Multiplication of a continuous and a binary variable

The expression $z = x \cdot \delta$ where $x$ is a continuous variable and $\delta$ is a binary variable can also be linearized fairly easily. The exact form depends on the bounds of $x$. Assume $x$ has lower bound $x^{lo}$ and upper bound $x^{up}$, then we can form the inequalities:

$$\boxed{\begin{split}\min\{0,x^{lo}\} &\le z \le \max\{0,x^{up}\} \\ x^{lo} \cdot \delta &\le z \le x^{up} \cdot \delta \\ x − x^{up} \cdot (1 − \delta) &\le z \le x − x^{lo} \cdot (1 − \delta)\end{split}}$$

The last restriction is essentially a big-M constraint:

$$\boxed{x − M_1 \cdot (1 − \delta) \le z \le x + M_2 \cdot (1 − \delta)}$$

where $M_1$, $M_2$ are chosen as tightly as possible while not excluding $z=0$ when $\delta=0$.

For the special case $x^{lo}=0$ this reduces to:

$$\boxed{\begin{split}& z \in [0,x^{up}]\\ & z ≤ x^{up} · \delta\\ & z ≤ x\\ & z ≥ x − x^{up} · (1 − \delta)\end{split}}$$

The construct $z=x· \delta$ can be used to model an OR condition: "$z = 0$ OR $z = x$".