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I have some constant values $c_i$ in $(0.5, 2)$. I also have binary variables $x_i$. For my integer program, for a particular constraint, I need to multiply only those $c_i$ when $x_i$ takes the value 1.

So basically, $\Pi_i \big[(c_i - 1)x_i + 1\big]$. I'm trying to linearize this. I can put a continuous variable $\gamma_i = (c_i - 1)x_i + 1$ and try to use piecewise linear functions, but since $(c_i - 1)x_i + 1$ can only take two values, i.e., $c_i$ and $1$, is there any way to use this property to have a more efficient linearization?

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    $\begingroup$ Your first paragraph describes $\prod_i c_i^{x_i}$, very far from the expression starting paragraph two. Which of the two is intended? $\endgroup$ Commented May 25, 2020 at 4:26
  • $\begingroup$ I'll try to clarify with an example. Let there be three values of i. So I have $c_1, c_2, c_3$ which are constants. I also have $x_1, x_2, x_3$ which are binary variables. Now, let $x_1=1$, $x_2=0$, and $x_3=1$. So I need $c_1c_3$. $\endgroup$ Commented May 25, 2020 at 4:30
  • $\begingroup$ Yes, you need $c_1^1 c_2^0 c_3^1 = c_1^{x_1} c_2^{x_2} c_3^{x_3}$. $\endgroup$ Commented May 25, 2020 at 4:31
  • $\begingroup$ Since $c_i$'s are constant terms, and I only know $i$ beforehand, I'm not clear how to index $c_i$ with the binary optimization variables $x_i$. The $c_i$ values are given known beforehand, and I also know how many $i$'s there are. $\endgroup$ Commented May 25, 2020 at 4:37
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    $\begingroup$ Depending on how you want to use the resulting product, you might be able to use a log transformation, which yields the linear function $\sum_i \log(c_i) x_i$. $\endgroup$ Commented May 25, 2020 at 5:29

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Depending on how you want to use the resulting product, you might be able to use a log transformation, which yields the linear function $\sum_i \log(c_i)x_i$.

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