#1
08-23-2012, 05:28 AM
 invis Senior Member Join Date: Jul 2012 Posts: 50

I never used QP in my practice. And I using Octave for solve HWs.
So there is built-in function for quadratic programming in Octave - qp. (example)

But I still dont know what parameters I should use to solve:

s.t.

Looks like H is (.* is element by element product)
But what is the other parameters
#2
08-23-2012, 09:09 AM
 invis Senior Member Join Date: Jul 2012 Posts: 50

And where is ?
#3
08-23-2012, 10:23 AM
 invis Senior Member Join Date: Jul 2012 Posts: 50

I am absolutly confused

Should I use QP to solve this inequality:

s.t.

OR to solve inequality in picture in previous message ?
#4
08-23-2012, 04:22 PM
 yaser Caltech Join Date: Aug 2009 Location: Pasadena, California, USA Posts: 1,477

Quote:
 Originally Posted by invis I am absolutly confused Should I use QP to solve this inequality: s.t. OR to solve inequality in picture in previous message ?
The version in the picture. It's the dual problem (equivalent) to the version that is in the quote.
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#5
08-24-2012, 03:40 AM
 invis Senior Member Join Date: Jul 2012 Posts: 50

Why sometimes it is impossible to find the solution and what to do with that ?
On 200 iterations with random line and random dots (#10) 81 times QP didnt find the solution (I plot the line and dots for this situations and all looks pretty clear).
This is the parameters for QP I use:

H = (Y*Y') .* (X*X');
A = Y';
q=-1*ones(n,1);
b=0;
lb=zeros(n,1); (lower bound)
ub=[]; (upper bound)

min 0.5 x'*H*x + x'*q
subject to

A*x = b
lb <= x <= ub

where x is our
#6
08-24-2012, 04:30 AM
 htlin NTU Join Date: Aug 2009 Location: Taipei, Taiwan Posts: 601

Quote:
 Originally Posted by invis Why sometimes it is impossible to find the solution and what to do with that ? On 200 iterations with random line and random dots (#10) 81 times QP didnt find the solution (I plot the line and dots for this situations and all looks pretty clear). This is the parameters for QP I use: H = (Y*Y') .* (X*X'); A = Y'; q=-1*ones(n,1); b=0; lb=zeros(n,1); (lower bound) ub=[]; (upper bound) min 0.5 x'*H*x + x'*q subject to A*x = b lb <= x <= ub where x is our
Sometimes the numerical condition of the optimization problem makes it hard for the solver to locate a good solution. One possibility is to set ub (upper bound) to a really large value. Hope this helps.
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#7
08-24-2012, 06:08 AM
 apinde Member Join Date: Jul 2012 Posts: 12

Can one use Solver in Excel for this problem? I've not used the other languages and platforms for quadratic programming packages and am desperately searching for a package that can be learned and used in a few days.

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