I am supposed to find the dual of
max $c^Tx$
subject to $a \le Ax \le b$
$l \le x \le u$.
In order to find the dual I think I have to write it in standard form, the standard form is:
max $Ax'$
subject to: $A'x'\le b'$
$x' \ge 0$.
The way I try to transform it is like this:
max $c^Tx$
subject to $Ax \le b$
$-Ax \le-a$
$x \le u$
$-x \le -l$
Any tips on how to proceed?, thanks.
Rewrite the problem into its canonical form.
\begin{align} \max z = c^T x & \\ \text{s.t. } \begin{bmatrix} A \\ -A \\ I \\ -I \end{bmatrix} x &\le \begin{bmatrix} b \\ -a \\ u \\ -l \end{bmatrix} \end{align}
Now, the standard form will be very clear.
\begin{align} \max z = c^T x & \\ \text{s.t. } Ax + s_1 &= b \\ -Ax + s_2 &= -a \\ x + s_3 &= u \\ -x + s_4 &= -l \\ s_i &\ge 0 \,\forall i = 1,\dots,4 \end{align}
Depending on the sign of $a,b,l,u$, you may need to multiply both sides of a constraint by $-1$ so that the RHS of these four equalities is positive.
Therefore, the dual is the following LPP.
\begin{align} \max z = \begin{bmatrix} b \\ -a \\ u \\ -l \end{bmatrix}^T \begin{bmatrix} t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} & \\ \text{s.t. } \begin{bmatrix} A \\ -A \\ I \\ -I \end{bmatrix}^T \begin{bmatrix} t_1 \\ t_2 \\ t_3 \\ t_4 \end{bmatrix} &= c \\ t_i &\ge 0 \,\forall i = 1,\dots,4 \end{align}
Simply the above LPP.
\begin{align} \max z = b^T t_1 - a^T t_2 + u^T t_3 - l^T t_4 & \\ \text{s.t. } A^T t_1 - A^T t_2 + t_3 - t_4 &= c \\ t_i &\ge 0 \,\forall i = 1,\dots,4 \end{align}
