1
$\begingroup$

I want to solve the following problem, \begin{equation} \begin{aligned} & \underset{\mathbf{x}}{\text{minimize}} & & \mathbf{x^T}\mathbf{Px} \\ & \text{subject to} & & \mathbf{A{x}}=\mathbf{0} \\ &&& \mathbf{{x}^{T}}\mathbf{{x}}=1\\ &&& \mathbf{{x}^{T}}\mathbf{S}\mathbf{{x}}= n_0\\ &&& \mathbf{{x}^{T}}\mathbf{T}\mathbf{{x}} \le k_0\\\end{aligned} \end{equation} Where $x\in \mathbb{R}^n, n_0,k_0 \in \mathbb{R^+}, P \in {S_{++}}(n) $

Is it possible to solve this problem ? If yes, then how?

$\endgroup$
  • $\begingroup$ If $n_0,k_0$ are fixed, why do you want both $x^tTx=n_0$ and $x^tTx\leq k_0$? $\endgroup$ – M Turgeon Mar 14 '14 at 14:04
  • $\begingroup$ Sorry! it is a typo! they are the different matrices! i have corrected :) Thanks :) $\endgroup$ – Saket Mar 14 '14 at 14:11
1
$\begingroup$

What about the matrices $T$ and $S$? Are they positive semi-definite?

In any case this problem is not convex due to the constraints $x^T x =1$ and $x^TSx=n_0$. So it is in general a difficult problem.

There might be some specialized algorithm since you are optimizing on the unitary sphere. In general, any nonlinear solver should give you at least a stationary point, if any (it might be infeasible). Otherwise you need to look to deterministic Branch-and-Bound (see Couenne for instance).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.