Berwin A Turlach berwin at maths.uwa.edu.au
Mon Sep 3 11:15:09 CEST 2007

```G'day Thomas,

On Mon, 3 Sep 2007 10:08:08 +0200
<thomas.schwander at mvv.de> wrote:

> What's wrong with my code?
>

> Dmat<-diag(1,7,7)
> # muss als quadratische Matrix eingegeben werden
> Dmat
> dvec<-matrix(0,7,1) # muss als Spaltenvektor eingegeben werden
> dvec
> mu<-0 # (in Mio. €)
> bvec<-c(1,mu,matrix(0,7,1)) # muss als Spaltenvektor eingegeben werden
> bvec
> mu_r<-c(19.7,33.0,0.0,49.7, 82.5, 39.0,11.8)
> Amat<-matrix(c(matrix(1,1,7),7*mu_r,diag(1,7,7)),9,7,byrow=T)
> # muss als Matrix angegeben werden, wie sie wirklich ist
> Amat
> meq<-2
> loesung<-solve.QP(Dmat,dvec,Amat=t(Amat),bvec=bvec,meq=2)

With the commands above, I get on my system:

> loesung<-solve.QP(Dmat,dvec,Amat=t(Amat),bvec=bvec,meq=2)
Error in solve.QP(Dmat, dvec, Amat = t(Amat), bvec = bvec, meq = 2) :
constraints are inconsistent, no solution!

Which is a bit strange, since Dmat is the identity matrix, so there
should be little room for numerical problems.

OTOH, it is known that the Goldfarb-Idnani algorithm can have problem
in rare occasions if the problem is "ill-scaled"; see the work of
Powell.

Changing the definition of Amat to

> Amat<-matrix(c(matrix(1,1,7),mu_r,diag(1,7,7)),9,7,byrow=T)

leads to a successful call to solve.QP.  And since mu=0, I really
wonder why you scaled up the vector mu_r by a factor of 7 when putting
it into Amat.... :)

> loesung<-solve.QP(Dmat,dvec,Amat=t(Amat),bvec=bvec,meq=2)

Now, with the solution I obtained the rest of your commands show:

> loesung\$solution %*% mu_r
[,1]
[1,] 6.068172e-15
> sum(loesung\$solution)
[1] 1
> for (i in 1:7){
a<-loesung\$solution[i]>=0
print(a)
}
[1] TRUE
[1] TRUE
[1] TRUE
[1] TRUE
[1] FALSE
[1] FALSE
[1] TRUE

But:

>  for (i in 1:7){
a<-loesung\$solution[i]>=- .Machine\$double.eps*1000
print(a)
}

[1] TRUE
[1] TRUE
[1] TRUE
[1] TRUE
[1] TRUE
[1] TRUE
[1] TRUE

>  for (i in 1:7) print(loesung\$solution[i])
[1] 0
[1] 0
[1] 1
[1] 0
[1] -4.669169e-17
[1] -1.436586e-17
[1] 8.881784e-16

This is a consequence of finite precision arithmetic.  Just as you
should not compare to numeric numbers directly for equality but rather
that the absolute value of their difference is smaller than an
appropriate chosen threshold, you should not check whether a number is
bigger or equal to zero, but whether it is bigger or equal to an
FAQ 7.31.

HTH.

Cheers,

Berwin