reliability - LOLE=0; disp('PL Pg Prob') for i=1:nl/2

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% HW#3 part 1 clc ng=3; % Gen Cap FOR Pg=[ 25 .05; 50 .05; 150 .15]; %case2 % ng=4 % Pg=[ 25 .05; % 25 .01 % 50 .05; % 150 .15]; %case3 % ng=4 % Pg=[ 25 .05; % 50 .05; % 75 .05 % 150 .15]; nl=8; % load peak prob load prob Pl=[100 .25 6/96; 120 .25 6/96; 175 .25 6/96; 140 .25 6/96; 80 .25 18/96; 70 .25 18/96; 50 .25 18/96; 80 .25 18/96]; % Genearation Availability P_gen=zeros(2^ng,2); for i=1:2^ng bin=zeros(ng,1); bin1=de2bi(i-1); [r c]=size(bin1); bin(1:c)=bin1; P_gen(i,2)=prod((ones(ng,1)-Pg(:,2)).*bin+Pg(:,2).*(ones(ng,1)-bin)); P_gen(i,1)=sum(Pg(:,1).*bin); end e %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Loss of Load Expectation (LOLE)
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Unformatted text preview: LOLE=0; disp('PL Pg Prob') for i=1:nl/2 disp('-------------------------') for j=1:2^ng prob=0; if P_gen(j,1)<Pl(i,1) LOLE=LOLE+P_gen(j,2)*Pl(i,2); prob=P_gen(j,2)*Pl(i,2); end fprintf( '%4.1f %4.1f %4.4f \n', [Pl(i) P_gen(j,1) prob]) end end fprintf( '\nLOLE is %4.2f days/yr \n', LOLE*365) f f %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Expected Unserved Energy (EUE) EUE=0; for i=1:nl for j=1:2^ng if P_gen(j,1)<Pl(i,1) EUE=EUE+(P_gen(j,2)*Pl(i,3))*(Pl(i,1)-P_gen(j,1)); end end end fprintf( '\nEUE is %4.2f MWh/yr \n', EUE*360*24)...
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This note was uploaded on 07/14/2011 for the course ECE 522 taught by Professor Tomsovic during the Summer '10 term at University of Florida.

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reliability - LOLE=0; disp('PL Pg Prob') for i=1:nl/2

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