# How to use conditional bounds for parameters with lsqcurvefit?

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Akhil on 25 Jan 2023
Commented: Akhil on 25 Jan 2023
I am trying to fit an exponential curve to a set of data (Exponential Cumulative Distribution Function).
I would like to solve for the parameters λ, , and , however there are certain conditions with which I need to bound the parameters.
"The and are non-negative values that can be both be 0 but both values cannot be greater than 0 at the same time" as well as "The natural logarithm of λ can be any value between zero and 4."
Is there a way in which I would be be able to define these conditions using lsqcurvefit to solve for the parameters? If not how should I approach fitting this curve to my data? Below is my attempt to use lsqcurvefit without the conditional between and .
xdata = 0:7;
ydata = 100*[0 0.112543130593672 0.814735805710535 1 1 1 1 1];
lb = [log(1), 0, 0];
ub = [log(4), 7, 100]; % How to define upper bound with conditions?
fun = @(x,xdata) (1 - exp(-x(1)*(xdata-x(2))) + x(3));
x = lsqcurvefit(fun,[0, 0, 0],xdata,ydata, lb, ub)
plot(xdata, 1*ydata, '.k', 'MarkerSize', 20)
hold on
plot(xdata, 1*fun(x, xdata))
Matt J on 25 Jan 2023
The natural logarithm of λ can be any value between zero and 4.
If 0<=log(lambda)<=4, then your bounds on lambda would be exp(0) <=lambda<=exp(4), whereas in your code, you have log(1) <=lambda<=log(4)

Matt J on 25 Jan 2023
Edited: Matt J on 25 Jan 2023
You must divide the problem into two cases: one case where X0 is fixed at 0 and the second when Y0 is fixed at zero.
xdata = 0:7;
ydata = 100*[0 0.112543130593672 0.814735805710535 1 1 1 1 1];
lb = [exp(0), 0, 0];
ub = [exp(4), 7, 100];
%Case 1: Y0=0
fun = @(x,xdata) (1 - exp(-x(1)*(xdata-x(2))) );
x = lsqcurvefit(fun,[0, 0],xdata,ydata, lb(1:2), ub(1:2));
Initial point is a local minimum. Optimization completed because the size of the gradient at the initial point is less than the value of the optimality tolerance.
sol{1}=[x(1),x(2),0];
%Case 2: X0=0
fun = @(x,xdata) (1 - exp(-x(1)*(xdata)) +x(2));
x = lsqcurvefit(fun,[0, 0],xdata,ydata, lb([1,3]), ub([1,3]) );
Local minimum possible. lsqcurvefit stopped because the final change in the sum of squares relative to its initial value is less than the value of the function tolerance.
sol{2}=[x(1) 0 x(2)];
Now test each solution and see which gives the best resnorm. Clearly, sol{2} is the best:
Fun = @(x) norm( 1 - exp( -x(1)*(xdata-x(2)) ) + x(3) -ydata);
Fun(sol{1})
ans = 235.7671
Fun(sol{2})
ans = 112.7009
Akhil on 25 Jan 2023
Thank you, definitely a much simpler solution than what I was expecting!

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