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QP solves for zero instead of min?
From: |
Juan Pablo Carbajal |
Subject: |
QP solves for zero instead of min? |
Date: |
Sat, 2 Aug 2014 19:38:20 +0200 |
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.
Q = - ones (size(H,1),1);
A = [ones(1,floor(length(Q)/2)) -ones(1,ceil(length(Q)/2))];
a0 = rand (size(H,2),1);
a = qp (a0, H, Q, A, 0); % I get all zeros here.
# Try with SQP
phi = @(x) 0.5*x.'*H*x + x.'*Q;
g = @(x) A*x;
a_sqp = sqp (a0, phi, g);
a_r = a_sqp / sqrt(sumsq (a_sqp));
phi(a) # gives a zero
g(a) # it solves the constraints
phi(a_r) # negative!
g(a_r) # solves the constratins to tolerance
- QP solves for zero instead of min?,
Juan Pablo Carbajal <=
- Re: QP solves for zero instead of min?, Olaf Till, 2014/08/03
- Re: QP solves for zero instead of min?, Juan Pablo Carbajal, 2014/08/03
- Re: QP solves for zero instead of min?, Olaf Till, 2014/08/03
- Re: QP solves for zero instead of min?, Juan Pablo Carbajal, 2014/08/03
- Re: QP solves for zero instead of min?, Olaf Till, 2014/08/03
- Re: QP solves for zero instead of min?, Juan Pablo Carbajal, 2014/08/03
- Re: QP solves for zero instead of min?, Olaf Till, 2014/08/03
- Re: QP solves for zero instead of min?, Juan Pablo Carbajal, 2014/08/04