Curve fitting by Genetic Algorithm

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Hi everybody, I have a very simple theoric problem but I don't know how to resolve it.
I just wanna do a curve fitting like this:
Captura.PNG
I want the red line to fit the red dashed one as much as possible in the [0.3 - 0.6] interval.
The red line is a 100x1 vector (x-axis) and 100x1 vector (y-axis).
The red dashed curve is given by this equation (much longer, cropped here):
function [ d2 ] = d2_current( alpha,beta,phi,N,V )
d2 = (4903985730770845.*N.*(((exp(alpha.*(phi - V.*(beta - 1))) + 1).*((alpha.^2.*beta.^2.*exp(alpha.*...
end
Where V is the dependent variable, the previous 100x1 vector (x-axis) corresponding to the [0.3 - 0.6] interval.
The equation depends on 4 independent parameters, each one of them may be included in these intervals:
beta; from to a 1
alpha; from 0 to 15
N; from 0 to 300
phi; from 0 to 3
I have to vary these parameters in order to obtain the best fitting, note that the parameters value must be the same for one single fitting.
The first thing I did was 4 different 'for' loops for each parameter and try to compare each single result with the red curve my means of the euclidean distance. (simplified code):
cont = 0;
for beta = 0:0.01:1
for alpha = 0:0.01:15
for N = 0:1:300
for phi = 0:0.01:3
distance_euclidean(cont,1) = norm(current_red_curve - current_red_dashed_curve);
cont = cont + 1;
end
end
end
end
minim_eucl = min(distance_euclidean);
The minimum euclidean distance would be the best fitting. For the graph example:
beta = 1;
alpha = 15;
N = 300;
phi = 0.45;
Everything is fine until this point, but there's a big problem: the computing time is just huge when the step in the 4 intervals decreases (needed because of the poor fitting).
Looking for better solutions I found the possible way to go: the genetic algorithm. I've been trying it on the Matlab optimization tool, but with no results.
Could somebody help me out?
Thank you very much for your help.
Regards,
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Accepted Answer

Star Strider
Star Strider on 18 Oct 2019
I can fit part of your data, however not all of it, because there is some noise in elements 19 through 24 that your function is likely to not able to account for. (I had to eliminate the commas from your data and translate them into decimal points. Thjat worked for the data you posted, however I do not know if it will work for your entire vector.)
Try this:
X = string([0,301067677000000
0,314157576000000
0,327247475000000
0,340337374000000
0,353427273000000
0,366517172000000
0,379607071000000
0,392696970000000
0,405786869000000
0,418876768000000
0,431966667000000
0,445056566000000
0,458146465000000
0,471236364000000
0,484326263000000
0,497416162000000
0,510506061000000
0,523595960000000
0,536685859000000
0,549775758000000
0,562865657000000
0,575955556000000
0,589045455000000
0,602135354000000]);
Y = string([0,0193930890000000
0,0221282684000000
0,0245255339000000
0,0265479770000000
0,0281674962000000
0,0293643273000000
0,0301265743000000
0,0304497403000000
0,0303362581000000
0,0297950211000000
0,0288409140000000
0,0274943436000000
0,0257807698000000
0,0237302359000000
0,0213768998000000
0,0187585647000000
0,0159162099000000
0,0128935212000000
0,00973642228000000
0,00649260510000000
0,00321106079000000
-5,83896230000000e-05
-0,00326556436000000
-0,00636039014000000]);
V = sscanf(sprintf('%s.%s\n',X.'),'%f');
y = sscanf(sprintf('%s.%s\n',Y.'),'%f');
d2 = @(beta,alpha,N,phi,V) (4903985730770845.*N.*(((exp(alpha.*(phi - V.*(beta - 1))) + 1).*((alpha.^2.*beta.^2.*exp(alpha.*...
(phi - beta.*V)))./(exp(alpha.*(phi - V.*(beta - 1))) + 1) - (alpha.^2.*exp(alpha.*(phi - V.*(beta - 1))).*...
(exp(alpha.*(phi - beta.*V)) + 1).*(beta - 1).^2)./(exp(alpha.*(phi - V.*(beta - 1))) + 1).^2 + ...
(2.*alpha.^2.*exp(2.*alpha.*(phi - V.*(beta - 1))).*(exp(alpha.*(phi - beta.*V)) + 1).*(beta - 1).^2)./...
(exp(alpha.*(phi - V.*(beta - 1))) + 1).^3 - (2.*alpha.^2.*beta.*exp(alpha.*(phi - beta.*V)).*exp(alpha.*...
(phi - V.*(beta - 1))).*(beta - 1))./(exp(alpha.*(phi - V.*(beta - 1))) + 1).^2))./(alpha.*(exp(alpha.*(phi - beta.*V))...
+ 1)) + (exp(alpha.*(phi - V.*(beta - 1))).*(beta - 1).*((alpha.*beta.*exp(alpha.*(phi - beta.*V)))./(exp(alpha.*(phi -...
V.*(beta - 1))) + 1) - (alpha.*exp(alpha.*(phi - V.*(beta - 1))).*(exp(alpha.*(phi - beta.*V)) + 1).*(beta - 1))./...
(exp(alpha.*(phi - V.*(beta - 1))) + 1).^2))./(exp(alpha.*(phi - beta.*V)) + 1) - (beta.*exp(alpha.*(phi - beta.*V)).*...
((alpha.*beta.*exp(alpha.*(phi - beta.*V)))./(exp(alpha.*(phi - V.*(beta - 1))) + 1) - (alpha.*exp(alpha.*(phi - V.*...
(beta - 1))).*(exp(alpha.*(phi - beta.*V)) + 1).*(beta - 1))./(exp(alpha.*(phi - V.*(beta - 1))) + 1).^2).*(exp(alpha.*...
(phi - V.*(beta - 1))) + 1))./(exp(alpha.*(phi - beta.*V)) + 1).^2))./63407003003326144512;
Ve = V(1:18);
ye = y(1:18);
ftns = @(b) norm(ye - d2(b(1),b(2),b(3),b(4),Ve));
PopSz = 500;
Parms = 4;
opts = optimoptions('ga', 'PopulationSize',PopSz, 'InitialPopulationMatrix',randi(1E+4,PopSz,Parms)*1E-2, 'MaxGenerations',2E3, 'PlotFcn',@gaplotbestf, 'PlotInterval',1);
t0 = clock;
fprintf('\nStart Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t0)
[B,fval,exitflag,output] = ga(ftns, Parms, [],[],[],[],[],[],[],[],opts)
t1 = clock;
fprintf('\nStop Time: %4d-%02d-%02d %02d:%02d:%07.4f\n', t1)
GA_Time = etime(t1,t0)
fprintf('\nElapsed Time: %23.15E\t\t%02d:%02d:%02d.%03d\n', GA_Time, hmsdv(GA_Time))
fprintf(1,'\tParameters:\n')
for k1 = 1:length(B)
fprintf(1, '\t\tB(%d) = %8.5f\n', k1, B(k1))
end
xv = linspace(min(Ve), max(Ve), 250);
figure
plot(Ve, ye, 'p')
hold on
plot(xv, d2(B(1),B(2),B(3),B(4),xv), '-r')
grid
producing these parameter estimates:
Parameters:
B(1) = 15.96980
B(2) = 0.63200
B(3) = 542.18665
B(4) = 7.56829
with the elements of ‘B’ being, in order: ‘beta’, ‘alpha’, ‘N’, and ‘phi’. They retain their full precision internally.
and this plot:
#Curve fitting by Genetic Algorithm.png
Plot images are not showing up with the ‘Picture’ icon for some reason, so I am also attaching the plot image as a separate file to be sure it is visible.
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More Answers (1)

Alex Sha
Alex Sha on 19 Oct 2019
Rerfer the results below:
Root of Mean Square Error (RMSE): 0.00353350868235726
Sum of Squared Residual: 0.00029965640659906
Correlation Coef. (R): 0.975244832242099
R-Square: 0.95110248281492
Adjusted R-Square: 0.946445576416342
Determination Coef. (DC): 0.902466880697908
Chi-Square: -0.0236084937698218
F-Statistic: 41.4027542316901
Parameter Best Estimate
---------- -------------
beta1 1
alpha 15
n 105.578188880725
phi 0.388542678989983
t1.jpg
  1 Comment
Juan Castillo
Juan Castillo on 19 Oct 2019
Many thanks for your answer, but apart from the parameters value I need to know the way you did it! hahah
I need to implement the fitting algorithm for thousands of curves, so the process is mandatory for me.
Regards,

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