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4. Primal to dual conversion example ( Enter your problem )
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2. Example-2

Find dual from primal conversion
MIN z = x1 - 3x2 - 2x3
subject to
3x1 - x2 + 2x3 <= 7
2x1 - 4x2 >= 12
-4x1 + 3x2 + 8x3 = 10
and x1,x2 >= 0 and x3 unrestricted in sign

Solution:
Primal is (Solution stpes of Primal by BigM method)

 MIN z_x =   x_1  -  3 x_2  -  2 x_3
subject to
  3 x_1  -   x_2  +  2 x_3 ≤ 7  2 x_1  -  4 x_2 ≥ 12  -  4 x_1  +  3 x_2  +  8 x_3 = 10
and x_1,x_2 >= 0; x_3 unrestricted in sign

all <= constraints can be converted to >= type by multipling both sides by -1

 MIN z_x =   x_1  -  3 x_2  -  2 x_3
subject to
  -  3 x_1  +   x_2  -  2 x_3 ≥ -7  2 x_1  -  4 x_2 ≥ 12  -  4 x_1  +  3 x_2  +  8 x_3 = 10
and x_1,x_2 >= 0; x_3 unrestricted in sign

The x_3 variable in the primal is unrestricted in sign, therefore the 3^(rd) constraint in the dual shall be equality.

Since 3^(rd) constraint in the primal is equality, the corresponding dual variable y_3 will be unrestricted in sign.

Dual is (Solution stpes of Dual by BigM method)

 MAX z_y =  -  7 y_1  +  12 y_2  +  10 y_3
subject to
  -  3 y_1  +  2 y_2  -  4 y_3 ≤ 1   y_1  -  4 y_2  +  3 y_3 ≤ -3  -  2 y_1  +  8 y_3 = -2
and y_1,y_2 >= 0; y_3 unrestricted in sign

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