How can i solve 3 parameters of Weibull distribution under with initial data? like nonliear equation? Or do we have any method for estimate 3 parameters distribution from fail

How can i solve 3 parameters of Weibull distribution under with initial data? like nonliear equation? Or do we have any method for estimate 3 parameters of Weibull distribution for number of failure strength?

clc; clear; close all;
syms m sigma sigma_0 sigma_th 



% Define failure strength data (30 values)
sigma_i = [137.48, 162.07, 162.24, 162.82, 164.01, 164.40, 165.87, ...
              167.72, 169.75, 171.39, 177.64, 180.30, 180.57, 181.05, ...
              182.19, 182.86, 187.90, 187.92, 188.07, 188.21, 188.25, ...
              189.46, 190.05, 190.82, 190.99, 191.17, 191.62, 214.00, ...
              218.37, 221.12];

N = length(sigma_i); % Number of failure strength values

% Equation 1: dL/dσ_th
    eq1 = -(m - 1) * sum(1 ./ (sigma_i - sigma_th)) + (m / sigma_0^m) * sum((sigma_i - sigma_th).^(m - 1));

% Equation 2: dL/dσ_0
    eq2 = -(n * m) / sigma_0 + (m / sigma_0^(m + 1)) * sum((sigma_i - sigma_th).^m);

% Equation 3: dL/dm
    z = (sigma_i - sigma_th) / sigma_0;
    logz = log(sigma_i - sigma_th);
   
    eq3 = (n / m) - n * log(sigma_0) + sum(logz) - sum(z.^m .* (logz - log(sigma_0)));

Answer 1

[Star Strider]

The Statistiics and Machine Learning Toolbox norrmally fits a two-parameter Weibull distribution, howeer fitting a three-parameter distru=ibution in described in Three=Parameter Weibull Distribution

Shamelessly copying large sections of that code —

sigma_i = [137.48, 162.07, 162.24, 162.82, 164.01, 164.40, 165.87, ...
              167.72, 169.75, 171.39, 177.64, 180.30, 180.57, 181.05, ...
              182.19, 182.86, 187.90, 187.92, 188.07, 188.21, 188.25, ...
              189.46, 190.05, 190.82, 190.99, 191.17, 191.62, 214.00, ...
              218.37, 221.12];

figure
histfit(sigma_i, 10, 'Weibull')

f = @(x,a,b,c) wblpdf(x-c,a,b);

try
    mle(sigma_i,'pdf',f,'Start',[1700 2 1500])
catch ME
    disp(ME)
end

  MException with properties:

    identifier: 'stats:mle:NonpositivePdfVal'
       message: 'Custom probability function returned negative or zero values.'
         cause: {}
         stack: [7x1 struct]
    Correction: []

statset('mlecustom')

ans = struct with fields:
          Display: 'off'
      MaxFunEvals: 400
          MaxIter: 200
           TolBnd: 1.0000e-06
           TolFun: 1.0000e-06
       TolTypeFun: []
             TolX: 1.0000e-06
         TolTypeX: []
          GradObj: 'off'
         Jacobian: []
        DerivStep: 6.0555e-06
      FunValCheck: 'on'
           Robust: []
     RobustWgtFun: []
           WgtFun: []
             Tune: []
      UseParallel: []
    UseSubstreams: []
          Streams: {}
        OutputFcn: []

opt = statset('FunValCheck','off');

params = mle(sigma_i,'pdf',f,'Start',[100 2 100],'Options',opt, ...
    'LowerBound',[0 0 -Inf],'UpperBound',[Inf Inf min(sigma_i)]);

fprintf('\nParameters —\n\ta = %12.7f\n\tb = %12.7f\n\tc = %12.7f\n',params)

Parameters —
	a =   65.3620433
	b =    3.6671814
	c =  122.5383564

x = linspace(0, 250);

figure
histogram(sigma_i,8,'Normalization','pdf')
hold on
plot(x,f(x,params(1),params(2),params(3)),'LineWidth',2)
hold off

.