Consider a linear optimization problem, with absolute values, of the following form:

\begin{equation} \begin{array}{rl} \text{minimize}\ &\mathbf{c'x}+\mathbf{d'y}\\ \text{subject to}\ &\mathbf{Ax}+\mathbf{By}\leq\mathbf{b}\\ &y_i=|x_i|, \end{array} \end{equation}

Assume that all entries of B and d are nonnegative.

I have to provide an example to show that if B has negative entries, the problem may have a local minimum that is not a global minimum, but I have really not idea how to it. Can you help me ?

ps: what will happen if the entries of c and A are negatives ?

  • $\begingroup$ I'd start with the case of scalar $x,y$. $\endgroup$ – hardmath Jul 9 '18 at 16:04
  • $\begingroup$ how do you define a local minimum in constrained optimization? $\endgroup$ – LinAlg Jul 9 '18 at 16:20


The set of $x\in\mathbb{R}$ which satisfy $1\leqslant|x|\leqslant2$ can be written as $[-2,-1]\cup[1,2]$. This constraint divides the feasible region into two disconnected components...

  • $\begingroup$ So if B has negative entries is possible to create disconnected feasible regions ? If yes, it's still not clear how it happens... $\endgroup$ – Qwerto Jul 11 '18 at 17:15
  • 1
    $\begingroup$ @Qwerto Yes, that’s what I’m hinting at here. We have the constraints $-|x|\leqslant-1$ and $|x|\leqslant2$, which is equivalent to $Ax+By\leqslant b$ with $A=0$, $B=(-1,1)^\text{T}$ and $b=(-1,2)^\text{T}$ $\endgroup$ – David M. Jul 11 '18 at 17:19
  • $\begingroup$ Now it's clear. Instead, what will happen if the entries of c and A are negatives ? $\endgroup$ – Qwerto Jul 11 '18 at 17:25
  • $\begingroup$ Lots of things could happen. You would have to be a little more specific. $\endgroup$ – David M. Jul 11 '18 at 17:26
  • $\begingroup$ To me, if c can assume negative values there is the risk of a unbounded solution. Instead do you think that if A has negative values similar problems can arise ? $\endgroup$ – Qwerto Jul 11 '18 at 17:30

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