Michael Griffiths wrote
>
> Dear List,
>
> I am using constrOptim to solve the following
>
> fr1 <- function(x) {
> b0 <- x[1]
> b1 <- x[2]
> ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3
> }
>
> As you can see, my objective function is
> ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would
> like to solve for both b0 and b1.
>
> If I were to use optim then I would derive the gradient of the function
> (grr) as follows:
>
> fr2 <-
> expression(((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3)
> grr <- deriv(fr2,c("b0","b1"), func=TRUE)
>
> and then simply use optim via
>
> optim(c(-5.2,0.22), fr1, grr)
>
> My problem is that I wish to place constraints (b0>=-0.2 and b1>= 0.1)
> upon
> the values of b0 and b1. I can set the constraints matrix and boundary
> values to
>
> ui=rbind(c(1,0),c(0,1)) and ci=c(-0.2,0.1), however, when I come to run
> constrOptim function via
>
>
> constrOptim(c(-0.1,0.2), fr1, grr, ui=rbind(c(1,0),c(0,1)),
> ci=c(-0.2,0.1))
>
> I get the following error message:
>
> "Error in .expr1 + b1 : 'b1' is missing"
>
> So, it seems to me that I am doing something incorrectly in my
> specification of grr in constrOptim.
>
grr is a function with two arguments. Do this
grr
and then you will see.
But the gradient function passed to constrOptim wants a function with a
vector argument.
So if you do
gradr <- function(x) {
b0 <- x[1]
b1 <- x[2]
grr(b0,b1)
}
and after testing with
gradr(c(-0.1,0.2))
this should work
constrOptim(c(-0.1,0.2), fr1, gradr, ui=rbind(c(1,0),c(0,1)),
ci=c(-0.2,0.1))
Berend
--
View this message in context:
http://r.789695.n4.nabble.com/constrOptim-and-problem-with-derivative-tp4217531p4217776.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.
