similar to: Re: non-linear model

AFAIK most model fitting techniques will only deal with additive errors, not multiplicative ones. You might want to try fitting: log(y-x) = a*x + e which is linear. Andy > From: Angelo Secchi > > Hi, > is there a way in R to fit a non linear model like > > y=x+exp(a*x)*eps > > where a is the parameter and eps is the error term? > Thanks > Angelo > >

I'm doing a non linear regression with 8 parameters to be fitted: J.Tl.nls<-nls(Gw~(a1/(1+exp(-a2*Tl+a3))+a4)*(b1/(1+exp(b2*Tl-b3))+b4),data=Enveloppe, start=list(a1=0.88957,a2=0.36298,a3=10.59241,a4=0.26308, b1=0.391268,b2=1.041856,b3=0.391268,b4=0.03439)) First, I fitted my curve on my data by guessing the parameters'

Hi, I'm trying to make a regression of the form : formula <- y ~ Asym_inf + Asym_sup * ( (1 / (1 + (n1 * (exp( (tmid1-x) / scal1) )^(1/n1) ) ) ) - (1 / (1 + (n2 * (exp( (tmid2-x) / scal2) )^(1/n2) ) ) ) ) which is a sum of the generalized logistic model proposed by richards. with data such as these: x <- c(88,113,128,143,157,172,184,198,210,226,240,249,263,284,302,340) y <-

Oups, I sent the email by error, as I was still writing my reply… Spencer, Le 22-avr.-09 à 03:33, spencerg a écrit : > Is your first model a special case of the second with eta1 = 0? > If yes, what about using 2*log(likelihood ratio) being approximately > chi-square? Yes, the first model is a special case of the second with eta1=0… Could you give me more explanation about

Hi, I'm trying to fit the following model to data using 'nls': y = alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x) and the call I've been using is: nls(y ~ alpha_1 * beta_1 * exp(-beta_1 * x) + alpha_2 * beta_2 * exp(-beta_2 * x), start=list(alpha_1=4, alpha_2=2, beta_1=3.5, beta_2=2.5), trace=TRUE, control=nls.control(maxiter =

Dear I have a data.frame, and want to fit a constrained non-linear model: data: x y -0.08 20.815 -0.065 19.8128 -0.05 19.1824 -0.03 18.7346 -0.015 18.3129 0.015 18.0269 0.03 18.4715 0.05 18.9517 0.065 19.4184 0.08 20.146 0 18.2947 model: y~exp(a)*(x-m)^4+exp(b)*(x-m)^2+const I try to use nls() and set start=list(a=1,b=1,c=1,m=1), but which always give me a error message that

Hello, I want to understand how to tell if a model is significant. For example I have vectX1 and vectY1. I seek first what model is best suited for my vectors and then I want to know if my result is significant. I'am doing like this: model1 <- lm(vectY1 ~ vectX1, data= d), model2 <- nls(vectY1 ~ a*(1-exp(-vectX1/b)) + c, data= d, start = list(a=1, b=3, c=0)) aic1 <- AIC(model1)

Hello there, I am using nls the first time for a non-linear regression with a logistic growth function: startparam <- c(alpha=3e+07,beta=4000,gamma=2) fit <- nls(dataset$V2~(( alpha / ( 1 + exp( beta - gamma * dataset$V1 ) ) ) ),data=dataset,start=startparam) Everything works fine and i get good results. Now I would like to improve the results using my DUMMY Variable (dataset$V6) the

Hi, I am using nls to fit a non linear function to some data but R keeps giving me "singular gradient matrix at initial parameter estimates" errors. For testing purposes I am doing this: ### R code ### x <- 0:140 y <- 200 / (1 + exp(17 - x)/2) * exp(-0.02*x) # creating 'perfect' samples with fitting model yeps <- y + rnorm(length(y), sd = 2) # adding noise # results

I am trying to plot a non-linear trend line along with my data. I am sure the solution is simple; any help greatly appreciated. Recent and historic attempts: fit = nls((sd~a+b*exp(-c/time)+d*geology), start=list(a=1, b=1, c=10, d=-1), model=TRUE) plot(time, sd, col=3, main = "Regression", sub=formula(fit)) lines(time, fitted.values(fit), lwd=2) #tt = seq(from=0, to=200, by=10)

I am trying to fit a curve to a cumulative mortality curve (logistic) where y is the cumulative proportion of mortalities, and t is the time in hours (see below). Asym. at 0 and 1 > y [1] 0.00000000 0.04853859 0.08303777 0.15201970 0.40995074 0.46444992 0.62862069 0.95885057 1.00000000 [10] 1.00000000 1.00000000 > t [1] 0 13 20 24 37 42 48 61 72 86 90 I tried to find starting values for

Dear R friends, I know that this topic has been mulled over before, and that there is a substantial difference between the convergence criteria for JMP and those for R. I apologize that this is somwehat raking cold coals. Summary: A model/data combination achieves convergence in JMP, and survives a reasonably rigorous examination (sensible parameter estimates, well-behaved surface,

Hi, I'm using a non linear model to fit experimental survival curves. This model describes the fraction of "still active" experiments as a function of time t as follows: f(t)=(1+exp(-etaD*cD)) / (1+exp(etaD(t-cD))) Moreover, when experiments are still active, they may change of state (from 0 to 1). But they may fall inactive before changing their state (their state still

I have the following data set, and I have to find the line of best fit using this equation, y = a*(1 - exp(-b*x)). samples = seq(1,20,by=1) species = c(5,8,9,9,11,11,12,15,17,19,20,20,21,23,23,25,25,27,27,27) plot(samples,species, main = "Accumulation Curve for Tree Species Richness", xlab = "Samples", ylab = "Number of Species") curve((y = 27*(1 -

Hi all, Like a lot of people I noticed that I get different results when I use nls in R compared to the exponential fit in excel. A bit annoying because often the R^2 is higher in excel but when I'm reading the different topics on this forum I kind of understand that using R is better than excel? (I don't really understand how the difference occurs, but I understand that there is a

I have a non-linear regression with 8 parameters to solve .... however it does not converge ... easily solves the excel ... including the initial estimates used in the R were found in the excel ... Another question is how to establish the increments of R by the parameters in the search ..

Hi all! i'm trying to use nls for fitting my data. I wrote this code to find some minimum, but it fails, returning 0 every time..... and i can't figure out the problem... any advice? grid <- expand.grid(A0 = seq(1000,10000,1000), A1 = seq(0,2,0.1)) exp.approx <- function(x,A0,A1) { A0 * exp(- x*A1) } ss <- function(p) { sum((durata.h.freq -

dear list members, my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as (I) expected. I'm posting this here since: (A) the problem might suggest a modification to the `m' component in the return argument of `nls' (making this post formally OK for this list) and (B) I got no

dear list members, I apologize in advance for posting a second time, but probably after one week chances are, the first try went down the sink.. my question concerns computation of confidence intervals in nonlinear fits with `nls' when weigthing the fit. the seemingly correct procedure does not work as expected, as is detailed in my original post below. any remarks appreciated. greetings

This data are kilojoules of energy that are consumed in starving fish over a time period (Days). The KJ reach a lower asymptote and level off and I would like to use a non-linear plot to show this leveling off. The data are noisy and the sample sizes not the largest. I have tried selfstarting weibull curves and tried the following, both end with errors. Days<-c(12, 12, 12, 12, 22, 22, 22,