Why MATLAB gives very bad fitting parameters?

Why MATLAB gives very bad fitting parameters when fittype and fit are used together. Below is the one example where I have a 3D matrix ( Auto) and in which I am fitting the Auto(p,q,:) elements using a for a loop. I am calculating one parameter ( D(p,q)=fit1.D) from the fitting. Sometimes I used catch and try to avoid the breakdown of the loop.

However, the point is that I think MATLAB is not efficient in curve fitting. I used ( Origin) which gave nice good fitting but the same data and the same equation does not give me expected parameters in MATLAB. I could use the curve fitting toolbox but I want to recursively fit the data and want to extract parameters in a loop which refrain me for using the toolbox.

clear all; close all; clc;
load('Auto.mat');

l2= size(Auto);
for p=1:l2(1)
    for q=1:l2(2)
        % This is part for the data which has values less than 0.6 in the
        % first two matrix in the AutoX3 matrix
        y=reshape(Auto(p,q,1:end),[],1);
        if y(1,:)<0.6 || y(2,:)<0.55
            y(:,:)=NaN;
        end
        x=(0:0.5:(length(y)-1)*0.5)'; % The X-axis
        % The fitting equation %
        f = fittype('(A./(sqrt(1+ ((x*D)./(0.16.^2)))))');

          % Try and catch to successfully continue the loop%
          try
              fit1 = fit(x,y,f,'StartPoint',[0.1 0.1]);
              RateCons(p,q)=fit1.D;
              Amplitude(p,q)=fit1.A;
          catch
              RateCons(p,q)=NaN;
              Amplitude(p,q)=NaN;
          end
      end
  end

[Walter Roberson]

Can you attach some sample data?

[Jaladhar Mahato]

Dear Walter Roberson,

Here I have attached one data set here and updated the code which I actually used to calculate the parameters. Thanks for your prompt reply.

Jaladhar

[Walter Roberson]

You have

        if y(1,:)<0.6 || y(2,:)<0.55
            y(:,:)=NaN;
        end

Why do you bother to fit in that case? With y being all NaN, you can never hope to get any coefficient other than NaN?

A lot of your results trigger the catch because you did not put in any boundary conditions.

[Jaladhar Mahato]

My question was about the reliability of fitting parameters. I want NaN coefficient values in those places where y is NaN. But where it does succeed, I want reliable parameters. Matlab very much depends on the 'StartPoint' values. I checked this with known parameters and then tried to fit this with the combination of fittype and fit.However, different 'StartPoint' results diffeerent parameters and sometime it is far from the known values. This is problematic when my parameters are unknown.I think my main point has gone while elaborating the problem. So the question is MATLAB good at fitting functions? Thanks

[Walter Roberson]

I sampled at a number of random locations, and in each case that I tested, the results from fit() in my code below were pretty good compared to the global minima -- off by less than 0.3%

[Walter Roberson]

For example, at p=32, q=33, then A and D are determined by fit to be 0.929878332309704, 0.00491910153972962 . When I convert the residue formula into symbolic form using rationals instead of floating point numbers, and use a symbolic package to minimize, I get .929869232279317, 0.00491889395158922 as the global best, which differ in positions from fit's results by about 1E-5 and 2E-7 respectively.

But the exact best position is subject to numeric roundoff problems, and working numerically in MATLAB I get slightly better residue at 0.929869232278125901, 0.00491889395270517921. We are talking about differences in residue on the order of 4E-17 .

The overall residue surfaces have a fairly consistent shape of being more or less a tilted parabola sheet outside the major basin of attraction; any minimizer that looked at the gradient would get centered towards the middle of the parabola pretty quickly, and would follow that line into the major basin of attraction. The basin of attraction is then an obvious paraboloid, so a gradient following minimizer would get pulled to a small distance from the global minima residue in a small number of iterations.

In the locations I tested, I saw no evidence at all of any important local minima: there was no competing minima that a reasonable algorithm might accidentally get caught in.

Which is not to say that fit() is immune to having problems with competing local minima: it does, and there have been a few cases discussed in the last month or so where fit could reasonably accidentally get stuck in the wrong minima, especially cases involving sum of exponentials. fit can end up with a pretty high confidence that it has found the right parameters, and yet be wrong because the other better set of parameters might be difficult to get close to. But this function is not one of those cases: there is an unambiguous area of global minima and in each trial with your data that I investigated, fit() got close to it without difficulty.

[Jaladhar Mahato]

Dear Walter, thanks for your elaboration. I think after setting the bound I should not get unwanted parameters as my function sometimes involve multiplication of only one exponential (not the sum of exponential).

Answer 1

[Walter Roberson]

%clear all; close all; clc;
load('Auto.mat');

%arbitrary constant as part of the equation. I do not know where it comes from
C = 0.16^2;

l2= size(Auto);

RateCons = NaN(l2(1), l2(2));
Amplitude = NaN(l2(1), l2(2));

x=(0:0.5:(l2(3)-1)*0.5)'; % The X-axis

LBA = -inf;
LBD = -C/x(end);    %below this, the calculation goes complex
UBA = inf;
UBD = inf;
LB = [LBA, LBD];
UB = [UBA, UBD];

% The fitting equation %
f = fittype(@(A,D,x) (A./(sqrt(1+ ((x*D)./C)))));
opts = fitoptions(f);
opts.Lower = LB;
opts.Upper = UB;
opts.StartPoint = [0.1 0.1];

for p=1:l2(1)
    for q=1:l2(2)
        % This is part for the data which has values less than 0.6 in the
        % first two matrix in the AutoX3 matrix
        y=reshape(Auto(p,q,1:end),[],1);
        if any(isnan(y(:))) || y(1,:)<0.6 || y(2,:)<0.55
            %we used to assign NaN here, but now we initialize everything to NaN
        else  
            % Try and catch to successfully continue the loop%
            try
                fit1 = fit(x,y,f,opts);
                RateCons(p,q)=fit1.D;
                Amplitude(p,q)=fit1.A;
            catch ME
                %we used to assign NaN here, but now we initialize everything to NaN
            end
        end
    end
end

[Jaladhar Mahato]

Thanks Walter Roberson. The output looks decent. I have to verify it with the known example. Thanks again.