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How to linearize difference of absolutes?

$$|a|\ge k|b|$$

where $a,b$ are variables and $k$ is a constant.

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    $\begingroup$ The feasible set isn’t convex so this can’t be done in linear programming. Do you want an answer involving integer variables? $\endgroup$ – Brian Borchers Apr 2 '20 at 2:25
  • $\begingroup$ Sure...I will try if I can convert my problem to integer variables. $\endgroup$ – Vinay Apr 2 '20 at 2:29
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    $\begingroup$ You can find helpful things in here: ibm.com/developerworks/community/forums/html/… $\endgroup$ – kur ag Apr 2 '20 at 9:57
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    $\begingroup$ Please see the question and answer here: or.stackexchange.com/questions/3/…. If that doesn't answer your question sufficiently, then please revise your question to indicate what's still unanswered. $\endgroup$ – LarrySnyder610 Apr 2 '20 at 20:42
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Create some extra variables

$AB_1, AB_2, X_1, X_2, X_3, X_4$, and you have $a,b,k$

\begin{align}a &= X_1-X_2\\AB_1 &= X_1+X_2\\b &= X_3-X_4\\AB_2 &= X_3+X_4\\AB_1 &\geq k \cdot AB_2\end{align}

Also, you have to minimize variable $X_1, X_2, X_3, X_4$ and $X_1, X_2, X_3, X_4 \geq 0$

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  • $\begingroup$ Is it possible to create constraints without objective function? I have another objective functions in my problem $\endgroup$ – Vinay Apr 2 '20 at 19:34
  • $\begingroup$ you can multiply it with an small value so that it doesn't have any impact on your objective function. Like this your current objective function + (x1+x2+x3+x4)*0.00000000001 some thing like this. You can also try it without adding objective function. $\endgroup$ – ooo Apr 3 '20 at 5:36
  • $\begingroup$ The small penalty approach is not entirely safe. If a solution with $|a|<k|b|$ produces a sizeable improvement in the objective function, the solver will "pad" $X_1$ and $X_2$ to make the solution look feasible and accept the small penalty. $\endgroup$ – prubin Apr 3 '20 at 21:33

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