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How to fit a custom equation?

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madhuri dubey
madhuri dubey on 9 Jul 2018
Answered: Alex Sha on 18 Feb 2020
My equation is y=a(1-exp(-b(c+x)) x=[0,80,100,120,150] y=[2195,4265,4824,5143.5,5329] When I am solving it in matlab, I am not getting a proper fit in addition, sse=6.5196e+05 and r square=0.899. Although the r square value is acceptable, the sse is too high. Therefore kindly help to get minimum sse. Further I have tried in curve fitting tool but I got same thing.

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Accepted Answer

Star Strider
Star Strider on 9 Jul 2018
I get good results with this:
yf = @(b,x) b(1).*(1-exp(-b(2)*(b(3)+x)));
B0 = [5000; 0.01; 10];
[Bm,normresm] = fminsearch(@(b) norm(y - yf(b,x)), B0);
SSE = sum((y - yf(Bm,x)).^2)
Bm =
6677.76372320411
0.0084077646869843
47.1622210493944
normresm =
195.173589996072
SSE =
38092.7302319547

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Star Strider
Star Strider on 10 Jul 2018
As always, my pleasure.
I would be tempted to use polyfit to get initial estimates of ‘a’, ‘c’, and ‘d’ (estimated as [-0.09, 34, 2200] when I did it) , then let your nonlinear parameter estimation routine (similar to my code) estimate them and ‘b’ (that I would initially estimate as 10).
I usually create my own initial population for ga. I would be tempted here to use a matrix of 500 individuals, defined as:
init_pop = randi(5000, 500, 4);
using the appropriate options function (linked to in the See Also section in the ga documentation) to define it as such. The ga function is efficient, however since it has to search the entire parameter space, it will take time for it to converge.
Note that you have 5 data pairs and you are now estimating 4 parameters.
madhuri dubey
madhuri dubey on 11 Jul 2018
When I use polyfit to get initial estimates of ‘a’, ‘c’, and ‘d’ , I got [-0.0001, 0.0346, 2.1807]. Why there is difference in constant values for the same data.
Star Strider
Star Strider on 11 Jul 2018
There isn’t. You’re ignoring the constant multiplication factor 1.0E+03. The full result:
p =
1.0e+03 *
-0.000088159928538 0.034559355475118 2.180742845451099

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More Answers (2)

Image Analyst
Image Analyst on 11 Jul 2018
For what it's worth, I used fitnlm() (Fit a non-linear model) because that's the function I'm more familiar with. You can see it gives the same results as Star's method in the image below. I'm attaching the full demo to determine the coefficients and plot the figure.

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Alex Sha
Alex Sha on 18 Feb 2020
If don't care the type of fitting function, try the below function, much simple but with much better result:
y = b1+b2*x^1.5+b3*x^3;
Root of Mean Square Error (RMSE): 18.0563068929128
Sum of Squared Residual: 1630.15109305524
Correlation Coef. (R): 0.999873897456616
R-Square: 0.999747810815083
Adjusted R-Square: 0.999495621630167
Determination Coef. (DC): 0.999747810815083
Chi-Square: 0.165495427247465
F-Statistic: 3964.27710070773
Parameter Best Estimate
---------- -------------
b1 2195.84396843687
b2 3.66989234203779
b3 -0.00107101963512847

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