I recently tried to solve a primal minimization problem using its maximization dual. In the optimal simplex tableau of the dual, there was a slack variable and only one dual variable in the basis.

So, am I right to interpret that since one of the dual variables is zero, therefore one of the constraint's limits of the primal is zero or is it that the value of the slack in dual's basis corresponds to the primal constraint?

The problem was solvable using the Big-M method, but I was unable to achieve the same result via dual. The original problem:

\begin{array}{l} \text{Minimize } z = 60 x_1 + 80 x_2\\ \text{subject to:} \\ 20x_1 + 30x_2 \geq 900\\ 40x_1 + 30x_2 \geq 1200\\ x_1, x_2 \in \mathbb{R}^+\\ \end{array}

  • 3
    $\begingroup$ Please include the mathematical formulations. $\endgroup$
    – Brannon
    Commented Feb 27 at 14:23
  • $\begingroup$ You have the primal formulation. The next step is to write out the dual formulation. $\endgroup$
    – RobPratt
    Commented Mar 2 at 22:55

1 Answer 1


As the problem contains only two decision variables, it would always be worth to investigate its schematic form. The feasible solution to the primal problem is as follows:

enter image description here

And the optimal solution to the problem is $2500$ with respect to $x_1 = 15$ and $x_2 = 20$. Also, the dual values are $c_1 = 2.33$ and $c_2 = 0.33$ respectively.

Now, consider its dual form:

enter image description here

Again, the value of the objective function would be $2500$ corresponding to $y_1 = 2.33$ and $y_2 = 0.33$.

Now, if you have achieved any other solutions, you will need to get back on your formulation and/or how to iterate pivoting in your tableau. Also, this link to show the relation between the primal and dual can be helpful.


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