If I have the following problem: y <= max(x1, x2) (make it linear)

Can I solve it as follows:

y <= x1 + M * z

and y <= x2 + M * (z-1)

or do I need the 2 inequalities described below too?

x1 >= x2 - M * z

and x2 >= x1 - M * (z-1)

asked 13 Jan '15, 13:51

Luckiiiz's gravatar image

Luckiiiz
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edited 13 Jan '15, 13:56


It depends...

> http://orinanobworld.blogspot.com/2011/01/max-and-min-functions-in-mip.html

P.S.: ...assuming that you meant to write \((1-z)\) -- not \((z-1)\).

link

answered 13 Jan '15, 14:46

fbahr's gravatar image

fbahr ♦
4.6k716
accept rate: 13%

edited 13 Jan '15, 15:13

Yes indeed (1-z)! So the last two inequalities are always irrelevant?

Because in the link there is only the possibility between my first two inequalities and using a slack variable that keeps track of the largest value.

(13 Jan '15, 15:29) Luckiiiz

You will need the (other) 2 inequalities iff [sic!], quoting @Paul Rubin", "you cannot be sure that the 'pressure' of the objective function will prevent inflated values of z from occurring, and you really do need the value of z to be correct."

(13 Jan '15, 15:40) fbahr ♦

Thank you very much!!

(13 Jan '15, 15:49) Luckiiiz

The link from @fbahr shows four constraints to enforce y = max(x1, x2). In this case, you want the inequality y <= max(x1, x2). Your first two constraints (with (1-z) instead of (z-1)) do just that. Your last two constraints are unnecessary.

link

answered 20 Jan '15, 12:33

optimizer's gravatar image

optimizer
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accept rate: 6%

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Asked: 13 Jan '15, 13:51

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