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Is the following problem NP-complete:

Input: A set of tuples $T = \left\{ t_i=(a_i,b_i) | 1\le i\le n \right\}$, an integer $k$ and numbers $C,D\in \mathbb Q_{\ge 0}$.

Question: Exists a subset $S\subseteq T$ with $|S|\le k$ and $$ \sum_{t_i\in S} a_i - C\cdot \prod_{t_i \in S} b_i \ge D. $$

There are some related threads that did not really help me:

As the problem seems to be quite fundamental, I am quite sure that there exists a source. But I have not so many good ideas what that could be called like. I have found somewhat similar but not close enough problems:

However all relations to Knapsack result in the issue that in the presented problem there are no weights to the tuples/elements. And connections to subset-sum fail because I require $\ge D$.

The most desirable variant for me would be $a_i\in \mathbb Q_{\ge 0}, b_i\in \mathbb Q\cap[0,1)$, but I would also be happy with an idea for the case where $a_i$s and $b_i$s are integers.

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  • $\begingroup$ Let's assume $a_i\ge 0, b_i \in [0,1)$. WLOG, we can assume $|S|=k$. Consider these two extremes: (1) If each $b_i$ is close to 1, then the problem might be easy: $\prod b_i \approx 1 - \sum (1-b_i)$, so the problem becomes approximately $\sum (a_i+C-Cb_i) \ge D+1$, which can be solved with a greedy algorithm. (2) If each $b_i$ is far from 1, the problem also might be easy: $\prod b_i$ is exponentially small (in $k$), so the problem is approximately $\sum a_i \ge D$, which can also be solved with a greedy algorithm. Any ideas for how to handle intermediate cases? $\endgroup$
    – D.W.
    Aug 29, 2022 at 17:32
  • $\begingroup$ The idea is actually quite interesting. Indeed when the $b_i$s are relatively close to each other (informally) then we can approximate $$ \prod_{t_i\in S} b_i \approx b_1^{|S|}. $$ And from that point we can go greedy about the $a_i$s. That generalizes the idea for "intermediate cases". So the the question is what about $b_i$s having quite values that are not so close to each other. $\endgroup$
    – xtyner
    Aug 30, 2022 at 9:31

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