I have the generic optimization problem:
$$ \max c^T|x|$$
$$ \text{s.t. } Ax \le b $$
$x$ is unrestricted
How do I convert it into a linear programming problem?
Online I read something about letting $x$ equal the difference of 2 positive numbers but I could not intuitively grasp why that worked. Plus the example applied only to minimization problems where the $c^T$ entries are all greater than $0$.
I'm sort of stuck
On
I think the question you are trying to ask is this: If we have a set of linear constraints involving a variable $x$, how can we introduce $|x|$ (the absolute value of $x$) into the objective function?
Here is the trick. Add a constraint of the form $$t_1 - t_2 = x$$ where $t_i \ge 0$. The Simplex Algorithm will set $t_1 = x$ and $t_2 = 0$ if $x \ge 0$; otherwise, $t_1 = 0$ and $t_2 = -x$. So $t_1 + t_2 =|x|$ in either case.
On the face of it, this trick shouldn't work, because if we have $x = -3$, for example, there are seemingly many possibilities for $t_1$ and $t_2$ other than $t_1 = 0$ and $t_2 = 3$; for example, $t_1 = 1$ and $t_2 = 4$ seems to be a possibility. But the Simplex Algorithm will never choose one of these "bad" solutions because it always chooses a vertex of the feasible region, even if there are other possibilities.
EDIT added Mar 29, 2019
For this trick to work, the coefficient of the absolute value in the objective function must be positive and you must be minimizing, as in
min $2(t_1+t_2)+\dots$
or the coefficient can be negative if you are maximizing, as in
max $-2(t_1+t_2)+\dots$
Otherwise, you end up with an unbounded objective function, and the problem must be solved by other methods, e.g. mixed integer-linear programming.
(If I knew this before, I had forgotten. Thanks to Discretelizard for pointing this out to me.)
On
I realize this is old, but I just ran into this issue. Please see: http://lpsolve.sourceforge.net/5.1/absolute.htm, which has a great explanation of the solution (both for minimization and maximization of an absolute value), and helped me a lot.
$$ \begin{array}{rcr} \min_{\mathbf{u},\mathbf{x}} \mathbf{1}^\mathrm{T}\mathbf{u} & \mathrm{s.t.} & \mathbf{x}-\mathbf{u} \preceq \mathrm{0} \\ & & -\mathbf{x}-\mathbf{u} \preceq \mathrm{0} \\ & & \mathbf{Ax} \preceq \mathrm{b} \\ \end{array} $$
where $\preceq$ is meant to signify an element-wise less-than relationship. I also left out $\mathbf{c}$ since you can just scale the elements of $\mathbf{x}$ to be $c_i x_i$.
As for implementation, some software may require you to combine $\mathbf{x}$ with the dummy vector $\mathbf{u}$. In which case, the problem might look like
$$ \begin{array}{rcr} \min_{\mathbf{u},\mathbf{x}} \left[ \mathbf{1} \,\, \mathbf{0}\right]^\mathrm{T}\left[ \mathbf{u} \,\, \mathbf{x} \right] & \mathrm{s.t.} & \mathbf{x}-\mathbf{u} \preceq \mathrm{0} \\ & & -\mathbf{x}-\mathbf{u} \preceq \mathrm{0} \\ & & \mathbf{Ax} \preceq \mathrm{b} \\ \end{array} $$
You would then just take the portion of the final answer that corresponds to $\mathbf{x}$.