CPLEX OPL 优化问题解决中运行时性能不一致

Question

我正在解决12.10版Cplex opl中的优化问题。该问题被表述为混合整数锥规划模型。我在一天 24 小时内分别执行代码，并相应地更改输入数据。优化过程随时间变化；对于某些时间，3 到 4 分钟内即可获得结果，而对于其他时间，最多需要 25 分钟才能获得最佳解决方案。

我的优化问题涉及确定具有 37 个开关的电力网络的最佳开关状态。主要目标是最大限度地减少网络损耗，同时保持网络的放射状结构，同时实施必要的重新配置。优化问题确定交换机的最佳状态。每小时应打开 5 个交换机。优化过程针对一天中的每个小时进行，考虑不同的负载条件以找出五个打开的开关。我观察到计算时间存在显着差异，有些小时在几分钟内即可产生结果，而另一些则需要 25 到 30 分钟

Answer 1

我鼓励您查看 CPLEX 文档中的“评估可变性”。

这将帮助您抵御好运气和坏运气。

这在 OPL 和其他 API 中可用

评估变异性

CPLEX 多次解决问题以评估可变性。

CPLEX 可以多次解决相同的混合整数问题，每次 使用不同的随机种子的时间。运用这个工具可以作为 模型随机扰动的替代品。事实上，锻炼身体 该工具可以帮助您评估类别的可变性 您的实例所属的问题。提示：你的问题一定是一个 以下类型的问题：

mixed integer linear program (MILP)
mixed integer quadratic program (MIQP)
mixed integer quadratically constrained program (MIQCP)

要调用此过程，请按照下列步骤操作：

First, enter or read your model, so that your problem is currently in memory and available to CPLEX.
Optionally, set CPLEX parameters in the usual way; these settings will be applied on every run.
Then invoke the following command, where n specifies the number of times that you want CPLEX to solve your problem for comparison.

注意：您还可以编写一些代码，例如 https://github.com/AlexFleischerParis/howtowithopl/blob/master/randomseedvariability.mod

    int n=30; // number of random seeds
    int d[1..n]; // duration
    int o[1..n]; // objective
    int iter[1..n]; // iterations
     
     // start of the model we want to test
     
     int Fixed        = 100;
    int NbWarehouses = 50;
    int NbStores     = 200;

    assert( NbStores > NbWarehouses );

    range Warehouses = 1..NbWarehouses;
    range Stores     = 1..NbStores;
    int Capacity[w in Warehouses] =
      NbStores div NbWarehouses +
      w % ( NbStores div NbWarehouses );
    int SupplyCost[s in Stores][w in Warehouses] =
      1 + ( ( s + 10 * w ) % 100 );
    dvar int Open[Warehouses] in 0..1;
    dvar float Supply[Stores][Warehouses] in 0..1;
    dexpr int TotalFixedCost = sum( w in Warehouses ) Fixed * Open[w];
    dexpr float TotalSupplyCost = sum( w in Warehouses, s in Stores )  SupplyCost[s][w] * Supply[s][w];
    minimize TotalFixedCost + TotalSupplyCost;

    subject to {
      forall( s in Stores )
        ctStoreHasOneWarehouse:
          sum( w in Warehouses )
            Supply[s][w] == 1;
      forall( w in Warehouses )
        ctOpen:
          sum( s in Stores )
            Supply[s][w] <= Open[w] * Capacity[w];
    }
     
    // end of the model we want to test


    main {

    thisOplModel.generate();


    var sum_d=0;
    var sum_o=0;;
    var sum_iter=0;

    writeln("seed objective iteration   runtime");

    for(var i=1;i<=thisOplModel.n;i++)
    {

        var opl=new IloOplModel(thisOplModel.modelDefinition);
        opl.generate();

        cplex.randomseed=i;

        
        
        
        var d1=new Date();
        cplex.solve();
        var d2=new Date();
        
        thisOplModel.d[i]=d2-d1;
        sum_d+=d2-d1;
        thisOplModel.d[i]=d2-d1;
        thisOplModel.o[i]=Opl.ftoi(Opl.round(cplex.getObjValue()));
        sum_o+=thisOplModel.o[i];
        thisOplModel.iter[i]=cplex.getNiterations();
        sum_iter+=thisOplModel.iter[i];
        writeln(i,"    ",thisOplModel.o[i],"     ",thisOplModel.iter[i],"       ",
        thisOplModel.d[i]/1000);
        cplex.clearModel();
    }    

    writeln("-----------------------------------------");
    writeln("average      ",sum_o/thisOplModel.n," ",
    sum_iter/thisOplModel.n," ",sum_d/thisOplModel.n/1000);
    writeln("std dev      ",Opl.standardDeviation(thisOplModel.o),"     ",
    Opl.standardDeviation(thisOplModel.iter),"     ",Opl.standardDeviation(thisOplModel.d)/1000);

    }