如何将 CPLEX 中的所有解决方案获取到 MIP

Question

更新：现在我得到了解决方案的数量，但是当我尝试将它们拿出来时，它只是给了我相同的时间表。我在下面添加了 Alex 建议的代码。例如，它给了我 4 种不同的解决方案，但是当我编写 X 矩阵时，它们是相同的。谁能帮我这个？我想要四种不同的解决方案。

我正在制定单循环锦标赛的时间表。 它在 CPLEX 中被建模为 MIP，在我的解决方案池中，目前有四个具有相同最佳目标值的解决方案。 我想获得这四种解决方案中的每一种，以便可以单独打印和检查它们。这可能吗？

// Create Parameters:
 {string} G1 = ...; // Set of teams in first group
 {string} G2 = ...; // Set of teams in second group
 
 {string} Teams = G1 union G2;
 
 tuple Match {string team1; string team2;}
 
 {Match} Matches_G1 = {<t1,t2>| ordered t1,t2 in G1};
 {Match} Matches_G2 = {<t1,t2>| ordered t1,t2 in G2};
 {Match} MD1 = ...;
 {Match} MD2 = ...;
 {Match} MD3 = ...;
 {Match} M = Matches_G1 union Matches_G2; //All matches for the two groups
 
 {Match} matchForTeam[t in Teams] = {m| m in M : m.team1 == t || m.team2 == t}; //List of all teams
 
 {string} S =...; //Set of stadiums
 
 {string} T = ...; //Set of kick off times
 
 {string} D = ...; //Set of kick off days
 
 int K[D][S][T] = ...; //Predetermined schedule between stadium and kickoff time
 
 float VT[M][T] = ...; //Value of match if played on Matchday M at Time T according to TV distribution

 // Decision Variables:
 
 dvar int X[M][S][T] in 0..1; // if match M is played at time T
 
 dvar int Dist; //Object function for distribution
 
 

 //////////// OBJECTIVE FUNCTION ///////////////
 
 maximize 
    Dist;
 
 //////////// CONSTRAINTS ///////////////
 
subject to{ 

Dist == sum(m in M, s in S, t in T) (VT[m][t])*X[m][s][t];


 //A match can only be played one time
 forall(m in M)
    sum(s in S, t in T) X[m][s][t] == 1;
    
//Simultaneous Kickoff on matchday 3
 sum(s in S)X[<"A1", "A4">][s]["22.00"] == sum(s in S)X[<"A2", "A3">][s]["22.00"];
     
//only one match on possible kick off times at matchday 1
 forall(t in T : t != "18.00")
   sum(s in S, m in MD1) X[m][s][t]==1;
//only one match on possible kick off times at matchday 2
 forall(t in T : t != "18.00")
   sum(s in S, m in MD2) X[m][s][t]==1;
//two matches per possible kick off times at matchday 3
 forall(t in T : t in {"18.00", "22.00"})
   sum(s in S, m in MD3) X[m][s][t]==2;
//One match per stadium on matchday 1
 forall(s in S)
    sum(m in MD1, t in T: t != "18.00") X[m][s][t] == 1;
 //One match per stadium on matchday 2
 forall(s in S)
   sum(m in MD2, t in T: t != "18.00") X[m][s][t] == 1;  
 //one match per stadium on matchday 3
forall(s in S)
 sum(m in MD3, t in T: t in {"18.00", "22.00"}) X[m][s][t] == 1;    
 
    
 //Each team can play at most two matches per stadium
forall(i in Teams, s in S)
 sum(t in T, m in matchForTeam[i]) X[m][s][t] <= 2;
  
 //Each team can play at most two matches per kickoff time
 forall(i in Teams, t in T)
  sum(s in S, m in matchForTeam[i]) X[m][s][t] <= 2;
  

 forall(s in S, t in T, m in MD1)
  X[m][s][t] <= K["1"][s][t];
 forall(s in S, t in T, m in MD2)
  X[m][s][t] <= K["2"][s][t];
 forall(s in S, t in T, m in MD3)
  X[m][s][t] <= K["3"][s][t];
  
 }
 
 
execute{
    
    writeln("schedule: ", X);
    var cd = new IloOplOutputFile("resbi2.txt");
        for(var m in M)
            for(var s in S)
                for(var t in T)
                    cd.writeln(thisOplModel.X[m][s][t]);
                    
            cd.close();
    
    
}
main{
  cplex.solnpoolintensity=4; 
  cplex.solnpoolagap=0;
  thisOplModel.generate(); 
  cplex.solve(); 
  
  if (cplex.populate()) {
          var nsolns = cplex.solnPoolNsolns; 
  writeln("number of solutions: ", nsolns);
 
  writeln("average object value: ", cplex.getSolnPoolMeanObjValue());
  writeln();
  
   for (var s=0; s<nsolns; s++) {
            thisOplModel.setPoolSolution(s); 
            
            var cd = new IloOplOutputFile("resAB" +s+".txt");
            cd.writeln(thisOplModel.X);
            cd.close();
            thisOplModel.postProcess();
            
          }
        }
}  
  
  

Answer 1

是的，在脚本中，您可以循环进入解决方案池中的所有解决方案。

见https://github.com/AlexFleischerParis/zooopl/blob/master/zooseveral.mod

    int nbKids=300;
    float costBus40=500;
    float costBus30=400;
     
    dvar int+ nbBus40;
    dvar int+ nbBus30;
     
    //minimize
     //costBus40*nbBus40  +nbBus30*costBus30;
     
    subject to
    {
     40*nbBus40+nbBus30*30>=nbKids;
    }

    execute
    {
    writeln("nbBus40 = ",nbBus40," and nbBus30 = ",nbBus30," and the cost is ",costBus40*nbBus40  +nbBus30*costBus30);
    }

     main {
    cplex.solnpoolintensity=4;

        thisOplModel.generate();
        cplex.solve();
        if (cplex.populate()) {
          var nsolns = cplex.solnPoolNsolns;
          
          
          writeln("Number of solutions found = ",nsolns);
          writeln();
          for (var s=0; s<nsolns; s++) {
            thisOplModel.setPoolSolution(s);
            thisOplModel.postProcess();
          }
        }
    }