Constructing Dual Problem
It has been stated earlier that every linear programming problem has its dual. the procedure of writing he dual to the given primal problem will be discussed.
If the primal problem is a maximization problem then for the prupose of writing its dual all constraints of the problem must be converted to less than or equal to type. In other words if primal problem (maximization) involves mixture of constraints i.e., involves greater then or equal to or equality constraints they all must be converted to 'less than or equal to type for writing the dual.
For a given maximization primal problem the following four cases may arise.
Case 1. All constraints are less than or equal to type.
Example 1. Write down the primal problem to the following dual problem:
Maximize w1 + w2 + w3
Subject to
2w1 + w2 + 2w3 < 2
4 w1 + 2w2 + w3 < 2
W1, w2, w3, > 0.
Solution. The primal to the given dual L.P problem is its dual.
Minimize 2x1 + 2x2
Subject to
2x1 + 4x2 > 1
X1 + 2x2 > 1
2x1 + x2 > 1
X1, x2 > 0
Case 11. One or more constraints are greater than or equal to type.
Example 3. Obtain the dual of :
Maximize 3x1 + 4x2
Subject to
2x1 + 3x2 < 16
5x1 + 2x2 > 20
X1 x2 > 0.
Solution. Note that the second inequality is of the form greater than or equal to so we first convert it to an equal to or less-than equality.
We multiply the second constraint on both sides by-1.
(-1) (5x1 + 2x2) > (-1) 20
-5x1 - 2x2 < -20
Or
And replace the second con.straint by this equivalent constraint. This yields the following equivalent model.
Maximize Z = 3x1 + 4x2
Subject to
2x1 + 3x2 < 16
-5x1 -2x2 < -20
X1 > 0, x2 > 0
The dual model of the problem can now be formulated as follows:
Minimize Z*= 16y1 - 20y2
Subject
2y1 - 5y2 > 3
3y1 - 2y2 > 4
Y1 > 0, y2 > 0.
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