Can I get the code for Economic Load Dispatch using Gradient Descent (matlab)?

11 Ansichten (letzte 30 Tage)
I'm not getting how to proceed after this!!
clc
a=[500;400;200];
b=[5.31;5.5;5.8];
g=[0.004;0.005;0.009];
lambda=input('enter vvalue of lambda');
iter=0;
Pd=800;
n=length(a);
delP=5;tcost=0;
tolerance=0.0001;
while abs(delP)>tolerance
iter=iter+1;
x=0;
for i=1:n
p(i)=(lambda-b(i))/(2*g(i));
x=x+(1/(2*g(i)));
cost=(a(i)+(b(i)*p(i))+(g(i)*p(i)*p(i)));
  11 Kommentare
mahesh kumar
mahesh kumar am 27 Jun. 2020
Bearbeitet: mahesh kumar am 27 Jun. 2020
try this::
B = [.00003 0 0;0 .00009 0 ; 0 0 .00012]; dB=diag(B); load=850;
x=max(b); dP=1;i=0;
z=randperm(10000,length(b))'; P=load*z/sum(z);
while abs(dP)>0.000001
i=i+1; disp(i);
dP=load+P'*B*P-sum(P); % loss=P'*B*P;
x=x+dP*2/(sum(1./a));
P=(x-b-2*(B*P-dB.*P))./(2*a+2*x*dB);
% plot(x,P,'x'); hold on; pause(1);
% C=c+b.*P+a.*P.*P;
% plot(x,sum(C)/100,'o');
end
Suzzane M
Suzzane M am 28 Jun. 2020
Yes, it solves my problem. Thankyou so much for the help. Was badly stuck in for days. Fianlly got some clarity. Thankyou again

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Antworten (1)

mahesh kumar
mahesh kumar am 29 Jun. 2020
Bearbeitet: mahesh kumar am 29 Jun. 2020

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