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4. Primal to dual conversion example ( Enter your problem )
 Rules & Example-1Example-2 Other related methods

1. Rules & Example-1

Rules
 In Primal Then in Dual 1. Objective function is maximum Objective function is minimum 2. x_1 unrestricted in sign 1^"st" constraint is = type 3. 1^"st" constraint is = type y_1 unrestricted in sign 4. constraint is <= type constraint is >= type 5. Objective function : total 3 variables (x_1,x_2,x_3) and coefficient c_1,c_2,c_3 constraints : total 3 constraints and right hand side constraint b_1,b_2,b_3.(c_1 becomes b_1, c_2 becomes b_2, c_3 becomes b_3) 6. constraints : total 2 constraints and right hand side constraint b_1,b_2 Objective function : total 2 variables (y_1,y_2) and coefficient c_1,c_2.(b_1 becomes c_1, b_2 becomes c_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 stpes of Primal by BigM 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;

Dual is (Solution stpes of Dual by BigM 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;

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