我正在解决一个最小化问题，用GLPK用轴平行线覆盖3d点。最初我设置了lb as 0和ub as 1。然后在获得GLPK结果后，我创建第一个fractional变量值0，然后将lb和ub值更改为：
i) lb will be 1 for the variable values 1
ii) ub will be 0 for the variable values 0
这是我的代码：
fid=fopen('output2.txt','w');
%input points are given below
numberOfPoints=size(points,1)
fprintf(fid,'numberofpoints=%d \n',numberOfPoints);
xy=unique(points(:,[1,2]),'rows');
yz=unique(points(:,[2,3]),'rows');
xz=unique(points(:,[1,3]),'rows');
L=size(xy,1)+size(yz,1)+size(xz,1); %number of lines
Axy=zeros(size(points,1),size(xy,1));
Ayz=zeros(size(points,1),size(yz,1));
Axz=zeros(size(points,1),size(xz,1));
for i=1:size(points,1)
for j=1:size(xy,1)
if(points(i,[1,2])==xy(j,:))
Axy(i,j)=1;
endif
end
end
Axy;
for i=1:size(points,1)
for j=1:size(yz,1)
if(points(i,[2,3])==yz(j,:))
Ayz(i,j)=1;
endif
end
end
Ayz;
for i=1:size(points,1)
for j=1:size(xz,1)
if(points(i,[1,3])==xz(j,:))
Axz(i,j)=1;
endif
end
end
Axz;
Acat=cat(2,Axy,Ayz);
A=cat(2,Acat,Axz); %constraint matrix
c=ones(1,size(A,2))'; %objective function co-efficients matrix
b=ones(1,size(A,1))'; %constraints' right hand side value matrix
lb=zeros(1,size(A,2)); %variable lower bound matrix
ub=ones(1,size(A,2)); %variable upper bound matrix
for i=1:size(A,1)
ctype(i)="L";
end
ctype;
for i=1:size(A,2)
vartype(i)="C";
end
vartype;
s = 1;
[xmin, fmin, status] = glpk (c, A, b, lb, ub, ctype, vartype, s);%GLPK function
fprintf(fid,'fmin=%0.8f \n',fmin);
fprintf(fid,'\n\n');
fprintf(fid,'status=%d \n',status);
fprintf(fid,'\n\n');
NumberofNodes = 1;
frac_value=intersect(find(xmin >0) , find(xmin < 1)) ; %positions of fractional variables
%fprintf(fid,'frac_value=%d \n',frac_value);
%fprintf(fid,'\n\n');
if (isempty(frac_value))
fprintf(fid,'no fractional value')
fprintf(fid,'\n\n');
break;
else
round_value=frac_value(1);
xmin_SP1=xmin;
ub1=ub;
fprintf(fid,'ub1=%d \n',ub1);
fprintf(fid,'\n');
lb1=lb;
fprintf(fid,'lb1=%d\n',lb1);
fprintf(fid,'\n');
xmin_SP1(round_value)=0; %making the first fractional value 0
NumberofNodes =NumberofNodes +1;
% fprintf(fid,'xmin_SP1=%0.8f \n',xmin_SP1);
% fprintf(fid,'\n\n');
one_value_SP1=find ( xmin_SP1== 1 ); %variables with value 1
% fprintf(fid,'one_value_SP1=%d \n',one_value_SP1);
% fprintf(fid,'\n\n');
zero_value_SP1=find ( xmin_SP1==0); %variables with value 0
% fprintf(fid,'zero_value_SP1=%d \n',zero_value_SP1);
% fprintf(fid,'\n\n');
ub1(zero_value_SP1)=0 ; %changing ub for 0 values
fprintf(fid,'ub1=%d \n',ub1);
fprintf(fid,'\n');
% fprintf(fid,'numberofub=%d \n',size(ub,2));
% fprintf(fid,'\n');
lb1(one_value_SP1)=1 ; %changing lb for 1 values
fprintf(fid,'lb1=%d\n',lb1);
fprintf(fid,'\n');
% fprintf(fid,'numberoflb=%d \n',size(lb,2));
% fprintf(fid,'\n');
[xminSP1, SP1min, status] = glpk (c, A, b, lb1, ub1, ctype, vartype, s) ; %GLPK function
%size(xmin_SP1,1);
fprintf(fid,'xmin_SP1=%0.8f \n',xminSP1);
fprintf(fid,'\n\n');
%fprintf(fid,'SP1min=%0.8f \n',SP1min);
%fprintf(fid,'\n\n');
fprintf(fid,'status=%d \n',status);
fprintf(fid,'\n\n');
LB1=SP1min
end
fclose(fid);
这是我的输入点和这些点的输出值：
points=
1 1 1
2 1 1
4 1 1
1 2 1
2 2 1
3 2 1
4 2 1
2 3 1
3 3 1
4 3 1
3 4 1
4 4 1
1 1 2
3 1 2
4 1 2
1 2 2
2 2 2
3 2 2
4 2 2
1 3 2
2 3 2
4 3 2
1 4 2
2 4 2
3 4 2
4 4 2
1 1 3
2 1 3
3 1 3
4 1 3
1 2 3
2 2 3
3 2 3
4 2 3
2 3 3
3 3 3
4 3 3
1 4 3
2 4 3
3 4 3
1 1 4
2 1 4
3 1 4
4 1 4
1 2 4
2 2 4
3 2 4
4 2 4
1 3 4
2 3 4
3 3 4
1 4 4
3 4 4
numberofpoints=53
fmin=16.00000000
status=0
initial ub:
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
initial lb:
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
changed ub for variable values 0:
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=0
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=1
ub1=0
ub1=1
ub1=0
ub1=1
ub1=1
ub1=0
ub1=1
ub1=0
ub1=1
ub1=0
ub1=1
ub1=1
ub1=0
ub1=1
ub1=1
ub1=0
ub1=0
ub1=1
ub1=0
ub1=0
ub1=1
ub1=1
ub1=0
ub1=0
ub1=0
ub1=0
ub1=1
ub1=0
ub1=1
ub1=0
ub1=1
ub1=0
changed lb for variable values 1:
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=0
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=1
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
lb1=0
variable values showing NA as status=10(no primal feasible solution)
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
xmin_SP1=NA
*******Result for SP1 node 2********
status=10
我真的不明白为什么在我更改no primal feasible solution status=10时显示1st fractional variable value 0。任何有关这方面的帮助将非常感谢。
答案 0 :(得分：0)
我可以解决我的问题。这是因为某些变量具有值0.000000000000000027773或0.99999999999999999。因此，如果我进行一些预处理，例如制作0.000000000000027773值0或0.99999999999999999 1，那么GLPK会提供可行的解决方案