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10021Does a convex objective subject to a polytope attain its optimum at a vertex?
https://www.asbestian.de/blog/index.php?/archives/1632-Does-a-convex-objective-subject-to-a-polytope-attain-its-optimum-at-a-vertex.html
Mathemathicshttps://www.asbestian.de/blog/index.php?/archives/1632-Does-a-convex-objective-subject-to-a-polytope-attain-its-optimum-at-a-vertex.html#commentshttps://www.asbestian.de/blog/wfwcomment.php?cid=16321https://www.asbestian.de/blog/rss.php?version=2.0&type=comments&cid=1632nospam@example.com (Sebastian Schenker)
<p>Today, while thinking about a work-related optimisation problem, I asked myself whether, given a domain corresponding to a nonempty polytope \(P\) and a convex objective, is there always an optimal solution which is a vertex of \(P\) ?</p>
<p>To warm up let's look at the <strong>linear programming case</strong> first. Given a nonempty polytope \(P = \{x \in \mathbb{R}^n: Ax \geq b \}\) and a linear objective \(c \in \mathbb{R}^n\) it's well known that there is a vertex \(v \in P\) which is optimal for \(\text{min } c \cdot x \text{ s.t. } x \in P\).
We can prove this as follows: Let \(Q\) be the set of all optimal solutions and let \(w \in \mathbb{R}\) be the optimal objective value. Then, \(Q = \{ x \in \mathbb{R}^n : Ax \geq b, c \cdot x = w \}\) is itself a polytope. Since \(Q \subset P\), it does not contain any lines implying it has at least one vertex. Let \(v \in Q\) be a vertex of \(Q\). We show that \(v\) is also a vertex of \(P\). Suppose \(v\) is not a vertex of \(P\). Then, there are \(x, y \in P\) such that \(x \neq v, y \neq v\) and some \(\lambda \in [0,1]\) such that \(v = \lambda x + (1-\lambda)y\). Moreover, we get $$w = c \cdot v = \lambda c \cdot x + (1-\lambda) c \cdot y$$. The latter can be used to show that \(x \in Q\) and \(y \in Q\) which contradicts that \(v\) is a vertex of \(Q\). </p>
<p>Given a <strong>convex</strong> objective function, it turns out that we have to distinguish between minimisation and maximisation unlike in the linear programming case.</p>
<p><strong>Minimisation:</strong> Let the polytope \(P \subset \mathbb{R}^2\) be given by the convex combination of the four vertices \( (1,0), (0,-1), (-1,0), (0,1) \) and let \(f : P \rightarrow \mathbb{R}\) be given by $$f(x) = x_1^2 + x_2^2 + x_1 x_2. $$ Consider \(\text{min } f(x) \text{ s.t. } x \in P\). It's not hard to see that none of the vertices corresponds to an optimal solution which is given by the interior point \((0,0) \in P\). In other words, if we minimise a convex function subject to a polytope, then we can generally not assume that an optimal solution is given by a vertex.</p>
<p><strong>Maximisation:</strong> Let \(P \subset \mathbb{R}^n\) be a polytope and let \(f : P \rightarrow \mathbb{R}\) be a convex function. Consider the optimisation problem \(\text{max } f(x) \text{ s.t. } x \in P\) and let \(x \in P\) be an optimal solution. Suppose \(x\) is not a vertex of \(P\). Then, by the equivalence of vertices and extreme points, we know that \(x\) can be written as \(x = \lambda y + (1-\lambda)z\) with \(\lambda \in (0,1)\) and \(y, z\) are vertices of \(P\). Now, by definition of a convex function we get \(f(x) = f(\lambda y + (1-\lambda)z) \leq \lambda f(y) + (1-\lambda) f(z)\). The latter implies that either \(f(x) \leq f(y)\) or \(f(x) \leq f(z)\) or both. In other words, either \(y\) or \(z\) or both will also be optimal.</p>
Wed, 20 Mar 2019 21:01:00 +0100https://www.asbestian.de/blog/index.php?/archives/1632-guid.htmlconvex optimisationlinear optimisation