``2``4/5``0``-2/5``0`
`R_2`(new)`= R_2`(old) - `2 R_3`(new)`6``0``6/5``0``2/5``1`


Iteration-2 `C_j``5``3``0``0``M`
`B``C_B``X_B``x_1``x_2``S_1``S_2``A_1`MinRatio
`(X_B)/(x_2)`
`S_1``0``8``0``(16/5)``1``2/5``0``(8)/(16/5)=5/2=2.5``->`
`A_1``M``6``0``6/5``0``2/5``1``(6)/(6/5)=5`
`x_1``5``2``1``2/5``0``-1/5``0``(2)/(2/5)=5`
`Z=6M+10` `Z_j``5``(6M)/(5)+2``0``(2M)/(5)-1``M`
`Z_j-C_j``0``(6M)/(5)-1``uarr``0``(2M)/(5)-1``0`


Positive maximum `Z_j-C_j` is `(6M)/(5)-1` and its column index is `2`. So, the entering variable is `x_2`.

Minimum ratio is `2.5` and its row index is `1`. So, the leaving basis variable is `S_1`.

`:.` The pivot element is `16/5`.

Entering `=x_2`, Departing `=S_1`, Key Element `=16/5`

`R_1`(new)`= R_1`(old) `xx5/16`
`R_1`(old) = `8``0``16/5``1``2/5``0`
`R_1`(new)`= R_1`(old) `xx5/16``5/2``0``1``5/16``1/8``0`


`R_2`(new)`= R_2`(old) - `6/5 R_1`(new)
`R_2`(old) = `6``0``6/5``0``2/5``1`
`R_1`(new) = `5/2``0``1``5/16``1/8``0`
`6/5 xx R_1`(new) = `3``0``6/5``3/8``3/20``0`
`R_2`(new)`= R_2`(old) - `6/5 R_1`(new)`3``0``0``-3/8``1/4``1`


`R_3`(new)`= R_3`(old) - `2/5 R_1`(new)
`R_3`(old) = `2``1``2/5``0``-1/5``0`
`R_1`(new) = `5/2``0``1``5/16``1/8``0`
`2/5 xx R_1`(new) = `1``0``2/5``1/8``1/20``0`
`R_3`(new)`= R_3`(old) - `2/5 R_1`(new)`1``1``0``-1/8``-1/4``0`


Iteration-3 `C_j``5``3``0``0``M`
`B``C_B``X_B``x_1``x_2``S_1``S_2``A_1`MinRatio
`(X_B)/(S_2)`
`x_2``3``5/2``0``1``5/16``1/8``0``(5/2)/(1/8)=20`
`A_1``M``3``0``0``-3/8``(1/4)``1``(3)/(1/4)=12``->`
`x_1``5``1``1``0``-1/8``-1/4``0`---
`Z=3M+25/2` `Z_j``5``3``-(3M)/(8)+5/16``(M)/(4)-7/8``M`
`Z_j-C_j``0``0``-(3M)/(8)+5/16``(M)/(4)-7/8``uarr``0`


Positive maximum `Z_j-C_j` is `(M)/(4)-7/8` and its column index is `4`. So, the entering variable is `S_2`.

Minimum ratio is `12` and its row index is `2`. So, the leaving basis variable is `A_1`.

`:.` The pivot element is `1/4`.

Entering `=S_2`, Departing `=A_1`, Key Element `=1/4`

`R_2`(new)`= R_2`(old) `xx4`
`R_2`(old) = `3``0``0``-3/8``1/4`
`R_2`(new)`= R_2`(old) `xx4``12``0``0``-3/2``1`


`R_1`(new)`= R_1`(old) - `1/8 R_2`(new)
`R_1`(old) = `5/2``0``1``5/16``1/8`
`R_2`(new) = `12``0``0``-3/2``1`
`1/8 xx R_2`(new) = `3/2``0``0``-3/16``1/8`
`R_1`(new)`= R_1`(old) - `1/8 R_2`(new)`1``0``1``1/2``0`


`R_3`(new)`= R_3`(old) + `1/4 R_2`(new)
`R_3`(old) = `1``1``0``-1/8``-1/4`
`R_2`(new) = `12``0``0``-3/2``1`
`1/4 xx R_2`(new) = `3``0``0``-3/8``1/4`
`R_3`(new)`= R_3`(old) + `1/4 R_2`(new)`4``1``0``-1/2``0`


Iteration-4 `C_j``5``3``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2`MinRatio
`x_2``3``1``0``1``1/2``0`
`S_2``0``12``0``0``-3/2``1`
`x_1``5``4``1``0``-1/2``0`
`Z=23` `Z_j``5``3``-1``0`
`Z_j-C_j``0``0``-1``0`


Since all `Z_j-C_j <= 0`

Hence, optimal solution is arrived with value of variables as :
`x_1=4,x_2=1`

Min `Z=23`

This material is intended as a summary. Use your textbook for detail explanation.
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7. Minimization example-1
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