Matlab實(shí)現(xiàn)Zoutendijk編程例子

minf(x)=x2+2x2+3x2+xx—2xx+xx—4x—6x1 2 3 12 13 23 1 2x+2x+x<4. 1 23x,x,x>0123取初始點(diǎn)x(i)=(0,0,0)t,通過(guò)兇@打@6編程實(shí)現(xiàn)求解過(guò)程。公用函數(shù)如下：1、function[f,x]=func%設(shè)置目標(biāo)函數(shù)symsx1x2x3;f=xr2+2*x2-2+3*x3-2+x1*x2-2*x1*x3+x2*x3-4*x1-6*x2；x=[x1,x2,x3];end2、functionf_val=fval(x0)%求目標(biāo)函數(shù)值x0=transpose(x0);[f,x]=func;f_val=subs(f,x,x0);end3、functions=diff_val(x0)%求目標(biāo)函數(shù)梯度[f,x]=func;grad=jacobian(f,x);s=subs(grad,x,x0);end4、functionh=fmin(x0,d0,vmax)%求函數(shù)最小值[f,x]=func;symsh;a=x0+h*d0;f_val=inline(subs(f,x,a));ifvmax==infmin_h=fminbnd(f_val,0,10000);elsemin_h=fminbnd(f_val,0,vmax);endh=min_h;endZoutendijk方法主函數(shù)function[X0,f_val]=zoutendijk(A,b,x0,Aeq,beq)%自定義函數(shù)diff_val(x0)作用是求所給函數(shù)在*0出的偏導(dǎo)數(shù)%自定義函數(shù)n21&0)作用是求所給函數(shù)在*0出的函數(shù)值formatlong;eps=;x0=transpose(x0)；%剛開(kāi)始給的x0為行向量[f,x]=func;sz=length(x0);[m,n]=size(A);%把人分解為人1,人2，其中人1為起作用約束fork=1:1:100A1=A;A2=A;b1=b;b2=b;fori=m:-1:1ifabs(A2(i,:)*x0-b2(i,:))<A2(i,:)=[];b2(i,:)=[];endendfori=m:-1:1ifabs(A1(i,:)*x0-b1(i,:))>=A1(i,:)=[];b1(i,:)=[];endendA1;A2;b1;b2;i2=rank(A2);AE=[A1;Aeq];[i1,j1]=size(AE);r=rank(AE);ifr<i1'行不滿秩'returnendifi2==0'無(wú)效'returnend%求解線性規(guī)劃問(wèn)題得到可行下降方向d0s=diff_val(x0);c=double(s);lb=-1*ones(sz,1);ub=ones(sz,1);k1=length(b1);k2=length(beq);p=zeros(k1,1);q=zeros(k2,1);[d0,mn,m1,m2,m3]=linprog(c,A1,p,Aeq,q,lb,ub);d0;mn;df=abs(s*d0);ifdf<'最優(yōu)解為'x0f_val=fval(x0)kreturnelse%進(jìn)行一維搜索，求f(x(k+1))的最小值b_=b2-A2*x0;d_=A2*d0;[dh,dl]=size(d_);ul=1;fori=1:1:dhifd_(i,:)>=0u=1;elseu=0;endul=ul*u;endul;b_;d_;vmax=inf;iful==0vmax=inf;elsefori=1:1:dhifd_(i,:)>0v=b_(i,:)/d_(i,:);ifv<vmaxvmax=v;endendendendendvmax;h=fmin(x0,d0,vmax);a=x0+h*d0;f_val=fval(a);x0=x0+h*d0;'s1^s1^s1^s1>%|>'X0=x0f_val=fval(x0)k^|cx|cx|cx|cx|cx|cx|cx|c^|c^|c^|c^|c^|cx^Cx^Cx^Cendend'TFt*toRATLAF?ht品thisVidti.■seeOiwas.睢rtad<rttti-neStarted?A=[L21：-]O0：0-10：00-L]k[H?？?]"：xQ=[Q0Q]:Aeq=[3.buq=U.label=L:>>zdllIendijkl&、)