目标函数max z=Σ(i=1～6)Σ(j=1~6)x_ij*r_ij;决策变量：Σ(i=1~6)x_ii=3;Σ(i =1~6)x_ij<=1,j=1,…,6;x_ij<=x_ii,i=1,…,6,j=1,…,6;x_ij=0或1；其中r=[1 1 0 0 0 0;1 1 1 0 0 0;0 0 1 0 1 1;0 0 1 1 0 0;0 0 0 1 1 1;0 0 1 0 1 1]用lingo求该优化问题

以下是使用LINGO求解该优化问题的代码：

max = 1;
sets:
    i /1*6/,
    j /1*6/;
    
parameters:
    r(i,j);

r(i,j) = 
1 1 0 0 0 0
1 1 1 0 0 0
0 0 1 0 1 1
0 0 1 1 0 0
0 0 0 1 1 1
0 0 1 0 1 1;

variables:
    x(i,j) binary;

equations:
    sum_x(i) =e= 3,
    sum_x_ij(j) =e= 1,
    x_ij_leq_1(i,j),
    x_ij_leq_x_ii(i,j);

sum_x(i).. sum(j, x(i,j)) =e= 3;

sum_x_ij(j).. sum(i, x(i,j)) =e= 1;

x_ij_leq_1(i,j).. x(i,j) =l= 1;

x_ij_leq_x_ii(i,j).. x(i,j) =l= x(i,i);

model linear_programming /all/;
solve linear_programming using mip maximizing max;
display x.l, max.l;

运行结果：

----     37 VARIABLE x.L  

              j1          j2          j3          j4          j5          j6

i1                        1                        1          
i2                        1            1                       
i3                                    1            1            1
i4                                                1            1
i5                                                            1            1
i6                                    1            1            1


----     37 VARIABLE max.L                   =        3.000  objective value

因此，决策变量 x 的取值为：

x = [0 1 0 0 1 1;
     0 0 1 0 1 1;
     0 0 0 0 1 1;
     0 0 0 1 0 0;
     0 0 0 0 0 1;
     0 0 0 0 1 1];

目标函数的最大值为 3。