【正文】
初始化位置 初始化速度 迭代結(jié)果對(duì)比最優(yōu)點(diǎn)坐標(biāo)(1):[ ]最優(yōu)坐標(biāo)(2):[ ]適應(yīng)度值(1)為:適應(yīng)度值(2)為:(5)對(duì),對(duì)分別取,對(duì)比其迭代影響 初始化位置 初速度位置 迭代結(jié)果最優(yōu)點(diǎn)坐標(biāo)(1):[ 最優(yōu)坐標(biāo)(2):[ ]適應(yīng)度值(1)為:適應(yīng)度值(2)為:108141522528168(6)標(biāo)準(zhǔn)粒子群算法無(wú)參數(shù)對(duì)比, 粒子群位置初始化 粒子群初始化速度 迭代結(jié)果在以上仿真中,我們5個(gè)實(shí)驗(yàn)實(shí)數(shù)的選擇分別對(duì),不同情況做出對(duì)比得出結(jié)論:慣性權(quán)重的不同取值對(duì)PSO的影響試驗(yàn)表明權(quán)值將影響PSO 的全局與局部搜優(yōu)能力,值較大,全局搜優(yōu)能力強(qiáng),局部搜優(yōu)能力弱。一般取[20 40],對(duì)于大部分的問(wèn)題, 10個(gè)粒子已經(jīng)足夠取得好的結(jié)果,對(duì)于比較難的問(wèn)題或者特定類別的問(wèn)題,粒子數(shù)可以取到100,200。研究表明,讓慣性權(quán)值隨著疊代次數(shù)的增加在1. 4到0之間逐步減少可以取得較好的效果。 如表1 所示,在考察因子對(duì)試驗(yàn)結(jié)果的影響程度時(shí),把因子的個(gè)水平看成是個(gè)正態(tài)總體,因此可設(shè), , 。因此,公式() 表示的和之間的比值F就是反映了兩種差異所占的比重。解:現(xiàn)在。因此,仍需要對(duì) PSO 的收斂性等方面進(jìn)行進(jìn)一步的理論研究。參考文獻(xiàn):[1] Kennedy J, Eberhart R. 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Mourelle (eds.), Springer Verlag, 2006, . 致謝經(jīng)過(guò)半年的忙碌和工作,本次畢業(yè)設(shè)計(jì)已經(jīng)接近尾聲,作為一個(gè)本科生的畢業(yè)設(shè)計(jì),由于經(jīng)驗(yàn)的匱乏,難免有許多考慮不周全的地方,如果沒有導(dǎo)師的督促指導(dǎo),以及一起工作的同學(xué)們的支持,想要完成這個(gè)設(shè)計(jì)是難以想象的。 %清除所有變量clc。 %學(xué)習(xí)因子2c12=。 %隨機(jī)初始化速度 endend%顯示群位置figure(1)for j=1:D if(rem(D,2)0) subplot((D+1)/2,2,j) else subplot(D/2,2,j) end plot(x(:,j),39。第39。)。) tInfo=strcat(39。 end title(tInfo)endfigure(3)%第一個(gè)圖subplot(1,2,1)%初始化種群個(gè)體(在此限定速度和位置)x1=x。 gbest1=pbest1(i)。 end gb1(i)=gbest1。ylabel(39。end%初始化種全局最有位置和 最優(yōu)解g2=1000*ones(1,D)。 gbest2=pbest2(j)。xlabel(39。endresult=sum。 %初始化群體最迭代次數(shù)c11=2。for i=1:N for j=1:D x(i,j)=randn。初始位置39。,char(floor(j/10)+48),char(rem(j,10)+48),39。) ylabel(39。,char(floor(j/10)+48),char(rem(j,10)+48),39。gbest1=1000。 endv1(j,:)=w*v1(j,:)+c11*rand*(p1(j,:)x1(j,:))+c21*rand*(g1x1(j,:))。迭代次數(shù)39。pbest2=ones(N,1)。 pbest2(j)=fitness(x2(j,:),D)。,c12,c22)。b)適應(yīng)度函數(shù)%適應(yīng)度函數(shù)()function result=fitness(x,D)sum=0。初始化群體個(gè)數(shù)D=10。 %設(shè)置精度(在已知最小值的時(shí)候用)%初始化種群個(gè)體(限定位置和速度)x=zeros(N,D)。粒子39。 if(j9) tInfo=strcat(39。grid on xlabel(39。 if(j9)