Very confusing code.
function [br]= Rate_optimum(k)
PL=3;
N=10^5;
MM=3;
si=1;
PT=10.^(5./10);
PR=10.^(0./10);
sign=10.^(0./10);sigd=10.^(0./10);sige=10.^(-7./10);
lamdap=0.1;
Rate_opt2 =0;
for kk=1:N
fprintf('Running %d per %d \n',kk,length(k));
w_opt=PR;
for bb = 1 : MM %relays
f= exprnd(1./lamdap);
h=exprnd(1./lamdap);
g= exprnd(1./lamdap);
alpha_opt=(k.*si.^(-1))./(PT.*f+(PR.*h.*(PT.*g+sign))+2.*sige);%k is an array so alpha_out will also be an array
SNR1_opt=PT.*f./(sige+sigd);
SNR2_opt=(w_opt.*h.*g.*PT)./(w_opt.*h.*sign+sige+sigd);%everything else is just constant and can be taken outside the loops
SNR_SD_opt=SNR1_opt+SNR2_opt;
Rate=0.5.*(1-alpha_opt).*log2(1+SNR_SD_opt);%Rate is also be an array
if (Rate > Rate_opt2 ) %does not make sense comparing an array to a scalar
Rate_opt2 = Rate ;
ID_max = bb % This is the selected relay from the set of relays we have
end
end
Rate_final=Rate_opt2;
Rate_opt2
end
br= Rate_final/N ;
br
end
How are you differentiating between the different relays? I believe there should be three values for each of your constants (different values for each relay). You never index with bb; therefore, all the relays have the same constants and the rates will be the same. You should be able to do this without any loops.
