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Home > Operation Research calculators > Primal to dual conversion example
4. Primal to dual conversion example ( Enter your problem )
( Enter your problem )
  1. Formulation and Rules
  2. Example-1
  3. Example-2
  4. Example-3
  5. Example-4
  6. Example-5
  7. Example-6
 		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
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3. Example-2


Find dual from primal conversion
MIN z = 3x1 - 2x2 + 4x3
subject to
3x1 + 5x2 + 4x3 >= 7
6x1 + x2 + 3x3 >= 4
7x1 - 2x2 - x3 <= 10
x1 - 2x2 + 5x3 >= 3
4x1 + 7x2 - 2x3 >= 2
and x1,x2,x3 >= 0

Solution:
Primal is (Solution steps of Primal by Simplex method)

MIN `z_x``=````3``x_1`` - ``2``x_2`` + ``4``x_3`
subject to
```3``x_1`` + ``5``x_2`` + ``4``x_3`≥`7`
```6``x_1`` + ````x_2`` + ``3``x_3`≥`4`
```7``x_1`` - ``2``x_2`` - ````x_3`≤`10`
`````x_1`` - ``2``x_2`` + ``5``x_3`≥`3`
```4``x_1`` + ``7``x_2`` - ``2``x_3`≥`2`
and `x_1,x_2,x_3 >= 0; `


Since objective function is minimizing, all `<=` constraints (3) can be converted to `>=` type by multipling both sides by -1

MIN `z_x``=````3``x_1`` - ``2``x_2`` + ``4``x_3`
subject to
```3``x_1`` + ``5``x_2`` + ``4``x_3`≥`7`
```6``x_1`` + ````x_2`` + ``3``x_3`≥`4`
` - ``7``x_1`` + ``2``x_2`` + ````x_3`≥`-10`
`````x_1`` - ``2``x_2`` + ``5``x_3`≥`3`
```4``x_1`` + ``7``x_2`` - ``2``x_3`≥`2`
and `x_1,x_2,x_3 >= 0; `


In primal, There are `3` variables and `5` constraints, so in dual there must be `3` constraints and `5` variables

In primal, The coefficient of objective function `c_1=3,c_2=-2,c_3=4` becomes right hand side constants in dual

In primal, The right hand side constants `b_1=7,b_2=4,b_3=-10,b_4=3,b_5=2` becomes coefficient of objective function in dual

In primal, objective function is minimizing, so in dual objective function must be maximizing

Let `y1,y2,y3,y4,y5` be the dual variables

Dual is (Solution steps of Dual by Simplex method)

MAX `z_y``=````7``y_1`` + ``4``y_2`` - ``10``y_3`` + ``3``y_4`` + ``2``y_5`
subject to
```3``y_1`` + ``6``y_2`` - ``7``y_3`` + ````y_4`` + ``4``y_5`≤`3`
```5``y_1`` + ````y_2`` + ``2``y_3`` - ``2``y_4`` + ``7``y_5`≤`-2`
```4``y_1`` + ``3``y_2`` + ````y_3`` + ``5``y_4`` - ``2``y_5`≤`4`
and `y_1,y_2,y_3,y_4,y_5 >= 0; `


This material is intended as a summary. Use your textbook for detail explanation.
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