f=xlsread('air04',2,'B1:B8904');
Aeq=zeros(823,8904);
M=xlsread('air04',4);
for k=1:size(M,1);
i=M(k,1);
j=M(k,2);
Aeq(i,j)=1;
end;
beq=ones(823,1);
lb=zeros(8904,1);
ub=ones(8904,1);
cvx_begin
variable x(8904) binary
minimize( f' * x )
subject to
Aeq * x == beq;
cvx_end
Name Size Bytes Class Attributes
f 8904x1 71232 double
Name Size Bytes Class Attributes
Aeq 823x8904 58623936 double
Name Size Bytes Class Attributes
beq 823x1 6584 double
Error using cvx_sdpt3>solve (line 124)
SDTP3 does not support integer variables.
Error in cvxprob/solve (line 427)
[ x, status, tprec, iters ] = shim.solve( At, b, c, cones,
quiet, prec, solv.settings );
Error in cvx_end (line 78)
solve( prob );
Error in air04 (line 17)
cvx_end
Should I use variable x(n) binary to describe the binary constraint? What’s wrong with my program?
Thanks a lot!