Re: QP solves for zero instead of min?

[Juan Pablo Carbajal]

On Sun, Aug 3, 2014 at 5:20 PM, Olaf Till <address@hidden> wrote:
> On Sun, Aug 03, 2014 at 04:42:46PM +0200, Juan Pablo Carbajal wrote:
>> On Sun, Aug 3, 2014 at 12:24 PM, Olaf Till <address@hidden> wrote:
>> > On Sat, Aug 02, 2014 at 07:38:20PM +0200, Juan Pablo Carbajal wrote:
>> >> Hi,
>> >>
>> >> It seems that when QP receives linear equality constrains it goes for
>> >> the zero of the cost function instead of a minimum.
>> >> Might somebody find my mistake? Here is my test problem
>> >>
>> >> x = randn (2,5);
>> >> H = x.'*x; #Cov matrix to make sure is a good one.
>> >
>> > H has no full rank, the result is not determined. Try x = randn (5);.
>> >
>> > Olaf
>> >
>> > --
>> > public key id EAFE0591, e.g. on x-hkp://pool.sks-keyservers.net
>>
>> That's not the issue, it was just an example. If rank deficient
>> matrices are a problem for the algorithm then at least QP should emit
>> a warning or ti should be commented on the docstring. Do not you
>> think?
>>
>> What is the source of this limitation for QP? I can't find a direct
>> reference to this problem, but just to the KKT matrix.
>
> This has nothing to do with a special algorithm, and nothing to do
> with a limitation of an algorithm. Your example problem had more than
> one solution. `qp' and `sqp' can come up with different solutions
> which are both correct.
>
> Olaf
>
> --
> public key id EAFE0591, e.g. on x-hkp://pool.sks-keyservers.net

With a small variation the example I provided is SVM. So far QP gives
the zero of the cost function and not the solution. Do you mean the
problem is not convex? How can that be?