# Tag Info

2

For a continuous function, all you need to do is prove that it's (i) non-convex, and (ii) monotonic. (i) can be shown using the eigenvalues of the hessian matrix, and (ii) using the gradient. However, in your case your domain is $\mathbb{Z}$, therefore derivatives are generally not defined, and neither is the concept of (pseudo)convexity. You can show ...

6

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

1

For the problem you describe, since the function is strongly convex and $x,y \in \mathbb{R}$ (so you don't have bound constraints), your solution is obtained by solving the system $\nabla f = 0$. Now, if you only have approximations to the gradient, the resulting error largely depends on the algorithm you chose to use to solve this. For instance, since your ...

5

Yes, because $\log$ is monotonic, it preserves inequalities. The tightness depends on your other constraints.

2

This is a convex problem under any definition. From Wikipedia: If $f$ is a convex function or if a minimum point of $f$ is being sought, then $f$ is called proper if there exists some point $x_0$ in its domain such that: \begin{aligned} f(x_0)<\infty\\ f(x) > -\infty \end{aligned} $x^3$ is convex for $x\geq 0$ and it also satisfies the above condition, ...

4

An objective function which is convex on only part of its unconstrained domain, but which is convex on the constraint set (i.e.., for any feasible point), can be 'convexified" by modifying the objective function to have value $\infty$ for infeasible points. This is essentially what is done by Disciplined Convex Programming (DCP) systems such as CVX, ...

5

With the natural binary decision variables $x_{s,u}$, $y_{u,g}$, and $z_{s,g}$ you had defined earlier, the problem is to maximize $\sum_u q_u$ subject to \begin{align} q_u &= \frac{\sum_s h_{s,u} x_{s,u}}{\sum_s h_{s,u} (1-x_{s,u})} &&\text{for all $u$} \tag1 \\ \sum_g y_{u,g} &= 1 &&\text{for all $u$} \tag2 \\ \sum_g z_{s,g} &= ...

1

Although the code that you shared shows that simple constraints have been generated, it does not show the constraints. A better way to generate these constraints in Pyomo is to use constraint lists as follow: model.cons1 = ConstraintList() and then in a for loop you will generate the expression for each constraint and add them to the list: model.cons1.add(...

2

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 general nonlinear form (and thus a general nonlinear solver), you might use a geometric programming form (and thus solve as a convex problem) if all your other ...

6

Maximize an auxiliary variable $z$ subject to the constraints $z\le \sum_{c=1}^C d_{u,c}x_{u,c}\ \forall u$.

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You can model this as a maxmin problem by introducing an auxiliary variable $\theta$: \begin{align} \max&\quad\theta &\\ \text{s.t.}&\quad\theta \leq \sum_{c=1}^C x_{uc}d_{uc} & \forall u=1,\dots,U \end{align} For future reference, if in contrast you had a minmax objective instead of a maxmin objective, you could apply the same trick: \begin{...

3

If I understand the problem correctly, the $y$ variables decide which components of $x$ are non-zero, and the rest is essentially some variant of a least-square problem. There are several ways you can prevent a solution $x^{*}, y^{*}$ from deviating too much from a previous solution $\bar{x}, \bar{y}$. Adding extra constraints Restrict the number of ...

2

Welcome to OR Stack Exchange. Your question is not clear. You are interested in special points in discrete optimization spaces. But you describe a mixed-variable problem involving both continuous variables $x$ and integer variables $y$. For a given $y$, the theory of calculus of variations applies to the continuous subproblem on $x$. But I don't think you ...

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