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I have this two problems and i only want to find the dual form:

$\begin{gather} max\hspace{.1cm}z =5x_1+6x_2\\ s.t\hspace{.1cm}x_1+2x_2=5\\ -x_1+5x_2 \ge 3\\ x_2 \ge 0\\ x_1\hspace{.1cm} unrestricted \end{gather}$

$\begin{gather} max\hspace{.1cm}z =x_1+x_2\\ s.t\hspace{.1cm}2x_1+x_2=5\\ 3x_1-x_2 =6\\ x_1, x_2 \hspace{.1cm}unrestricted \end{gather}$

So, in dual form i have this:

$\begin{gather} min\hspace{.1cm}z' =5y_1-3y_2\\ s.t\hspace{.1cm}y_1+3y_2\ge 5\\ -2y_1-5y_2 \ge 6\\ y_2 \ge 0\\ \end{gather}$

$\begin{gather} min\hspace{.1cm}z' =5y_1+6y_2\\ s.t\hspace{.1cm}2y_1+3y_2\ge 1\\ y_1-y_2 \ge 1\\ \end{gather}$

But I do not know how to create constraints on unrestricted variables

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1 Answer 1

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I have designed the dual problems according to the attached table.

First dual problem

$\begin{gather} \color{blue}{min}\hspace{.1cm} \ 5y_1+3y_2\\ s.t.\hspace{.1cm} \ \ y_1-y_2= 5\\ \ \ 2y_1+5y_2 \ge 6\\ y_1 \ \text{free}, \ y_2 \le 0 \end{gather}$

Second dual problem

$\begin{gather} \color{blue}{min}\hspace{.1cm} \ 5y_1+6y_2\\ s.t.\hspace{.1cm} \ \ 2y_1+3y_2 = 1\\ y_1-y_2 = 1 \\y_1,y_2 \ \text{free}\\ \end{gather}$

If you have any questions regarding the dual formulations or the table, please feel free to ask.

enter image description here

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  • $\begingroup$ that means that the unrestricted variable produces an equality in the existent constraint. I thought it generated a new equality constraint.. $\endgroup$
    – LP0
    Oct 8, 2014 at 20:08
  • $\begingroup$ If you have n (in-)equalities and m variables at the primal problem, than you have m (in-)equalities and n variables at the dual problem. Thus you "add" nothing. $\endgroup$
    – callculus42
    Oct 8, 2014 at 20:21

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