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4. Primal to dual conversion example ( 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

1. Formulation and Rules
(Previous example)
3. Example-2
(Next example)

2. Example-1





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


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

MAX `z_x``=``````x_1`` - ````x_2`` + ``3``x_3`
subject to
`````x_1`` + ````x_2`` + ````x_3``10`
```2``x_1`` - ````x_2`` - ````x_3``2`
```2``x_1`` - ``2``x_2`` - ``3``x_3``6`
and `x_1,x_2,x_3 >= 0; `


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

In primal, The coefficient of objective function `c_1=1,c_2=-1,c_3=3` becomes right hand side constants in dual

In primal, The right hand side constants `b_1=10,b_2=2,b_3=6` becomes coefficient of objective function in dual

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

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

Dual is (Solution steps of Dual by Simplex method)

MIN `z_y``=````10``y_1`` + ``2``y_2`` + ``6``y_3`
subject to
`````y_1`` + ``2``y_2`` + ``2``y_3``1`
`````y_1`` - ````y_2`` - ``2``y_3``-1`
`````y_1`` - ````y_2`` - ``3``y_3``3`
and `y_1,y_2,y_3 >= 0; `



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1. Formulation and Rules
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