I get the Primal LP in non-standard form
$$\begin{align}
\operatorname{max} & \sum^{n}_{j = 1} {c_j x_j} \\
\operatorname{s. t.} & \sum^{n}_{j = 1} {a_{ij} x_j} = b_i & \forall i = 1, \cdots, m \\
& x_j \ge 0 & \forall j = 1, \cdots, n
\end{align}$$
And I am asked to derive the Dual LP. So I standardize the Primal LP:
$$\begin{align}
\operatorname{max} & \sum^{n}_{j = 1} {c_j x_j} \\
\operatorname{s. t.} & \sum^{n}_{j = 1} {a_{ij} x_j} \le b_i & \forall i = 1, \cdots, m \\
& -\sum^{n}_{j = 1} {a_{ij} x_j} \le -b_i & \forall i = 1, \cdots, m \\
& x_j \ge 0 & \forall j = 1, \cdots, n
\end{align}$$
And now I use the transformation to the Dual LP:
$$\begin{align}
\operatorname{min} & \sum^{m}_{i = 1} {b_i y_i} – \sum^{m}_{i = 1} {b_i y_i}\\
\operatorname{s. t.} & \sum^{m}_{i = 1} {a_{ji} y_i} – \sum^{m}_{i = 1} {a_{ji} y_i} \ge c_j & \forall j = 1, \cdots, n \\
& y_i \ge 0 & \forall i = 1, \cdots, m
\end{align}$$
I don't know if I'm hallucinating or something but this to me can be simplified as
$$\begin{align}
\operatorname{min} & \ \ 0\\
\operatorname{s. t.} & \ \ 0 \ge c_j & \forall j = 1, \cdots, n \\
\end{align}$$
Any help is greatly appreciated.
Best Answer
Let's rewrite your problem as (P):
$$ \begin{aligned} & \text{max} & c^Tx \\ & \text{s.t.} & Ax = b, \\ && x \geq 0. \end{aligned} $$
The dual of this is simply (D):
$$ \begin{aligned} & \text{min} & w^T b \\ & \text{s.t.} & w A \geq c, \\ && w \text{ unrestricted}. \end{aligned} $$
The only place in the dual that we need to take the equality $Ax = b$ in the primal into consideration is in determining the sign of the entries in $w$.
To see this, rewrite the primal as you originally did, as problem (P'):
$$ \begin{align} & \text{max} & c^T x \\ & \text{s.t.} & Ax \leq b, \\ && -Ax \leq -b, \\ && x \geq 0. \end{align} $$
Let $w = [w_1 \ \ w_2]$ be the vector of decision variables for the dual to this problem, where $w_1$ corresponds the $b$ and $w_2$ corresponds to $-b$. We then have a new dual (D'):
$$ \begin{align} & \text{min} & (w_1 - w_2)^T b \\ & \text{s.t.} & (w_1 - w_2) A \geq c, \\ && w_1, w_2 \geq 0. \end{align} $$
Letting $w = w_1 - w_2$, we see that $w$ is unrestricted, so (D') is the same as (D). Notice that $w_1 = w^+$ and $w_2 = w^-$ are just the positive and negative parts of $w$.