qp() in Octave 3.0.0 returns result egregiously violating input constraints

[Joshua Redstone]

Hi Gabriele,
My mistake, The l2-norm of eVrH is 1.  I didn't realize Z was a matrix and
so was printing too many entries of eVrH.
So I'm back to trying to figure out why 'p' has such small values in it
(1e-17)....
Are those small values in 'p' indeed problematic?
It's a bit hard tracking down this bug because I don't have a good idea of
the math behind the code so don't know what to expect.
Josh

On Thu, Apr 10, 2008 at 1:33 AM, Gabriele Pannocchia <
g.pannocchia at ing.unipi.it> wrote:

> Hi Josh,
>
> > I put in the expressions you suggested into qp.m right before the
> > __qp__() call.  The output I get is:
> > for  norm(A*x0-b):
> >    ans =  5.7220e-16
> > and for
> >       all((Ain*x0) > bin -
> > 1e-15)
> > I got 1.  (I got zero if I put a tolerance of 1e-18)
>
> OK.
>
> > I have a bit more debugging information.
> > So when __qp__ runs, it starts off with a feasible solution.  Then it
> > varies between thinking the problem is convex and not convex.
> > That is, it varies between setting info to 0 and 1, and I think this
> > depends on the set of active constraints.
>
> Loosely speaking, yes. Depending on which inequality constraints are in
> the current active set, the reduced Hessian could be positive definite
> or not. Your Hessian is actually indefinite, so the reduced Hessian
> could be itself indefinite.
>
> The reduced Hessian is checked in line 240 of __qp__.cc, and pR tells us
> if it is positive definite or not. If it is, i.e. info=0, the Cholesky
> factors are computed and the inverse of the reduced Hessian computed.
>
> If instead the reduced Hessian is not positive definite, i.e. info=1,
> the "most negative curvature" is found from an eigenvalue/eigenvector
> decomposition of rH.
> The most negative eigenvalue is evaluated in line 277, the corresponding
> index is found in the subsequent lines and finally eVrH is the
> eigenvector associated to the minimum eigenvalue.
>
> By definition (as far as I know), eigenvectors should be normalized to
> length 1 (i.e. the 2-norm of eVrH should be 1), so that if you report
> values in the vector eVrH that exceeds 1 (in magnitude), then the
> problem may be with the library that makes the eigenvalue decomposition.
>
> Can you confirm that the 2-norm of eVrH is huge?
>
> Thanks,
>         Gabriele
>
>
>
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