# Re: [R] formula for nls

From: Jarosław Jasiewicz <jarekj_at_amu.edu.pl>
Date: Mon, 28 Jan 2008 09:18:07 +0100

Christian Ritz pisze:
> 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
>
> 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
yes, I must only check if coefs are expected

but...
fortunalty I found that my problem is solved by package HydroMe

thanks again
Jarek

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