you could re-parameterize, e.g.,
EM <- c(4.5000,6.0000,10.5000,5.0000,27.0000,20.7500,16.7500,23.6666,38.7500)
W <- array(EM, c(3,3))
d <- c(10, 20, 20)
##############33
fn <- function(x){
x <- exp(x) / sum(exp(x))
r <- W%*%x - d
crossprod(r, r)[1,1]
}
opt <- optim(rnorm(3), fn)
res <- exp(opt$par) / sum(exp(opt$par))
res
I hope it helps.
Best, Dimitris
---- Dimitris Rizopoulos Ph.D. Student Biostatistical Centre School of Public Health Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium Tel: +32/16/336899 Fax: +32/16/337015 Web: http://www.med.kuleuven.ac.be/biostat/ http://www.student.kuleuven.ac.be/~m0390867/dimitris.htm
----- Original Message ----- From: "Stefaan Lhermitte" <[EMAIL PROTECTED]>
To: <[email protected]>
Sent: Thursday, March 17, 2005 3:13 PM
Subject: [R] Optimization of constrained linear least-squares problem
Dear R-ians,
I want to perform an linear unmixing of image pixels in fractions of pure endmembers. Therefore I need to perform a constrained linear least-squares problem that looks like :
min || Cx - d || � where sum(x) = 1.
I have a 3x3 matrix C, containing the values for endmembers and I have a 3x1 column vector d (for every pixel in the image). In theory my x values should all be in the (0,1) interval but I don't want to force them so I can check the validity of my solution. I just want to calculate the x values. Can anyone help me with this problem? I've been checking the optim, optimize, constrOptim and nlm help files, burt I don't understand it very well. Wich function should I use for my problem? I did a first test using optim:
# Make my C matrix
EM<- c(4.5000,6.0000,10.5000,5.0000,27.0000,20.7500,16.7500,23.6666,38.7500)
C <- array(EM, c(3,3))
# Take an arbitrary d d<-c(10, 20, 20)
# Define the function fr <- function(x) { C[1,]*x=d C[2,]*x=d C[3,]*x=d sum(x)=1}
# Perform the optimization optim(c(0.25,0.25,0.25),fr)
But it did not work. I got the eror couldn't find function. Can anyone tell me what functyion I should use for my problem and how should I program it?
Thanx in advance, Stef
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