fsolve

Question

Doug Hates Squirrels 2011-4-26

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链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/6220-fsolve

回答： Lidianne Mapa 2018-1-14

I'm working on solving for the values of a series of parameters that are from a set of equations. While I have tried following every example I can find for fsolve, none have been particularly helpful. I've past the code from my m file below. Any help is appreciated. THANKS!

function F=calibrate(X)
%known variables;
nm = 0.031;
se = 0.108;
ne = 0.862;
c = 0.091;
ke = 0.545;
km = 0.040;
m = 0.097;
l = 0.333;
phi = 0.045;
A = 0.017;
B = 0.048;
%unknown parameters;
delta=X(1);
alpha=X(2);
gamma=X(3);
tau=X(4);
theta=X(5);
zeta1=X(6);
zeta2=X(7);
zeta3=X(8);
eta=X(9);
omegae = X(10);
omegam = X(11);
beta = X(12);
F(1) = A*((ke^alpha)*(ne^(1-alpha)))^(1-gamma) * m^gamma - c - ke*phi - km*phi +(ke+km)*(1 - delta);
F(2) = B*(theta * (km^tau) + (1-theta)*(nm^tau))^(1/tau) -m;
F(3) = omegae*se + omegam*nm - phi;
F(4) = 1 - ne - nm - se -l;
F(5) = zeta1 * ke^(alpha*(1-gamma))* ne^(-alpha - gamma*(1- alpha))*m^gamma - c;
F(6) = (gamma/eta)*A*B*(ke^(alpha*(1-gamma)))*(ne^((1-alpha)*(1-gamma)))*(nm^(tau-1))*(m^(gamma - tau)) + c*(omegae /omegam) - c;
F(7) = zeta2 * (ke^((alpha -1)-(gamma*alpha)))*(ne^((1-alpha)*(1-gamma)))*(m^gamma)+ zeta3 - phi;
F(8) = zeta3 - beta * gamma * theta * A* B*(ke^(alpha*(1-gamma)))*(ne^((1-alpha)+(1-gamma)))*(km^(tau-1))*(m^(gamma-tau)) - phi;
F(9) = beta*omegae*(1-l) - phi;
F(10) = (A*(1-gamma)*(1 - alpha))/eta - zeta1;
F(11) = beta*A*(1-gamma)*alpha - zeta2;
F(12) = beta*(1-delta)- zeta3;

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Doug Hates Squirrels 2011-4-26

Here's my solution when I try to run it, but I know it's not correct because when I plug the answers into the my system of equations they don't add up

EDU>> fsolve('calibrate',[1 1 1 1 1 1 1 1 1 1 1 1])

Solver stopped prematurely.

fsolve stopped because it exceeded the function evaluation limit,

options.MaxFunEvals = 1200 (the default value).

ans =

0.7982 1.0012 1.1022 0.9963 1.0038 0.0085 0.0001 0.0397 0.9970 0.4465 0.7347 0.1712

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Answer 1

Walter Roberson 2011-4-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/6220-fsolve#answer_8616

在 MATLAB Online 中打开

fsolve(@calibrate, ones(1,12), optimset('MaxFunEvals', 3000, 'FunValCheck', 'on', 'PlotFcns', @optimplotfval))

After your first run, when you have verified that it isn't trying to work with invalid values and have verified that the minimization is going well, you would likely want to remove the last two pairs of options.

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bym 2011-4-26

good thing the question didn't come from "Doug Hates Raccoons"

Paulo Silva 2011-4-26

proecsm that was a nice joke, thanks :D

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Answer 2

Lidianne Mapa 2018-1-14

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链接

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https://ww2.mathworks.cn/matlabcentral/answers/6220-fsolve#answer_300053

在 MATLAB Online 中打开

Hello dear, I'm with similar problem. This is my function:

 if y==1
 N1{y}=[(diff(Beta{y}(:,1),qsol(1))) diff(Beta{y}(:,1),qsol(2)) diff(Beta{y}(:,1),qsol(3));
        (diff(Beta{y}(:,2),qsol(1))) diff(Beta{y}(:,2),qsol(2)) diff(Beta{y}(:,2),qsol(3));
        (diff(Beta{y}(:,3),qsol(1))) diff(Beta{y}(:,3),qsol(2)) diff(Beta{y}(:,3),qsol(3))]
N2{y}=[(diff(Alfa{y}(:,1),qsol(1))) diff(Alfa{y}(:,1),qsol(2)) diff(Alfa{y}(:,1),qsol(3));
      (diff(Alfa{y}(:,2),qsol(1))) diff(Alfa{y}(:,2),qsol(2)) diff(Alfa{y}(:,2),qsol(3));
      (diff(Alfa{y}(:,3),qsol(1))) diff(Alfa{y}(:,3),qsol(2)) diff(Alfa{y}(:,3),qsol(3))]
else
    N1{y}=[(diff(Beta{y}(:,1),qsol(4))) diff(Beta{y}(:,1),qsol(5)) diff(Beta{y}(:,1),qsol(6));
      (diff(Beta{y}(:,2),qsol(4))) diff(Beta{y}(:,2),qsol(5)) diff(Beta{y}(:,2),qsol(6));
      (diff(Beta{y}(:,3),qsol(4))) diff(Beta{y}(:,3),qsol(5)) diff(Beta{y}(:,3),qsol(6))]
N2{y}=[(diff(Alfa{y}(:,1),qsol(4))) diff(Alfa{y}(:,1),qsol(5)) diff(Alfa{y}(:,1),qsol(6));
      (diff(Alfa{y}(:,2),qsol(4))) diff(Alfa{y}(:,2),qsol(5)) diff(Alfa{y}(:,2),qsol(6));
      (diff(Alfa{y}(:,3),qsol(4))) diff(Alfa{y}(:,3),qsol(5)) diff(Alfa{y}(:,3),qsol(6))]
end
end

KNL1=[N1{1},zeros(size(TMgmod{1},1),size(TMgmod{1},2)); zeros(size(TMgmod{1},1),size(TMgmod{1},2)) N1{2}];

KNL2=[[N2{1},zeros(size(TMgmod{1},1),size(TMgmod{1},2)); zeros(size(TMgmod{1},1),size(TMgmod{1},2)) N2{2}]] KTOTAL=1/2*KNL1+1/3*KNL2+Ktotal Forca=[10;10;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0;0] TFORCA=Tcbacop'*Forca; x0=inv(Ktotal)*TFORCA fun_sym =rootedo(qsol,KTOTAL,TFORCA); fun = matlabFunction(fun_sym, 'vars', {qsol.'}); fsolve(fun,x0, optimset('MaxFunEvals', 3000,'MaxIter', 800, 'FunValCheck', 'on', 'PlotFcns', @optimplotfval))

function [F]=rootedo(qsol,KTOTAL,TFORCA) F=KTOTAL*qsol'-TFORCA

This error:

fsolve stopped because the relative size of the current step is less than the default value of the step size tolerance squared, but the vector of function values is not near zero as measured by the default value of the function tolerance.

criteria details

ans = 1.0e-08 * 0.5447 0.1506 0.0561 0.5447 0.1506 0.0561

Can you help me please?