# Converting from primal to dual linear problem

If a primal problem has an unrestricted variable then the corresponding dual constraint is an equality.

This is what I've been told.

However, the following online primal to dual conversion software states the same fact but doesn't use it in the solution.

Could someone clarify this for me:

Should it not be: $y_1 = 6$ ?

Yes, it should. I think you typed incorrect data. This is what I got. Note the difference in the primal problem representation. In your case it says $$x_1,x_2\ge 0,\ X_1\text{ unrestricted}$$ but $X_1$ (capital) is not a variable there.
• Not really, the dual set is empty, because it is not possible for $y_1$ to be $6$ and $\le 1$ at the same time. The solution in the primal problem does not exist either, because the objective function is unbounded below. Take $x_1=3-2x_2$ (in the set) then the objective function is $18-10x_2\to \color{red}{-\infty}$ as $x_2\to +\infty$. – A.Γ. Apr 29 '17 at 7:43