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Home > Operation Research calculators > Dual Simplex method example
5. Dual simplex method example ( Enter your problem )
( Enter your problem )
  1. Algorithm
  2. Example-1
  3. Example-2
  4. Example-3
 		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
2. Example-1
(Previous example)		4. Example-3
(Next example)

3. Example-2


Find solution using dual-simplex method
MIN Z = 2x1 + 3x2 + 0x3
subject to
2x1 - x2 - x3 >= 3
x1 - x2 + x3 >= 2
and x1,x2,x3 >= 0

Solution:
Problem is
Min `Z``=````2``x_1`` + ``3``x_2`
subject to
```2``x_1`` - ````x_2`` - ````x_3`≥`3`
`````x_1`` - ````x_2`` + ````x_3`≥`2`
and `x_1,x_2,x_3 >= 0; `


In order to apply the dual simplex method, convert Min Z to Max Z and all `>=` constraint to `<=` constraint by multiply -1.

Problem is
Max `Z``=`` - ``2``x_1`` - ``3``x_2`
subject to
` - ``2``x_1`` + ````x_2`` + ````x_3`≤`-3`
` - ````x_1`` + ````x_2`` - ````x_3`≤`-2`
and `x_1,x_2,x_3 >= 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`

After introducing slack variables
Max `Z``=`` - ``2``x_1`` - ``3``x_2`` + ``0``x_3`` + ``0``S_1`` + ``0``S_2`
subject to
` - ``2``x_1`` + ````x_2`` + ````x_3`` + ````S_1`=`-3`
` - ````x_1`` + ````x_2`` - ````x_3`` + ````S_2`=`-2`
and `x_1,x_2,x_3,S_1,S_2 >= 0`


Iteration-1 `C_j``-2``-3``0``0``0`
`B``C_B``X_B``x_1``x_2``x_3``S_1``S_2`
`S_1``0``-3``(-2)``1``1``1``0`
`S_2``0``-2``-1``1``-1``0``1`
`Z=0` `Z_j``0``0``0``0``0`
`Z_j-C_j``2``3``0``0``0`
`"Ratio"=(Z_j-C_j)/(S_1,j)`
and `S_1,j<0`		`-1``uarr`------------


Minimum negative `X_B` is `-3` and its row index is `1`. So, the leaving basis variable is `S_1`.

Maximum negative ratio is `-1` and its column index is `1`. So, the entering variable is `x_1`.

`:.` The pivot element is `-2`.

Entering `=x_1`, Departing `=S_1`, Key Element `=-2`

`R_1`(new)`= R_1`(old) `-: (-2)`
`R_1`(old) = `-3``-2``1``1``1``0`
`R_1`(new)`= R_1`(old) `-: (-2)``3/2``1``-1/2``-1/2``-1/2``0`


`R_2`(new)`= R_2`(old) + `R_1`(new)
`R_2`(old) = `-2``-1``1``-1``0``1`
`R_1`(new) = `3/2``1``-1/2``-1/2``-1/2``0`
`R_2`(new)`= R_2`(old) + `R_1`(new)`-1/2``0``1/2``-3/2``-1/2``1`


Iteration-2 `C_j``-2``-3``0``0``0`
`B``C_B``X_B``x_1``x_2``x_3``S_1``S_2`
`x_1``-2``3/2``1``-1/2``-1/2``-1/2``0`
`S_2``0``-1/2``0``1/2``(-3/2)``-1/2``1`
`Z=-3` `Z_j``-2``1``1``1``0`
`Z_j-C_j``0``4``1``1``0`
`"Ratio"=(Z_j-C_j)/(S_2,j)`
and `S_2,j<0`		------`-0.6667``uarr``-2`---


Minimum negative `X_B` is `-1/2` and its row index is `2`. So, the leaving basis variable is `S_2`.

Maximum negative ratio is `-0.6667` and its column index is `3`. So, the entering variable is `x_3`.

`:.` The pivot element is `-3/2`.

Entering `=x_3`, Departing `=S_2`, Key Element `=-3/2`

`R_2`(new)`= R_2`(old) `xx(-2/3)`
`R_2`(old) = `-1/2``0``1/2``-3/2``-1/2``1`
`R_2`(new)`= R_2`(old) `xx(-2/3)``1/3``0``-1/3``1``1/3``-2/3`


`R_1`(new)`= R_1`(old) + `1/2 R_2`(new)
`R_1`(old) = `3/2``1``-1/2``-1/2``-1/2``0`
`R_2`(new) = `1/3``0``-1/3``1``1/3``-2/3`
`1/2 xx R_2`(new) = `1/6``0``-1/6``1/2``1/6``-1/3`
`R_1`(new)`= R_1`(old) + `1/2 R_2`(new)`5/3``1``-2/3``0``-1/3``-1/3`


Iteration-3 `C_j``-2``-3``0``0``0`
`B``C_B``X_B``x_1``x_2``x_3``S_1``S_2`
`x_1``-2``5/3``1``-2/3``0``-1/3``-1/3`
`x_3``0``1/3``0``-1/3``1``1/3``-2/3`
`Z=-10/3` `Z_j``-2``4/3``0``2/3``2/3`
`Z_j-C_j``0``13/3``0``2/3``2/3`
Ratio---------------


Since all `Z_j-C_j >= 0` and all `X_(Bi) >= 0` thus the current solution is the optimal solution.

Hence, optimal solution is arrived with value of variables as :
`x_1=5/3,x_2=0,x_3=1/3`

Max `Z=-10/3`

`:.` Min `Z=10/3`

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


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