lingo 解0-1规划问题

简要复述要求:1、20个33元一次不等式,1个等式的可能结果。33个变量结果只能是0或1,且可以赋值某些变量,减少后期结果的数量。
lingo代码如下
model: sets: row/1..20/: b; !20个目标约束; col/1..33/: x; !33个变量; link(row,col):a; endsets !数据申明; data: b=4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4; a=0 1 1 0 1 0 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 1 0 1 1 0 1 0 0 0 1 0 1 0 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 0 1 0 0 1 1 1 0 1 1 0 0 1 1 0 0 1 0 1 0 0 1 1 0 0 1 0 1 0 1 1 0 0 0 1 0 1 1 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 0 0 1 1 1 0 1 0 1 0 0 0 1 1 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 0 1 1 1 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 1 0 0 1 0 0 0 0 1 1 0 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 1 0 1 1 0 0 0 0 1 0 1 0 1 1 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1 0 0 1 0 0 0 1 1 1 0 0 0 1 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 0 1 0 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 1 1 0 0 1 0 1 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 1 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 1 1 0 0 1 0 1 0 0 0 1 0 0 1 0 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0; enddata !目标函数; 6=@sum(col:x); @for(row(i):@sum(col(j):a(i,j)*x(j))<=b(i)); @for(col:@bin(x)); end
需要lingo破解软件的可以在我的邮箱留言:liuronghong123@foxmail.com
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