Ceiling function for objective function in MILP optimization

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m c 2015-11-19

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评论： m c 2015-11-24

Hi.

I am using intlinprog to optimise a room allocation schedule for an event. My variables are all integers.

I need to minimise the cost of running this event. Cost is incurred for multiples of 28 people in a room. For example having a room with 0-28 people costs 15$, a room with 29-56 people costs 30$.

I figured I could use the a ceiling function to round up (# of people / 28) to the closest integer, but I cont find a way to incorporate this into the objective function.

Is it possible to use ceiling, or is there a workaround?

Any ideas would be appreciated.

Thanks for reading my question. M

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Answer 1

Alan Weiss 2015-11-20

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You can use an extra variable that keeps track of the number of segments of people/28. Here's how.

Suppose that n is the number of people (this might be a sum of some other variables in your problem). Introduce a new variable y that has the following constraints:

y is integer-valued
28*y >= n (you can represent this as a linear inequality constraint)
The (linear) cost function includes y as one term

Then the minimum of the cost function will occur when y is minimized, meaning when it is as small as it can be but still above n/28. Then you can write your cost in terms of y.

Alan Weiss

MATLAB mathematical toolbox documentation

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m c 2015-11-24

I have been trying to work through the example, but I am still stuck on one thing.

I see now I have to append the n variables to the end of the x vector.

However, I dont know how to relate the n part of the x vector to the part of the x vector that controls room allocations. I imagine I would have to include part of the x vector in the beq vector.

How do I do this?

Thanks again M

m c 2015-11-24

Never mind. I figured it out.

I just had to modify the A matrix and b vector.

And now I have my optimised schedule!

Thanks for your help.

M

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