Is it necessary to consider $\epsilon$ when it come to strict inequality linearization?

Question

My decisions variables are all binary. One the constraints I'd like to include into a model is in the from of $\sum_{(i,j) \in E} \delta_{i,j}^t \ge 1 \implies \theta^t =1 \qquad \forall t$

I wanted reverse the statement and write it as $\theta^t =0 \implies \sum_{(i,j) \in E} \delta_{i,j}^t < 1 $ and this leads to introducing $\epsilon$ which I try to avoid. Is there any way to circumvent it?

Answer 1

For a sum of integers, $<1$ is equivalent to $\le 0$. You can enforce the indicator constraint $$\theta^t =0 \implies \sum_{(i,j) \in E} \delta_{i,j}^t \le 0$$ via linear big-M constraints $$\sum_{(i,j) \in E} \delta_{i,j}^t \le |E| \theta^t. \tag1\label1$$ Because all variables are binary, you can obtain a stronger formulation by disaggregating \eqref{1} as follows: $$\delta_{i,j}^t \le \theta^t \quad \text{for all $(i,j) \in E$}.$$

Answer 2

Try:
$ M(\sum_{(i,j) \in E} \delta_{i,j}^t - 1) \le \theta^t - 1$
$ \theta^t \le M\sum_{(i,j) \in E} \delta_{i,j}^t $\