我正在製定單人循環賽的時間表。它在 CPLEX 中建模為 MIP，並且在我的解決方案池中，目前有四個解決方案具有相同的最佳目標值。我想獲得這四個解決方案中的每一個，以便可以單獨打印和檢查它們。那可能嗎？
// 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];
}
main{
thisOplModel.generate();
cplex.populate();
writeln("number of solutions: " + cplex.getSolnPoolNsolns());
writeln("Average objective value is: " + cplex.getSolnPoolMeanObjValue());
}
是的，在腳本編寫中，您可以循環到解決方案池中的所有解決方案。
見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();
}
}
}
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