An embedded and charset-unspecified text was scrubbed... Name: not available URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20111220/98ec4445/attachment.pl>
constrOptim and problem with derivative
3 messages · Michael Griffiths, Berend Hasselman
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.
Berend Hasselman wrote
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)
}
This is incorrect.
The gradient function should return a vector. It was returning a scalar with
attributes.
The gradient function should be
# Correct
gradr <- function(x) {
b0 <- x[1]
b1 <- x[2]
g <- grr(b0,b1)
attr(g,"gradient")
}
and this looks better
gradr(c(-0.1,0.2))
str(gradr(c(-0.1,0.2)))
constrOptim(c(-0.1,0.2), fr1, gradr, ui=rbind(c(1,0),c(0,1)),
ci=c(-0.2,0.1))
I'm puzzled why constrOptim or optim didn't issue an error message in the
original case.
Berend
--
View this message in context: http://r.789695.n4.nabble.com/constrOptim-and-problem-with-derivative-tp4217531p4218207.html
Sent from the R help mailing list archive at Nabble.com.