Re: [Help-glpk] How to linearize a weighted average with a decision variable?

[Michael Hennebry]

On Tue, 24 Apr 2018, Matt wrote:

> *max sum(i) { enabled[i] * value[i] * weight[i] } / sum(i) { enabled[i] *
> weight[i] }*
>
> *s.t. sum (i) enabled[i] = M*
>
> - *value* is a vector of decimal numbers in [0, 1] (precomputed)
> - *weight* is a vector of decimal numbers in [0, 1] (precomputed)
> - *enabled* is a vector of either 0 or 1 (decision variable)

For linear constraints, there is a tranformation to an LP:
https://en.wikipedia.org/wiki/Linear-fractional_programming#Transformation_to_a_linear_program
It does not convert an integer problen to an integer problem.
My suggestion is to use it to get an LP-based bound, call it q.
Then maximize numerator - q*denominator as an IP.
If it's zero, you are done.
If it's negative, the true objective gives you another q.
If it's positive, you made a mistake.

You might need to explicitly bound the denominator.

--
Michael   address@hidden
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