資源簡介
通過pso進行簡單的單目標搜索案例,體現了簡潔的MATLAB 代碼,解決問題,通過七個數據內容進行存儲速度u,v,的速度更新
代碼片段和文件信息
%%?https://www.youtube.com/watch?v=OQ3T575sbI8
%find?minimum?of?the?objective?function
%%?initialization
clear?
clc
iterations=30;
inertia=1.0;
correction_factor=2.0;
swarms=50;
%------initial?swarm?position--------
swarm=zeros(507);
step=1;
for?i=1:50
????swarm(step1:7)=i;
????step=step+1;
end
swarm(:7)=1000;?????%Greater?than?maximum?possible?value
swarm(:5)=0;???????????%initial?velocity
swarm(:6)=0;???????????%initial?velocity
%%?Iterations
for?iter=1:iterations
????
????%------position?of?swarm------
????for?i=1:swarms
????????swarm(i1)=swarm(i1)+swarm(i5)/1.2??????%update?u?position
????????swarm(i2)=swarm(i2)+swarm(i6)/1.2??????%update?v
????????u=swarm(i1)
????????v=swarm(i2)
????????
????????value=(u-20)^2+(v-10)^2?????????????????????????%objective?function
????????
????????if?value
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