C++ 中的优化问题不会产生与 CPLEX 中相同的结果-解网

问：

到目前为止，我直接在 IBM ILOG CPLEX Optimization Studio 中对优化问题进行了建模。

现在，我想尝试在使用 CPLEX API 的同时使用 C++ 来表述模型。

为此，我从一个非常简单的问题开始，只是为了学习 C++ 并感受如何使用 CPLEX API。

问题在 CPLEX 中如下所示：

range j = 1..3;

dvar int+ x[i];
dvar float+ y[j];

maximize -3*x[1] - 2*x[2] + 4*x[3] + y[1] - 3*y[2] - 2*y[3];

subject to {
  c1: -2*x[1] + 4*x[2] - x[3] + 3*y[1] - 2*y[2] + 3*y[3] <= -4;
  c2: -x[1] - 2*x[2] + 4*x[3] + 2*y[1] + 4*y[2] - 5*y[3] <= 2;
  c3: forall(i in i) x[i] <= 6;
}

现在，我只想用 C++ 编写相同的模型。这是我当前的代码：

#include <iostream>
#include "ilcplex//ilocplex.h";
using namespace std;

int main() {

    IloEnv env;
    IloModel Model(env);

    IloIntVar x1(env, 0, IloInfinity, "x1");
    IloIntVar x2(env, 0, IloInfinity, "x2");
    IloIntVar x3(env, 0, IloInfinity, "x3");
    IloNumVar y1(env, 0, IloInfinity, ILOFLOAT, "y1");
    IloNumVar y2(env, 0, IloInfinity, ILOFLOAT, "y2");
    IloNumVar y3(env, 0, IloInfinity, ILOFLOAT, "y3");

    Model.add(IloMaximize(env, -3 * x1 - 2 * x2 + 4 * x3 + y1 - 3 * y2 - 2 * y3));

    Model.add((-2 * x1 + 4 * x2 - x3 + 3 * y1 - 2 * y2 + 3 * y3) <= -4);
    Model.add((-x1 - 2 * x2 + 4 * x3 + 2 * y1 + 4 * y2 - 5 * y3) <= 2);
    Model.add(x1 <= 6);
    Model.add(x2 <= 6);
    Model.add(x3 <= 6);

    IloCplex cplex(Model);

    if (!cplex.solve()) {
        env.error() << "Optimization failed" << endl;
        throw(-1);
    }

    double obj = cplex.getObjValue();
    cout << "\n\n\t objective value: " << obj << endl;  

    cout << "x1 = " << cplex.getValue(x1) << endl;
    cout << "x2 = " << cplex.getValue(x2) << endl;
    cout << "x3 = " << cplex.getValue(x3) << endl;
    cout << "y1 = " << cplex.getValue(y1) << endl;
    cout << "y2 = " << cplex.getValue(y2) << endl;
    cout << "y3 = " << cplex.getValue(y3) << endl;
}

老实说，我已经看了一段时间了，我无法弄清楚为什么这两个程序不能产生相同的解决方案。

CPLEX 程序的目标值为 2.429，这是正确的解决方案。 C++ 程序打印出优化失败，这意味着它没有找到解决方案。

有人看到我犯的错误吗？

C++ 数学优化 CPLEX 混合整数规划运筹学

IloIntVar x1(env, 0, 6, "x1");
IloIntVar x2(env, 0, 6, "x2");
IloIntVar x3(env, 0, 6, "x3");
IloNumVar y1(env, 0, IloInfinity, ILOFLOAT, "y1");
IloNumVar y2(env, 0, IloInfinity, ILOFLOAT, "y2");
IloNumVar y3(env, 0, IloInfinity, ILOFLOAT, "y3");

Model.add(IloMaximize(env, -3 * x1 - 2 * x2 + 4 * x3 + y1 - 3 * y2 - 2 * y3));

Model.add((-2 * x1 + 4 * x2 - x3 + 3 * y1 - 2 * y2 + 3 * y3) <= -4);
Model.add((-x1 - 2 * x2 + 4 * x3 + 2 * y1 + 4 * y2 - 5 * y3) <= 2);
//Model.add(x1 <= 6);
//Model.add(x2 <= 6);
//Model.add(x3 <= 6);

即更改是在声明这些 x 变量时将它们的边界设置为 6，而不是通过约束单独设置。以下是修改后的模型的输出日志：

Version identifier: 20.1.0.0 | 2020-11-10 | 9bedb6d68
Found incumbent of value -6.000000 after 0.00 sec. (0.00 ticks)
Tried aggregator 1 time.
Reduced MIP has 2 rows, 6 columns, and 12 nonzeros.
Reduced MIP has 0 binaries, 3 generals, 0 SOSs, and 0 indicators.
Presolve time = 0.00 sec. (0.00 ticks)
Tried aggregator 1 time.
Reduced MIP has 2 rows, 6 columns, and 12 nonzeros.
Reduced MIP has 0 binaries, 3 generals, 0 SOSs, and 0 indicators.
Presolve time = 0.00 sec. (0.00 ticks)
MIP emphasis: balance optimality and feasibility.
MIP search method: dynamic search.
Parallel mode: deterministic, using up to 16 threads.
Root relaxation solution time = 0.00 sec. (0.00 ticks)

        Nodes                                         Cuts/
   Node  Left     Objective  IInf  Best Integer    Best Bound    ItCnt     Gap

*     0+    0                           -6.0000                           0.00%
*     0+    0                           -1.9524                           0.00%
      0     0        4.0000     1       -1.9524        4.0000        3  304.88%
*     0+    0                            1.9231        4.0000           108.00%
*     0+    0                            2.2857        4.0000            75.00%
      0     0        2.5000     2        2.2857    MIRcuts: 1        5    9.37%
*     0+    0                            2.4286        2.5000             2.94%
      0     0        cutoff              2.4286                      5     ---
Elapsed time = 0.05 sec. (0.05 ticks, tree = 0.01 MB, solutions = 5)

Mixed integer rounding cuts applied:  1

Root node processing (before b&c):
  Real time             =    0.05 sec. (0.05 ticks)
Parallel b&c, 16 threads:
  Real time             =    0.00 sec. (0.00 ticks)
  Sync time (average)   =    0.00 sec.
  Wait time (average)   =    0.00 sec.
                          ------------
Total (root+branch&cut) =    0.05 sec. (0.05 ticks)


         objective value: 2.42857
x1 = 4
x2 = -0
x3 = 5
y1 = 0.142857
y2 = 0
y3 = 2.85714

我原以为这应该在 CPLEX 的预解析之后生成一个非常相似的模型并给出相同的结果，但它显然做了一些不同的事情。我会将此报告给 IBM 进行进一步调查。

编辑：更多挖掘。看起来它不喜欢用 IloInfinity 的上限声明 IloIntVars。如果我将模型导出为 LP 文件，我会在这些变量上获得意想不到的边界，将它们全部设置为零：

\ENCODING=ISO-8859-1
\Problem name: IloCplex

Maximize
 obj1: - 3 x1 - 2 x2 + 4 x3 + y1 - 3 y2 - 2 y3
Subject To
 c1: - 2 x1 + 4 x2 - x3 + 3 y1 - 2 y2 + 3 y3 <= -4
 c2: - x1 - 2 x2 + 4 x3 + 2 y1 + 4 y2 - 5 y3 <= 2
 c3: x1 <= 6
 c4: x2 <= 6
 c5: x3 <= 6
Bounds
      x1 = 0
      x2 = 0
      x3 = 0
Generals
 x1  x2  x3 
End

如果我将声明更改为以下内容，它似乎可以正常工作：

IloNumVar x1(env, 0, IloInfinity, ILOINT, "x1");
IloNumVar x2(env, 0, IloInfinity, ILOINT, "x2");
IloNumVar x3(env, 0, IloInfinity, ILOINT, "x3");

C++ 中的优化问题不会产生与 CPLEX 中相同的结果

Optimization problem in C++ does not yield the same result as in CPLEX

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