我想使用列式建模解决cplex中的一个简单问题。 这是问题,
maximize 2x + 3y
subject to x<= 5
y<=2
x,y >=0
以下是我必须编写的代码来解决它:
public static void Model_1() {
try {
//create new model
IloCplex cplex = new IloCplex();
//define variables
IloNumVar x;
IloNumVar y;
IloObjective objective;
objective = cplex.addMaximize();
IloRange cons01;
IloRange cons02;
cons01 = cplex.addRange(0, 5, "c1");
cons02 = cplex.addRange(0, 2, "c1");
IloColumn new_col = cplex.column(objective, 2);
IloColumn new_col2 = cplex.column(objective,3);
new_col = new_col.and(cplex.column(cons01,1));
new_col2 = new_col2.and(cplex.column(cons02,1));
x = cplex.numVar(new_col, 0, Double.MAX_VALUE);
y = cplex.numVar(new_col, 0, Double.MAX_VALUE);
//solve model
if (cplex.solve()) {
System.out.println("obj = "+cplex.getObjValue());
System.out.println("x = "+cplex.getValue(x));
System.out.println("y = "+cplex.getValue(y));
}
else {
System.out.println("Model not solved");
}
cplex.end();
}
catch (IloException exc) {
exc.printStackTrace();
}
}
但我没有得到正确的解决方案。我在编写代码时犯了什么错误吗?
答案 0 :(得分:0)
尝试调试此类问题时,将模型导出为LP格式以确保正确生成它总是有用的。您可以在调用cplex.solve
之前添加以下代码行:
cplex.exportModel("model.lp");
如果你这样做,model.lp的内容如下所示:
Maximize
obj: 2 x1 + 2 x2
Subject To
c1: x1 + x2 - Rgc1 = 0
c1: - Rgc1 = 0
Bounds
0 <= Rgc1 <= 5
0 <= Rgc1 <= 2
End
这揭示了你在程序中犯的两个错别字。即,您应该替换:
cons02 = cplex.addRange(0, 2, "c1");
与
cons02 = cplex.addRange(0, 2, "c2");
而且,您应该替换:
y = cplex.numVar(new_col, 0, Double.MAX_VALUE);
与
y = cplex.numVar(new_col2, 0, Double.MAX_VALUE);
完成这两个更改后,model.lp看起来像:
Maximize
obj: 2 x1 + 3 x2
Subject To
c1: x1 - Rgc1 = 0
c2: x2 - Rgc2 = 0
Bounds
0 <= Rgc1 <= 5
0 <= Rgc2 <= 2
End
并且,您从程序中获得以下输出:
obj = 16.0
x = 5.0
y = 2.0