primal problem is: $$\min z = 4x_1-3x_2+5x_3$$
$$7x_1+6x_2+24x_3\le16$$
$$2x_1+5z_2+3x_3\le10$$
$$x_i\ge0$$ the optimal solution is: $(0,2,0), z = -6$
The dual problem is : $$ \max g = 16w_1+10w_2$$
$$7w_1+2w_2\le4$$
$$6w_1+5w_2\le-3$$
$$24w_1+3w_2\le5$$
$$w_1,w_2\le0$$ I get the optimal solution $g=0$ which is wrong because of the duality theorem, $z(opt)=g(opt)$. What's wrong with it? Thanks.
$\endgroup$2 Answers
$\begingroup$The linear program you give as the dual is correct. However, the optimal solution isn't $g=0$, but rather $g=-6$ at $(w_1,w_2)=\left(0,-\frac{3}{5}\right)$.
Notice that $g=0$ isn't a possibility because if $g=0$ then we have $w_1=w_2=0$ which then does not satisfy the constraint $$6w_1+5w_2\le-3$$ You can also notice that this is the only nontrivial constraint in the dual program - the other constraints are satisfied merely by the $w_1,w_2\le 0$ requirement.
$\endgroup$ 3 $\begingroup$i think this will help you
MIN zx = x1 + 2 x2 subject to - 2 x1 - 4 x2 ≥ -160 x1 - x2 = 30 x1 ≥ 10 and x1,x2≥0;
Since 2nd constraint in the primal is equality, the corresponding dual variable y2 will be unrestricted in sign.
Dual is (Solution stpes of Dual by BigM method)
MAX zy = - 160 y1 + 30 y2 + 10 y3 subject to - 2 y1 + y2 + y3 ≤ 1 - 4 y1 - y2 ≤ 2 and y1,y3≥0;y2 unrestricted in sign
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