LINPROG requires the following inputs to be of data type double: 'f'.
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%% Discrete modelling
Ae=(1-(K*ts)/Cth);
B=(ts*COP*HP)/Cth;
D=ts*K/Cth;
u=zeros(1,i);
Tin=zeros(1,i);
x=zeros(3,i);
cost=[gamma;zeros(1,1008);zeros(1,1008)];
xcost=zeros(3,i);
for j=1:i
if j>1
Tin(j)=Ae*Tin(j-1)+B*u(j-1)+C*Tout(j-1);
x=[u(j-1);Tin(j-1);gamma(j-1)];
else
Tin(j)=0;
u(j)=0;
x(j)=0;
end
% x=[u(j);Tin(j);gamma(j)];
cost=[gamma(j);0;0];
f=@(x) sum(cost(j).'*x(j));
%Starting point%
x0=0;
%Constraints%
%Inequality const%
A=[1 0 0;-1 0 0;0 1 0;0 -1 0;0 0 1;0 0 -1];
b=[umin; -umax; Tmin; -Tmax; gammamin; -gammamax];
options = optimoptions(@linprog,'Display', 'off');
[x(j),xcost(j)]=linprog(f,x0,A,b);
end
Antworten (1)
Ameer Hamza
am 18 Okt. 2020
Bearbeitet: Ameer Hamza
am 18 Okt. 2020
linprog() does not require you to multiply 'f' with x. It will do that internally. Just directly give the vector 'f'
f = cost; % instead of f = @(x) sum(cost(j).'*x(j));
2 Kommentare
Ameer Hamza
am 18 Okt. 2020
The dimensions of A and b seem to be fine? Can you add a breakpoint and see if the dimensions are correct when the issue happens?
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