What varialbes enter the $\min/\max$ in dual problem?

Question

Having the following linear program:

\begin{cases} \max & -x_1 & -2 x_2&+x_3\\ & -3 x_1 &+x_2 & &\le -1\\ & x_1 &-x_2 & &\ge 1\\ &-2x_1 & +7 x_2 & & \le 6\\ & -5x_1& &+2x_3 &=-3\\ & 7x_1 &-3x_2 &&\le6\\ &x_1\ge0\\ &x_3\le0 \end{cases}

Why do we have:

\begin{cases} \min &-y_1 &-y_2 &+6y_3 & +6y_4& &-3y_6\\ &-3y_1&-y_2 & -2y_3& +7y_4 & & -5y_6&\ge-1\\ &y_1 &+y_2 &+y_3 &-3y_4&&&=-1\\ &&&&&y_5&+2y_6&=1\\ \forall i\in[[1,5]],y_i\ge0 \end{cases}

And not

\begin{cases} \min &-y_1 &-y_2 &+6y_3 & -3y_4& &+6y_5\\ &-3y_1&-y_2 & -2y_3& +7y_4 & & -5y_6&\ge-1\\ &y_1 &+y_2 &+y_3 &-3y_4&&&=-1\\ &&&&&y_5&+2y_6&=1\\ \forall i\in[[1,5]],y_i\ge0 \end{cases}

I just read the $b_i$ in the vertical way and put them at the place of the $c_i$ in the dual problem.

Answer 1

They changed the order of the inequalities:

\begin{cases} \max & -x_1 & -2 x_2&+x_3\\ & -3 x_1 &+x_2 & &\le -1 \color{blue}{\qquad (y_1)}\\ & x_1 &-x_2 & &\ge 1 \color{blue}{ \qquad (y_2)}\\ &-2x_1 & +7 x_2 & & \le 6\qquad \color{blue}{ (y_3)}\\ & 7x_1 &-3x_2 &&\le6\qquad \color{blue}{(y_4)}\\ & x_3\leq 0 \qquad \qquad \color{blue}{(y_5)}\\ & -5x_1& &+2x_3 &=-3\qquad \color{blue}{(y_6)}\\ &x_1\geq 0\\ \end{cases}

The second constraint has to be multiplied by 1.

Note that the second constraint of the dual program has to be

$y_1+y_2+7y_3-3y_4+2y_6=\color{red}{-2} \quad (x_2)$

The corresponding coefficient in the objective function is -2 ($-2x_2$).

The others constraints are:

$-3y_1-y_2-2y_3+7y_4-5y_6\geq -1 \quad (x_1)$

$y_5+2y_6=1 \quad (x_3)$