資源簡介
采用PSO算法,以成本最小為目標函數,對蓄電池和超級電容的容量進行了優化
代碼片段和文件信息
%%?funm
function?y=funm(pop)
Ew=[277.6??238.5??243.4??240.4??238.5??222.2??208.8??205.8??205.7??236.3??265.4??310.7];
Es=[31.3??37.8??54.8??60.63??69.93??67.07??65.03??62.02??59.92??43.6??31.47??26.74];
El=[294.5??266??285??273??294.5??283??295??281??282??294??285??299];
yitac=0.95;
delE=(Ew+Es)*yitac-El;
Eb=zeros(112);
Ec=zeros(112);
Elps=0;
for?k=1:12
????if?delE(k)>0
????????if?k?==?1
???????????[EbtEct]=pro1(delE(k)pop00);
????????else
????????????[EbtEct]=pro1(delE(k)popEb(k-1)Ec(k-1));
????????end
????????Eb(k)=Ebt;
????????Ec(k)=Ect;
????else
????????delE(k)=-1*delE(k);
????????if?k?==?1
????????????[EbtEctElps]=pro2(delE(k)popElps00);
????????else
????????????[EbtEctElps]=pro2(delE(k)popElpsEb(k-1)Ec(k-1));
????????end
????????Eb(
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????1096??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\funm.asv
?????文件????????1096??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\funm.m
?????文件?????????735??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\pro1.asv
?????文件?????????723??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\pro1.m
?????文件?????????981??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\pro2.asv
?????文件?????????964??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\pro2.m
?????文件????????2204??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\PSO_R.asv
?????文件????????2324??2020-09-22?06:53??PSO_R╕─╜°╚?╓╪\PSO_R.m
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