Re: [Help-glpk] Absolute value constraint

[Andrew Makhorin]

> Hi, is it possible to express 'abs(f(x)) >= m' as a linear constraint?
 
Yes, using a mip approach.

There are two alternatives:

   f(x) >= 0 that gives f(x) >= m

   f(x) <  0 that gives -f(x) >= m or, eqiuvalently, f(x) <= -m.

Let z be a binary variable such that:

   z = 1 means m <= f(x) <= +M

   z = 0 means -M <= f(x) <= -m

where M is a "big" constant. Then the constraint can be formulated
as follows:

m * z - M * (1 - z) <= f(x) <= M * z - m * (1 - z)

(under assumption that f(x) is a linear function of variables).