# primal to dual solution conversion ??

i have an optimization problem

$$\text{ maximize } z=3x+4y$$ $$\text{ such that: } x+y ≤ 450 \text{ and } 2x+y ≤ 600$$

the optimal solution to this problems comes to be $x=0$; $y=450$; $p=150$ (the slack variable)

consequently, the dual problem becomes :

$$\text{ minimize } 450a+600b$$ $$\text{ such that } a+2b ≥ 3 \text{ and }a+b ≥ 4;$$

the optimal solution of dual becomes $a=4$; $b=0$; $c=1$ (surplus variable)

my doubt is that when i apply the strong duality theorem on the primal solution, i'm unable to get the dual solution. that is:

(C transpose) multiplied by (b inverse) $C^Tb^{-1}$= {4,0}*{{1,0},{-1,1}}={4,0} which is not correct since we should get the dual solution. where am i going wrong?

• Should the primal problem not have two slack variables? Are there any positivity conditions? – Lutz Lehmann Feb 22 '14 at 16:27
• the first slack variable in primal comes to be zero. and the second surplus variable in dual also becomes zero. yes, the variables x,y,a,b are all non-negative – mathlover Feb 22 '14 at 17:03
• I do not understand your problem. The strong duality theorem holds as the primal and dual optimal solution coincide. – guanglei Jan 1 '17 at 11:21

In the primal, for that solution, you need non-negativity constraints on $x$ and $y$.

So, the primal is: $$\text{ maximize } z=3x+4y$$ $$\text{ such that: } x+y ≤ 450 \text{ and } 2x+y ≤ 600$$ $$x,y\geq0$$

which is equivalent to:

$$\text{ minimize } z=-3x-4y$$ $$\text{ such that: } x+y ≤ 450 \text{ and } 2x+y ≤ 600$$ $$x,y\geq0$$

Which gives the answer $(x,y)=(0,450)$ and a primal optimal solution value of $-1800$.

Now, the dual becomes:

$$\text{ maximize } 450a+600b$$ $$\text{ such that } a+2b \leq -3 \text{ and }a+b \leq -4;$$ $$a,b\leq0$$

Which when solved gives the answer $(a,b)=(-4,0)$ which leads to optimal dual value of $-1800$.