Asked by Saber_phy
on 15 Jun 2018

Hi

I am a student enrolled in the 2nd year physics Master's degree, i am a beginner in Matlab and i have the same problem like Mattias, can you help me and is it was possible to get your e-mail.

My problem is to make a fit of a nonlinear equation to data with the genetic algorithm and to minimize R, my model function is

R*(1-exp(-t/rate))

The industrial data Rexp and t are known: these are two vectors that give the variation of Rexp as a function of time (t)

rate also is known and equal to 40

I declared my function (fitnes function) as follows:

function S=ga_Newton(R)

t=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28];

Rexp=[1 .7987e-05 3.07996e-05 3.90531e-05 1.89562e-05 2.82794e-05 5.55377e-05

4.75723e-05 6.2447e-05 5.35282e-05 5.16775e-05 4.67144e-05 4.63215e-05 6.62753e-05

3.72551e-05 3.54374e-05];

for i=1:15

S(i) =sum((Rexp(i)-R*(1-exp(-t(i)./40)))^2);

end

The function

S (i) = sum ((Rexp (i) -R * (1-exp (-t (i) ./ 40))) ^ 2)

I use it in the methods of Gauss Newton and Levenberg marquardt and I got a good result of R which is in the order of 0.000059 but with the genetic algorithm I found no result.

Answer by Walter Roberson
on 15 Jun 2018

Accepted Answer

Replace

for i=1:15

S(i) =sum((Rexp(i)-R*(1-exp(-t(i)./40)))^2);

end

with

S = sum((Rexp-R*(1-exp(-t./40))).^2);

Saber_phy
on 16 Jun 2018

Walter Roberson
on 16 Jun 2018

I have attached the exact files I used. Execution requires less than 3 seconds.

More generally, the approach I would use would be to use the Symbolic Toolbox to prepare the function to be minimized:

t=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28];

Rexp=[1.7987e-05 3.07996e-05 3.90531e-05 1.89562e-05 2.82794e-05 5.55377e-05 4.75723e-05 6.2447e-05 5.35282e-05 5.16775e-05 4.67144e-05 4.63215e-05 6.62753e-05 3.72551e-05 3.54374e-05];

syms R

prediction = R*(1-exp(-t./40));

actual = Rexp;

residue = sum((actual - prediction).^2);

F = matlabFunction(residue);

nvars = 1;

A = []; b = [];

Aeq = []; beq = [];

lb = []; ub = [];

nonlcon = [];

options = gaoptimset('PlotFcn', {@gaplotbestf @gaplotbestindiv @gaplotexpectation @gaplotstopping}, ...

'TolFun', 1e-13, 'Generations', 1000);

[R, fval, exitflag, output] = ga(F, nvars, A, b, Aeq, beq, lb, ub, nonlcon, options);

fprintf('best R: %g\n', R);

fprintf('best residue: %g\n', fval);

Note: you will not get especially close to the true minima. The true minima is at approximately 0.000128381339932429 which can be quickly determined:

[R, fval] = fminunc(F, 0)

Saber_phy
on 16 Jun 2018

thank you very much :) :)

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Answer by Saber_phy
on 10 Sep 2018

Hello;

My problem is to make a fit of a nonlinear equation to data with the genetic algorithm and to minimize ‘R’, and ‘rate’ at the same time, my model function is R*(1-exp(-t/rate)) The industrial data Rexp and t are known: these are two vectors that give the variation of Rexp as a function of time (t), first of all I will test my code by a single value of R and rate I declared my function (fitnes function) as follows:

function residue =fitness_XY(x)

t=[0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120];

Rexp= 0.000157*(1-exp(-t/70));

residue=sum((Rexp-x(1)*(1-exp(-t./x(2)))).^2) ;

%genetic code

F=@fitness_XY

nvars = 2;

A= []; b = [];

Aeq = []; beq = [];

lb = []; ub = [];

nonlcon = [];

%Options

options=gaoptimset('PlotFcn',{@gaplotbestf}, 'CrossoverFraction',0.01,'TolFun',1e-13,'Generations',300) ;

[x,fval,exitflag,output]=ga(F, nvars, A, b, Aeq, beq, lb, ub, nonlcon, options) ;

fprintf('best R and rate: %g\n', x) ;

fprintf('The number of generations was : %d\n', output.generations);

fprintf('The number of function evaluations was : %d\n', output.funccount);

fprintf('The best function value found was(best residue) : %g\n', fval);

at the end the code must return the exact value of R=0.000157 and rate=70 but it still running more than 30mn and return a false value, I don’t know where is the problem exactly

Saber_phy
on 12 Sep 2018

Walter Roberson
on 13 Sep 2018

"ga does not give an acceptable result within an acceptable time" is a valid conclusion.

You can use ga options to pass in a starting point: look at the initial population matrix parameter.

If it is reasonable to use a lower bound or upper bound on the values, you should do so. Even if it is just lower bound 0.

Saber_phy
on 13 Sep 2018

i already use the ga option and i used a lower and upper bound on the values and i got reasonable value.

thank you very much :)

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Answer by Saber_phy
on 4 Oct 2018

Edited by Saber_phy
on 4 Oct 2018

hello

I have a problem in tracing curves( plot) with the results obtained by genetic algorithme(i can't extract the result obtained by GA in each generation, i obtain just the result of final generation), my fitness function has been already declared:

function residue =fitness_XY(x)

t=[0;2;4;6;8;10;12;14;16;18;20;22;24;26;28;30;32;34;36;38;40;42;44;46;48;50;52;54;56;58;60;62;64;66;68;70;72;74;76;78;80;82;84;86;88;90;92;94;96;98;100;102];

Rexp=[-2.52708010767266e-14;5.32064895666630e-06;5.42159073721499e-06;7.75678296093904e-06;1.67256311050823e-05;1.23031850182854e-05;1.09794849995505e-05;1.20652240252192e-05;1.24698346918211e-05;2.33361919229394e-05;1.82336903162199e-05;2.33101153576165e-05;2.45945311647171e-05;1.69556354173215e-05;1.75494764515605e-05;2.36697072906705e-05;2.01220931647325e-05;2.94548337297039e-05;1.93821061390601e-05;2.03102518112670e-05;2.60712020709046e-05;2.20680494493528e-05;2.37415365853562e-05;1.95054548844394e-05;2.67173634397363e-05;3.21106195109085e-05;2.63163846059201e-05;2.80977713622924e-05;3.08339673608630e-05;4.27002816931244e-05;3.96175206240119e-05;2.90039717124409e-05;4.73335574049721e-05;3.45245585232448e-05;3.57778734384076e-05;4.21295733024959e-05;3.75669338617277e-05;4.87572144151742e-05;3.77461296632742e-05;3.58914963798983e-05;4.30056378514206e-05;3.84680097772286e-05;3.82897377331970e-05;5.62266325833374e-05;3.80177616775395e-05;4.43695039815036e-05;4.00476713092609e-05;3.88913016586448e-05;4.49264984228608e-05;4.49264984228608e-05;4.49264984228608e-05;4.49264984228608e-05];

Rth=x(1)*(1-exp(-t./x(2)));

residue=sum((Rexp-Rth).^2) ;

how i can plot:

-my fitness function(or any function) in each generation, to see the evolution of my code.

-Rexp and Rth in each generation.

-how my function converge to the best result in different generation.

Walter Roberson
on 4 Oct 2018

For plotting the fitness value at each generation (not each iteration) then modify

options = gaoptimset('PlotFcn', {@gaplotbestf @gaplotbestindiv @gaplotexpectation @gaplotstopping}, ...

'TolFun', 1e-13, 'Generations', 1000);

to suit your needs.

Rth and Rexp are vectors in each evaluation of the fitness_XY() function. Are you sure it makes sense to plot them?

Saber_phy
on 4 Oct 2018

sorry you don't understand me, i don't like to see the default plot function in matlab(like @gaplotbest, @gaplotbestindiv...).

i want to get a custom function, in the xlabel(generation) and in the ylabel(an ather parameter in differents figure)

or [xlabel(x(1)) and ylabel(x(2)) at different generation] to show how the populatios of individus converg to the best individu from generation to an ather.

thank you very much.

Walter Roberson
on 4 Oct 2018

Use nested functions with shared variables to take a record of all the values you want plotted. Provide your own plot function that you name in the 'PlotFcn' options.

ga does not appear to track individuals as to whether they are converging to the best of the generation. A work around might be for you to use an extra variable in your vector that encoded a unique identity, and code custom mutation and cross-over functions that somehow encoded identity as well, and then provide your own custom plot function that analyzed the Population state to show whatever it is you are trying to plot.

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