如何在CPLEX中将多维数组模型转换为元组结构？

Question

我想使用一个大型数据集(3100个需求位置)来解决我的设施选址问题。

约束之一是距离矩阵的大小。如果我使用二维数组来表示位置之间的距离，我会存储大量不必要的数据。(比如我不会在模型中使用的长距离，所以我添加了另一个约束，比如<= maxdist )

我没有使用二维数组，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)，我得不到解决方案？

谢谢你的建议...

    {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

将解决方案从一个问题移动到另一个答案：

我的新代码作为解决方案：

元组圆弧{ 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 ={ | in Distance :FROM in Supply & TO in Demand & MILES<=maxDist}；setof( ARC ) DCD_Arcs ={ < FROM，TO> | 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) XARCp*G2DIST*Ci;