similar to: R-beta: nonlinear fitting

GENERAL REFERENCE ON NONLINEAR OPTIMIZATION? What are your favorite references on nonlinear optimization? I like Bates and Watts (1988) Nonlinear Regression Analysis and Its Applications (Wiley), especially for its key insights regarding parameter effects vs. intrinsic curvature. Before I spent time and money on several of the refences cited on the help pages for "optim",

Dear all, I just know how to solve an eaquation by using optim() or nlm(). But, now, I have three nonlinear equations, how could we use optim() or nlm() to solve  a system of nonlinear equations in R?  Thank you so much. Sincerely, Joe ___________________________________________________ 您的生活即時通 － 溝通、娛樂、生活、工作一次搞定！ [[alternative HTML version deleted]]

I am trying to fit a nonlinear model using nlm(). My application of nlm() is a bit complicated. Here is the story behind the model being fit: The observer is trying to detect a signal corrupted by noise. On each trial, the observer gets stim=signal+rnorm(). In the simulation below I have 500 trials. Each row of stim is a new trial. On each trial, if the cross-correlation between the stim and the

I tried to ?ms, ?nls and apparently these aren't implemented on R yet. However I seem to remember postings on this list having to do with fitting nonlinear models (no I don't mean GLM type fits, I have a REAL nonlinear model: y=ax^b + c). So please tell me if it is possible to fit nonlinear models in R (by least squares or ML). Thanks! Bill Simpson

After many months, I am now banging my head against the wall because I can't find a solution to this seemingly trivial problem.  Any help would be appreciated: I am trying to apply a nonlinear fitting routine to a 3D MR image on a voxel-by-voxel basis.  I've tested the routine using simulated data and things went well.  As for the real data, the fitting routine

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

Simple-mindedly I tried getting MLE and SE for one-parameter model in the same way as for multi-param models. out<-nlm(fn,p=c(2),hessian=T) But sqrt(diag(solve(out$hessian))) gives the answer 1. The Hessian has only one entry, not really a matrix. diag(x) gives 1 if x is just a single number. Is this what I should be doing to get SE for MLE? sqrt(solve(out$hessian)) Thanks very much for

This surprised me. Is it the way it is supposed to be? > x<-c(1,2,3,4,5) > y<-c(1.2,1.3,1.4,1.5,NA) > x-y [1] -0.2 0.7 1.6 2.5 NA > (x-y)^2 [1] 0.04 0.49 2.56 6.25 NaN >>>so NA^2 = NaN? Why not still NA? > sum((x-y)^2) [1] NaN >>>yes that is reasonable So if you ever have a data set with missing observations (NAs), you can't do any nlm() least

Hi, I have been brought back to the "R-Side" from MatLab. I have used R in graduate econometrics but only for statistics and regression (linear and nonlinear). But now I need to run general nonlinear optimization. I know about the add-in quadprog but my problem is not QP. My problem is a general nonlinear (obj funct) with linear constraints.I know about the "ms" and

I'm a very recent user of R. I have been adapting my Splus programmes and I found only one (important) problem. There exists no function "nlmin" in R and its substitute, "nlm", does not work well with my kind of problems, sometimes no achieving convergence, other tines "converging" to impossible values. My models are highly nonlinear and are to be estimated by

I'm looking for an example of a simple R script that impliments a contrained nonlinear function using nlm or optim. I'm not exactly sure how to impliment the constraints within the objective function that is passed to nlm/optim. obj.func <- function( p ) { x(p) <- unconstrained obj function value if( constraint1 > something ) { obj.func <- x(p) } else {

Dear r-helpers, This might be an elementary question, but I have a hard time getting my head around it, so all help is much appreciated. I am working on a nonlinear regression model of the form if z > 0 y = f1(x,y), else y = f2(x,t) . In other words, the functional form of f(.) changes according to some criteria z. Natural approach would be to fit two models, i.e. model1 <- nlm(y ~ ...,

Hi, I'm trying to solve the following system of nonlinear equations P1 - F2 = x[1] + (1/2) * x[3] * x[1]^2 P2 - F2 = x[2] + (1/2) * x [3] * x[2]^2 F1 - F2 = -(1/2) * x[1] - (1/2) * x[2] + (1/8) * x [3] * (x[1] + x[2])^2 B1 - F2 = (1/4) * x[1] - (1/4) * x[2] + (1/16) * x[3] * (x[1] - x[2])^2 B2 - F2 = (1/4) * x[1] + (1/4) * x[2] + (1/16) * x[3] * (x[1]

Hello colleagues, I am attempting to determine the nonlinear least-squares estimates of the nonlinear model parameters using nls. I have come across a common problem that R users have reported when I attempt to fit a particular 3-parameter nonlinear function to my dataset: Error in nls(r ~ tlm(a, N.fix, k, theta), data = tlm.data, start = list(a = a.st, : step factor 0.000488281

Dear R users, I?m trying to optimize simultaneously two binomials inequalities (used in acceptance sampling) which are nonlinear solution, so there is no simple direct solution. Please, let me explain shortly the the problem and the question as following. The objective is to obtain the smallest value of 'n' (sample size) satisfying both inequalities: (1-alpha) <= pbinom(c, n, p1)

I have a question about passing arguments to the function f that nlm minimizes. I have no problems if I do this: x<-seq(0,1,.1) y<-1.1*x + (1-1.1) + rnorm(length(x),0,.1) fn<-function(p) { yhat<-p*x+(1-p) sum((y-yhat)^2) } out<-nlm(fn,p=1.5,hessian=TRUE) But I would like to define fn<-function(x,y,p) { yhat<-p*x+(1-p) sum((y-yhat)^2) } so

Hi It is preferable to echo your posts to r-help, you usually get more answers and some definitelly superb to mine. It is also better to start a new mail if your question has nothing to do with original subject "Maura E Monville" <maura.monville at gmail.com> napsal dne 01.10.2007 17:44:43: > Unluckily I do not have the privilege of practising with R all day > long. I

Admittedly I am using an old version 1.7.1, but can anyone tell if this is or was a problem. I can only get nlm (nonlinear minimization) to adjust the first three components of function variable. No gradient or hessian is supplied. E.G.; fnoise function(y) { y[5]/(y[4]*sp2) * exp(-((x[,3]-y[1]-y[2]*x[,1]-y[3] *x[,2])/y[4])^2/2) + (1-y[5])/(y[9]*sp2) * exp(-((x[,3]-y[6]-y[7]*x[,1]-y[8]

Dear R People: I am trying to duplicate the example from Dennis and Schnabel's "Numerical Methods for Unconstrained Optimization and Nonlinear Equations", which starts on page 149. My reason for doing so: to try to understand the "nlm" function. Here is the function: >mfun1 function(x) { z <- matrix(0,nrow=2,ncol=1) z[1,1] <- x[1]^2 + x[2]^2 -

Hi R users I'm trying to fit a model y=ax^b. I know if I made ln(y)=ln(a)+bln(x) this is a linear regression. But I obtain differente results with nls() and lm() My commands are: nls(CV ~a*Est^b, data=limiares, start =list(a=100,b=0), trace = TRUE) for nonlinear regression and : lm(ln_CV~ln_Est, data=limiares) for linear regression Nonlinear