1
$\begingroup$

I am dealing with a knapsack-like problem with one difference from the conventional problem: the “weights” can be positive or negative and the constraint is $\sum w_i x_i \ge 0$ instead of $\sum w_i x_i \le W$. The "values" can also be positive or negative.

Can this be transformed to a knapsack problem or is it some other type of combinatorial optimization problem?

$\endgroup$
6
  • $\begingroup$ This is better suited to or.stackexchange.com $\endgroup$ Commented Oct 3, 2021 at 15:08
  • $\begingroup$ Are the values all positive? If so, we should certainly take all items with positive weights, and then the question of the best subset of negative-weight items seems to be classical knapsack again. $\endgroup$ Commented Oct 4, 2021 at 2:34
  • $\begingroup$ Good point! Forgot to mention that values can also be negative. $\endgroup$ Commented Oct 4, 2021 at 11:47
  • $\begingroup$ What type of variable is $x$? $\endgroup$ Commented Oct 4, 2021 at 15:11
  • $\begingroup$ Binary indeed as you noted $\endgroup$ Commented Oct 5, 2021 at 9:13

1 Answer 1

1
$\begingroup$

Assuming $x_i$ is binary, perform a change of variables $\bar{x}_i:=1-x_i$ where $w_i>0$: \begin{align} \sum_i w_i x_i &= \sum_{i:w_i>0} w_i x_i + \sum_{i:w_i<0} w_i x_i \\ &= \sum_{i:w_i>0} w_i(1-\bar{x}_i) + \sum_{i:w_i<0} w_i x_i \\ &= \sum_{i:w_i>0} w_i + \sum_{i:w_i>0} (-w_i)\bar{x}_i + \sum_{i:w_i<0} w_i x_i \end{align} So $\sum_i w_i x_i \ge 0$ is equivalent to $$\sum_{i:w_i>0} w_i\bar{x}_i + \sum_{i:w_i<0} (-w_i) x_i \le \sum_{i:w_i>0} w_i$$

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.