I want to find the dual problem of the following LP:
$\min c'x$ s.t. $Ax=b,x\ge a$ where a>0.
I'm considering substitute $y=x-a$ so that it becomes:
$\min c'y+c'a$ s.t. $Ax=b-Aa,y\ge 0$.
I know I can drop the constant $c'a$ and then write the dual problem. I'm wondering is there a way to write the dual of it without dropping the constant? Any advice will be greatly appreciated.
Yes, there are two other approaches. The first one is using the Lagrangian. The second is by writing the problem as: \begin{align} \min \quad & c^Tx \\ \text{s.t.} \quad & \begin{pmatrix}A \\ -A \\ I\end{pmatrix}x \geq \begin{pmatrix}b \\ -b \\ a\end{pmatrix} \\ & x \geq 0. \end{align} The dual is: \begin{align} \max \quad & \begin{pmatrix}b \\ -b \\ a\end{pmatrix}^Ty \\ \text{s.t.} \quad & \begin{pmatrix}A \\ -A \\ I\end{pmatrix}^T y \leq c \\ & y \geq 0. \end{align} You could simplify this to: \begin{align} \max \quad & b^Ty_1 + a^T y_2 \\ \text{s.t.} \quad & A^Ty_1 + y_2 \leq c \\ & y_2 \geq 0. \end{align}