# Convex optimization problem to quadratic programming problem

Briefly, have the following problem: $$\sum_{i = 0}^n a_i \ (max [ F_i( \bar x ), 0 ] )^2 \rightarrow min, \\\\ s.t.\\\\ A \bar x \leq b$$ where $F( \bar x )$ is a linear function, $a_i \gt 0$, $n$ is huge comparing to the size of $x$.

It is possible to write an equal Quadratic Programming problem, such as

$$\sum_{i=0}^n a_i \ ( G_i )^2 \rightarrow min \\\\ s.t. \\\\ G_i \geq {\bf 0}, \quad i = 0..n \\\\ G_i \geq F_i( \bar x ) \quad i = 0..n \\\\ A \bar x \leq b$$

which can be solved very efficiently with an appropriate numerical method.

Unfortunately in my particular case such conversion doesn't work: it adds a lot of new restrictions, and that appropriate numerical method doesn't converge.

I tried to figure out another equal QPP, which adds fewer new constraints, but nothing came across my mind. Is there another way?

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If you are talking about $\min[\max(F( \bar x), 0)]^2$ where $\bar{x}$ is unconstrained, you don't need any software to do it: as $F$ is linear, the answer is zero, and global minimum is attained at $\bar{x}=0$. If there are additional constraints, you may try to specify them in your question. –  user1551 Dec 7 '12 at 22:27
@user1551 there are standart linear constraints. I edited the question. –  Artem Pyanykh Dec 7 '12 at 22:47

## 1 Answer

This is not really an answer. I just want to say that your optimization problem can be converted into a linear programming problem:

$\min F(\bar{x})$ subject to $A\bar{x}\le b$.

If the minimum found is $m$ and the minimizer is $x_0$, then the minimum for your original problem is $\max(m,0)^2$ and the minimizer is $x_0$.

Reason: If $F(x_0)=m<0$, then $\max(F(x_0), 0)^2=\max(m, 0)^2=0$, which is the least possible value of $\max(F(\bar{x}), 0)^2$ over the whole space. Hence $x_0$ is a feasible and global minimizer.

If $F(x_0)=m\ge0$, then $F(\bar{x})\ge F(x_0)\ge0$ for every $\bar{x}\in D=\{\bar{x}: A\bar{x}\le b\}$. Hence $\max(F(\bar{x}), 0) = F(\bar{x})\ge0$ for every $\bar{x}\in D$. Therefore $$\min_{\bar{x}\in D} \max(F(\bar{x}), 0)^2 = \min_{\bar{x}\in D} F(\bar{x})^2 = \left(\min_{\bar{x}\in D} F(\bar{x})\right)^2 = m^2 = \max(m,0)^2.$$

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Yes, that's right. I apologize for being not enough specific. Please, see the edited question. –  Artem Pyanykh Dec 8 '12 at 14:14