CPLEX OPL错误RUNTIME:未键入lloType

Question

这很令人困惑；我不知道问题出在哪里。有人知道问题是什么吗？我得到了这个错误，"；opl不键入运行时错误"；，当我在CPLEX中运行代码时，我的目标函数中的一些表达式。我正在写一篇关于调度问题设计的论文，以便它适用于A2是空载跳闸，我遇到了一个错误，我无法用字符串输出结果，请帮我复习代码，我将非常感激。我的代码如下：

`int numLocomotive=...;
range RangeLocomotive=1..numLocomotive;
{string} Pij=...;
{string} Node=...;
{string} Locomotive=...;{string} Vd=...;
{string} Va=...;
{string} Vs=...;
{string} Vf=...;
{string} Exp= Va union Vs;
{string} Exp= Va union Vs;
{string} Exp= Va union Vs;
{string} V= Vd union Va union Vs;
int ckm=...;
int cloc=...;
tuple A{
string I;
string j;
}
{A} A2 ={<i,j> | ordered i,j in Node};
int dis=...;
dvar boolean x[<i,j> in A2];
dvar boolean s[Locomotive];
dvar boolean q[Node][Locomotive];

execute PRE_SETUP
{
cplex.epgap = 0.001;
cplex.tilim = 21600;
}
dexpr float Totalcost = sum(<i,j> in A2)(dis*x[<i,j>])*ckm + sum(k in Locomotive)s[k]*cloc;
minimize Totalcost;
subject to{
ct01:
forall(j in Vd)
sum(<i,j> in A2) x[<i,j>] == 1;
ct02:
forall(i in Exp)
sum(<i,j> in A2) x[<i,j>] == 1;
ct03:
forall(j in Vf)
sum(<i,j> in A2) x[<i,j>] == numLocomotive;   
ct04:
forall(<i,j> in A2, k in Locomotive)
s[k] + x[<i,j>] == 1;
ct05:
forall(i in Node, k in Locomotive)
q[i][k] == 1;
ct06:
forall(i in V)
sum(k in Locomotive, i in Node) q[i][k] == 1;
ct07:
forall(i in Node, j in Node, k in Locomotive: k in Pij)
q[i][k] == q[j][k];
forall(i in Node, j in Node, k in Locomotive: k not in Pij)  
q[i][k] + q[j][k] == 0;  
ct08:
forall(<i,j> in A2, k in Locomotive)
q[j][k] >= q[i][k] - (1-x[<i,j>]);
}    

我的数据

`numLocomotive=2;
ckm=10;
cloc=300;
Node={"S","K","V","Vi","Va","B","E"};
Locomotive={"L1","L2"};
Vd={"S","K","V"};
Va={"V","Vi","Va","B"};
Vs={"S";"K"};
Vf={"E"};
Pij={<L1 L2> <L1 L2> <L1 L2> <L1 L2>};

A1
i  S   K   V   S 
j  V   Vi  Va  B
A2
i  S   S   S    k   k   k
j  S   Vi   K   S   K   V
Dis   50  45  40  56  50   59
ckm=10;
cloc=300;  ` 

Answer 1

.mod

int numLocomotive=...;
range RangeLocomotive=1..numLocomotive;
{string} Node=...;
{string} Locomotive=...;{string} Vd=...;
{string} Va=...;
{string} Vs=...;
{string} Vf=...;
{string} Exp= Va union Vs;
{string} V= Vd union Va union Vs;
int ckm=...;
int cloc=...;
tuple A{
string I;
string j;
}
{A} Pij=...;
{string} ij={k.I | k in Pij} union {k.j | k in Pij};

{A} A2 ={<i,j> | ordered i,j in Node};
int dis=...;
dvar boolean x[<i,j> in A2];
dvar boolean s[Locomotive];
dvar boolean q[Node][Locomotive];

execute PRE_SETUP
{
cplex.epgap = 0.001;
cplex.tilim = 21600;
}
dexpr float Totalcost = sum(<i,j> in A2)(dis*x[<i,j>])*ckm + sum(k in Locomotive)s[k]*cloc;
minimize Totalcost;
subject to{
ct01:
forall(j in Vd)
sum(<i,j> in A2) x[<i,j>] == 1;
ct02:
forall(i in Exp)
sum(<i,j> in A2) x[<i,j>] == 1;
ct03:
forall(j in Vf)
sum(<i,j> in A2) x[<i,j>] == numLocomotive;   
ct04:
forall(<i,j> in A2, k in Locomotive)
s[k] + x[<i,j>] == 1;
ct05:
forall(i in Node, k in Locomotive)
q[i][k] == 1;
ct06:
forall(i in V)
sum(k in Locomotive, i in Node) q[i][k] == 1;
ct07:
forall(i in Node, j in Node, k in Locomotive: k in ij)
q[i][k] == q[j][k];
forall(i in Node, j in Node, k in Locomotive: k not in ij)  
q[i][k] + q[j][k] == 0;  
ct08:
forall(<i,j> in A2, k in Locomotive)
q[j][k] >= q[i][k] - (1-x[<i,j>]);
} 

.dat

numLocomotive=2;
ckm=10;
cloc=300;
Node={"S","K","V","Vi","Va","B","E"};
Locomotive={"L1","L2"};
Vd={"S","K","V"};
Va={"V","Vi","Va","B"};
Vs={"S","K"};
Vf={"E"};
Pij={<L1 L2> <L1 L2>, <L1 L2> <L1 L2>};
dis=2;

工作良好