gurobi bigM constraint vs. epsilon

Question

I am new to mathematical programming and I am trying to implement case specific constrains in Gurobi with Python.

I am wondering about how I can implement my constraints in the fastest or most common way.

There are three variables: $$ w_i,w_j \in \mathbb{N}, y_\text{i,j} \in \{0,1\} $$ The indices $i$ and $j$ referes to fiction tasks since this model takes part of a task scheduling problem. Variable $w$ is a weight of the task and $y$ shows if task $i$ precedes $j$. So there are some other constraints to provide some preceding rules.

I need to implement different constraints (let's say $C_1, C_2, C_3$) depending to those variables:
$$ 1: (w_i > w_j) \land (y_\text{i,j} \lor y_\text{j,i}) \rightarrow C_1 \\ 2: (w_i = w_j) \land (y_\text{i,j} \lor y_\text{j,i}) \rightarrow C_2 \\ 3: (w_i < w_j) \land (y_\text{i,j} \lor y_\text{j,i}) \rightarrow C_3 $$

To implement this constraint I need introduce new binary variables to the model (let's say $a_1,a_2,a_3$) wich are only equal $1$, when the refered case is true. Since Gurobi does not implement strict less or greater operators, I need to model a bigM constraints for the variables $a$.

$$ w_i \ge w_j + \epsilon - M(1-a_1) \\ w_i \le w_j + M \times a_1 \\ a_1 \in {0,1} \\ \epsilon << w_i \tag{1} $$

Since there is a constraint that says, that either $y_\text{i,j}$ or $y_\text{j,i}$, but not both, are equal 1, I plan to implement a constraint for each case with a new binary variable $b$ to represent the logical condition: $$ b_1 = a_1 \times (y_\text{i,j} + y_\text{j,i}) $$

Finally, I am able to implement each case with an indicator constraint like $b_x \rightarrow C_x; x\in {1,2,3}$

At first, I am wondering if the second equation of EQ:1 can also expressed with epsilon instead of M as $w_i \le (w_j - \epsilon) \times a_1$.

What is the difference?

Second, Is this the correct way to implement a problem like this or is there a better way?

Answer 1

Changing the second inequality in (1) as proposed would mean that $w_i \notin (w_j - \epsilon, w_j + \epsilon).$ It would shrink the feasible region in a way that might be detrimental to the solution of the problem.

To get the split you want (into three cases), you could employ three binary variables $z_1,z_2,z_3$ together with the following constraints: $$ w_i \ge w_j + \epsilon - M(z_2 + z_3)$$ $$ w_j - Mz_2 \le w_i \le w_j + Mz_2$$ $$ w_i \le w_j - \epsilon +M(z_1 + z_2)$$ $$ z_1 + z_2 + z_3 = 1.$$

The last equation can be used to eliminate one of the binary variables. (I would leave it to the presolver to decide whether to do that.) The three $z$ variables correspond to your cases 1, 2 and 3.