Feels like you are asking two things, tractability of convex problems and convexity of integer problems. A first order approximation is that convex programs are tractable, .i.e., most problems you ...

I suspect you want to minimize $\sum\limits_i \left[k_1\max(e_i,0)^2 + k_2\max(-e_i,0)^2\right]$. For instance, $a=0.5$ would correspond to $k_1 = 0.25$ and $k_2 = 2.25$). This can be formulated as a ...

Non-convex means not convex, which could mean concave but also neither convex nor concave, such as a bilinear term $xy$.

Introduce four binary variables indicating which region you are in and the function value. \begin{align} \delta_1 &\rightarrow x_2\geq c, ~x_3\geq d, ~x_1 = a\\ \delta_2 &\rightarrow x_2\...

As all have mentioned, the problem is intrinsically hard, as it effectively involves a strict inequality. One way to represent it, using no strict inequalities or magic constants, is to use the ...

In YALMIP you can use arbitrary black-box operators to circumvent modelling n = 5; c = randn(3*n,1); A = randn(10*n,3*n); b = rand(10*n,1); % MILP model x = sdpvar(2*n,1); Domain= [0 <= x <= 1];...

You are maximizing a convex quadratic (the monotonic log is irrelevant) so the maximum is attained at the border, i.e. either $0$ or $\min(1,\sqrt{1-\text{constant}})$.

You can ask Gurobi to do this for you https://www.gurobi.com/documentation/9.1/refman/preqlinearize.html Whether a linearization leads to better or worse performance is almost impossible to know ...

My interpretation is that you want $y$ to be $i$ if $p_i=1$. You can do that with a simple multiplication $y=c^Tp$ where the constant vector $c$ is given by $c_i=i$.

No, it is an intrinsically non-convex constraint. Just take a diagonal matrix, and the feasible set would be the coordinate axes, i.e. nonconvex and highly ill-conditioned as the feasible set has ...

Besides simply adding a large bound (which can cause numerical issues and lead to poor branching) or presolve from constraints involving the unbounded variable, the solver might be able to derive ...

Consider maximizing $\sum_{i=1}^n x_i$ subject to $x$ binary and $\sum_{i=1}^n i x_i \leq 2$ Solve relaxation and it gives $x_1 = 1, x_2 = 1/2$ with remaining variables 0. Branch on $x_2 = 0$ and the ...

You just seem to have hidden a long list of constraints of the form $(x_i=j) \Rightarrow \text{equalities}_{ij}$ Introduce a binary matrix $C_{ij}$ with $\sum_j C_{ij}= 1$ and $C_{ij} \Rightarrow \{... View answer 5 votes By Farkas lemma, infeasibility of$Ax\leq b$is equivalent to feasibility of$A^Ty = 0, y^Tb < 0, y\geq 0$, or more practically useful$A^Ty=0, y^Tb \leq -1, y\geq 0$. Unfortunately, this will lead ... View answer Accepted answer 5 votes Although focused on implementing the model in YALMIP instead of CVX (converting the code should be trivial), precisely this case is described in the following tutorial https://yalmip.github.io/... View answer Accepted answer 4 votes Introduce a binary variable$\delta_t$to represent which case it is and$z_t$to represent the modelled product, and your MILP model of the piecewise-affine dynamics would be${EP}_t\ =\ \sum_{i=1}^{...

Are you talking about doing it in some particularly clever way? If you simply intend to solve the problem in the original dual space (i.e. seeing $X$ as a matrix parameterized by its elements and ...

Extending Robs answer slightly, taking into account that you asked about ILP (which I interpret as mixed-integer linear program), the constraint is MILP-representable as long as $f$ is MILP-...

You are not going to be able to add these logs and quadratic terms to the model via simple double-sided big-M constraints, as they generate non-convex use of convex quadratics and logs, and CVX does ...

No, the product is indefinite so it can neither be bounded from above using an tight convex epigraph representation, nor from below using a convex hypograph representation. If you cannot accept a ...

Introduce a term $y_m$ to replace and lower bound the terms inside the logarithm. Those lower bound constraints simplify to $\sum_{n=1}^{N}(1-x_{m,n})\omega_{m,n}+z\ge y_m(\sum_{n=1}^{N}x_{m,n}\omega_{... View answer 3 votes When you call optimize without any options set, the default values will be used, and those are created by the function baronset. View answer Accepted answer 2 votes Integrating a piecewise linear function means you end up with a piecewise quadratic function. Unless there is come convex structure in the resulting piecewise quadratic function (i.e. the PWL is non-... View answer 2 votes You are asking about representations of$\sqrt{a^2 + b^2} \leq b$. It is trivially only feasible for$a = 0$and$b\geq 0$. The constraint$b\geq 0$term will be problematic in your new model as it ... View answer Accepted answer 2 votes Define$\mu_i = \sqrt{\lambda_i}$and the problem is a convex quadratically constrained problem in$(b,\mu)\$