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There's something I don't understand about CVXPY's example on its MIQP use. It says that the algorithm returns a solution $x \in \mathbb{Z}^n$ but I thought in general the point of MIQP algorithms was to return a solution $x$ such as

$$\forall i, x_{i} \in [m,M] \cup \{0\}$$

given $m,M\in \mathbb{R}$. Am I missing some parameters here?

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What you described is a problem for which every variable is semicontinuous. In mixed integer programming, the variables are $(x,y)\in\mathbb{Z}^{n_1} \times \mathbb{R}^{n_2}$. For (pure) integer programming, take $n_2=0$.

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  • $\begingroup$ Okay, and let's say I want to solve an optimization problem where the variables are the $x_{i}, i=1,\ldots,N$, can I consider it as a mixed integer programming problem with $2*N$ variables, the first N being the vector $(x_{i})_{i=1,\ldots,N}$ and the other N are indicator variables $y \in \{0,1\}$ such as $\forall i, m*y_{i} \leq x_{i} \leq M*y_{i}$ ? Or do you know a more efficient way to solve a semi continous quadratic optimization in Python ? $\endgroup$
    – FredNgu
    Aug 31, 2020 at 18:18
  • $\begingroup$ @FredNgu see this thread, where several solvers are mentioned. $\endgroup$
    – RobPratt
    Aug 31, 2020 at 18:27
  • $\begingroup$ Thank you, they mentioned the lp_solver but unfortunately it seems like it only solves linear programming. Another solution they mentioned is the big-M formulation that is very similar to the approach in my last comment. I am going to try this approach, however I'm afraid the algorithm will face numerical errors. For example if for one binary variable $y_{i}=0$, we will have for that $i$ : $0 \leq x_{i} \leq 0$ but the algorithm will still affect a non-zero value (although very low) to it. What do you think about that ? $\endgroup$
    – FredNgu
    Aug 31, 2020 at 20:01

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