call function in fmicon optimization failing
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Hi guys, I have 4 functions where each one is derived from previous one. When I call the last one (revenue.m) I got an error using fmincon. Any clue? I want to optimize in respect to x0 that is present in each function. Many Thanks! Here is my code:
%% Total out flow equation T = zeros([N,M]); S=zeros([N,M]); for i = 1:M
R(:,i)=revenue(price(:,1),HH(:,i),ro,g,eff,x0(:,i),N,k1);
options = optimset('MaxFunEvals',Inf,'MaxIter',50,...
'Algorithm','interior-point','Display','iter');
tic
[x(:,i),fval(:,i)] = fmincon(@revenue,x0(1:2*N,i),A,b(:,i),Aeq,beq(:,i),LB(:,i),UB(:,i),[],options); %good
toc
end
with revenue fun being:
function R=revenue(price,HH,ro,g,eff,x0,N,k1)
R= -sum(price.*((HH*ro*g*eff).*x0(1:N))/k1);
end
% saved as revenue.m
Error msg:
Not enough input arguments.
Error in revenue (line 5)
R= -sum(price.*((HH*ro*g*eff).*x0(1:N))/k1);
Error in fmincon (line 535)
initVals.f = feval(funfcn{3},X,varargin{:});
Caused by:
Failure in initial objective function evaluation. FMINCON cannot continue.
end
4 Comments
Accepted Answer
Stephen23
on 25 May 2018
Edited: Stephen23
on 25 May 2018
The fmincon documentation clearly states that the first input " fun is a function that accepts a vector or array x and returns a real scalar f, the objective function evaluated at x". So it only accepts one input, which can be a vector or matrix. How many inputs does revenue have?
You need to parameterize the function, e.g. using an anonymous function and assuming that all of HH, ro, etc. are defined in that workspace, and the first input price is that one that you want to optimize:
fun = @(x) revenue(x,HH,ro,g,eff,x0,N,k1);
fmincon(fun,...)
5 Comments
Torsten
on 25 May 2018
Edited: Torsten
on 25 May 2018
The solution inside fmincon is passed as a one-dimensional vector.
Writing
... = fmincon(@(x0)revenue(price(:,i),HH(:,i),ro,g,eff,x0(:,i),N,k1),...
assumes that x0 inside fmincon is a matrix from which you want to pass the i'th column to "revenue". But this i'th column of a matrix does not exist unless i=1.
Best wishes
Torsten.
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