Question

我正在尝试使用MATLAB fmincon函数解决问题。我有一个如下所示的等式，我已经使用一些时间点生成了测试数据。我想基于生成的测试数据，通过使用优化来估计参数x（1），x（2）和x（3）。使用fmincon当前估计的参数不接近我用于生成数据的初始参数。任何帮助将受到高度赞赏。

测试数据; 时间点= [10：10：300,500,700,1000]; x = [0.1,0.5,0.3]; ％感兴趣的参数数据= x（1）* sin（x（2）。*时间点）+ log（x（3）。*时间点）; ％使用测试方程生成数据

% 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

提供代码示例时，至少要确保它有效，并包含定义测试数据的代码。

经过适当修改以使其正常工作后，以下代码显示代码生成的解决方案非常好。你有一个非常非线性的函数，因此解决方案是局部最小值而不是全局最小值（即原始解决方案）并不奇怪。根据初始条件（即x0），你会得到略微不同的结果。

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

提供代码示例时，至少要确保它有效，并包含定义测试数据的代码。

经过适当修改以使其正常工作后，以下代码显示代码生成的解决方案非常好。你有一个非常非线性的函数，因此解决方案是局部最小值而不是全局最小值（即原始解决方案）并不奇怪。根据初始条件（即x0），你会得到略微不同的结果。

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);

fmincon - 结果不接近原始参数

2 个答案: