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))
Related
When I type help gmres in MATLAB I get the following example:
n = 21; A = gallery('wilk',n); b = sum(A,2);
tol = 1e-12; maxit = 15;
x1 = gmres(#(x)afun(x,n),b,10,tol,maxit,#(x)mfun(x,n));
where the two functions are:
function y = afun(x,n)
y = [0; x(1:n-1)] + [((n-1)/2:-1:0)'; (1:(n-1)/2)'].*x+[x(2:n); 0];
end
and
function y = mfun(r,n)
y = r ./ [((n-1)/2:-1:1)'; 1; (1:(n-1)/2)'];
end
I tested it and it works great. My question is in both those functions what is the value for x since we never give it one?
Also shouldn't the call to gmres be written like this: (y in the #handle)
x1 = gmres(#(y)afun(x,n),b,10,tol,maxit,#(y)mfun(x,n));
Function handles are one way to parametrize functions in MATLAB. From the documentation page, we find the following example:
b = 2;
c = 3.5;
cubicpoly = #(x) x^3 + b*x + c;
x = fzero(cubicpoly,0)
which results in:
x =
-1.0945
So what's happening here? fzero is a so-called function function, that takes function handles as inputs, and performs operations on them -- in this case, finds the root of the given function. Practically, this means that fzero decides which values for the input argument x to cubicpoly to try in order to find the root. This means the user just provides a function - no need to give the inputs - and fzero will query the function with different values for x to eventually find the root.
The function you ask about, gmres, operates in a similar manner. What this means is that you merely need to provide a function that takes an appropriate number of input arguments, and gmres will take care of calling it with appropriate inputs to produce its output.
Finally, let's consider your suggestion of calling gmres as follows:
x1 = gmres(#(y)afun(x,n),b,10,tol,maxit,#(y)mfun(x,n));
This might work, or then again it might not -- it depends whether you have a variable called x in the workspace of the function eventually calling either afun or mfun. Notice that now the function handles take one input, y, but its value is nowhere used in the expression of the function defined. This means it will not have any effect on the output.
Consider the following example to illustrate what happens:
f = #(y)2*x+1; % define a function handle
f(1) % error! Undefined function or variable 'x'!
% the following this works, and g will now use x from the workspace
x = 42;
g = #(y)2*x+1; % define a function handle that knows about x
g(1)
g(2)
g(3) % ...but the result will be independent of y as it's not used.
need to find a set of optimal parameters P of the system y = P(1)*exp(-P(2)*x) - P(3)*x where x and y are experimental values. I defined my function
f = #(P) P(1)*exp(-P(2)*x) - P(3)*x
and
guess = [1, 1, 1]
and tried
P = fminsearch(f,guess)
according to Help. I get an error
Subscripted assignment dimension mismatch.
Error in fminsearch (line 191)
fv(:,1) = funfcn(x,varargin{:});
I don't quite understand where my y values would fall in, as well as where the function takes P from. I unfortunately have no access to nlinfit or optimization toolboxes.
You should try the matlab function lsqnonlin(#testfun,[1;1;1])
But first make a function and save in an m-file that includes all the data points, lets say your y is A and x is x like here below:
function F = testfun(P)
A = [1;2;3;7;30;100];
x = [1;2;3;4;5;6];
F = A-P(1)*exp(-P(2)*x) - P(3)*x;
This minimized the 2-norm end gives you the best parameters.
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 created a function in matlab that returns a vector like
function w = W_1D(x,pos,h)
w=zeros(1,length(x));
if (h~=0)
xmpos = x-pos;
inds1 = (-h <= xmpos) & (xmpos < 0);
w(inds1) = xmpos(inds1)./h + 1;
inds2 = (0 <= xmpos) & (xmpos <= h);
w(inds2) = -xmpos(inds2)./h + 1;
else
error('h shouldn't be 0')
end
end
Thus, in the end, there is a vector w of size length(x).
Now i created a second function like
function f = W_2D(x,y,pos_1,pos_2,h)
w_x = W_1D(x,pos_1,h);
w_y = W_1D(y,pos_2,h);
f = w_x'*w_y;
end
where length(x)=length(y). Thus, the function W_2D obviously returns a matrix.
But when I now try to evaluate the integral over a rectangular domain like e.g.
V = integral2(#(x,y) W_2D(x,y,2,3,h),0,10,0,10);
matlab returns some errors:
Error using integral2Calc>integral2t/tensor (line 242)
Integrand output size does not match the input size.
Error in integral2Calc>integral2t (line 56)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 10)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 107)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
I also tried to vary something in the W_2D-function: instead of f = w_x'*w_y;
I tried f = w_x.'*w_y;
or w_y = transpose(w_y); f = kron(w_x,w_y);, but there is always this error with the Integrand output size-stuff.
Can anyone explain, where my fault is?
EDIT: After Werner's hint with the keyboard debugging method, I can tell you the following.
The first step returns w_x of type <1x154 double>, w_y is <1x192 double>, x and y are both <14x14 double>. In the next step, f appears with a value of <154x192 double>. Then everything disappears, except x and y and the matlab-function integral2Calc.m appears in the editor and it jumps to the Function Call Stack integral2t/tensor and after some more steps, the error occurs here
Z = FUN(X,Y); NFE = NFE + 1;
if FIRSTFUNEVAL
if ~isfloat(Z)
error(message('MATLAB:integral2:UnsupportedClass',class(Z)));
end
% Check that FUN is properly vectorized. This is important here
% because we (otherwise) always pass in square matrices, which
% reduces the probability of the user generating an error by
% using matrix functions instead of elementwise functions.
Z1 = FUN(X(VTSTIDX),Y(VTSTIDX)); NFE = NFE + 1;
if ~isequal(size(Z),size(X)) || ~isequal(size(Z1),size(VTSTIDX))
% Example:
% integral2(#(x,y)1,0,1,0,1)
error(message('MATLAB:integral2:funSizeMismatch'));
end
Hope that information is detailed enough...I have no idea what happenes, because my example is exact as it is given on the mathworks site about integral2, isn't it?
Maybe I should precise a bit more, what I wanna do: since W_2D gives me a surface w(x,y) of a compactly supported 2-dimensional hat-function, stored in a matrix w, I want to calculate the volume between the (x,y)-plane and the surface z=w(x,y)...
EDIT2: I still do not understand how to handle the problem, that integral2 creates matrices as inputs for my W_1D-functions, which are called in W_2D and intended to have a <1xn double>-valued input and return a <1xn double> output, but at least I can simply use the following to solve the integration over the tensor product by using two one-dimensional integral-calls, that is
V = integral(#(x)integral(#(y)W_1D(y,3,h),0,10).*W_1D(x,2,h),0,10);
This first function is quite wrong. You are not indexing the array positions while you are doing w = x inside for.
Besides, if that would work, you are returning a line vector, that is, size 1xlength(x) and when you do w_x'*w_y you are doing length(x)x1 times 1xlength(y), which would give you a matrix length(x)*length(y).
Consider correcting your function:
function w = W_1D(x,pos)
w = zeros(length(x),1); % Allocate w as column vector, so that the product gives a scalar (as I suppose that it is what you want.
for ii=1:length(x) % Here, so that is indexes w and x elements as you need
w(ii)=x(ii) - pos; % I changed your code to something that makes sense, but I don't know if that is what you want to do, you have to adapt it to work correctly.
end
end
You may also want to debug your functions, consider adding keyboard before your operations and check what they are returning using dbstep. I.e:
function f = W_2D(x,y,pos_1,pos_2)
w_x = W_1D(x,pos_1);
w_y = W_1D(y,pos_2);
keyboard
f = w_x'*w_y;
end
Execution will stop at keyboard, then you can check w_x size, w_y size, and do dbstep to go after f = w_x'*w_y and see what it returned. After you finish debug, you can do dbcont so that it will continue execution.
This answer is a draft as it is quite difficult to help you with the information you have provided. But I think you can start working the things out with this. If you have more doubts feel free to ask.
I want to solve two nonlinear equations in MATLAB so i did the following:
part of my script
c=[A\u;A\v];
% parts of code are omitted.
x0=[1;1];
sol= fsolve(#myfunc,x0);
the myfunc function is as follows
function F = myfunc(x)
F=[ x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj];
end
i have two unknowns x(1) and x(2)
my question is How to pass a values(c,ii,jj) to myfunc in every time i call it?
or how to overcome this error Undefined function or method 'c' for input arguments of type 'double'.
thanks
Edit: The previous answer was bogus and not contributing at all. Hence has been deleted. Here is the right way.
In your main code create a vector of the coefficients c,ii,jj and a dummy function handle f_d
coeffs = [c,ii,jj];
f_d = #(x0) myfunc(x0,coeffs); % f_d considers x0 as variables
sol = fsolve(f_d,x0);
Make your function myfunc capable of taking in 2 variables, x0 and coeffs
function F = myfunc(x, coeffs)
c = coeffs(1:end-2);
ii = coeffs(end-1);
jj = coeffs(end);
F(1) = x(1)*c(1)+x(2)*c(2)+x(1)*x(2)*c(3)+c(4)-ii;
F(2) = x(1)*c(5)+x(2)*c(6)+x(1)*x(2)*c(7)+c(8)-jj;
I think that should solve for x0(1) and x0(2).
Edit: Thank you Eitan_T. Changes have been made above.
There is an alternative option, that I prefer, if a function handle is not what you are looking for.
Say I have this function:
function y = i_have_a_root( x, a )
y = a*x^2;
end
You can pass in your initial guess for x and a value for a by just calling fsolve like so:
a = 5;
x0 = 0;
root = fsolve('i_have_a_root',x0,[],a);
Note: The [] is reserved for fsolve options, which you probably want to use. See the second call to fsolve in the documentation here for information on the options argument.