Question

我想使用大型数据集（3100个需求位置）来解决我的设施位置问题。

其中一个限制因素是距离矩阵的大小。如果我使用二维数组来获取位置之间的距离，我会存储大量不必要的数据。（就像我在模型中不会使用的长距离，所以我添加了另一个约束，如＆lt; = maxdist）

我没有使用2D数组，而是尝试使用以下元组，但是如果我不使用完整的距离矩阵（转换为元组）我没有得到解决方案？

感谢您的建议......

    {string} Supply = ...;      // Supply locations
    {string} DC = ...;          // Candidate facility locations
    {string} Demand = ...;      // Demand locations

    tuple Dist_Tup{
    string FROM;
    string TO;
    float MILES;
    };

    setof(Dist_Tup) DistanceTmp=...;
    setof(Dist_Tup) Distance = { <FROM,TO,MILES> | <FROM,TO,MILES> in DistanceTmp : FROM in Supply || FROM in DC};

    dexpr float TransportCost1 = sum(i in Supply , a in Alt , j in DC , p in Period, BB in Distance : BB.TO==j && BB.FROM==i) X[i][a][j][p]*Dist[BB]*C[i][j];

    //dexpr float TransportCost1 = sum(i in Supply , a in Alt , j in DC , p in Period) X[i][a][j][p]*G[i][a][j]*C[i][j];

Answer 1

将解决方案从问题转移到答案：

我的新代码作为解决方案：

tuple Arc { 
 string FROM;
 string TO;
} 
tuple Dist_Tup{
 string FROM;
 string TO;
 float MILES;
};
setof(Dist_Tup) Distance=...;
setof(Arc) SDC_Arcs = { <FROM,TO> | <FROM,TO,MILES> in Distance : FROM in Supply && TO in DC};
setof(Arc) SD_Arcs =  { <FROM,TO> | <FROM,TO,MILES> in Distance : FROM in Supply && TO in Demand && MILES<=maxDist};
setof(Arc) DCD_Arcs = { <FROM,TO> | <FROM,TO,MILES> in Distance : FROM in DC && TO in Demand && MILES<=maxDist};

dexpr float TransportCost1 = sum(i in Supply , a in Alt , j in DC , p in Period, DIST in Distance : DIST.FROM==i && DIST.TO==j, ARC in SDC_Arcs : ARC.FROM==i && ARC.TO==j) X[ARC][a][p]*G2[DIST]*C[i][j];

如何将多维数组模型转换为CPLEX中的元组结构？

1 个答案: