You have merely found a little used
entry point into Circle 1 of 'The R Inferno'.
The third element of your answer is zero
to the precision of the QP algorithm. So it
is obeying the non-negative constraints that
you put on the problem to the best of its
ability. You should not expect numerical
exactness.
You might consider using the 'zapsmall'
function on your 'w' vector.
On 11/04/2010 16:31, li li wrote:
Hi,
thanks!
I added meq=1 and it did not seem to work. The result is the same as before.
x<- 2*c(1,0.5,0.8,0.5,1,0.9, 0.8,0.9,1)
Dmat<- matrix(x, byrow=T, nrow=3, ncol=3)
dvec<- numeric(3)
Amat<- matrix(0,3,4)
Amat[,1 ]<- c(1,1,1)
Amat[,2:4 ]<- t(diag(3))
bvec<- c(3,0,0,0)
solve.QP(Dmat,dvec,Amat,bvec=bvec, meq=1)
$solution
[1] 1.500000e+00 1.500000e+00 -8.881784e-16
$value
[1] 6.75
$unconstrained.solution
[1] 0 0 0
$iterations
[1] 3 0
$Lagrangian
[1] 4.5 0.0 0.0 0.6
$iact
[1] 1 4
2010/4/11 Gabor Grothendieck<ggrothendi...@gmail.com>
Add meq=1 to the arguments.
On Sun, Apr 11, 2010 at 9:50 AM, li li<hannah....@gmail.com> wrote:
Hi, thank you very much for the reply!
Consider minimize quadratic form w'Aw with A be the following matrix.
Dmat/2
[,1] [,2] [,3]
[1,] 1.0 0.5 0.8
[2,] 0.5 1.0 0.9
[3,] 0.8 0.9 1.0
I need to find w=(w1,w2,w3), a 3 by 1 vector, such that sum(w)=3, and
wi>=0
for all i.
Below is the code I wrote, using the function solve.QP , however, the
solution for w still have a
negtive component. Can some one give me some suggestions?
Thank you very much!
x<- 2*c(1,0.5,0.8,0.5,1,0.9, 0.8,0.9,1)
Dmat<- matrix(x, byrow=T, nrow=3, ncol=3)
dvec<- numeric(3)
Amat<- matrix(0,3,4)
Amat[,1 ]<- c(1,1,1)
Amat[,2:4 ]<- t(diag(3))
bvec<- c(3,0,0,0)
solve.QP(Dmat,dvec,Amat,bvec=bvec)
$solution
[1] 1.500000e+00 1.500000e+00 -8.881784e-16
$value
[1] 6.75
$unconstrained.solution
[1] 0 0 0
$iterations
[1] 3 0
$Lagrangian
[1] 4.5 0.0 0.0 0.6
$iact
[1] 1 4
2010/4/10 Gabor Grothendieck<ggrothendi...@gmail.com>
Check out the quadprog package.
On Sat, Apr 10, 2010 at 5:36 PM, li li<hannah....@gmail.com> wrote:
Hi, thanks for the reply.
A will be a given matrix satisfying condition 1. I want to find the
vector w that minimizes the
quadratic form. w satisfies condition 2.
2010/4/10 Paul Smith<phh...@gmail.com>
On Sat, Apr 10, 2010 at 5:13 PM, Paul Smith<phh...@gmail.com>
wrote:
I am trying to minimize the quardratic form w'Aw, with certain
constraints.
In particular,
(1) A=(a_{ij}) is n by n matrix and it is symmetric positive
definite,
a_{ii}=1 for all i;
and 0<a_{ij}<1 for i not equal j.
(2) w'1=n;
(3) w_{i}>=0
Analytically, for n=2, it is easy to come up with a result. For
larger
n, it
seems
difficult to obtain the result.
Does any one know whether it is possible to use R to numerically
compute
it?
And your decision variables are? Both w[i] and a[i,j] ?
In addition, what do you mean by "larger n"? n = 20 is already large
(in your sense)?
Paul
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