How to find good start values for nls function?

Question

I don\'t understand why I can\'t have a nls function for these data. I have tried with a lot of different start values and I have always the same error.

Here is what I h

Answer 1

This can be written with two linear parameters (.lin1 and .lin2) and one nonlinear parameter (b) like this:

a*(1-exp(-x/b)) + c  
= (a+c) - a * exp(-x/b)
= .lin1 + .lin2 * exp(-x/b)

where .lin1 = a+c and .lin2 = -a (so a = - .lin2 and c = .lin1 + .lin2) This lets us use "plinear" which only requires specification of a starting value for the single nonlinear parameter (eliminating the problem of how to set the starting values for the other parameters) and which converges despite the starting value of b=75 being far from that of the solution:

nls(y ~ cbind(1, exp(-x/b)), start = list(b = 75), alg = "plinear")

Here is the result of a run from which we can see from the size of .lin2 that the problem is badly scaled:

> x <- c(77.87,87.76,68.6,66.29)
> y <- c(1,1,0.8,0.6)
> nls(y ~ cbind(1, exp(-x/b)), start = list(b = 75), alg = "plinear")
Nonlinear regression model
  model:  y ~ cbind(1, exp(-x/b)) 
   data:  parent.frame() 
         b      .lin1      .lin2 
 3.351e+00  1.006e+00 -1.589e+08 
 residual sum-of-squares: 7.909e-05

Number of iterations to convergence: 9 
Achieved convergence tolerance: 9.887e-07 
> R.version.string
[1] "R version 2.14.2 Patched (2012-02-29 r58660)"
> win.version()
[1] "Windows Vista (build 6002) Service Pack 2"

EDIT: added sample run and comment on scaling.