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I'm using the non linear least squares Levenburg Marquardt algorithm in java to fit a number of exponential curves (A+Bexp(Cx)). Although the data is quite clean and has a good approximation to the model the algorithm is not able to model the majority of them even with a excessive number of iterations(5000-6000). For the curves it can model, it does so in about 150 iterations. 

LeastSquaresProblem problem = new LeastSquaresBuilder()
        .start(start).model(jac).target(dTarget)
        .lazyEvaluation(false).maxEvaluations(5000)
        .maxIterations(6000).build();

LevenbergMarquardtOptimizer optimizer = new LevenbergMarquardtOptimizer();
LeastSquaresOptimizer.Optimum optimum = optimizer.optimize(problem);}

My question is how would I define a convergence criteria in apache commons in order to stop it hitting a max number of iterations?

alto125
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1 Answers1

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I don't believe Java is your problem. Let's address the mathematics.

This problem is easier to solve if you change your function.

Your assumed equation is:

y = A + B*exp(C*x)

It'd be easier if you could do this:

y-A = B*exp(C*x)

Now A is just a constant that can be zero or whatever value you need to shift the curve up or down. Let's call that variable z:

z = B*exp(C*x)

Taking the natural log of both sides:

ln(z) = ln(B*exp(C*x))

We can simplify that right hand side to get the final result:

ln(z) = ln(B) + C*x

Transform your (x, y) data to (x, z) and you can use least squares fitting of a straight line where C is the slope in (x, z) space and ln(B) is the intercept. Lots of software available to do that.

duffymo
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  • great tip, however A is very important to what I'm looking for, how do you get it back from the transformation? – alto125 Feb 24 '16 at 14:43
  • I think it's a translation of the fitted curve up or down on the y-axis to match the data. Should that constant be the y value at x = 0 of your data? The asymptotic value? An average? Maybe a second fit using the B and C values you get to determine A? I don't know. Why not try this approach and see what you get? – duffymo Feb 24 '16 at 14:45
  • Thanks for the help - I'll give all of the above a go! – alto125 Feb 24 '16 at 14:56
  • If nothing else, it'll give you a better starting guess for your non-linear optimization on the original equation. Maybe you can take the best results you get out of it and stuff those into the iterative solution as your starting guess. Non-linear methods converge faster if you happen to guess a starting point that's not far from the true solution. – duffymo Feb 24 '16 at 15:50