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

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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);
}