# How to couple a binary variable to a continuous variable to indicate values greater 0

I have a continuous variable $$x_t$$. A binary variable $$b_t$$ should be coupled to $$x_t$$ such that $$b_t$$ has the value $$1$$ if $$x_t$$ has a value greater than $$0$$ and $$b_t$$ has the value $$0$$ if $$x_t$$ has the value $$0$$. Any idea how I can do that without including the term $$b_t$$ in the objective function?

• This is a FAQ -- frequently asked here and frequently answered here.
– prubin
Commented May 17, 2021 at 20:22
• You will have to choose a tolerance as to how close to zero should be considered zero, taking into account solver tolerance, wherein a variable for which the true optimum is exactly zero will not be exactly zero as computed in and returned by the solver. Commented May 18, 2021 at 0:14
• @prubin I agree this is a FAQ, here is quite a complete answer. But it may be worth answering $x_t=0\; \Rightarrow \; b_t = 0$ (?), which is the hard part, I think. Commented May 18, 2021 at 6:53
• @MarkL.Stone Thanks for your answer Mark L. Stone. How can I formulate this with equations? Commented May 18, 2021 at 9:04
• @prubin: Thanks for your comment. Would you mind sending me 2 or 3 links to such questions? I'd highly apprciate this. Commented May 18, 2021 at 9:16

## 3 Answers

I'm only aware of a mechanism that works if there is an upper bound for the continuous variable.

\begin{align}x_{t, \max}\cdot b_t &\geq x_t\\ m\cdot x_t &\geq b_t\end{align}

I used this in answering this question. I'm not aware of a way to solve it for unbounded $$x_t$$, unless the solver handles floating points infinities correctly. This would need to be tested by solvers. The second equation implements the $$x_t < \frac{1}{m}\implies b_t=0$$. If the trueness of $$b_t$$ is penalized in the objective the second term is unnecessary.

• Thanks worldsmithhelper for your answer. Basically the continous variable has an upper bound of 1 and a lower bound of 0. Nonetheless I think your approach does not work and is wrong. If x(t) has the value 0.2, then your second equations forces b(t) to have the value 0 which is wrong. Commented May 20, 2021 at 9:01
• Mhh you are right. You could fix that by multiplying $x_t$ by a sufficiently large numbers. I edited my answer. Commented May 20, 2021 at 9:03
• Thanks for your answer and effort worldsmithhelper. This solution might work. I will try it. How would your suggested solution change, if the lower bound of x is let's say 0.25 (while the upper bounds remains 1)? Commented May 20, 2021 at 9:12
• $x_{t,max} = 1$ and $m=4$ Commented May 20, 2021 at 9:16
• Thanks worldsmithhelper for your answer. I upvoted and accepted your answer. I really appreciate your tremendous help. Commented May 20, 2021 at 10:22

Enforcing $$b_t$$ to take value $$1$$ when $$x_t$$ is positive is done with $$x_t \le b_t$$, assuming $$x_t \le 1$$.

For the second part, quoting @MarkL.Stone:

You will have to choose a tolerance as to how close to zero should be considered zero

Let $$\epsilon$$ be this tolerance. So you want to enforce $$x_t < \epsilon\implies b_t = 0$$

Now referring to this link (careful as $$b$$ in the link is a constant $$\neq$$ your $$b_t$$):

To enforce "if $$x < b$$ then $$y=1$$": $$b - x \le My,$$ where $$M$$ is a large constant. The logic is that if $$b - x > 0$$, then $$y$$ must equal 1, and otherwise it may equal 0.

Given that $$x_t \le 1$$, and that you want the binary variable to take value $$0$$ (and not $$1$$), the constraint becomes: $$\epsilon-x_t \le 1-b_t$$

It is easy to see that if $$b_t$$ takes value $$1$$, then it implies $$x_t \ge \epsilon$$, which is the contrapositive of what you want.

• Thanks Kuifje for your answer (I upvoted it). The shorter (and simpler) answer given by worldsmithhelper seems to work. I have implement it and not experienced any problems so far. Do you see any problems with that answer? Commented May 20, 2021 at 10:10
• Both approaches are similar and deal with a tolerance ($\epsilon$ or $1/m$), but the way the tolerance is handled is a bit different. Commented May 20, 2021 at 10:18
• I would say try both and reward the one giving you better performance in your problem. Commented May 20, 2021 at 10:22

Many solvers and APIs allow logical constraints.

For instance with OPL CPLEX you may write

dvar float+ x;
dvar boolean b;

subject to
{

b==!(x==0);
}

• Thanks for your answerAlex How can I formulate this with equations? I do not use OPL CPLEX and I would like to have an answer that is independent from the modelling language and the solver. Commented May 18, 2021 at 9:05
• Commented May 18, 2021 at 9:44
• Thanks for the link Alex. I think that the answer there does not quite answer my question. There it is stated for continous variables that there are 2 cases: Case 1 "if x>b then y=1" --> this also holds for my example and is in line with what I want. In my case b is just 0. However, the Case 2: "if x=b then y=1" is not in line with what I want. if x=b (and b is 0 in my case) then y should be also 0. Any idea how I can formulate this with equations? Commented May 18, 2021 at 10:49