I am asking a follow-up question of this problem. For simplicity, let's say we have a constraint, $I(-a_i x\leq b) \geq m$.
We reformulate it as $$y_i \geq m,\,-a_i x \le b + A(1-y_i) \text{ such that } y_i \in \{0,1\}.$$
As I understood, if $-a_ix-b>0$, then $y_i$ has to be $0$ in the inequality above, else our inequality will not be satisfied. But if $-a_ix-b \leq 0$, I don't see a reason why $y_i$ must be $1$. So, only one condition of the indicator function is satisfied when we construct the inequality. Could someone please tell me what am I missing here?
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Cherryblossoms
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You are missing two things. First, both of your $\ge m$ constraints should have a $\sum_i$ on the left hand side. Second, the (big-M) constraint $-a_i x \le b + A(1-y_i)$ enforces $y_i = 1 \implies -a_i x \le b$ but not its converse. When $-a_i x \le b$ is satisfied, $y_i$ is free to take either value.
RobPratt
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Regarding your second comment; so we are relaxing the indicator constraint and not equivalently writing it, right? – Cherryblossoms Apr 04 '23 at 07:47
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Correct, and this relaxation suffices because the indicator appears only in the $\ge m$ constraint. – RobPratt Apr 04 '23 at 11:48
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Key is $m \le \sum_{i}^M y_i $ which means more than one $y$s have to take $1$ as value unless $-a_ix \gt b$ holds
Sutanu Majumdar
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