1
$\begingroup$

Let $N_i=\{0,1,\dots,\bar{n}_i\}$ and define $N=N_1\times \dots \times N_I$. I want to maximize $f$ on $N$. $f$ has the following form $$ f(n) = \sum_i A_i n_i -\sum_i \sum_{j\neq i} B_{ij} (n_i-n_j)^2 $$ where $A_i\geq 0$ and $B_{ij}\geq 0$ for all $i,j$. Note that $f$ is a quadratic and concave function when defined on $\mathbb{R}^I$.

I know that integer programing is in general a hard problem but I would like to find nontrivial conditions on $A$ and $B$ such that this problem is easy to solve. By easy to solve I mean that there is a procedure that is fast and that is guaranteed to find the global maximum.

What I have tried

I have tried the following way of solving the problem. Define the mapping $T$ for which the $i$th element is defined as $$(Tn)_i=\arg\max_{\tilde{n}_i\in N_i} f\left( \left\{ n_1,\dots,\tilde{n}_i,\dots,n_I\right\}\right)$$ and iterate on $T$ until convergence. This is essentially a coordinate ascent algorithm but in a discrete space.

A few remarks:

  1. We can look at the fixed points of $T$. The vector $n^*$ that maximizes $f$ is obviously a fixed point of $T$ (can't improve by deviating in any dimensions) but there might be multiple fixed points.
  2. The mapping $T$ is monotone in the sense that $n\geq n'$ implies $T(n)\geq T(n')$. As a result we can use Tarski's fixed point theorem to show that the (non empty) set of fixed point is a lattice. We can find a lower bound of that set by iterating on $T$ from $(0,\dots,0)$ and an upper bound by iterating from $(\bar{n}_1,\dots,\bar{n}_I)$. If both procedures end up on the same vector then there is a unique fixed point and it must be $n^*$.
  3. When it works that procedure is quite fast numerically but I don't know what conditions on $A$ and $B$ are needed for it to work.
  4. Under some conditions, $T$ might be a contraction mapping which would guarantee uniqueness but I haven't made much progress along that path yet.

Any help is appreciated! Also, if somebody has a good reference on these problems that would be useful.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.