I wanted to have a fast rule for computing dual LP at hand.. I always seem to struggle to form the dual not in terms of the objective function or the coefficients of constraints but I am struggling to decide whether the constrains should be $\leq, \geq , = $ and also sign constraints..
I found this https://www.usna.edu/Users/math/dphillip/sa305.s12/sob.pdf
And it has this table :
I applied it to a couple of LP's and got the duals right . But then I stumbled on this one and I get it wrong:
The primal is
$$ \min 3x1 +x2 + x3 + x4$$
$$ \text{s.t } 3x1 -x2 -7x3 + x4 = -3 $$
$$ 3x1 - 4x3 + x4 = -1 $$
$$ x3 \leq 3 $$
$$x1, x2, x3, x4 \geq 0 $$
To compute the dual :
$$ \max -3y1 -y2 +3y3 $$
$$ \text{s.t } 3y1 + 3y2 \; ?\; 3 $$
$$ -y1 \; ? \; $$
$$ -7y1 -4y2 + y3 \; ?\; 1 $$
$$ y1 + y2 \; ? \; 1 $$
$$ ? $$
So to my understanding, according to the rule , the type of constraints is defined by the sign constraint of the corresponding primal variable .. So since all of them are positive we introduce only $\leq $ constraints.
And the sign constraints are defined by the type of constraints in the primal . The $=$ constraints let $y1, y2$ to be free and the $\leq$ constraints is according to the pdf a "bizarre" one and it implies that $y3 \leq 0 $
$$ \max -3y1 -y2 +3y3 $$
$$ \text{s.t } 3y1 + 3y2 \; \leq 3 $$
$$ -y1 \leq 1$$
$$ -7y1 -4y2 + y3 \leq 1 $$
$$ y1 + y2 \leq 1 $$
$$ y3 \leq 0 $$
But by putting my LP into an online calculator my result seems to be wrong..
Can you help me work out my mistake ?
In my view you are on the right track. The dual is
$$ \max\ -3y_1 -y_2 +3y_3 $$ $$ \textrm{s.t } 3y_1 + 3y_2 \leq 3$$
$$ -y_1 \leq 1$$ $$ -7y_1 -4y_2 + y_3\leq 1 $$ $$ y_1 + y_2 \leq 1 $$
The last (third) constraint of the primal min-problem is a $\leq$-constraint.Thus $y_3\leq 0$. I don't see any mistake in your model.
Remark:
I've calculated the optimal solutions with this calculator. The optimal solutions are
primal: $(x_1^*,x_2^*,x_3^*,x_4^*)=\left(0,\frac54, \frac{1}{4},0\right)$
dual: $(y_1^*,y_2^*y_3^*)=\left(-1,\frac32,0\right)$
In both cases the optimal value of the objective function is $\frac32$ and hence the strong duality theorem is satisfied.