Hello,

You cannot change the numerical accuracy, it's a built-in constant. To see it use

?.Machine
.Machine$double.eps  # smallest value different from zero


Actually, .Machine$double.eps is the "the smallest positive floating-point number x such that 1 + x != 1"

You can try the following function

is.zero <- function(x, eps = .Machine$double.eps^0.5) abs(x) < eps

is.zero(your_value)  # TRUE?


Or even try ?all.equal

all.equal(your_value, 0)


Hope this helps,

Rui Barradas

Em 10-04-2013 22:13, nntx escreveu:
Rui, thanks for your reply. You meant that it is the issue of accuracy? So if
I change the numerical accuracy, my results can be output? Thanks a lot!



--
View this message in context: 
http://r.789695.n4.nabble.com/Optimization-problem-tp4663821p4663928.html
Sent from the R help mailing list archive at Nabble.com.

______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Reply via email to