It is confusing; I do not what the problem is. Does anyone know what the problem is?I get this error,"opl not to type runtime error", for some expression in my objective function when I am running the code in CPLEX.i'm working on a thesis with a scheduling problem design so that it's suitable for A2 is deadhead trip, I'm getting an error that i can't output the result with string, please help me review the code, I'll be very grateful My code is as follow:
`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>]);
}
My data
`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; `
CodePudding user response:
.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;
work fine