# linear inequality constrains based on absolute values

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D D am 6 Okt. 2022
Kommentiert: Walter Roberson am 7 Okt. 2022
x+y <= 0.1*abs(x)
x+y >= -0.1*abs(x)
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### Akzeptierte Antwort

Matt J am 6 Okt. 2022
Bearbeitet: Matt J am 6 Okt. 2022
The constraints correspond to a non-convex region in (as Walter's second plot shows). You would have to break it into two regions and optimize over each one separately:
Region 1:
x<=0
x+y <= 0.1*(-x)
x+y >= -0.1*(-x)
Region 2:
x>=0
x+y <= 0.1*(x)
x+y >= -0.1*(x)
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Torsten am 7 Okt. 2022
The one that gives the lowest value of the objective, I guess.
help min
Walter Roberson am 7 Okt. 2022
There is no point which is not in one of the regions or the other, so solving separately and looking for the best between the two is going to get you the same result as if you had no constraint.
It would make more sense if the conditions were "and" and you processed the intersection of the constraints in two pieces, one for negative x and the other for non-negative x, and took the best between the two of those.

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### Weitere Antworten (1)

Walter Roberson am 6 Okt. 2022
x = linspace(-0.005, 0.005, 100);
y = linspace(-0.005, 0.005, 101).';
M1 = x + y <= 0.1*abs(x);
M2 = x + y >= -0.1*abs(x);
[r1, c1] = find(M1);
[r2, c2] = find(M2);
plot(x(c1), y(r1), 'k.', x(c2), y(r2), 'ro');
As you can see from the plot, there is nowhere which is not part of one of the regions or the other, so nothing is constrained out.
Matters would be different if the constraints were "and".
x = linspace(-0.0003, 0.0003, 1000);
y = linspace(-0.0003, 0.0003, 1001).';
M1 = x + y <= 0.1*abs(x);
M2 = x + y >= -0.1*abs(x);
[r1, c1] = find(M1 & M2);
plot(x(c1), y(r1), 'k.');
I think the small gap is a matter of resolution.
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