Hi Jarek, an alternative approach is to provide more precise starting values!
It pays off to realise that it's possible to find quite good guesses for some of the parameters in your model function: t ~ tr+ (ts-tr)/ ((1+ (a*h)^n)^(1-(1/n))) The parameters tr and ts correspond to the response t for h equal to infinity and h=0. Therefore by looking at the plot of your data I would set: tr = 0 ts = 15000 The parameter n must be above 1 in order to achieve a decreasing function. We will not think more about this at the moment. For n somewhat larger than 1 (in which case I would approximate the exponent 1-(1/n) by 1) the parameter a is approximately equal to the reciprocal of the h value resulting in a response halfway between tr and ts. Therefore (again from looking at the plot) I would set a=10. Using the above starting values and simply increasing n in steps of 1 eventually results in a useful model fit: ## Fails tmp.m1<-nls(t ~ tr+ (ts-tr)/ ((1+ (a*h)^n)^(1-(1/n))), data = tmp, start = list(a=10, n=1, tr=0, ts=15000)) ## Fails tmp.m1<-nls(t ~ tr+ (ts-tr)/ ((1+ (a*h)^n)^(1-(1/n))), data = tmp, start = list(a=10, n=2, tr=0, ts=15000)) ## Works!! tmp.m1<-nls(t ~ tr+ (ts-tr)/ ((1+ (a*h)^n)^(1-(1/n))), data = tmp, start = list(a=10, n=3, tr=0, ts=15000)) summary(tmp.m1) plot(t ~ h, data = tmp) lines(tmp$h, predict(tmp.m1)) I hope you can use this explanation?! Christian ______________________________________________ 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.
