Re: lsqcurvefit bounds influence fitting result

Without analyzing your particular function, I can say that many, perhaps most, realistic fitting problems have many local minima. Changing the bounds causes lsqcurvefit to reach different local answers (minima). If you want a global minimum, you have to take extra steps.

If you have a Global Optimization Toolbox license, you can use MultiStart to search for a global minimum very easily. Even without MultiStart, you can write a program of your own to take a variety of start points and collect the results.

Good luck,

Alan Weiss
MATLAB mathematical toolbox documentation

On 3/17/2011 1:37 PM, Philipp Steffen wrote:
Hallo,
I have been using lsqcurvefit for a while to fit measured data to a
gaussian function with an initial constant part to extract parameters
that describe the constant part as well as the gaussian shape of the curve.
Now I realized, that this is quite sensitive to the upper and lower
bound. I need to set the lower bound to be >0. In principle I dont need
a upper bound but the results of the fit vary quite drastically
depending on if I put ub=[], ub=[1,1,1] or ub=[10,10,10].
In my dataset for example, ub=[10,10,10] yields the parameter
beta=0.0000 0.1995 1.3786 (resnorm=0.1626)
when I put ub=[1,1,1] i get:
beta=0.4395 0.2434 1.0000 (resnorm=0.2531)
and with ub=[]:
beta=2.6643 0.4244 0.6430 (resnorm=1.4990)
Clearly the fit using ub=[10,10,10] is the best. Why is this? Why do the
boundaries influence the fit in this way?
I appreciate any hints and comments and hope that someone can help me to
solve this issue. Basically, I don't want to fix the bounds to a value,
since I would like to keep the function as flexible as possible.

Here is my code:

radius: 48 double values
RP: 48 double values

beta0(1)=0.4;
beta0(2)=0.79;
beta0(3)=0.8;
beta = lsqcurvefit(@func,beta0,radius,RP,[realmin,realmin,realmin],[]);

Function:
function [yhat] = func(b, radius) for i=1:length(radius),
if (radius(i) < b(1)),
yhat(i) = b(2);
else
yhat(i) = 1 - ((1-b(2))*exp(-0.5*(((radius(i)-b(1))^2)/(b(3)^2))));
end
end;
yhat=yhat';
end


Thanks a lot,

Philipp

.

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