# IF X = 0 THEN Y = 1, IF X > 0 THEN Y => 0

I'm trying to model the following

IF $$tS = 0$$ THEN $$Y = 1$$, IF $$tS \gt 0$$ THEN $$Y \ge 0$$

$$tS$$ is a positive real number and $$Y$$ is binary.

I tried the following: $$tS - \epsilon \ge -M Y$$ but this doesn't work.

The optimiser always sets $$ts = \epsilon$$ and $$Y = 0$$

• If tS is a positive real number, as you say, then the IF tS = 0 THEN Y = 1 part is irrelevant because positive numbers are never zero. – RocketNuts Mar 20 '20 at 19:51
• He meant nonnegative. – RobPratt Mar 21 '20 at 13:47

Your second if-then statement is always true because $$Y$$ is binary. For your first if-then statement, rewrite as its contrapositive $$Y=0 \implies tS \ge \epsilon$$. The following big-M constraint enforces that: $$\epsilon - tS \le MY$$ This is equivalent to what you tried. Note that $$(tS,Y)=(\epsilon,0)$$ is feasible, so if the solver always returns it, maybe it is optimal. As a sanity check, you could fix $$tS$$ to $$0$$ and see what happens.

Update based on a lower bound of $$0$$ for $$tS$$: you can take $$M=\epsilon-0=\epsilon$$, yielding $$tS\ge\epsilon(1-Y)$$

• I am failing to see why the big-M is necessary and why epsilon - tS <= Y is not sufficient. Am I missing something obvious? – Renaud M. Mar 20 '20 at 8:31
• Hi Rob $\\$ Valid is, $0 \le tS \le \alpha, \alpha$ not known a priori; anything within that range is feasible. For these values, according to the problem's logic, $Y$ should be $1$. However, $Y=0$ doesn't make the problem infeasible. The "funny" thing is, why does CPLEX prefer to set $ts=\epsilon$, for whatever $\epsilon \le \alpha$ and $Y=0$ and not $tS=0$ and $Y=1$. $\\$ As for Renaud's question, I think he is right. $M=1$ should be fine. – Clement Mar 20 '20 at 9:53
• I updated the answer with a recommended $M$ just now. – RobPratt Mar 20 '20 at 12:44
• The original version, ϵ−tS≤MY, is actually better. The case M=e creates problems, when tS<ϵ. Then, after trasformation we get Y ≥ 1 - (tS/ϵ), which leads to Y=1 but we should have Y ≥ 0. When M = BigM, then we get Y ≥ (ϵ - tS) / M. If in this case tS < ϵ, the (ϵ - tS) / M gets past the tolerance for 0 and thus Y ≥ 0. As an example, M = ϵ = 1E-08 drives the problem infeasible using the default settings of CPLEX.. – Clement Mar 21 '20 at 12:12
• Better not to divide by $\epsilon$. – RobPratt Mar 21 '20 at 13:49

$$1 - M \cdot ts \leq Y$$

when $$ts = 0$$ then $$1 \leq Y$$

when $$ts \gt 0$$ then $$-M \leq Y$$ and $$0 \leq Y$$ (because $$Y$$ is binary).

Here the value of $$M$$ must be chosen carefully by taking into consideration the decimal precision of $$ts$$, i.e., $$\forall ts \in (0,1] \quad ts\cdot M > 1$$

• This is algebraically equivalent to what I suggested if you set your $M$ to $1/\epsilon$. But dividing by a small number will cause numerical difficulties, so it is better to leave it in the form I had, with $\epsilon$ as a coefficient. – RobPratt Mar 21 '20 at 13:54
• yes that is better. – ooo Mar 21 '20 at 17:28