Hi 晓辉,
I understand from your question that even though you specify the type of each element to be integer for the matrix 'x', you get decimal values in the solution.
This happens because when you set the data type constaint to integer, the output actually has a tolerance value by which the solution can deviate from an integer value. Simply put, if the tolerance value is 't', and your solution variable is 's', then the value of 's' can range from 'i' +/- 't', where 'i' is any integer.
This tolerance value is specified by the parameter 'IntegerTolerance', which can be set or changed using 'optimoptions'. The default value of 'IntegerTolerance' is 1e-5, and it can be set from 1e-6 to 1e-3 in MATLAB R2023b. I have updated your code by setting this tolerance to 1e-6, and the output shows that most of the values are now integers. Also, I have taken a random matrix ranging from -10 to 10 as 'a'.
a = -10 + 20*rand(14,14);
a(isnan(a))=0; %把不确定值NaN替换为0
M=10^7; w=ones(14)*M;
for i=1:14
for j=1:14
if i~=j, w(i,j)=sum(a(:,i).*a(:,j)); end
end
end
n=15; w(n,n)=M; prob=optimproblem;
x=optimvar('x',n,n,'Type','integer','LowerBound',0,'UpperBound',1);
u=optimvar('u',n,'LowerBound',0) %序号变量
prob.Objective=sum(sum(w.*x));
prob.Constraints.con1=[sum(x,2)==1; sum(x,1)'==1; u(1)==0];
con2 = optimconstr(n*(n-1)); k=0;
for i=1:n
for j=2:n
k=k+1; con2(k)=u(i)-u(j)+n*x(i,j)<=n-1;
end
end
prob.Constraints.con2 = con2;
prob.Constraints.con3 = [1<=u(2:end)'; u(2:end)'<=14];
options = optimoptions(prob);
options.IntegerTolerance = 1e-06;
[sol, fval, flag]=solve(prob)
xx=sol.x; [i,j]=find(xx);
fprintf('xij=1对应的行列位置如下：\n')
ij=[i'; j'];
Here is the screenshot of the output:
You can learn more about 'optimoptions' and other parameters available for solver of type 'intlinprog' from the below documentations:
Furthermore, MATLAB R2024a has an updated range for 'IntegerTolerance' from 1e-10 to 1e-3, so you can consider upgrading your version if you need more precision. Here is the documentation for the same.
Hope this helps!
