fmincon -结果与原始参数不接近

Question

我试图用MATLAB的fmincon函数来解决一个问题。下面显示了一个等式，为此我使用一些时间点生成了测试数据。我希望使用优化方法，根据生成的测试数据估计参数x(1)、x(2)和x(3)。当前使用fmincon估计的参数与用于生成数据的初始参数不接近。任何帮助都将不胜感激。

测试数据.时间点= 10:10:300,500,700,1000；x= 0.1，0.5，0.3；兴趣数据%参数=使用测试方程生成数据的x(1)*sin(x(2).*Timepoints)+log(x(3).*Timepoints)；%

% Parameters used to run the fmincon
x0 = [0, 0.1, 0.1]; % initial guess
lb = zeros(1, length(x0)); % lower bound of parameters
ub = ones(1, length(x0)); % upper bound of parameters

[x, fval, exitflag, output] = fmincon(@modelA1, x0, [], [], [], [], lb, ub, [], options, Timepoints, Data); 


function fvalues = modelA(x, Timepoints, fvals) 
Fvalues = zeros(1, length(Timepoints)); 
PreFvalues = zeros(1, length(Timepoints)); 

for Temp = 1:length(Timepoints)
tempY = x0(1)*sin(x0(2).*Timepoints(Temp))+log(x0(3).*Timepoints(Temp));
PreFvalues(Temp) = (fvals(Temp)-tempY)^2; 
end 
fvalues=sqrt(sum(PreFvalues)); 

Answer 1

在提供代码示例时，至少要确保其工作正常，并包括定义测试数据的代码。

经过适当的修改使其工作，下面的代码显示，您的代码生成的解决方案是相当好的。你有一个相当非线性的函数，所以它的解是一个局部极小，而不是一个整体极小(即原始解)，这并不奇怪。根据初始条件(即x0)，您将得到稍微不同的结果。

对于这个问题，我还建议使用lsqcurvefit而不是fmincon。这方面的一个例子也在代码中。

function test

Timepoints = [10:10:300, 500, 700, 1000]; x = [0.1, 0.5, 0.3]; 
Data = x(1)*sin(x(2).*Timepoints)+log(x(3).*Timepoints);

options = optimoptions('fmincon');

% Parameters used to run the fmincon
x0 = [0, 0.1, 0.1]; % initial guess
lb = zeros(1, length(x0)); % lower bound of parameters
ub = ones(1, length(x0)); % upper bound of parameters

[x, fval, exitflag, output] = fmincon(@modelA, x0, [], [], [], [], lb, ub, [], options, Timepoints, Data); 

% Using lsqcurvefit
F = @(x,xData)x(1)*sin(x(2).*xData)+log(x(3).*xData);
x = lsqcurvefit(F,x0,Timepoints,Data,lb,ub);

plot(...
    Timepoints,Data,...
    Timepoints,x(1)*sin(x(2).*Timepoints)+log(x(3).*Timepoints),...
    Timepoints,F(x,Timepoints));
legend({'Original','fmincon','lsqcurvefit'});

function fvalues = modelA(x, Timepoints, fvals) 

tempY = x(1)*sin(x(2)*Timepoints)+log(x(3)*Timepoints);
PreFvalues = (fvals-tempY).^2; 

fvalues=sum(PreFvalues); 

Answer 2

在提供代码示例时，至少要确保其工作正常，并包括定义测试数据的代码。

经过适当的修改使其工作，下面的代码显示，您的代码生成的解决方案是相当好的。你有一个相当非线性的函数，所以它的解是一个局部极小，而不是一个整体极小(即原始解)，这并不奇怪。根据初始条件(即x0)，您将得到稍微不同的结果。

对于这个问题，我还建议使用lsqcurvefit而不是fmincon。这方面的一个例子也在代码中。

function test

Timepoints = [10:10:300, 500, 700, 1000]; x = [0.1, 0.5, 0.3]; 
Data = x(1)*sin(x(2).*Timepoints)+log(x(3).*Timepoints);

options = optimoptions('fmincon');

% Parameters used to run the fmincon
x0 = [0, 0.1, 0.1]; % initial guess
lb = zeros(1, length(x0)); % lower bound of parameters
ub = ones(1, length(x0)); % upper bound of parameters

[x, fval, exitflag, output] = fmincon(@modelA, x0, [], [], [], [], lb, ub, [], options, Timepoints, Data); 

% Using lsqcurvefit
F = @(x,xData)x(1)*sin(x(2)*xData)+log(x(3)*xData);
x = lsqcurvefit(F,x0,Timepoints,Data,lb,ub);

plot(...
    Timepoints,Data,...
    Timepoints,x(1)*sin(x(2).*Timepoints)+log(x(3).*Timepoints),...
    Timepoints,F(x,Timepoints));
legend({'Original','fmincon','lsqcurvefit'});

function fvalues = modelA(x, Timepoints, fvals) 

tempY = x(1)*sin(x(2)*Timepoints)+log(x(3)*Timepoints);
PreFvalues = (fvals-tempY).^2; 

fvalues=sum(PreFvalues);