我正在尝试使用OPL和IBM ILOG CPLEX中的软时间窗来设置一个旅行商问题来解决。毕竟我有Probelm作为安装的小版本，但找不到我的错误，所以我很无奈。
请原谅我在编写OPL时可能有点复杂的方法，我从未做过这样的事情。
TSP - 问题涉及1个仓库（x [1] [1]）和3个客户。从仓库的Start开始，必须访问所有三个客户，然后返回到仓库。
因此可以超过时间窗口（a [i]，b [i]）并低于此值。
w是早到的输出时间，m是迟到的输出时间。
基础数据位于.dat文件中。
我认为问题在于等式4c和4d，它应该消除子层。
因为我不幸完全无能为力。
非常感谢任何帮助。
提前致谢
迈克尔
OPL-Programm（.mod）：
//number of Nodes
int V=...;
// initialization driving Time between two nodes
int u[1..V][1..V]=...;
//initialization Time-Windows
int a[2..V]=...;
int b[2..V]=...;
//initialization panalty-costs (not used in objective function)
int c1[2..V]=...;
int c2[2..V]=...;
// Helping Variable for Linearization
int M=51000;
//initialization decision variable
dvar boolean x [1..V][1..V];
//initialization of d,w, m
//arrival Time
dvar int d[1..V];
// come too late
dvar int m[2..V];
//come too early // waitingtime
dvar int w[2..V];
//objective function
minimize
sum( i in 1..V, j in 1..V)
u[i][j]*x[i][j]
// + sum(i in 2..V )
// c1[i]*w[i]
// + sum(i in 2..V)
// c2[i]*m[i];
;
//constraints
subject to{
//1)
// in each account only one edge must arrive TSP
forall(i in 2..V) sum(j in 1..V)x[i][j]==1;
//2) 3)
//Ensures that the depot (1) is left and is approached.
sum(j in 1..V)x[1][j]==1;
sum(i in 1..V)x[i][1]==1;
//4)
//Ensures that every customer is visited and leave afterwards.
forall(l in 2..V)
sum(i in 1..V)x[i][l]-sum(j in 1..V)x[l][j] ==0;
//4c) und 4d)
//prevents subtours
forall(i in 2..V, j in 1..V)
(w[i]+d[i]+u[i][j])-(M*(1-x[i][j]))<=d[j];
forall(j in 2..V)
x[1][j]*u[1][j]+w[j]<=d[j];
//4a)
//earliest arrival time z [i] equal is greater than the lower limit of the customer time window.
forall(i in 2..V)
a[i]-w[i]<=d[i];
//4b)
//earliest arrival z [i] is equal to less than the upper limit of the customer's time slot.
forall(i in 2..V )
b[i]+m[i]>=d[i];
forall(i in 1..V)
d[i]>=0;
forall(i in 2..V)
w[i]>=0;
forall(i in 2..V)
m[i]>=0;
}
数据文件（.dat）：
//number of Nodes
V = 4;
// Traveltimes
u=[ [1000, 3, 5, 30] [3, 1000, 30, 9] [5, 30, 1000, 4] [30, 9, 4, 1000] ];
//Time window lower limit // [Customer 1 Customer 2 Customer 3]
a=[ 23 , 3, 10];
//Time window limit // [Customer 1 Customer 2 Customer 3]
b=[50, 4, 30];
//penalty costs // [Customer 1 Customer 2 Customer 3]
c1=[1, 1, 1];
//penalty costs // [Customer 1 Customer 2 Customer 3]
c2=[1, 1, 1];