整数规划分支定界法MATLAB 程序1.这种方法绝对能都解出答案，而且答案正确function[x,val]=fzdj(n,f,a,b,aeq,beq,lb,ub) x=zeros(n,1);x1=zeros(n,1);m1=2;m2=1;[x1,val1]=linprog(f,a,b,aeq,beq,lb,ub);if (x1==0) x=x1;val=val1;elseif (round(x1)==x1) x=x1;val=val1;else e1={0,a,b,aeq,beq,lb,ub,x1,val1}; e(1,1)={e1};zl=0; zu=-val1;while (zu~=zl)for c=1:1:m2if (m1~=2)if (cell2mat(e{m1-1,c}(1))==1) e1={1,[],[],[],[],[],[],[],0}; e(m1,c*2-1)={e1};e(m1,c*2)={e1}; continue;end;end; x1=cell2mat(e{m1-1,c}(8)); x2=zeros(n,1);s=0;s1=1;s2=1; lb1=cell2mat(e{m1-1,c}(6)); ub1=cell2mat(e{m1-1,c}(7)); lb2=cell2mat(e{m1-1,c}(6)); ub2=cell2mat(e{m1-1,c}(7));for d=1:1:nif (abs((round(x1(d))-x1(d)))>0.0001)&(s==0) s=1;lb1(d)=fix(x1(d))+1;if (a*lb1<=b)s1=0;end;ub2(d)=fix(x1(d));if (a*lb2<=b)s2=0;end;end;end;e1={s1,a,b,aeq,beq,lb1,ub1,[],0};e2={s2,a,b,aeq,beq,lb2,ub2,[],0};e(m1,c*2-1)={e1};e(m1,c*2)={e2};end;m1=m1+1;m2=m2*2;for c=1:1:m2if (cell2mat(e{m1-1,c}(1))==0)[x1,val1]=linprog(f,cell2mat(e{m1-1,c}( 2)),cell2mat(e{m1-1,c}(3)),cell2mat(e{m1-1,c}(4)),cell2 mat(e{m1-1,c}(5)),cell2mat(e{m1-1,c}(6)),cell2mat(e{m1-1,c}(7)));e1={cell2mat(e{m1-1,c}(1)),cell2mat(e{m1-1,c}(2)),cell2mat(e{m1-1,c}(3)),cell2mat(e{m1-1,c}( 4)),cell2mat(e{m1-1,c}(5)),cell2mat(e{m1-1,c}(6)),cell2mat(e{m1-1,c}(7)),x1,val1};e(m1-1,c)={e1};end;z=val1;if ((-z)<(-zl))e1={1,[],[],[],[],[],[],[],0};e(m1-1,c)={e1};elseif (abs(round(x1)-x1)<=0.0001)zl=z;end;end;for c=1:1:m2if (cell2mat(e{m1-1,c}(1))==0)zu=cell2mat(e{m1-1,c}(9));end;end;for c=1:1:m2if (-cell2mat(e{m1-1,c}(9))>(-zu))zu=cell2mat(e{m1-1,c}(9));end;end;end;for c=1:1:m2if (cell2mat(e{m1-1,c}(1))==0)&(cell2mat(e{m1-1,c}(9))==zu) x=cell2mat(e{m1-1,c}(8)); end;end;val=zu;end;2.这种方法是课本上的程序，但是不能解题，希望高手能将它改进function [x,y]=IntLp(f,G ,h,Geq,heq,lb,ub,x,id,options)%整数线性规划分支定界法，可求解全整数线性或混合整数线性规划%y=min f'*x subjectto:G*x<=h Geq*x=heq x 为全整数%数或混合整数列向量%用法% [x,y]=IntLp(f,G ,h)% [x,y]=IntLp(f,G ,h,Geq,heq)[x,y]=IntLp(f,G ,h,Geq,heq,lb,ub)[x,y]=IntLp(f,G ,h,Geq,heq,lb,ub,x)% [x,y]=IntLp(f,G ,h,Geq,heq,lb,ub,x,id)% [x,y]=IntLp(f,G ,h,Geq,heq,lb,ub,x,id,options) %参数说明% x:最优解列向量；y:目标函数最小值；f:目标函数系数列向量% G :约束不等式条件系数矩阵；h :约束不等式条件右端列向量% Gep：约束等式条件系数矩阵；heq:约束等式条件右端列向量% lb:解的下界列向量(Default:」nf)% ub:解得上界列向量(Default:inf)% x:迭代初值列向量；% id:整数变量指标列向量，1—整数，0 —实数(default")% options 的设置请参见optimset 或linprog%例min z=x1+4x2%s.t. 2x1+x2<=8% x1+2x2>=6% x1,x2>=0 且为整数%先将x1+x2>=6 化为-x1-2x2<=-6 %[x,y]=IntLp([1;4],[2 1;-1 -2],[8;-6],[],[],[0;0])global upper opt c x0 A b Aeq beq ID options;if nargin<10,options=optimset({});options.Display='off';rgeScale='off';endif nargin<9,id=ones(size(f));endif nargin<8,x=[];endif nargin<7|isempty(ub),ub=inf*ones(size(f));endif nargin<6|isempty(lb),lb=zeros(size(f));endif nargin<5,heq=[];endif nargin<4,Geq=[];end upper=inf;c=f;x0=x;A=G;b=h;Aeq=Geq;beq=heq;ID=id;ftemp=IntLp(lb(:),ub(:)); x=opt;y=upper;%以下为子函数function ftemp=IntLp(vlb,vub) global upper opt c x0 A b Aeq beq ID options; [x,ftemp,how]=linprog(c,A,b,Aeq,beq,vlb,vub,x0,options); if how<=0return;end;if ftemp-upper>0.00005%in order to avoid error return;end;if max(abs(x.*ID-round(x.*ID)))<0.00005if upper-ftemp>0.00005%in order to avoid error opt=x';upper=ftemp;return;else opt=[opt;x']; return;end;end; notintx=find(abs(x-round(x))>0.00005);%in order to avoid error intx=fix(x);tempvlb=vlb;tempvub=vub;if vub(notintx(1,1),1)>=intx(notintx(1,1),1)+1tempvlb(notintx(1,1),1)=intx(notintx(1,1),1)+1; ftemp=IntLp(tempvlb,vub);end;if vlb(notintx(1,1),1)<=intx(notintx(1,1),1) tempvub(notintx(1,1),1)=intx(notintx(1,1),1);ftemp=IntLp(vlb,tempvub);end;3.这种方法是我网上搜的，也不能解题，希望高手改进function [y,fval]=BranchBound(c,A,b,Aeq,beq)NL=length(c);UB=inf;LB=-inf;FN=[0];AA(1)={A};BB(1)={b};k=0;flag=0;while flag==0;[x,fval,exitFlag]=linprog(c,A,b,Aeq,beq);if (exitFlag == -2) | (fval >= UB)FN(1)=[];if isempty(FN)==1flag=1;elsek=FN(1);A=AA{k};b=BB{k};endelsefor i=1:NLif abs(x(i)-round(x(i)))>1e-7 kk=FN(end); FN=[FN,kk+1,kk+2];temp_A=zeros(1,NL); temp_A(i)=1; temp_A1=[A;temp_A]; AA(kk+1)={temp_A1};b1=[b;fix(x(i))]; BB(kk+1)={b1}; temp_A2=[A;-temp_A];AA(kk+2)={temp_A2}; b2=[b;-(fix(x(i))+1)]; BB(kk+2)={b2}; FN(1)=[];k=FN(1); A=AA{k}; b=BB{k}; break;endendif (i==NL) & (abs(x(i)-round(x(i)))<=1e-7)UB=fval;y=x;FN(1)=[];if isempty(FN)==1 flag=1;elsek=FN(1);A=AA{k};b=BB{k};endendendend y=round(y); fval=c*y;以上程序都是我网上搜到的，还有自己从课本上自己敲的，第后两个不能，也许是程序有问题，也许是我不会调用，希望高手指点，应用这三个程序！！！个程序能成功解出答案，帮助更多朋友们能够。