I'm wondering if there is any way to convexify, approximate, and/or simplify the following problem.
$\max. \sum_{k \in K} \prod_{i \in I} (a_{ik} x_{ik} + b_{ik})$
s.t. $x_{ik} \in [0,1]$
where the data structure is such that $a_{ik} \in (-1,1)$, $b_{ik} \in (0,1)$, and $a_{ik} x_{ik} + b_{ik} \in (0,1)$.
Any help or references will be greatly appreciated.
It's actually close to a convex optimization problem. If $n=|I|$, the number of indices in $I$, then this is a convex problem:
$$\begin{array}{ll} \text{maximize} & \sum_{k\in K} \left( \prod_{i\in I} a_{ik}+b_{ik} \right)^{1/n} \\ \text{subject to} & a_{ik} x_{ik} + b_{ik} \in [0,1] ~ \forall i,k \\ & x_{ij} \in [0,1] ~ \forall i,k \end{array}$$ The $1/n$ exponents turn the products into concave geometric means.
If you need the case without the exponents, this suggests a heuristic. Consider the weighted case $$\begin{array}{ll} \text{maximize} & \sum_{k\in K} w_k \left( \prod_{i\in I} a_{ik}x_{ik}+b_{ik} \right)^{1/n} \\ \text{subject to} & a_{ik} x_{ik} + b_{ik} \in [0,1] ~ \forall i,k \\ & x_{ij} \in [0,1] ~ \forall i,k \end{array}$$ where $w_k\geq 0$. First, solve this with $w_k\equiv 1$, the unweighted version. Then take that current solution and define $$w_k \triangleq \left(\prod_{i\in I} a_{ik} x_{ik} + b_{ik} \right)^{1-1/n}$$ and now solve the weighted problem. Iterate until convergence. I'm not saying this is guaranteed of course...