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Considering the first given objective function, the problem is trivial to solve. The decision variables X1 and X2 are positive. Because polynomial with positive coefficients only, the objective function is thus monotonically increasing over the space of feasible solutions. The last constraint over X1 and X2 is inactive because dominated by the previous ones. ...


NP-hardness is an asymptotic result of increasing problem dimensions. It is not a property of one fixed problem instance. So to ask whether your problem class is NP-hard, you would have to explain how the problem would grow with increasing numbers of $X$ variables. As far as solving your specific instance, a number of nonlinear solvers could probably do it, ...


The notion of NP-hardness relates to whether one class of problems can be solved by a solver for another class of problems where the translation overhead is negligible. The problem you presented is member of many classes of problems. However NP-hardness is the property of a class of problems. So it is impossible to answer whether this particular instance is ...


There are different solvers in Excel that are suited for different types of problems: Simplex LP GRG Nonlinear Evolutionary The Simplex LP solver is the only solver that can guarantee an optimal solution, however, the problem size is somewhat limited and all of the constraints as well as the objective function have to be linear. Your problem is small, ...


Your question isn't entirely clear and isn't really an OR question, but I think what you are trying to do is the following: for j in n: for k in r: o += xsum(w[k] *y[k][jj] for jj in n if jj <= j) >= xsum (a[i][k]* x[i][j] for i in p)


\begin{align}\text{diff}^2&=c^2+4\left(\left(\sum s_jx_j\right)^2-c\sum s_jx_j\right)\\&=c^2+4\left(\sum s_j^2x_j^2+\sum_{\rm cyc}s_ks_\ell x_kx_\ell-c\sum s_jx_j\right)\\&=c^2+4\left(\sum x_j\boldsymbol{s_j(s_j-c)}x_j+\sum_{\rm cyc}x_k\boldsymbol{s_ks_\ell}x_\ell\right)=c^2+4x^\top Qx\end{align}


Yes, a real PSD matrix $M$ is a symmetric matrix with $$x^TMx\ge 0$$ for any $x$ (see e.g. However, this is not a real restriction. (We have two meanings of "real" here). We can form $$M' = \frac{M+M^T}{2}$$ Now $M'$ is symmetric and we have $$x^TM'x = x^TMx$$ for any $x$. So you can make $M$ ...


I would advise Bayesian Optimization. The benefits imho are that they don’t require a gradient, work for a wide variety of optimization problems and are made for when we are dealing with functions that are hard or slow to evaluate.


Excluding the constant portions of the objective function is perfectly fine. The optimal solutions will be unchanged and it should have no impact on how long the solver needs to solve the model.

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