Timeline for Quasiconvex Constraints in Maximisation
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 28, 2023 at 21:48 | comment | added | Amit | We need to compare all the local optimums to determine the global optimum in cases where either $f$ is not quasi-concave or $g$ is not quasi-convex. | |
| Feb 27, 2023 at 15:00 | comment | added | CormJack | I see, so we need it as a second oder condition to ensure globally? So we can still find local optimums if both constraint and objective function ore concave (or both convex) but we won't be assure that this is a global optimum? correct? Thanks for the input! | |
| Feb 15, 2023 at 0:29 | vote | accept | CormJack | ||
| Feb 15, 2023 at 0:27 | comment | added | Amit | However if $f$ is quasi concave and $g$ is quasi-convex, local optimality guarantees optimality. This is certainly a desirable property if you want to write efficient programs to determine optimal solutions. | |
| Feb 15, 2023 at 0:27 | comment | added | Amit | Consider the following modified problem: $\displaystyle\max_{(x,y)\in\mathbb{R}^2_+} 2x+y \text{ s.t. } \sqrt{x} + \sqrt{y} \leq 1$. This problem has a unique solution which is $(1,0)$. However, $(0,1)$ is also a "local" optimum i.e. If you're at $(0,1)$ and you try to move along the constraint boundary, the value of the objective will fall initially. So local optimality does not guarantee optimality. | |
| Feb 14, 2023 at 19:14 | comment | added | CormJack | Hi Amit, good to see you again so soon! Thanks for providing another answer, I'm not too clear on this one....With your final example you provide a solution which is not a convex set. But why do we need it to be a convex set? I'm still not clear intuitively what goes wrong if we have both a concave constraint and a concave objective function? | |
| Feb 14, 2023 at 18:08 | history | answered | Amit | CC BY-SA 4.0 |