Can solve the problem that the number of nodes is within 20.
The number of nodes is out of 20 and cannot be solved.
No problem with the model.
I feel that it is mainly used in =>.
Use in the model in load weight constraints and time window constraints =>.
I feel that there may be some problems in this logic.
MOD as below:
/*********************************************
* OPL 12.9.0.0 Model
* Author: 86131
* Creation Date: 2019年5月24日 at 下午10:17:34
*********************************************/
int n = …;
int K = …;
{int} A = asSet(1..n);//配送网点与客户点
{int} N = asSet(2..n);
{int} Trucks = asSet(1..K);//一组同质的车辆(人员)
float T =…;//单次收派的最大工作时间
float Z =…;//一个极大正数
float Q[Trucks] = …;//载重量上限
float d[N] = …;//收件重量
float v[N] = …;//派件重量
float CC[1..n*n*K] = …;
float c[i in 1..n, j in 1..n,k in 1..K] = CC[k+K*(j-1)+n*K*(i-1)];//行走成本（时间）
float t[N][Trucks] =…;//客户 i的服务时间
float a[N] = …;//客户 i要求的最早到达时间
float b[N] = …;//客户 i要求的最晚到达时间*/
dvar boolean x[A][A][Trucks];//车辆 k 是否访问过客户 i 和客户 j 所构成的弧
dvar float+ q[A][Trucks];//车辆 k 离开客户i时所装载重量
dvar float+ w[A][Trucks];//收派员 k 达到客户 i的时间,原文文中是k，这里符号冲突了
dexpr float time[k in Trucks] =sum(i,j in A:j!=i)(c[i][j][k]*x[i][j][k])+sum(i in A,j in N:j!=i)(t[j][k]*x[i][j][k]);
/************************************************/
// Objective
minimize sum (i,j in A:i!=j,k in Trucks) c[i][j][k]*x[i][j][k];
subject to {
forall(i in N)
ct1:sum(j in A:j!=i,k in Trucks)(x[i][j][k])==1;
forall(j in N)
ct2:sum(i in A:i!=j,k in Trucks)(x[i][j][k])==1;
forall(k in Trucks)
ct3:sum(j in N)(x[1][j][k])==1;
forall(k in Trucks)
ct4:sum(i in N)(x[i][1][k])==1;
forall(j in N)
ct5:sum(k in Trucks)(x[1][j][k])<=1;
forall(i in N)
ct6:sum(k in Trucks)(x[i][1][k])<=1;
forall(k in Trucks,h in N)
ct7:sum(i in A:i!=h)(x[i][h][k])-sum(j in A:j!=h)(x[h][j][k])==0;//改了A
forall(k in Trucks)
ct8:time[k]<=T;
forall(k in Trucks)
ct9:q[1][k]==sum(j in N)(v[j]*sum(i in A:i!=j)(x[i][j][k]));
forall(i in A,j in N:j!=i,k in Trucks)
ct10:x[i][j][k]==1=>(q[j][k]-q[i][k])==-(d[j]+v[j]);//不用考虑回原点
//forall(i in A,j in N:j!=i,k in Trucks)
//ct10:(q[j][k]-q[i][k])>=-(d[j]+v[j])-Z*(1-x[i][j][k]);//不用考虑回原点
forall(i in A,j in A:j!=i,k in Trucks)
ct11:x[i][j][k]==1=>q[i][k]<=Q[k];//之前有问题，q意思是离开客户i的载重量
forall(i in N,j in A:j!=i,k in Trucks)
ct12:x[i][j][k]==1=>maxl(0,-d[i])<=q[i][k];
forall(j in N,k in Trucks)
ct13:x[1][j][k]==1=>w[j][k] >= w[1][k]+c[1][j][k];//这里 w[1][k]表示离开原点1的时间
forall(i in N,j in N:j!=i,k in Trucks)
ct14:x[i][j][k]==1=>w[j][k] >= w[i][k]+t[i][k]+c[i][j][k];
//forall(j in N,k in Trucks)
//ct13:w[j][k] >= w[1][k]+c[1][j][k]-(1-x[1][j][k])*Z;//这里 w[1][k]表示离开原点1的时间
//forall(i in N,j in N:j!=i,k in Trucks)
//ct14:w[j][k] >= w[i][k]+t[i][k]+c[i][j][k]-(1-x[i][j][k])*Z;//这里 w[1][k]表示离开原点1的时间
forall(i in A,j in N:j!=i,k in Trucks)
ct15:x[i][j][k]==1=>a[j]<=w[j][k];
forall(i in A,j in N:j!=i,k in Trucks)
ct16:x[i][j][k]==1=>w[j][k]<=b[j];
ct18:sum(i in A,k in Trucks)x[i][i][k]==0;
}
Personal feeling problems mainly appear in the load weight constraint ct10-ct12 and time window constraint ct13-ct16.
I don’t know if => is correct, if not, how to represent the above constraints.
In addition, with similar constraints such as adding an infinite number Z, the model does not apply.
Thanks for solving my problem.
Source: mTSP with time window