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I'm trying to solve a system of non linear odes in Matlab as follows.
editparams %parameters
Tend = 50;
Nt = 100;
% Define RHS functions
RHS = #(t,x) ModelRHS(t,x,param); %anonymous function defining the set of equations
%Execution-----------------------------------------------------------------
x0 =[0.04;0.75;0.85]; %Initial condition
t = linspace(0,Tend,Nt); %TSPAN
[t x] = ode45(RHS, t, x0);
Now, I need to find the steady state of the system and I'm trying to create a function for this. I thought I'd calculate the steady state using the Jacobian. My equations are in an anonymous function which is defined as f in the code below. However, I realised that jacobian does not work for anonymous functions (or may be there is a way to do with anonymous functions) . so I thought I would convert the anonymous function to a symbolic function and try it. But
i still have a difficulty in getting that done. So any help is really appreciated!
function SS = SteadyState(param, E)
f = #(t,x)ModelRHS(t,x,param,E); %set of odes
SymbolicSystem = sym(f); %trying to make the anonymous function symbolic
SymbolicJacobian = jacobian(SymbolicSystem',x); %jacobian
Jacob = matlabFunction(SymbolicJacobian,x);
end
Also if there is any other way apart from finding the Jacobian, kindly let me know about that too.
I tried using 'fsolve' to calculate the steady-state as follows:
f = #(t,x)ModelRHS(t,x,param);
x0 =[0.04;0.75;0.85]';
options = optimoptions('fsolve','Display','iter'); % Option to display output
SS = fsolve(f,x0,options); % Call solver
but it returned an error
Not enough input arguments.
Error in #(t,x)ModelRHS(t,x,param)
Error in fsolve (line 242)
fuser = feval(funfcn{3},x,varargin{:});
Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.
I am trying to use mle() function in MATLAB to estimate the parameters of a 6-parameter custom distribution.
The PDF of the custom distribution is
and the CDF is
where Γ(x,y) and Γ(x) are the upper incomplete gamma function and the gamma function, respectively. α, θ, β, a, b, and c are the parameters of the custom distribution. K is given by
Given a data vector 'data', I want to estimate the parameters α, θ, β, a, b, and c.
So, far I have come up with this code:
data = rand(20000,1); % Since I cannot upload the acutal data, we may use this
t = 0:0.0001:0.5;
fun = #(w,a,b,c) w^(a-1)*(1-w)^(b-1)*exp^(-c*w);
% to estimate the parameters
custpdf = #(data,myalpha,mybeta,mytheta,a,b,c)...
((integral(#(t)fun(t,a,b,c),0,1)^-1)*...
mybeta*...
igamma(myalpha,((mytheta/t)^mybeta)^(a-1))*...
(mytheta/t)^(myalpha*mybeta+1)*...
exp(-(mytheta/t)^mybeta-(c*(igamma(myalpha,(mytheta/t)^mybeta)/gamma(myalpha)))))...
/...
(mytheta*...
gamma(myalpha)^(a+b-1)*...
(gamma(myalpha)-igamma(myalpha,(mytheta/t)^mybeta))^(1-b));
custcdf = #(data,myalpha,mybeta,mytheta,a,b,c)...
(integral(#(t)fun(t,a,b,c),0,1)^-1)*...
integral(#(t)fun(t,a,b,c),0,igamma(myalpha,(mytheta/t)^mybeta)^mybeta/gamma(myalpha));
phat = mle(data,'pdf',custpdf,'cdf',custcdf,'start',0.0);
But I get the following error:
Error using mlecustom (line 166)
Error evaluating the user-supplied pdf function
'#(data,myalpha,mybeta,mytheta,a,b,c)((integral(#(t)fun(t,a,b,c),0,1)^-1)*mybeta*igamma(myalpha,((mytheta/t)^mybeta)^(a-1))*(mytheta/t)^(myalpha*mybeta+1)*exp(-(mytheta/t)^mybeta-(c*(igamma(myalpha,(mytheta/t)^mybeta)/gamma(myalpha)))))/(mytheta*gamma(myalpha)^(a+b-1)*(gamma(myalpha)-igamma(myalpha,(mytheta/t)^mybeta))^(1-b))'.
Error in mle (line 245)
phat = mlecustom(data,varargin{:});
Caused by:
Not enough input arguments.
I tried to look into the error lines but I can't figure out where the error actually is.
Which function lacks fewer inputs? Is it referring to fun? Why would mle lack fewer inputs when it is trying to estimate the parameters?
Could someone kindly help me debug the error?
Thanks in advance.
exp() is a function, not a variable, precise the argument
exp^(-c*w) ---> exp(-c*w)
The starting point concerns the 6 parameters, not only one
0.1*ones(1,6)
In custcdf mle requires the upper bound of the integral to be a
scalar, I did some trial and error and the range is [2~9]. for the
trial some values lead to negative cdf or less than 1 discard them.
Then use the right one to compute the upper bound see if it's the
same as the one you predefined.
I re-write all the functions, check them out
The code is as follow
Censored = ones(5,1);% All data could be trusted
data = rand(5,1); % Since I cannot upload the acutal data, we may use this
f = #(w,a,b,c) (w.^(a-1)).*((1-w).^(b-1)).*exp(-c.*w);
% to estimate the parameters
custpdf = #(t,alpha,theta,beta, a,b,c)...
(((integral(#(w)f(w,a,b,c), 0,1)).^-1).*...
beta.*...
((igamma(alpha, (theta./t).^beta)).^(a-1)).*...
((theta./t).^(alpha.*beta + 1 )).*...
exp(-(((theta./t).^beta)+...
c.*igamma(alpha, (theta./t).^beta)./gamma(alpha))))./...
(theta.*...
((gamma(alpha)).^(a+b-1)).*...
((gamma(alpha)-...
igamma(alpha, (theta./t).^beta)).^(1-b)));
custcdf = #(t,alpha,theta,beta, a,b,c)...
((integral(#(w)f(w,a,b,c), 0,1)).^-1).*...
(integral(#(w)f(w,a,b,c), 0,2));
phat = mle(data,'pdf',custpdf,'cdf',custcdf,'start', 0.1.*ones(1,6),'Censoring',Censored);
Result
phat = 0.1017 0.1223 0.1153 0.1493 -0.0377 0.0902
I am working on an optimization problem where I want to maximize utility while searching over variables lx_init and kx_init. Each time I solve an issue with fmincon another one pops up. The error right now is..
"Not enough input arguments."
"Failure in initial objective function evaluation. FMINCON cannot continue."
I have tried to trace back the error in debug mode but tracing back my error is proving difficult. The function is able to be evaluated on its own and spit back a value so I know the error is in the fmincon function. My Matlab experience is very minimal. I am sure there are other syntax errors I am guilty of and too naive to see. I will provide all variables so the code can be replicated.
A second question: I want to maximize my utility value here but also find the values of x and y given once the utility is maximized. However, we are also maximizing over lx_init and kx_init. How do I get the function to return x and y? Right now it is just returning the utility value.
Here is my function I am optimizing
function [utility, x, y] = utility_with_prod(gamma_X, gamma_Y, alpha_LX, alpha_KX, alpha_LY, alpha_KY, alpha_uX, sigma_U, sigma_X, sigma_Y, L_bar, K_bar, lx_init, kx_init)
% X's production function
x = gamma_X*((alpha_LX*lx_init^((sigma_X-1)/sigma_X)) + ((alpha_KX)*kx_init^((sigma_X-1)/sigma_X)))^(sigma_X/(sigma_X-1));
% Y's production function
y = gamma_Y*((alpha_LY*(L_bar-lx_init)^((sigma_Y-1)/sigma_Y)) + ((alpha_KY)*(K_bar-kx_init)^((sigma_Y-1)/sigma_Y)))^(sigma_Y/(sigma_Y-1));
% utility function with nested production function
utility = -(((alpha_uX*x^((sigma_U-1)/sigma_U)) + ((1-alpha_uX)*y^((sigma_U-1)/sigma_U)))^(sigma_U/(sigma_U-1)));
end
Here are my initial values
sigma_U= 0.5038;
sigma_X= 0.5029;
sigma_Y= 0.5029;
alpha_uX= 0.000236017865342454;
alpha_LX= 0.180813595922536;
alpha_KX= 0.819186404077464;
gamma_X= 1.768587207836113;
alpha_LY= 0.505368332690592;
alpha_KY= 0.494631667309408;
gamma_Y= 1.999942066647923;
lx_init = 1;
kx_init = 2;
L_bar = 3;
K_bar = 3;
x0 = [lx_init, kx_init];
and the optimization
utility = #(lx_init, kx_init)utility_with_prod(gamma_X, gamma_Y, alpha_LX, alpha_KX, alpha_LY, alpha_KY, alpha_uX, sigma_U, sigma_X, sigma_Y, L_bar, K_bar, lx_init, kx_init)
[optimal_lx_kx, utility_max, exitflag] = fmincon(utility, x0, [],[])
x0 in fmincon is a vector, that's an n by 1 matrix or 1 by n,
here 1 by 2 ---> x0 = [lx_init, kx_init];
Function handle #(lx_init, kx_init) is different from #([lx_init, kx_init]).
#([lx_init, kx_init])accepts only one input.
#(lx_init, kx_init) accepts only two inputs, no more, no less
Also input variable should not be predefined value
Change #(lx_init, kx_init) to #(x) where x(1) = lx_init and x(2) = kx_init
To sum up
utility = #(lx_init, kx_init)utility_with_prod(gamma_X, gamma_Y, alpha_LX,alpha_KX, alpha_LY, alpha_KY, alpha_uX, sigma_U, sigma_X, sigma_Y, L_bar, K_bar, lx_init, kx_init)
is replaced by
utility = #(x)utility_with_prod(gamma_X, gamma_Y, alpha_LX, alpha_KX, alpha_LY, alpha_KY, alpha_uX, sigma_U, sigma_X, sigma_Y, L_bar, K_bar, x(1), x(2))
i have some experimental data and a theoretical model which i would like to try and fit. i have made a function file with the model - the code is shown below
function [ Q,P ] = RodFit(k,C )
% Function file for the theoretical scattering from a Rod
% R = radius, L = length
R = 10; % radius in Å
L = 1000; % length in Å
Q = 0.001:0.0001:0.5;
fun = #(x) ( (2.*besselj(1,Q.*R.*sin(x)))./...
(Q.*R.*sin(x)).*...
(sin(Q.*L.*cos(x)./2))./...
(Q.*L.*cos(x)./2)...
).^2.*sin(x);
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
end
with Q being the x-values and P being the y-values. I can call the function fine from the matlab command line and it works fine e.g. [Q,P] = RodFit(1,0.001) gives me a result i can plot using plot(Q,P)
But i cannot figure how to best find the fit to some experimental data. Ideally, i would like to use the optimization toolbox and lsqcurvefit since i would then also be able to optimize the R and L parameters. but i do not know how to pass (x,y) data to lsqcurvefit. i have attempted it with the code below but it does not work
File = 30; % the specific observation you want to fit the model to
ydata = DataFiles{1,File}.data(:,2)';
% RAdius = linspace(10,1000,length(ydata));
% LEngth = linspace(100,10000,length(ydata));
Multiplier = linspace(1e-3,1e3,length(ydata));
Constant = linspace(0,1,length(ydata));
xdata = [Multiplier; Constant]; % RAdius; LEngth;
L = lsqcurvefit(#RodFit,[1;0],xdata,ydata);
it gives me the error message:
Error using *
Inner matrix dimensions must agree.
Error in RodFit (line 15)
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
Error in lsqcurvefit (line 199)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Caused by:
Failure in initial user-supplied objective function evaluation. LSQCURVEFIT cannot continue.
i have tried i) making all vectors/matrices the same length and ii) tried using .* instead. nothing works and i am giving the same error message
Any kind of help would be greatly appreciated, whether it is suggestion regading what method is should use, suggestions to my code or something third.
EDIT TO ANSWER Osmoses:
A really good point but i do not think that is the problem. just checked the size of the all the vectors/matrices and they should be alright
>> size(Q)
ans =
1 1780
>> size(P)
ans =
1 1780
>> size(xdata)
ans =
2 1780
>> size([1;0.001]) - the initial guess/start point for xdata (x0)
ans =
2 1
>> size(ydata)
ans =
1 1780
UPDATE
I think i have identified the problem. the function RodFit works fine when i specify the input directly e.g. [Q,P] = RodFit(1,0.001);.
however, if i define x0 as x0 = [1,0.001] i cannot pass x0 to the function
>> x0 = [1;0.001]
x0 =
1.0000
0.0010
>> RodFit(x0);
Error using *
Inner matrix dimensions must agree.
Error in RodFit (line 15)
P = (integral(fun,0,pi/2,'ArrayValued',true))*k+C;
The same happens if i use x0 = [1,0.001]
clearly, matlab is interpreting x0 as input for k only and attempts to multiplay a vector of length(ydata) and a vector of length(x0) which obviously fails.
So my problem is that i need to code so that lsqcurvefit understands that the first column of xdata and x0 is the k variable and the second column of xdata and x0 is the C variable. According to the documentation - Passing Matrix Arguments - i should be able to pass x0 as a matrix to the solver. The solver should then also pass the xdata in the same format as x0.
Have you tried (that's sometimes the mistake) looking at the orientation of your input data (e.g. if xdata & ydata are both row/column vectors?). Other than that your code looks like it should work.
I have been able to solve some of the problems. One mistake in my code was that the objective function did not use of vector a variables but instead took in two variables - k and C. changing the code to accept a vector solved this problem
function [ Q,P ] = RodFit(X)
% Function file for the theoretical scattering from a Rod
% R = radius, L = length
% Q = 0.001:0.0001:0.5;
Q = linspace(0.11198,4.46904,1780);
fun = #(x) ( (2.*besselj(1,Q.*R.*sin(x)))./...
(Q.*R.*sin(x)).*...
(sin(Q.*L.*cos(x)./2))./...
(Q.*L.*cos(x)./2)...
).^2.*sin(x);
P = (integral(fun,0,pi/2,'ArrayValued',true))*X(1)+X(2);
with the code above, i can define x0 as x0 = [1 0.001];, and pass that into RodFit and get a result. i can also pass xdata into the function and get a result e.g. [Q,P] = RodFit(xdata(2,:));
Notice i have changed the orientation of all vectors so that they are now row-vectors and xdata has size size(xdata) = 1780 2
so i thought i had solved the problem completely but i still run into problems when i run lsqcurvefit. i get the error message
Error using RodFit
Too many input arguments.
Error in lsqcurvefit (line 199)
initVals.F = feval(funfcn_x_xdata{3},xCurrent,XDATA,varargin{:});
Caused by:
Failure in initial user-supplied objective function evaluation. LSQCURVEFIT cannot continue.
i have no idea why - does anyone have any idea about why Rodfit recieves to many input arguments when i call lsqcurvefit but not when i run the function manual using xdata?
I have a 1000x2 data file that I'm using for this problem.
I am supposed to fit the data with Acos(wt + phi). t is time, which is the first column in the data file, i.e. the independent variable. I need to find the fit parameters (A, f, and phi) and their uncertainties.
My code is as follows:
%load initial data file
data = load('hw_fit_cos_problem.dat');
t = data(:,1); %1st column is t (time)
x = t;
y = data(:,2); %2nd column is y (signal strength)
%define fitting function
f = fittype('A*cos(w*x + p)','coefficients','A','problem',{'w','p'});
% check fit parameters
coeffs = coeffnames(f);
%fit data
[A] = fit(x,y,f)
disp('confidence interval/errorbars');
ci = confint(A)
which yields 4 different error messages that I don't understand.
Error Messages:
Error using fit>iAssertNumProblemParameters (line 1113)
Missing problem parameters. Specify the values as a cell array with one element for each problem parameter
in the fittype.
Error in fit>iFit (line 198)
iAssertNumProblemParameters( probparams, probnames( model ) );
Error in fit (line 109)
[fitobj, goodness, output, convmsg] = iFit( xdatain, ydatain, fittypeobj, ...
Error in problem2 (line 14)
[A] = fit(x,y,f)
The line of code
f = fittype('A*cos(w*x + p)','coefficients','A','problem',{'w','p'});
specifies A as a "coefficient" in the model, and the values w and p as "problem" parameters.
Thus, the fitting toolbox expects that you will provide some more information about w and p, and then it will vary A. When no further information about w and p was provided to the fitting tool, that resulted in an error.
I am not sure of the goal of this project, or why w and p were designated as problem parameters. However, one simple solution is to allow the toolbox to treat A, w, and p as "coefficients", as follows:
f = fittype('A*cos(w*x + p)','coefficients', {'A', 'w', 'p'});
In this case, the code will not throw an error, and will return 95% confidence intervals on A, w, and p.
I hope that helps.
The straightforward answer to your question is that the error "Missing problem parameters" is generated because you have identified w and p as problem-specific fixed parameters,
but you have not told the fit function what these fixed values are.
You can do this by changing the line
[A] = fit(x,y,f)
to
[A]=fit(x,y,f,'problem',{100,0.1})
which supplies the values w=100 and p=0.1 in the fit. This should resolve the errors you specified (all 4 error messages result from this error)
In general specifying some of the quantities in your fit equation as problem-specific fixed parameters might be a valid thing to do - for example if you have determined them independently and have good reason to believe the values you obtained to be reliable. In this case, you might know the frequency w in this way, but you most probably won't know the phase p, so that should be a fit coefficient.
Hope that helps.