how to further reduce residual error in an unconstrained levenberg-marquardt optimisation

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Anuj Anand
Anuj Anand 2014-2-14
评论： Anuj Anand 2014-2-18
I am running an optimisation function using unconstrained levenberg-marquardt. The algorithm converges and gives a residual of 412.192 in the 11th iteration. Is there a way in which I can futher reduce this residual error?? Many Thanks.
if true
First-Order Norm of
Iteration Func-count Residual optimality Lambda step
0 3 35865.3 2.85e+04 0.01
1 6 1450.89 2.76e+03 0.001 137.848
2 9 415.676 229 0.0001 458.723
3 12 412.195 2.5 1e-05 4.34834
4 15 412.192 0.221 1e-06 1.42228
5 18 412.192 0.0141 1e-07 0.0867978
6 21 412.192 0.000922 1e-08 0.0057553
7 24 412.192 4.6e-05 1e-09 0.000382006
8 27 412.192 7.44e-06 1e-10 2.42769e-05
9 35 412.192 3.43e-05 1e-05 4.1357e-06
10 41 412.192 0.000123 0.01 1.21969e-05
11 51 412.192 0.000117 100000 1.66032e-11
end

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回答（2 个）

Alan Weiss
Alan Weiss 2014-2-14
The Lambda parameter climbs to a very high value at the end. I wonder if your function is smooth.
Did you try starting at various start points? Did you formulate your problem correctly, passing in the components of your objective vector, or did you erroneously pass in the sum of squares? Without more details it is hard to know what to advise you to do.
Alan Weiss
MATLAB mathematical toolbox documentation
  1 个评论
Anuj Anand
Anuj Anand 2014-2-17
Here is the function. i also tried plotting the range of values over which I expect the correct solution (i.e. k(1)=[3000 6000] and k(2)=[0.01 5]) and plotted this function (see attached). It probably does not look like if there are more solution. But perhaps, you can suggest better. Many Thansk
if true
function [ a , b , sse ] = measureMT( signal , alpha , Rin )
Rin = 0.00314;
signal=[168.818176,168,167.969696,168.030304,168.212128,349.363647,349.84848,349.757568,349.878784,350.363647,433.818176,433.060608,432.909088,433.545441,432.727264,319.84848,318.727264,319.727264,318.84848,319.121216];
alpha =[2,2,2,2,2,5,5,5,5,5,10,10,10,10,10,18.59634,18.59634,18.59634,18.59634,18.59634];
format long
ahat = zeros(2,2); sse = zeros(1,2);
options = optimoptions(@lsqcurvefit,... 'Algorithm','levenberg-marquardt','Display','iter',... 'FunValCheck','on','ScaleProblem','Jacobian',... 'TolFun',1e-10,'TolX',1e-10,'MaxFunEvals',1e10,'MaxIter',1e10);
for i=1:1:2 x0 = [5000/i i/3.92] ; [ahat(:,i), sse(i)] = ... lsqcurvefit(@(x,xdata) cal(x,alpha,Rin),x0,alpha,signal,[],[],options);
figure, plot(alpha,signal,'o'), hold on
plot(alpha,cal(ahat,alpha,Rin),'-r')
end
a=ahat(1,:); b=ahat(2,:);
function F = cal(k,xdata,Rin) % MODEL
F = k(1)*sin(xdata*pi/180)*(1-exp(-Rin*k(2)))./(1-(cos(xdata*pi/180)*exp(-Rin*k(2))));
end
end
end

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Anuj Anand
Anuj Anand 2014-2-14
编辑：Matt J 2014-2-14
Dear Alan, Many Thanks again. I did try various start points, but it still gives very large residual. However, i cannot do too much with it as i have to use un-constrained levenberg-marquardt. So the initial condition that I am giving is not very far from the solution anyways (i think). I dont understand what do you mean by if i am formulating my problem correctly? I mean there is an non-linear equation which act as a model defined in function 'cal' which I have copy pasted from a paper so it is hard to fo wrong with it. Many Thanks for your help again.
function [ a , b , sse ] = measureMT( signal , alpha , Rin )
Rin = 0.00314;
signal=[168.818176,168,167.969696,168.030304,168.212128,349.363647,349.84848,349.757568,349.878784,350.363647,433.818176,433.060608,432.909088,433.545441,432.727264,319.84848,318.727264,319.727264,318.84848,319.121216];
alpha =[2,2,2,2,2,5,5,5,5,5,10,10,10,10,10,18.59634,18.59634,18.59634,18.59634,18.59634];
format long
ahat = zeros(2,2); sse = zeros(1,2);
options = optimoptions(@lsqcurvefit,... 'Algorithm','levenberg-marquardt','Display','iter',... 'FunValCheck','on','ScaleProblem','Jacobian',... 'TolFun',1e-10,'TolX',1e-10,'MaxFunEvals',1e10,'MaxIter',1e10);
for i=1:1:2 x0 = [5000/i i/3.92] ; [ahat(:,i), sse(i)] = ... lsqcurvefit(@(x,xdata) cal(x,alpha,Rin),x0,alpha,signal,[],[],options);
figure, plot(alpha,signal,'o'), hold on
plot(alpha,cal(ahat,alpha,Rin),'-r')
end
a=ahat(1,:); b=ahat(2,:);
function F = cal(k,xdata,Rin) % MODEL
F = k(1)*sin(xdata*pi/180)*(1-exp(-Rin*k(2)))./(1-(cos(xdata*pi/180)*exp(-Rin*k(2))));
end
end
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Matt J
Matt J 2014-2-17
编辑：Matt J 2014-2-17
Your attachment didn't make it, but if the figure shows that you are at the global minimum, doesn't that mean the problem is resolved? Your original concern was that you weren't sure if you had fully minimized the function.
Anuj Anand
Anuj Anand 2014-2-18
Many Thanks again for your replly. I just wanted to be sure and that is why I am asking. I am not very experiences with it and I was surprised to know that if my solution has reached global minima then why should this residual error be so high?

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