I have the next primal problem.I have to solve It obtaining the Dual problem and up to here everything is nice,but the problem appears when I have to apply the Complementary Slackness:

Primal problem is: Maximize $2x_1+3x_2-4x_3$,
Subject to: ,
$3x_1+5x_2+2x_3=15$ ,
$2x_1+3x_2-4x_3=8$ ,
$x_1\geq0, x_2\geq0,x_3\geq0$

I obtain the Next Dual Problem: Minimize $15\pi_1 +8\pi_2$

Subject to:
$x_2:\ 5\pi_1+3\pi_2\geq3$ ,
$x_3:\ 2\pi_1-4\pi_2\geq-4$

When I solve this dual problem I have that the Dual solution is (0,1).The issue here is when I put it in the equations the condition of $>$ doesnt satisfy,It only satiesfies the condition of =,so I cannot apply Complementary Slackness and say that any primal variable is equal to zero.If anybody knows how to proceed now,I'd very greatful.


1 Answer 1


The best way to deal with your problem is to use the strong duality. It says that the the primal optimal objective is equal to the dual optimal objective: $f(x^*)=g(\pi^*)=8$. That means you have three equations with 3 unknowns.

\begin{eqnarray} &2x_1+3x_2-4x_3=&8 \\ & 3x_1+5x_2+2x_3=&15\\ & {2x_1+3x_2-4x_3}=&8\end{eqnarray}

It is obvious that one equation is redundant. So for the optimal solution you have to express two variables by the remaining variable, i.e. $x_1$. And also you have to regard the non-negativity condition. Can you proceed?

  • $\begingroup$ I understand what you are saying,but the problem says that i should find the primal solutions applying optimality conditions to the dual problem so i cannot use your method :( $\endgroup$ Oct 9, 2021 at 17:48
  • $\begingroup$ And the condition I mentioned is part of the optimality conditions. I would even say that it is the optimality condition. $\endgroup$ Oct 9, 2021 at 17:52
  • $\begingroup$ ok,I understand.But i have a little doubt about this.If I do what you are saying,i will not get an unique solution, but infinite.But the optimal primal solution its supposed to be unique,right?or not?Sorry,my first year with this subject :/ $\endgroup$ Oct 9, 2021 at 18:06
  • $\begingroup$ @Aymar21 But the optimal primal solution its supposed to be unique,right? Not necessarily.The objective function and the second constraint are identical, which means at the optimal solution the obj. function lies on the second constraint, if it is active. An then you have a solution on a (constrained) line, which is not a point. I've tried an explanation here, with a 2D-graph. $\endgroup$ Oct 9, 2021 at 18:12
  • $\begingroup$ @Aymar21 Your are welcome. I would be pleased, if you accepted the answer. $\endgroup$ Oct 9, 2021 at 18:19

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