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following spreadsheet中。我正在解决经典的 A x = b 问题,但需要注意的是 x 的所有组件都必须是非负的。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。我尝试使用Microsoft Solver Foundation或Solver SDK在C#中复制此内容。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,并且使用Solver SDK我总是回到状态“最佳”和所有0的解决方案,这绝对不是本地的最小。
以下是我的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这样的代理。
答案 0 :(得分:2)
另一种方法是将其表述为LP问题:
最小化 x
中元素的总和受 Ax&gt; = b
的约束根据其中一个LP样本,使用Solver Foundation制定这个应该相当简单。
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。使用Microsoft Solver Foundation,和最小化差异的平方和,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
如果我没有弄错,与GRG不同,Solver Foundation不能支持开箱即用的一般非线性约束,我相信你需要额外的插件来处理这些。对于您的问题,这当然不是问题。
为了完整性,要改为制定LP问题,请使用以下代码在 START NLP SECTION 和 END NLP SECTION 之间交换代码:
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));
将产生以下输出(请注意,在两种情况下目标函数不同,因此f
中的差异很大):
f = 2125502.27815564
x[0] = 2017159
x[1] = 75302.7580022821
x[2] = 27215.9247379241
x[3] = 5824.5954154355
x[4] = 0