Using min/max operator in integer programming

Question

I am trying to optimize an objective function using integer programming, I have to use Max operator in my function, I want to know is there any way to deal with that?

Actually my question is similar to Using min/max within an Integer Linear Program but is different in some aspects:

All variables are binary.
Note that x4 and x5 are presented in two place.
One possible solution is using auxiliary variables like the answer of similar question, but I am confused when using this solution for my example.

Example:

Minimize (c1 * x1) + (c2 * x2) + (c3 * x3) + Max(c4 * x4, c5 * x5) + (c6 * x4) + (c7 * x5)

subject to
some equality and inequality constraints

It's refreshing when someone posts in the beginning why it differs from similar questions. — Dennis Meng, Dec 29 '13 at 23:09
@Dennis Meng, sorry, I can't understand what you mean, my question isn't clear enough? — oMiD, Dec 29 '13 at 23:15
Not at all. I was remarking on how you mentioned at the beginning why this isn't exactly like the other question. It's a good thing. :) — Dennis Meng, Dec 29 '13 at 23:17
It seems to me that you can use the technique in the other answer which you linked, are you running into problems? — Alexander Torstling, Dec 29 '13 at 23:24
@AlexanderTorstling, I can't understand how to use it, can u explain me plz? — oMiD, Dec 29 '13 at 23:29

score 7 · Accepted Answer · answered Dec 29 '13 at 23:36

Use the approach in the question you linked. The expression

Max(c4 * x4, c5 * x5)

can be replaced by a variable x6, provided that you add the following additional constraints:

x6 >= c4 * x4
x6 >= c5 * x5

So you total set becomes:

Minimize (c1 * x1) + (c2 * x2) + (c3 * x3) + x6 + (c6 * x4) + (c7 * x5)

subject to:

some equality and inequality constraints

and the new requirements:

x6 >= c4 * x4
x6 >= c5 * x5

This works since Max(c4 * x4, c5 * x5) will either take the value c4 * x4 or c5 * x5. The introduced variable x6 will always be larger or equal to both of these expressions, and so will always be larger or equal to the total max-expression. When properly minimized, x6 will bottom out at the value of the max-expression. So when minimized, the two forms are equivalent.

thanks for your very clear explanation, I totally get that :) — oMiD, Dec 29 '13 at 23:42

Using min/max operator in integer programming

1 Answers1

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