Is there a way to formulate the following Linear Program in a mixed-integer LP that I could solve with most linear programs in R/Python that support Mixed Integer Linear Programs (MILP)?
So my question is: How can I use a combination of integer, binary and continuous variables to reformulate the constraints (1) below?
Constants: $C_i$ (factor exposure), $x_i^a$ (initial weight)
Decision variables: $x_i$ (portfolio weight)
Portfolio Maximization:
$\max_{x_{i}}\sum_{i=1}^{N}x_{i}\cdot C_{i}$
subject to:
(1) $\boldsymbol{1}_{\left\{ x_{i}\geq x_{i}^{a}\right\} }\left(x_{i}-x_{i}^{a}\right)\in\{0\}\cup\left[0.025,\infty\right],\forall i$ (Minimum purchase size of 0.025)
where
$\boldsymbol{1}_{\left\{ x_{i}\geq x_{i}^{a}\right\} }=\begin{cases}
1 & \text{if } x_{i} \geq x_{i}^{a}\\
0 & \text{otherwise}
\end{cases}$
Best Answer
Introduce a small constant tolerance $\epsilon > 0$, binary decision variables $y_i^1$ and $y_i^2$, and linear constraints \begin{align} y_i^1 + y_i^2 &\le 1 &&\text{for all $i$} \tag1\\ (0-x_i^a) y_i^1 + 0.025 y_i^2 \le x_i - x_i^a &\le -\epsilon y_i^1 + (1-x_i^a)y_i^2 &&\text{for all $i$} \tag2\\ \end{align} Constraints $(1)$ and $(2)$ enforce $x_i - x_i^a \le -\epsilon \lor x_i - x_i^a = 0 \lor x_i - x_i^a \ge 0.025$. Explicitly, the three cases are \begin{align} (y_i^1,y_i^2)=(1,0): && -x_i^a \le x_i - x_i^a &\le -\epsilon \\ (y_i^1,y_i^2)=(0,0): && 0 \le x_i - x_i^a &\le 0 \\ (y_i^1,y_i^2)=(0,1): && 0.025 \le x_i - x_i^a &\le 1-x_i^a \\ \end{align}
More simply, introduce a binary decision variable $z_i$ and linear constraints $$-x_i^a (1-z_i) + 0.025 z_i \le x_i - x_i^a \le (1-x_i^a)z_i \text{ for all $i$}$$
If $z_i=0$, the constraint implies $-x_i^a \le x_i - x_i^a \le 0$, so $x_i \le x_i^a$.
If $z_i=1$, the constraint implies $0.025 \le x_i - x_i^a \le 1-x_i^a$, so $x_i \ge x_i^a + 0.025$.