# [R] Increasing precision of rgenoud solutions

Jasjeet Singh Sekhon sekhon at berkeley.edu
Thu May 10 21:35:38 CEST 2007

```Hi Paul,

I see.  You want to increase the population size (pop.size)
option---of lesser importance are the max.generations,
wait.generations and P9 options.  For more details, see
http://sekhon.berkeley.edu/papers/rgenoudJSS.pdf.

For example, if I run

a <- genoud(myfunc, nvars=2,
Domains=rbind(c(0,1),c(0,1)),max=TRUE,boundary.enforcement=2,solution.tolerance=0.0000001,
pop.size=6000, P9=50)

options("digits"=12)

I obtain:

#approx analytical solution
sum(c(0.707106781186548,0.707106781186548))
[1] 1.41421356237

#genoud solution
#a\$value
[1] 1.41421344205

#difference
a\$value-sum(c(0.707106781186548,0.707106781186548))

[1] -2.91195978441e-09

If that's not enough precision, increase the options (and the
run-time).  This would be faster with analytical derivatives.

Cheers,
Jas.

=======================================
Jasjeet S. Sekhon

Associate Professor
Travers Department of Political Science
Survey Research Center
UC Berkeley

http://sekhon.berkeley.edu/
V: 510-642-9974  F: 617-507-5524
=======================================

Paul Smith writes:
>Thanks, Jasjeet, for your reply, but maybe I was not enough clear.
>
>The analytical solution for the optimization problem is the pair
>
>(sqrt(2)/2,sqrt(2)/2),
>
>which, approximately, is
>
>(0.707106781186548,0.707106781186548).
>
>The solution provided by rgenoud, with
>
>solution.tolerance=0.000000001
>
>was
>
>\$par
>[1] 0.7090278 0.7051806
>
>which is not very precise comparing with the values of the
>(analytical) solution. Is it possible to increase the degree of
>closeness of the rgenoud solutions with the analytical ones?
>
>Paul
>
>Paul Smith writes:
> > Dear All
> >
> > I am using rgenoud to solve the following maximization problem:
> >
> > myfunc <- function(x) {
> >   x1 <- x[1]
> >   x2 <- x[2]
> >   if (x1^2+x2^2 > 1)
> >     return(-9999999)
> >   else x1+x2
> > }
> >
> > genoud(myfunc, nvars=2,
> > Domains=rbind(c(0,1),c(0,1)),max=TRUE,boundary.enforcement=2,solution.tolerance=0.000001)
> >
> > How can one increase the precision of the solution
> >
> > \$par
> > [1] 0.7072442 0.7069694
> >
> > ?
> >
> > I have tried solution.tolerance but without a significant improvement.
> >
> > Any ideas?
> >