# How can I simplify this quadratic optimization?

I have no experience in the field of optimization, so I have no idea how hard or naive it is. I got no response on math.stackexchange so I am posting it here, though I doubt it is research-level.

I want to minimize $x^t P x + q^t x$ subject to the following constraint:

For all $b \in B$, $|x^b| \le C \sum_{b' \in B} |x^{b'}|$

where $B = \{1, ..., n\}$ and $x^b$ is the $b$th component of the $n$-dimensional column vector $x$. $C$ is some positive constant which, to avoid triviality, should satisfy $1/|B| \le C \le 1$.

The only way I know how to do this is to do $2^{|B|}$ optimizations over the convex cone given by:

For all $b \in B$, $x^b \ge 0$ and $x^b \le C \sum_{b' \in B} x^{b'}$

and its reflections. Is there a more efficient way to solve this problem?

For my purposes let's say $C = 1/5$ and $n = 100$. I'm not sure I have much choice in the structure of $P$ and $q$, so an efficient solution for general $P$ and $q$ is desirable. [EDIT: $P$ is positive semidefinite] (Perhaps an approximate solution is much easier to find. Help with that would be appreciated too.)

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Is there a reason for $\sum_{b'\in B}$ as opposed to $\displaystyle\sum_{b=1}^n$ ? (which you would get with \displaystyle\sum_{b=1}^n ) – Ricky Demer Jul 26 '11 at 7:00
@Ricky: no particular reason. That's just the notation that I happened to settle on. – Tom Ellis Jul 26 '11 at 8:05
I have thought about this some more. An algorithm solving my problem would be similar to an algorithm solving the Closest Vector Problem, which I believe is known to be NP-hard, so perhaps a subexponential algorithm is asking too much! (Still, I don't have experience in this area so a confirmation of this point from someone knowledgeable would be welcome!). – Tom Ellis Jul 26 '11 at 10:03
So is $P$ an arbitrary symmetric matrix, or it is semidefinite? – Suvrit Jul 26 '11 at 17:46
@Suvrit: All the eigenvalues are positive, although in the case I'm considering the smallest are $10^7$ times smaller than the largest. – Tom Ellis Jul 27 '11 at 6:42

Thanks, I've edited the statemet to say that $P$ is positive semidefinite. – Tom Ellis Jul 27 '11 at 6:44