Relationship between Primal and Dual problems

Question

Considering the following program:

\begin{cases} \max & 8x_1 & + 3x_2\\ & x_1 &-6x_2&\ge2\\ & 5x_1 +&7x_2&=-4\\ &x_1&&\le 0\\ && x_2&\ge 0 \end{cases}

Why do we have

\begin{cases} \min & 2w_1 & -4 w_2\\ & w_1 &+5x_2&\ge8\\ & -6w_1 &+7w_2&\ge 3\\ &w_1&&\le 0\\ && w_2 \mbox{ unrestricted} \end{cases}

And not

\begin{cases} \min & 2w_1 & -4 w_2\\ & w_1 &+5x_2& +w_3&\le8\\ & -6w_1 &+7w_2&&\le 3\\ & 6w_1 &-7w_2&&\le -3\\ &w_1&&&\le 0\\ && w_2 &&\le 0 \end{cases}

I thought having this from the table 6.1 given in Mokthar S.Bazara and John J.Jarvis, Linear Programing and Network Flows p241

Answer 1

When you write the constraints of the dual problem, you should proceed in 3 steps:

1. Transpose the matrix of constraints, excluding the constraints of type $x_i\ge\mbox{or}\le 0$ (which you did not do, as you have 3 constraints in your dual matrix). The right hand term are the costs of the objective function. For the moment, the signs $\ge\mbox{or}\le \mbox{or} =$ are undefined: \begin{cases} &\omega_1+5\omega_2 &\mbox{ ? } 8\\ &-6\omega_1+7\omega_2 &\mbox{ ? } 3\\ \end{cases}
2. Now, determine the signs $\ge\mbox{or}\le \mbox{or} =$. To do so, check how variables $x_1$ and $x_2$ are constrained:$x_1$ should be negative, which means the first constraint should have a $\le$ sign. $x_2$ is positive, so the second constraint should have a $\ge$ sign: \begin{cases} &\omega_1+5\omega_2 &\le 8\\ &-6\omega_1+7\omega_2 &\ge 3\\ \end{cases}
3. The last step is to determine how variables $\omega_1$ and $\omega_2$ are constrained. Since the first primal constraint has a $\ge$ sign, $\omega_1$ should be negative, and since the second one has a $=$ sign, $\omega_2$ is unconstrained: \begin{cases} &\omega_1+5\omega_2 &\le 8\\ &-6\omega_1+7\omega_2 &\ge 3\\ &\omega_1 \le 0\\ &\omega_2 \in \mathbb{R} \end{cases}