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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
1. Algorithm
(Previous example)		3. Example-2
(Next example)

2. Example-1


1. Find solution using dual-simplex method
MAX Z = -2x1 - x2
subject to
-3x1 - x2 <= -3
-4x1 - 3x2 <= -6
-x1 - 2x2 <= -3
and x1,x2 >= 0

Solution:
Problem is
Max `Z``=`` - ``2``x_1`` - ````x_2`
subject to
` - ``3``x_1`` - ````x_2`≤`-3`
` - ``4``x_1`` - ``3``x_2`≤`-6`
` - ````x_1`` - ``2``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 add slack variable `S_3`

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


Iteration-1 `C_j``-2``-1``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2``S_3`
`S_1``0``-3``-3``-1``1``0``0`
`S_2``0``-6``-4``(-3)``0``1``0`
`S_3``0``-3``-1``-2``0``0``1`
`Z=0` `Z_j``0``0``0``0``0`
`Z_j-C_j``2``1``0``0``0`
`"Ratio"=(Z_j-C_j)/(S_2,j)`
and `S_2,j<0`		`-0.5``-0.3333``uarr`---------


Minimum negative `X_B` is `-6` and its row index is `2`. So, the leaving basis variable is `S_2`.

Maximum negative ratio is `-0.3333` and its column index is `2`. So, the entering variable is `x_2`.

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

Entering `=x_2`, Departing `=S_2`, Key Element `=-3`

`R_2`(new)`= R_2`(old) `-: (-3)`
`R_2`(old) = `-6``-4``-3``0``1``0`
`R_2`(new)`= R_2`(old) `-: (-3)``2``4/3``1``0``-1/3``0`


`R_1`(new)`= R_1`(old) + `R_2`(new)
`R_1`(old) = `-3``-3``-1``1``0``0`
`R_2`(new) = `2``4/3``1``0``-1/3``0`
`R_1`(new)`= R_1`(old) + `R_2`(new)`-1``-5/3``0``1``-1/3``0`


`R_3`(new)`= R_3`(old) + `2 R_2`(new)
`R_3`(old) = `-3``-1``-2``0``0``1`
`R_2`(new) = `2``4/3``1``0``-1/3``0`
`2 xx R_2`(new) = `4``8/3``2``0``-2/3``0`
`R_3`(new)`= R_3`(old) + `2 R_2`(new)`1``5/3``0``0``-2/3``1`


Iteration-2 `C_j``-2``-1``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2``S_3`
`S_1``0``-1``(-5/3)``0``1``-1/3``0`
`x_2``-1``2``4/3``1``0``-1/3``0`
`S_3``0``1``5/3``0``0``-2/3``1`
`Z=-2` `Z_j``-4/3``-1``0``1/3``0`
`Z_j-C_j``2/3``0``0``1/3``0`
`"Ratio"=(Z_j-C_j)/(S_1,j)`
and `S_1,j<0`		`-0.4``uarr`------`-1`---


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

Maximum negative ratio is `-0.4` and its column index is `1`. So, the entering variable is `x_1`.

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

Entering `=x_1`, Departing `=S_1`, Key Element `=-5/3`

`R_1`(new)`= R_1`(old) `xx(-3/5)`
`R_1`(old) = `-1``-5/3``0``1``-1/3``0`
`R_1`(new)`= R_1`(old) `xx(-3/5)``3/5``1``0``-3/5``1/5``0`


`R_2`(new)`= R_2`(old) - `4/3 R_1`(new)
`R_2`(old) = `2``4/3``1``0``-1/3``0`
`R_1`(new) = `3/5``1``0``-3/5``1/5``0`
`4/3 xx R_1`(new) = `4/5``4/3``0``-4/5``4/15``0`
`R_2`(new)`= R_2`(old) - `4/3 R_1`(new)`6/5``0``1``4/5``-3/5``0`


`R_3`(new)`= R_3`(old) - `5/3 R_1`(new)
`R_3`(old) = `1``5/3``0``0``-2/3``1`
`R_1`(new) = `3/5``1``0``-3/5``1/5``0`
`5/3 xx R_1`(new) = `1``5/3``0``-1``1/3``0`
`R_3`(new)`= R_3`(old) - `5/3 R_1`(new)`0``0``0``1``-1``1`


Iteration-3 `C_j``-2``-1``0``0``0`
`B``C_B``X_B``x_1``x_2``S_1``S_2``S_3`
`x_1``-2``3/5``1``0``-3/5``1/5``0`
`x_2``-1``6/5``0``1``4/5``-3/5``0`
`S_3``0``0``0``0``1``-1``1`
`Z=-12/5` `Z_j``-2``-1``2/5``1/5``0`
`Z_j-C_j``0``0``2/5``1/5``0`
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=3/5,x_2=6/5`

Max `Z=-12/5`

This material is intended as a summary. Use your textbook for detail explanation.
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1. Algorithm
(Previous example)		3. Example-2
(Next example)


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