On 21.12.2011 18:08, Michael Griffiths wrote:
Dear List,
I have the code below, where I am using the constrained optimisation
package, 'constrOptim.nl' to find the values of two values, b0 and b1.
I have no problems when I enter further variable information DIRECTLY into
the functions, fn, and heq. In this instance I require fn to have -0.0075
appended to it, and in the case of heq, h[1] has -0.2.
library(alabama)
fn<-function(x)
(((1/(1+exp(-x[1]+x[2]))+(1/(1+exp(-x[1])))+(1/(1+exp(-x[1]-x[2])))))/3)-0.0075
heq<- function(x) {
h<- rep(NA,2)
h[1]<- (1-(1/(1+exp(-x[1]))/(1/(1+exp(-x[1]-x[2])))))-0.2 #Constraint
1: diff = 20%
h[2]<- -log((1-0.0075)/0.0075)-x[1] #Constraint 2: set
b0
h
}
p0<-c(-4,0.3)
ans<-constrOptim.nl(par=p0,fn=fn,heq=heq)
This code does indeed work as expected and the results have been verified
elsewhere.
However, I do have problems when I wish to enter this information
indirectly via the constrOptim function, e.g.
fn<-function(x,RR)
(((1/(1+exp(-x[1]+x[2]))+(1/(1+exp(-x[1])))+(1/(1+exp(-x[1]-x[2])))))/3)-RR
heq<- function(x,Diff) {
h<- rep(NA,2)
h[1]<- (1-(1/(1+exp(-x[1]))/(1/(1+exp(-x[1]-x[2])))))-Diff #Constraint
1: diff = 20%
h[2]<- -log((1-RR)/RR)-x[1]
#Constraint 2: set b0
h
}
p0<-c(-4,0.3)
ans<-constrOptim.nl(par=p0,fn=fn,heq=heq, RR=0.0075, Diff=0.2)
In this instance I get the following error,
'Error in heq(theta, ...) : unused argument(s) (RR = 0.0075)'
I do not appear to be able to input further arguments in this manner, any
help from the list would be gratefully received.
Reason is that you need to specify fn and heq in a way that will accept
these arguments!
So the defitionions should start with
fn <- function(x,RR,Diff)
and
heq <- function(x,RR,Diff)
respectively.
Uwe Ligges
Regards
Mike Griffiths
On Tue, Dec 20, 2011 at 11:53 AM, Michael Griffiths<
griffi...@upstreamsystems.com> 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.
If the list could help, I would be most grateful.
Regards
Mike Griffiths
--
*Michael Griffiths, Ph.D
*Statistician
*Upstream Systems*
8th Floor
Portland House
Bressenden Place
SW1E 5BH
<http://www.google.com/url?q=http%3A%2F%2Fwww.upstreamsystems.com%2F&sa=D&sntz=1&usg=AFrqEzfKYfaAalqvahwrpywpJDL9DxUmWw>
Tel +44 (0) 20 7869 5147
Fax +44 207 290 1321
Mob +44 789 4944 145
www.upstreamsystems.com<http://www.google.com/url?q=http%3A%2F%2Fwww.upstreamsystems.com%2F&sa=D&sntz=1&usg=AFrqEzfKYfaAalqvahwrpywpJDL9DxUmWw>
*griffi...@upstreamsystems.com<einst...@upstreamsystems.com>*
<http://www.upstreamsystems.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.
