Minimize the sum of squared errors between the experimental and predicted data in order to estimate two optimum parameters
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In my research work, I want to do Vehicle survival fraction, survival rates were calculated using the log-logistic survival function, in equation two unknown parameters a, b is to obtain by minimize the sum of squared errors between the experimental and modeled.
equation of the model is:
S = [1 + (t/a)^b]^(-1)
age of the vehicle t = [0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20];
registered vehicles of respective age t, N = [6801342;6364669;6104616;5849372;5426898;4969076;4549439;4117610;3714272;3324896;2980512;2652830;2320935;2041282;1818527;1594733;1335572;1053197;800212;566590;376620];
experimental value u_exp = [406183941.2;270789294.1;812367882.4;541578588.2;406183941.2;1353946471;406183941.2;135394647.1;135394647.1;135394647.1;676973235.3;270789294.1;406183941.2;0;270789294.1;406183941.2;0;135394647.1;0;0;135394647.1];
u_mod = @(P) ((1+(t./P(1)).^P(2)).^(-1)).*N;
modeled value = S*N
and I want to calculate the parameters “a” and “b” by minimizing the sum of squared errors between “n exp” and “n model”.
Someone here can help me please?
Thank you already for your help!
9 个评论
Sam Chak
2024-4-1
Hi @Vikas Meena
Below, I have overlaid the Log-Logistic Distribution (LLD) with your normalized experimental data. It is worth noting that it is possible to find a unique LLD that corresponds to each non-zero data point. Given that there are 4 zeros out of the total 21 data points, it implies that you can determine 17 or fewer unique LLDs. I would appreciate it if you could clarify whether this is your desired outcome from a purely mathematical perspective.
t = linspace(0, 20, 2001);
a = 2.5*[1:8]; % alpha: 8 intersection points
b = 2.^[1/2 1 3/2 2 5/2 3 3.5]; % beta : 7 curve characteristics
for i = 1:numel(a)
for j = 1:numel(b)
LLD = 1./(1 + 1./((t/a(i)).^b(j)));
plot(t, LLD), hold on
end
end
z = [0;1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20];
u_exp = [4061839.412;2707892.941;8123678.824;5415785.882;4061839.412;13539464.71;4061839.412;1353946.471;1353946.471;1353946.471;6769732.353;2707892.941;4061839.412;0;2707892.941;4061839.412;0;1353946.471;0;0;1353946.471];
stem(z, u_exp/max(u_exp), 'k', 'linewidth', 1.5)
hold off, grid on
xlabel('t')
ylabel('LLD')
title('Overlay Plots')
回答(2 个)
the cyclist
2024-3-30
Unless I'm making a mistake in my thinking, though, that is not a very good function to fit your data. As t approaches zero, S approaches 1, regardless of the parameters.
rng default
% Define the data to be fit
t = [0; 1; 2; 3; 4; 5; 6; 7; 8; 9; 10; 11; 12; 13; 14; 15; 16; 17; 18; 19; 20];
N = [3; 2; 6; 4; 3; 10; 3; 1; 1; 1; 5; 2; 3; 0; 2; 3; 0; 1; 0; 0; 1];
% Tabulate the data
tbl = table(t,N);
% Define function that will be used to fit data
% (p is a vector of fitting parameters)
S = @(p,t) (1 + (t./p(1)).^p(2)).^(-1);
% Fit the model
beta0 = [5 5]; % Initial parameter guess
mdl = fitnlm(tbl,S,beta0)
% Calculate the model values at the empirical x
y_predicted = predict(mdl,t);
% Plot the data and fit
figure
plot(t,N,'*',t,y_predicted,'g');
legend('data','fit')
1 个评论
the cyclist
2024-3-30
This solution was written prior to a significant re-write of the question, and is no longer applicable as a specific solution to the question.
I'm only leaving it here for OP's later reference, if they wish.
Sam Chak
2024-4-2
编辑:Sam Chak
2024-4-2
Hi @Vikas Meena
Taking into account the information provided, this solution is likely the most suitable option I can propose.
%% Experimental Data
t = [0:20];
uex = [4061839.412; 2707892.941; 8123678.824; 5415785.882; 4061839.412; 13539464.71; 4061839.412; 1353946.471; 1353946.471; 1353946.471; 6769732.353; 2707892.941; 4061839.412; 0; 2707892.941; 4061839.412; 0; 1353946.471; 0; 0; 1353946.471]';
y = uex/max(uex); % normalized data
%% Determine the parameters
idx1= find(y == 1);
idx2= find(y == 0);
tt = t; tt(1) = 1e-6; % data processing in t
yy = y; yy(idx1) = 1 - 1e-6; yy(idx2) = 1e-6; % data processing in y
b = 3; % fix beta
a = ((- (tt.^b).*(yy - 1)).^(1/b))./(yy.^(1/b)); % find alpha
%% Log-Logistic Distribution model
LLD = 1./(1 + 1./((tt./a).^b));
%% Sum of squared errors
Err = y - LLD;
sse = sum(Err.^2)
%% Plot result
stem(t , y), axis([-1 21 0 1.1]), hold on
stem(tt, LLD, 'markersize', 12), hold off, grid on
xlabel('t')
legend('Normalized Data', 'Prediction', 'fontsize', 12)
title('Prediction by Log-Logistic Distribution Model')
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