Hi, Michael,
I think the SPSS answer is wrong. Your starting values are way off.
Look at this plot for verification:
con <- textConnection("time bod
1 1 0.47
2 2 0.74
3 3 1.17
4 4 1.42
5 5 1.60
6 7 1.84
7 9 2.19
8 11 2.17")
mydata <- read.table(con, header = TRUE)
close(con)
beta <- c(3, -0.1) # your initial values
beta <- c(2.4979, -2.02456) # SPSS answer
mycurve <- function(x) {
beta[1]/(1 - exp(beta[1] * x))
}
curve(mycurve, from = 1, to = 11,
ylim = range(mydata$bod, mycurve(mydata$time)))
points(mydata$time, mydata$bod)
You might want to look at package nls2 which allows a brute force grid
search to find some starting values. Or rethink the equation you're
trying to fit.
HTH,
--sundar
On Tue, Aug 25, 2009 at 9:10 AM, Michael Pearmain<mpearm...@google.com> wrote:
> Hi All,
>
> I'm trying to run nls on the data from the study by Marske (Biochemical
> Oxygen Demand Interpretation Using Sum of Squares Surface. M.S. thesis,
> University of Wisconsin, Madison, 1967) and was reported in Bates and Watts
> (1988).
>
> Data is as follows, (stored as mydata)
>
> time bod
> 1 1 0.47
> 2 2 0.74
> 3 3 1.17
> 4 4 1.42
> 5 5 1.60
> 6 7 1.84
> 7 9 2.19
> 8 11 2.17
>
> I then run the following;
> #Plot initial curve
> plot(mydata$time, mydata$bod,xlab="Time (in days)",ylab="biochemical oxygen
> demand (mg/l) ")
>
> model <- nls(bod ~ beta1/(1 - exp(beta2*time)), data =
> mydata, start=list(beta1 = 3, beta2 = -0.1),trace=T)
>
> The start values are recommended, (I have used these values in SPSS without
> any problems, SPSS returns values of Beta1 = 2.4979 and Beta2 = -2.02 456)
>
> but return the error message,
> Error in nls(bod ~ beta1/(1 - exp(beta2 * time)), data = mydata, start =
> list(beta1 = 3, : singular gradient
>
> Can anyone offer any advice?
>
> Thanks in advance
>
> Mike
>
>
>
>
>
>
>
>
>
> --
> Michael Pearmain
> Senior Analytics Research Specialist
>
> “Statistics are like women; mirrors of purest virtue and truth, or like
> whores to use as one pleases”
>
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