如何在C#中模拟Microsoft Excel的Solver功能(GRG Nonlinear)？

Question

我有约束的非线性优化问题.它可以使用Solver加载项在Microsoft Excel中解决,但我无法在C#中复制它.

我的问题显示在以下电子表格中.我正在解决经典的A x = b问题,但需要注意的是x的所有分量都必须是非负的.因此,不使用标准线性代数,而是使用具有非负约束的Solver,最小化平方差的总和,并获得合理的解.我试图使用Microsoft Solver Foundation或Solver SDK在C#中复制它.然而,我似乎无法与他们在任何地方,因为有了MSF,我无法弄清楚如何定义目标和使用Solver SDK我总是回到状态"最佳"和所有0的解决方案,这绝对不是一个本地最小.

这是我的Solver SDK代码:

static double[][] A = new double[][] { new double[] { 1, 0, 0, 0, 0 }, new double[] { 0.760652602, 1, 0, 0, 0 }, new double[] { 0.373419404, 0.760537565, 1, 0, 0 }, new double[] { 0.136996731, 0.373331934, 0.760422587, 1, 0 }, new double[] { 0.040625222, 0.136953801, 0.373244464, 0.76030755, 1 } };
static double[][] b = new double[][] { new double[] { 2017159 }, new double[] { 1609660 }, new double[] { 837732.8125 }, new double[] { 330977.3125 }, new double[] { 87528.38281 } };

static void Main(string[] args)
{
    using(Problem problem = new Problem(Solver_Type.Minimize, 5, 0))
    {
        problem.VarDecision.LowerBound.Array = new double[] { 0.0, 0.0, 0.0, 0.0, 0.0 };
        problem.VarDecision.UpperBound.Array = new double[] { Constants.PINF, Constants.PINF, Constants.PINF, Constants.PINF, Constants.PINF };

        problem.Evaluators[Eval_Type.Function].OnEvaluate += new EvaluateEventHandler(SumOfSquaredErrors);

        problem.ProblemType = Problem_Type.OptNLP;

        problem.Solver.Optimize();

        Optimize_Status status = problem.Solver.OptimizeStatus;

        Console.WriteLine(status.ToString());
        foreach(double x in problem.VarDecision.FinalValue.Array)
        {
            Console.WriteLine(x);
        }
    }
}

static Engine_Action SumOfSquaredErrors(Evaluator evaluator)
{
    double[][] x = new double[evaluator.Problem.Variables[0].Value.Array.Length][];
    for(int i = 0; i < x.Length; i++)
    {
        x[i] = new double[1] { evaluator.Problem.Variables[0].Value.Array[i] };
    }

    double[][] b_calculated = MatrixMultiply(A, x);

    double sum_sq = 0.0;
    for(int i = 0; i < b_calculated.Length; i++)
    {
        sum_sq += Math.Pow(b_calculated[i][0] - b[i][0], 2);
    }
    evaluator.Problem.FcnObjective.Value[0] = sum_sq;

    return Engine_Action.Continue;
}

static double[][] MatrixMultiply(double[][] left, double[][] right)
{
    if(left[0].Length != right.Length)
    {
        throw new ArgumentException();
    }

    double[][] sum = new double[left.Length][];
    for(int i = sum.GetLowerBound(0); i <= sum.GetUpperBound(0); i++)
    {
        sum[i] = new double[right[i].Length];
    }

    for(int i = 0; i < sum.Length; i++)
    {
        for(int j = 0; j < sum[0].Length; j++)
        {
            for(int k = 0; k < right.Length; k++)
            {
                sum[i][j] += left[i][k] * right[k][j];
            }
        }
    }

    return sum;
}

我没有Microsoft Solver Foundation的任何代码,因为我不认为目标函数可以写在一行中,并且它不允许像Solver SDK这样的委托.

Answer 1

一种替代方法是将其表述为 LP 问题：

最小化x中元素的总和

服从Ax >= b

使用 Solver Foundation 基于 LP 样本之一来制定这一点应该相当简单。

7 月 5 日更新

上述方法看起来也过于复杂，但这可能是由于 Frontline Solver API 造成的。使用 Microsoft Solver Foundation并最小化平方差之和，执行以下程序：

private static void Main(string[] args)
{
    var solver = SolverContext.GetContext();
    var model = solver.CreateModel();

    var A = new[,]
        {
            { 1, 0, 0, 0, 0 }, 
            { 0.760652602, 1, 0, 0, 0 }, 
            { 0.373419404, 0.760537565, 1, 0, 0 },
            { 0.136996731, 0.373331934, 0.760422587, 1, 0 },
            { 0.040625222, 0.136953801, 0.373244464, 0.76030755, 1 }
        };
    var b = new[] { 2017159, 1609660, 837732.8125, 330977.3125, 87528.38281 };

    var n = A.GetLength(1);
    var x = new Decision[n];
    for (var i = 0; i < n; ++i)
        model.AddDecision(x[i] = new Decision(Domain.RealNonnegative, null));

    // START NLP SECTION
    var m = A.GetLength(0);
    Term goal = 0.0;
    for (var j = 0; j < m; ++j)
    {
        Term Ax = 0.0;
        for (var i = 0; i < n; ++i) Ax += A[j, i] * x[i];
        goal += Model.Power(Ax - b[j], 2.0);
    }
    model.AddGoal(null, GoalKind.Minimize, goal);
    // END NLP SECTION

    var solution = solver.Solve();
    Console.WriteLine("f = {0}", solution.Goals.First().ToDouble());
    for (var i = 0; i < n; ++i) Console.WriteLine("x[{0}] = {1}", i, x[i].GetDouble());
}

生成以下解决方案，该解决方案应与链接的 Excel 工作表中的解决方案一致：

f = 254184688.179922
x[0] = 2017027.31820845
x[1] = 76226.6063397686
x[2] = 26007.3375581303
x[3] = 1.00650383558278E-07
x[4] = 4.18546775823669E-09

如果我没记错的话，与 GRG 不同，Solver Foundation 无法支持开箱即用的一般非线性约束，我相信您将需要额外的插件来处理这些。对于你的问题来说，这当然不是问题。

为了完整起见，要表述 LP 问题，请使用以下代码交换START NLP SECTION和END NLP SECTION之间的代码：

    var m = A.GetLength(0);
    var constraints = new Term[m];
    for (var j = 0; j < m; ++j)
    {
        Term Ax = 0.0;
        for (var i = 0; i < n; ++i) Ax += A[j, i] * x[i];
        model.AddConstraint(null, constraints[j] = Model.GreaterEqual(Ax, b[j]));
    }
    model.AddGoal(null, GoalKind.Minimize, Model.Sum(x));

这将产生以下输出（请注意，两种情况下的目标函数不同，因此 中的差异很大f）：

f = 2125502.27815564
x[0] = 2017159
x[1] = 75302.7580022821
x[2] = 27215.9247379241
x[3] = 5824.5954154355
x[4] = 0