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I have this following relation

$$N = \dfrac{(113834700(3000-c)(1-e^{-(Xn)/(3000-c)})}{n}$$

I also have a set of values for $N$ and $X$, these are vectors. So, we have a scatter plot $N$ vs $X$.

We have to fit the data ($N$ and $X$) with the above relation and thereby have to find $c$ and $n$, which are scalars. How do I do this?

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  • $\begingroup$ At first you need to understand which type of distance will satisfy you. Is Euclidean distance OK? $\endgroup$
    – German Demidov
    Commented Apr 24, 2016 at 14:58
  • $\begingroup$ I wanted to give one answer, but there are so many pitfalls about it so I would better recommend you to take a look at this page en.wikipedia.org/wiki/Mathematical_optimization and choose method appropriate for your data. $\endgroup$
    – German Demidov
    Commented Apr 24, 2016 at 15:01
  • $\begingroup$ Non Linear Regression might do the trick here. If you use R, try the nls function. $\endgroup$
    – Greenparker
    Commented Apr 24, 2016 at 15:32
  • $\begingroup$ @Greenparker I tried but kept getting some error messages in R. Could you please help me out with the R syntax for solving this? $\endgroup$
    – Debdatta Sinha Roy
    Commented Apr 24, 2016 at 15:34
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    $\begingroup$ I suspect that noticing that there is really only one parameter, (3000-c)/n, is part of the assignment. $\endgroup$
    – JimB
    Commented Apr 24, 2016 at 17:25

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As @JimBaldwin noted in a comment, there really is only 1 free parameter to fit in your equation, $(3000-c)/n$. If your code for nls or another nonlinear curve fitting routine included both $c$ and $n$ as separate parameters, then it would not be surprising for the routine to run with errors: there is an infinite set of $c$ and $n$ paired values that could "fit." So solve that problem by defining another variable, call is $a$:

$$a=\frac{3000-c}{n}.$$

It also can be easier to avoid accidental errors by avoiding unusual or large constant terms. I'd simply divide all the $N$ values by the constant 113834700 to start, so that there's less chance for error in writing the formula to submit to the fitting routine. Let's say:

$$B= \frac{N}{113834700}.$$

Then you are fitting a simple relation:

$$B=a\left(1-e^{-\frac{X}{a}}\right),$$

which will minimize the chance of errors both in your coding and in the routine.

As with any nonlinear fitting, it's important to start with a good initial estimate for your unknown parameter $a$, and to be prepared to test carefully to make sure that you have found a global minimum for residual squared error rather than some local minimum. I have some examples of R code for nls on this page in a similar context, which illustrates some of the considerations.

Also, make sure that the value of 113834700 is truly a constant and not a fitted value from someone else's analysis of another data set, for which you should be estimating a new value from your data set. That type of problem arose in the question I linked in the previous paragraph. If you do only have 1 free parameter, you could use the optim function instead of nls in R.

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  • $\begingroup$ I only see your answer after I have posted my answer at stats.stackexchange.com/questions/209068/… . Your final paragraph matches with what I concluded using the data posted in that other question. I concluded his model is lousy, and that hard-coded multiplicative constant "ain't cutting it". I introduced another parameter in its place and got a much better fit,. $\endgroup$
    – Mark L. Stone
    Commented Apr 24, 2016 at 23:03

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