How to use conditional bounds for parameters with lsqcurvefit?

조회 수: 2 (최근 30일)
Akhil
Akhil 2023년 1월 25일
댓글: Akhil 2023년 1월 25일
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))
  댓글 수: 1
Matt J
Matt J 2023년 1월 25일
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)

댓글을 달려면 로그인하십시오.

채택된 답변

Matt J
Matt J 2023년 1월 25일
편집: Matt J 2023년 1월 25일
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
  댓글 수: 1
Akhil
Akhil 2023년 1월 25일
Thank you, definitely a much simpler solution than what I was expecting!

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Nonlinear Optimization에 대해 자세히 알아보기

제품


릴리스

R2021b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by