我正在Java应用程序中使用Cplex建立一个多目标优化模型。但是,当它求解模型时,它会过度补偿一个变量,而其余变量则为0。我可以用什么来获得更加分散的解决方案?
int[] x = {49,43,43,44,48,49,51,54,51,52,57,59};
double[] y = {10, 12, 13.2, 22.7, 17.1, 16.5, 14.87, 12, 16.5, 14.8, 12, 11.5};
int[] z = {59, 59, 57, 57, 53, 53, 52, 51, 51, 50, 50, 50};
int totalVacations = 73;
try {
IloCplex model = new IloCplex();
int size = 12;
IloNumVarType varType = IloNumVarType.Int;
double[] lb = new double[size];
double[] ub = new double[size];
IloNumVarType[] varTypes = new IloNumVarType[size];
for (int i = 0; i < lb.length ; i++) {
lb[i] = 0.0;
ub[i] = Double.MAX_VALUE;
varTypes[i] = varType;
}
IloNumVar[] varUsed = model.numVarArray(size, lb, ub, varTypes);
for (int i = 0; i < varUsed.length; i++) {
model.add(varUsed[i]);
}
IloNumExpr[] objArray = new IloNumExpr[size];
for (int i = 0; i < objArray.length; i++) {
IloObjective next = model.maximize();
IloNumExpr exprVar = varUsed[i];
double setValue = z[i] - x[i] - y[i];
next.setExpr(model.diff(setValue, exprVar));
next.setSense(IloObjectiveSense.Maximize);
objArray[i] = next.getExpr();
}
model.add(model.maximize(model.staticLex(objArray)));
model.addEq(totalVacations, model.sum(varUsed), "c1");
if (model.solve()) {
double[] results = model.getValues(varUsed);
for (int i = 0; i < results.length; i++) {
System.out.println(results[i]);
}
}
System.out.println(model.toString());
model.end();
} catch (IloException e) {
System.err.println("Concert exception caught: " + e);
}
这些值是 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 73.0
当我希望它更加分散时(虽然不均匀)。在这种情况下有什么建议吗?
答案 0 :(得分:2)
让我使用OPL来向您展示如何做:
您写了Java的等价物
int totalVacation=73;
range r=1..12;
int x[r] = [49,43,43,44,48,49,51,54,51,52,57,59];
float y[r] = [10, 12, 13.2, 22.7, 17.1, 16.5, 14.87, 12, 16.5, 14.8, 12, 11.5];
float z[r] = [59, 59, 57, 57, 53, 53, 52, 51, 51, 50, 50, 50];
dvar int varUsed[r] in 0..totalVacation;
maximize sum(i in r) (z[i] - x[i] - y[i]-varUsed[i]);
subject to
{
c1:sum(i in r) varUsed[i]==totalVacation;
}
给出
varUsed = [73 0 0 0 0 0 0 0 0 0 0 0];
因为客观上没有关于公平的事情。
但是如果您将目标从...更改
maximize sum(i in r) (z[i] - x[i] - y[i]-varUsed[i]);
到
maximize staticLex(sum(i in r) (z[i] - x[i] - y[i]-varUsed[i]),
-max(j in r) varUsed[j]+min(j in r) varUsed[j]);
然后您添加第二个字典目的以最大程度地提高公平性,然后您将得到
varUsed = [7 6 6 6 6 6 6 6 6 6 6 6];