Algorithm (SciPy?) for Solving Optimization Problem with Absolute Values

Asked May 08 '18 at 01:12

Active May 08 '18 at 01:14

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I am at a point in n-dimensional space $X^0$ and it costs me to move to $X_i$ with varying cost depending on direction $\Sigma_i c_i |X_i-X^0_i|$.

But, it's good for me to move closer to a target point $X^*$ because I'm penalized quadratically for being away from the target $d \Sigma_i |X_i-X^*_i|^2$.

I know the absolute values make this problematic but is there a way to globally minimize the total cost function for $X_i$? (algorithmically?)

$Cost =\Sigma_i c_i |X_i-X^0_i| + d \Sigma_i |X_i-X^*_i|^2$

edited May 08 '18 at 01:14

asked May 08 '18 at 01:12

rhaskett

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It seems like you're asking an algorithms question - in which case you'd get better answers over at SO instead. – B. Mehta May 08 '18 at 01:14
Sure I'll give it a shot. – rhaskett May 08 '18 at 01:16
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The cost function I see is plainly sum of functions each depending on a single (and unique) coordinate. If that is it, it may perfectly well be optimized coordinate-wise. And, by the way, why this is tagged 'non-convex optimization'? ;) – metamorphy May 08 '18 at 01:40
@rhaskett: You might also want to take a look at this. – VHarisop May 08 '18 at 01:52
@metamorphy I think I see. If I shift the problem to where it is along a line between X^0 and X^* it becomes a 1D quadratic problem and I can split the space where X>X^0 and X<X^0 to deal with the absolute value. – rhaskett May 08 '18 at 02:17
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Minimizing absolute value is expressible as a linear problem: $x\geq |y|$ is the same as $x\geq y \geq -x$. The square can be handled with a second-order cone. – Michal Adamaszek May 08 '18 at 06:54
@MichalAdamaszek only if $c_i \geq 0$. – LinAlg May 08 '18 at 17:44
Yes, it is a cost so $c_i \geq 0$. – rhaskett May 08 '18 at 19:41
Do you have a link where I can learn about that transformation? – rhaskett May 08 '18 at 19:42

Algorithm (SciPy?) for Solving Optimization Problem with Absolute Values

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