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I have a variable (alpha) which depends on some other binary variables, denoted as X_i. So, for some combination of other variables, alpha may take a value (Beta_j). I added some auxillary variables (b_j) for the sake of clarity. The variable alpha is defined as follows:

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I think one way of linearizing these functions is to use big-M for each function. But, I am looking for another way to efficiently linearize this function in terms of computational time.

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  • $\begingroup$ You say $\alpha = \beta_1$ for a certain combination of $x$ values, which seems to contradict the expression of $\alpha$ as a linear combination of $b$ variables. Do you mean that $x_1 + x_2 = 2 \iff b_1 = 1?$ If so, do you have a constraint limiting at most one $b$ variable to taking the value 1? $\endgroup$
    – prubin
    Commented Apr 9 at 15:37
  • $\begingroup$ Yes, each b corresponds to the right hand side of the alpha function. The point is that any of those combinations can happen. That's why each of b_j can take the value of 1. So, the sum of b_j can be greater or equal to 1. $\endgroup$
    – Sam
    Commented Apr 9 at 22:53
  • $\begingroup$ I think I can rephrase my questions in another way. It si more clear I guess. I just edit the question. I hope it is more clear now. $\endgroup$
    – Sam
    Commented Apr 10 at 14:37

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I assume here that the $\beta_j$ are all nonnegative. You can linearize the definition of $b_1$ as follows: $$b_1 \le \beta_1 x_1$$ $$b_1 \le \beta_1 x_2$$ $$b_1 \ge \beta_1 (x_1 + x_2 - 1).$$ The other $b_i$ are handled similarly.

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  • $\begingroup$ many thanks for your help ! $\endgroup$
    – Sam
    Commented Apr 11 at 12:00
  • $\begingroup$ May I ask you what would be the situation if Beta_j would be a negative value. Indeed, beta could be both positive and negative value. $\endgroup$
    – Sam
    Commented Apr 11 at 15:05
  • $\begingroup$ If $\beta_j < 0$ then $b_j$ must be declared as a nonpositive variable. The three inequalities I listed would have their directions reversed in this case. $\endgroup$
    – prubin
    Commented Apr 11 at 15:55
  • $\begingroup$ Thanks for the comment ans your help ! $\endgroup$
    – Sam
    Commented Apr 11 at 19:58

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