> 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). _______________________________________________ Help-glpk mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-glpk
