Maximize $z = 5x_1 – 2x_2 + 3x_3$

Subject to $ 2x_1 + 2x_2 – x_3 ≥ 2 \\ 3x_1 – 4x_2 ≤ 3 \\ X_1 + 3x_3 ≤ 5 \\ X_1, x_2, x_3 ≥ 0$

Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 06

Year : MAY 2015

when primal is of maximization type, the constraints should be of “≤” type.

So multiplying first constraint by ‘-1’ we have,

$-2x_1-2x_2+x_3 ≤ -2$

Primal:

Maximize $z=5x_1-2x_2+3x_3$

constraints

$-2x_1+2x_2+1x_3 \leq -2; \\ 3x_1 -4x_2+0x_3 \leq 3; \\ 1x_1+0x_2+3x_3 \leq 5 ; \\ x_1,x_2,x_3 \geq 0$

The dual of given primal is

Minimize $w=-2y_1-3y_2+5y_3$

Constraints:

$-2y_1+3y_2+1y_3 \geq 5 \\ -2y_1-4y_2+0y_3 \geq -2 \\ i.e.\\ 2y_1+4y_2 00y_3 \geq 2; \\ 1y_1+0y_2 + 3y_3 \geq 3; \\ y_1,y_2,y_3 \geq 0;$

Dual in standard form,

Maximize

$w'=-2=2y_1-3y_2-5y_3; \\ i.e\\ w'=2y_1-3y_2-5y_3+0s_1+0s_2+0s_3 -MA_1-MA_3;$

Constraints:

$-2y_1+3y_2+1y_3-1s_1+0s_2+0s_3+1A_1+0A_3= 5; \\ 2y_1+4y_2- 0y_3+0s_1+1s_2+0s_3 +0A_10A_3=2; \\ 1y_1+0y_2 + 3y_3 + 0s_1+0s_2 -1s_3 +0A_1+ 1A_3=3 \\ y_1,y_2,y_3,s_1,s_2,s_3,A_1,A_2 \geq 0$

Since all values of ∆j are negative or zero and Artificial variable (M) remain in solution the dual of given LPP has infeasible solution.

So, primal of given LPP has unbounded solution