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I want to solve a linear program but with a subset of the variables taken from a unit sphere. That is, given fixed $\textbf{c} \in \mathbb{R}^{n}$, $\textbf{A} \in \mathbb{R}^{m \times (n+k)}$, I want to find variables $\left[ \begin{array}{c} \textbf{x} \\ \textbf{y} \end{array} \right]$ with $\textbf{x} \in \mathbb{R}^k$ and $\textbf{y} \in \mathbb{R}^n$ so that

$\min~~\textbf{c}^T \textbf{y}$

$s.t.~~\textbf{A} \left[ \begin{array}{c} \textbf{x} \\ \textbf{y} \end{array} \right] \geq 0$

$and~~~~\|\textbf{x}\|_2 = 1$.

Introducing the equality constraint on the norm of $\textbf{x}$ makes the problem a (non-convex) quadratically-constrained linear program. My understanding is that problems with non-convex constraints are not easy to solve in the general case (e.g., solving a general QCQP is an NP-hard problem).

That said, this problem has some structure to it --- the objective function is linear, and in particular the only quadratically constrained variables $\textbf{x}$ do not (explicitly) appear in the objective function.

With that in mind, I have two questions:

(1) Is this problem type a well-studied sub-problem of some sort? It seems like the unit-vector constraint is common enough that the problem type might be studied in its own right (e.g. linear program, over a unit sphere). Can this be converted to a well-studied problem type (e.g. a semidefinite program) that has an efficient solution?

(2) Regardless of the answer to (1), is there a good approach to solving this optimization problem? An efficient means of solving it or obtaining a good approximation?

I'd be grateful for any insight. Thank you!

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    $\begingroup$ If you replace $\|\mathbf{x}\|_2 = 1$ with $\|\mathbf{x}\|_2 \leq 1$, you can rewrite as a (convex) semidefinite program. $\endgroup$ – Rodrigo de Azevedo Jul 29 '19 at 8:30
  • $\begingroup$ I had known about this technique but didn't mention it because for my specific problem, the optimum over the unit-ball does not lie on the unit sphere, and I specifically need $\textbf{x}$ on the sphere itself. Is there a way to use this relaxation to find optima on the sphere itself? Perhaps by modifying the cost function to penalize smaller $\|\textbf{x}\|$ ? Something like $\textbf{c}^T \textbf{y} - \gamma \| \textbf{x} \|_2$ for some $\gamma$ ? $\endgroup$ – kklosteer Jul 29 '19 at 21:25
  • $\begingroup$ Seems related: google.com/url?sa=t&source=web&rct=j&url=http://… $\endgroup$ – user35593 Apr 19 '20 at 10:48
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Going by the first comment, the optimal solution to the convex problem (= replaced by $\leq$) must give a solution on the unit sphere. Firstly, Since $\{0,0\}$ is a feasible point, the optimal value cannot be positive. It can either be $0$ or negative. If its zero, an optimal solution can be scaled to lie on the unit sphere. So it suffices to check for negative case.

Now suppose $||x^*||=\delta<1$ for an optimal solution $\{x^*, y^*\}$. One can then choose the feasible point $\{\frac{1}{\delta}x^*,\frac{1}{\delta}y^*\}$, which has a smaller cost. This is a contradiction to the assumption of optimality.

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  • $\begingroup$ Nice! Great solution, thank you. $\endgroup$ – kklosteer Apr 19 '20 at 22:50

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