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I have this minimization problem

$\min \sum_{i=1}^k\alpha_i {X_i\choose r}$

s.t. $\sum_{i=1}^kX_i=m$

and $0<X_i<N$, $\alpha_i>0$, $1<k<N$, $m=(k-1)N$, and $1<r<N$

where $N$, $m$, $r$, and $\alpha_i$ are constants. ($N$ is the number of nodes in the problem)

But I have no idea how to approach it. Any ideas will help greatly.

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  • $\begingroup$ $\forall i, X_i > r$ ? Are there more bounds even relative ? $\endgroup$ Commented Sep 13, 2016 at 17:28
  • $\begingroup$ In practice, $X_i$ can be less than $r$. But then $X_i\choose r$ will be zero. $\endgroup$ Commented Sep 13, 2016 at 20:03
  • $\begingroup$ then it is a very general and complicate question ... $\endgroup$ Commented Sep 13, 2016 at 20:08
  • $\begingroup$ Typical size of problem, and typical value on r and m? $\endgroup$ Commented Sep 14, 2016 at 13:20
  • $\begingroup$ There is not a typical limit on these variables and they can become large. But I'll add some ranges in relation to other variables in the problem. $\endgroup$ Commented Sep 14, 2016 at 18:09

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Combinatorics in integer programming, pretty cool...

You are facing a typical case of integer programming with a non-linear cost function. What you can do (like always in those cases), is to define

$ \forall i \in [1,k], \forall n \in [0,m], Y_{i,m} $ the binary variable equal to $1$ iff $X_i = n$.

Your problem then becomes

$\min \sum_{i \in [1,k]} \sum_{n \in [0,m]} \alpha_i \binom{n}{r} Y_{i,n} $

s.t. $ \sum_{i \in [1,k]} \sum_{n \in [0,m]} n \cdot Y_{i,n} = m $

$\forall i \in [1,k], \sum_{n \in [0,m]} Y_{i,n} = 1$

Which is linear

I hope for you that this trick doesn't make an unsolvable MIP (I don't know the size of your problem). This kind of linearization trick is usually expensive computation-wise.

I sincerely wish you find a trick requiring less variables (for instance, if you are lucky enough so that $\binom{X_i}{r}$ is always convex, you don't need all of this). Good luck

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  • $\begingroup$ Thank you, I still don't know if this will solve my problem but it gets me on the right path. $\endgroup$ Commented Sep 14, 2016 at 21:42

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