| written 9.4 years ago by | • modified 9.4 years ago |
Max $z = 2x_1+ x_2$
Subject to $2x_1 – x_2 ≤ 2\\ X_1+ x_2 ≤ 4 \\ X_1 ≤ 3 \\ X_1, x_2 ≥ 0 $
Mumbai University > COMPS > Sem 4 > Applied Mathematics 4
Marks : 06
Year : DEC 2014
| written 9.4 years ago by | • modified 9.4 years ago |
Max $z = 2x_1+ x_2$
Subject to $2x_1 – x_2 ≤ 2\\ X_1+ x_2 ≤ 4 \\ X_1 ≤ 3 \\ X_1, x_2 ≥ 0 $
Mumbai University > COMPS > Sem 4 > Applied Mathematics 4
Marks : 06
Year : DEC 2014
| written 9.4 years ago by |
When primal of maximization type, constraint should be of “ ≤” type.
So the primal is
$Max\space \space z=2x_1+x_2$
Constraints:
$2x_1-1x_2 ≤ 2; \\ 1x_1+1x_2 ≤ 4 \\ 1x_1+0x_2 ≤ 3 \\ x_1 , x_2 ≥ 0$
Dual of above primal is
Minimize $w=2y_1+4y_2+3y_3$
Constraints:
$ 2y_1+1y_2+1y_3 ≥ 2; \\ -1y_1+1y_2+0y_3 ≥ 1; \\ y_1, y_2, y_3 ≥ 0$
the dual in standard form
maximize $W’=-W \\ =-2y_1+4y_2+3y_3+0s_1+0s_2-MA_1-MA_2$
Constraints
$2y_1+1y_2+1y_3-1s_1+0s_2+1A_1+0A_2=2 \\ -1y_1+1y_2+0y_3+0s_1-0s_2+0A_1+1A_2=1 \\ y_1, y_2, y_3, s_1, s_2, A_1, A_2 ≥ 0 $

Iteration-1

Iteration-2

For Maximization of dual, $W_{max}=-6 \\ \therefore W_{min}=-W_{max}=6$
from iteration 2 and in $W_j$ row $S_1=2; S_2=2$
for minimization of primal , $X_1=2; X_2=2; Z_{max}=6$