Displaying 20 results from an estimated 7000 matches similar to: "Optimization Question"
2011 Sep 22
1
nlm's Hessian update method
Hi R-help!
I'm trying to understand how R's nlm function updates its estimate of the Hessian matrix. The Dennis/Schnabel book cited in the references presents a number of different ways to do this, and seems to conclude that the positive-definite secant method (BFGS) works best in practice (p201). However, when I run my code through the optim function with the method as "BFGS",
2011 Aug 13
3
optimization problems
Dear R users
I am trying to use OPTIMX(OPTIM) for nonlinear optimization.
There is no error in my code but the results are so weird (see below).
When I ran via OPTIM, the results are that
Initial values are that theta0 = 0.6 1.6 0.6 1.6 0.7. (In fact true vales
are 0.5,1.0,0.8,1.2, 0.6.)
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2010 Jun 22
1
Subject: Re ZINB by Newton Raphson??
I have not included the previous postings because they came out very strangely on my mail
reader. However, the question concerned the choice of minimizer for the zeroinfl()
function, which apparently allows any of the current 6 methods of optim() for this
purpose. The original poster wanted to use Newton-Raphson.
Newton-Raphson (or just Newton for simplicity) is commonly thought to be the
2017 Dec 31
1
Order of methods for optimx
Dear R-er,
For a non-linear optimisation, I used optim() with BFGS method but it
stopped regularly before to reach a true mimimum. It was not a problem
with limit of iterations, just a local minimum. I was able sometimes to
reach better minimum using several rounds of optim().
Then I moved to optimx() to do the different optim rounds automatically
using "Nelder-Mead" and
2011 Sep 02
5
Hessian Matrix Issue
Dear All,
I am running a simulation to obtain coverage probability of Wald type
confidence intervals for my parameter d in a function of two parameters
(mu,d).
I am optimizing it using "optim" method "L-BFGS-B" to obtain MLE. As, I
want to invert the Hessian matrix to get Standard errors of the two
parameter estimates. However, my Hessian matrix at times becomes
2009 Dec 06
5
optim with constraints
Hi, dear R users
I am a newbie in R and I wantto use the method of meximum likelihood
to fit a Weibull distribution to my survival data. I use "optim" as
follows:
optim(c(1, 0.25),weibull.like,mydata=mydata,method="L-BFGS-B",hessian
= TRUE)
My question is: how do I setup the constraints so that the two
parametrs of Weibull to be pisotive? Or should I use other function
2023 Jan 05
1
R 'arima' discrepancies
Rob J Hyndman gives great explanation here
(https://robjhyndman.com/hyndsight/estimation/) for reasons why results
from R's arima may differ from other softwares.
@iacobus, to cite one, 'Major discrepancies between R and Stata for
ARIMA'
(https://stackoverflow.com/questions/22443395/major-discrepancies-between-r-and-stata-for-arima),
assign the, sometimes, big diferences from R
2012 Nov 15
1
hessian fails for box-constrained problems when close to boundary?
Hi
I am trying to recover the hessian of a problem optimised with
box-constraints. The problem is that in some cases, my estimates are very
close to the boundary, which will make optim(..., hessian=TRUE) or
optimHessian() fail, as they do not follow the box-constraints, and hence
estimate the function in the unfeasible parameter space.
As a simple example (my problem is more complex though,
2011 Sep 27
2
Error in optim function.
I'm trying to calculate the maximum likelihood estimate for a binomial
distribution. Here is my code:
y <- c(2, 4, 2, 4, 5, 3)
n <- length(y)
binomial.ll <- function (pi, y, n) { ## define log-likelihood
output <- y*log(pi)+(n-y)*(log(1-pi))
return(output)
}
binomial.mle <- optim(0.01, ## starting value
binomial.ll,
2008 Jun 24
1
Hessian in box-constraint problem - concern OPTIM function
Hello all useRs,
I am using the OPTIM function with particular interest in the method
L-BFGS-B,
because it is a box-constraint method.
I have interest in the errors estimates too.
I make:
s.e. <- sqrt( diag( solve( optim(...,method='L-BFGS-B',
hessian=TRUE)$hessian )))
but in help say:
"Note that this is the Hessian of the unconstrained problem even if the
box constraints
2012 Oct 10
1
"optim" and "nlminb"
#optim package
estimate<-optim(init.par,Linn,hessian=TRUE, method=c("L-BFGS-B"),control =
list(trace=1,abstol=0.001),lower=c(0,0,0,0,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf,-Inf),upper=c(1,1,1,1,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf,Inf))
#nlminb package
estimate<-nlminb(init.par,Linn,gr=NULL,hessian=TRUE,control =
2007 Nov 10
1
polr() error message wrt optim() and vmmin
Hi,
I'm getting an error message using polr():
Error in optim(start, fmin, gmin, method = "BFGS", hessian = Hess, ...) :
initial value in 'vmmin' is not finite
The outcome variable is ordinal and factored, and the independant variable
is continuous. I've checked the source code for both polr() and optim()
and can't find any variable called
2009 Jan 28
3
initial value in 'vmmin' is not finite
Dear r helpers
I run the following code for nested logit and got a message that
Error in optim(c(0, 0, 0, 0, 0.1, -2, -0.2), fr, hessian = TRUE, method = "BFGS") : initial value in 'vmmin' is not finite
What does this mean? and how can I correct it?
Thank you
June
> yogurt = read.table("yogurtnp.csv", header=F,sep=",")> attach(yogurt)>
2008 Jun 24
2
L-BFGS-B needs finite values of 'fn'
Hi,
When I run the following code,
r <- c(3,4,4,3,5,4,5,9,8,11,12,13)
n <- rep(15,12)
x <- c(0, 1.1, 1.3, 2.0, 2.2, 2.8, 3.7, 3.9, 4.4, 4.8, 5.9, 6.8)
x <- log10(x)
fr <- function(c, alpha, beta) {
P <- c + (1-c) * pnorm(alpha + beta * x)
P <- pmax(pmin(P,1),0)
-(sum(log(choose(n,r))) + sum(r * log(P)) + sum((n -r)* log(1-P)))
}
fit <- mle((fr), start = list(c
2011 Oct 21
2
Arima Models - Error and jump error
Hi people,
I´m trying to development a simple routine to run many Arima models result
from some parâmeters combination.
My data test have one year and daily level.
A part of routine is:
for ( d in 0:1 )
{ for ( p in 0:3 )
{ for ( q in 0:3 )
{ for ( sd in 0:1 )
{ for ( sp in 0:3 )
{ for ( sq in 0:3 )
{
2007 Apr 09
1
R:Maximum likelihood estimation using BHHH and BFGS
Dear R users,
I am new to R. I would like to find *maximum likelihood estimators for psi
and alpha* based on the following *log likelihood function*, c is
consumption data comprising 148 entries:
fn<-function(c,psi,alpha)
{
s1<-sum(for(i in 1:n){(c[i]-(psi^(-1/alpha)*(lag(c[i],-1))))^2*
(lag(c[i],-1)^((-2)*(alpha+1))
)});
s2<- sum(for(m in 1:n){log(lag(c[m],-1)^(((2)*alpha)+2))});
2013 Nov 15
1
optimization
x1<-c(5.548,4.896,1.964,3.586,3.824,3.111,3.607,3.557,2.989,18.053,3.773,1.253,2.094,2.726,1.758,5.011,2.455,0.913,0.890,2.468,4.168,4.810,34.319,1.531,1.481,2.239,4.204,3.463,1.727)
y<-c(2.590,3.770,1.270,1.445,3.290,0.930,1.600,1.250,3.450,1.096,1.745,1.060,0.890,2.755,1.515,4.770,2.220,0.590,0.530,1.910,4.010,1.745,1.965,2.555,0.770,0.720,1.730,2.860,0.760)
2000 Mar 17
1
optim: problem with additional arguments (PR#493)
The R function "optim" fails when a function requires additional
arguments, if the option "hessian=T" is specified. I am using R Version
1.0.0 on Windows 98.
Here is an example of the Rosenbrock Banana function from the optim help
example, with the function and gradient modified to take an additonal
argument. Note that the call to optim works fine unless a hessian is
2009 May 03
3
Optim function in the loop
Hi all,
I wrote the following lines of codes try to do some iterations to find the
global optimal values, but the function does not execute properly. Every
time codes stop after one iteration right after executing the optim()
function. Does anyone could have me to take a look? Thanks.
if (count>0){
k=k+0.05;
mu0=c(83+k,0,0)
Sigma0= diag(0.4,3)
initpar=c(.1+10*k,10*k,10*k,10*k) # initial
2006 May 01
1
Problem with optim()
I am having a problem with optim() using the "L-BFGS-B" method. When I
set the lower limit for the third parameter equal to zero I get an
error message:
> low.lim.3 <- 0
> phi_opt <- optim(phi_, model_lik, NULL, method = "L-BFGS-B", lower=c(0.2, -100, low.lim.3, 0), upper= c(10, 100, 10, 10), control = list(maxit = 1000, parscale = c(0.2, u1, 0.002, 0.002), trace =