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我正尝试在C#中使用极限优化例程创建风险平价投资组合。在C#中使用极端优化的风险均值投资组合优化
我主要是在我买他们之前先试试他们是否喜欢他们(我是个学生,所以钱很紧)。
我的想法是实施这种称为风险平价的新型投资组合优化。它基本上说,为了使您的投资组合多样化,您应该为每个组件分配相同的风险。
运行np1.Solve()时出现空错误,我不明白为什么。我认为其他一切都是由Extreme Optimization计算的。
1.我做错了什么?
2.有没有更快的方法来做这个优化,我不知道?
3.如果您不了解EO库,但是可以用C#中的其他东西来实现此功能,请您注释一下如何解决此问题?
顺便说一句,关于投资组合构造的细节在距离函数的评论中,以防您感兴趣。
最好的问候,
爱德华
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using Extreme.Statistics;
using Extreme.Mathematics;
using Extreme.Mathematics.Optimization;
namespace TestingRiskParityOptimization
{
class Program
{
static void Main(string[] args)
{
NonlinearProgram np1 = new NonlinearProgram(2);
Func<Vector, double> distance = DistanceFunction;
np1.ObjectiveFunction = distance;
np1.InitialGuess = Vector.CreateConstant(2, 1.0/((double)2));
np1.AddNonlinearConstraint(x => x[0] + x[1], ConstraintType.GreaterThanOrEqual, 0);
Vector solution = np1.Solve();
Console.WriteLine("Solution: {0:F6}", solution);
Console.WriteLine("Optimal value: {0:F6}", np1.OptimalValue);
Console.WriteLine("# iterations: {0}", np1.SolutionReport.IterationsNeeded);
Console.Write("Press Enter key to exit...");
Console.ReadLine();
}
private static double DistanceFunction(Vector Weights)
{
Matrix Sigma = Matrix.Create(new double[,] {
{0.1, 0.2},
{0.2, 0.4}
});
// if VarP = Weights' * CovarMatrix * Weights and VolP = sqrt(VarP)
// Then the marginal contribution to risk of an asset is the i-th number of
// Sigma*Weights*VolP
// And thus the contribution to risk of an asset is simply Weights . (Sigma*Weights/VarP)
// we need to find weights such that Weights (i) * Row(i) of (Sigma*Weights/VarP) = 1/N
// that is we want to minimize the distance of row vector (Weights (i) * Row(i) of (Sigma*Weights/VarP)) and vector 1/N
double Variance = Vector.DotProduct(Weights, Sigma * Weights);
Vector Beta = Sigma * Weights/Variance;
for (int i = 0; i < Beta.Length; i++)
{
// multiplies row of beta by weight to find the percent contribution to risk
Beta[i] = Weights[i] * Beta[i];
}
Vector ObjectiveVector = Vector.CreateConstant(Weights.Length, 1.0/((double)Weights.Length));
Vector Distance = Vector.Subtract(Beta, ObjectiveVector);
return Math.Sqrt(Vector.DotProduct(Distance, Distance));
}
}
}
我认为[email protected]对你来说是一个更好的地方。 – 2012-07-15 22:22:55
乍一看,该功能看起来不错。如果您使用任意向量调用函数本身,它会返回一个double值还是抛出? – 2012-07-16 17:01:49
它抛出:/我做了其他的东西,我用了Nelder-mead的算法,我只是想在短期内接受它...后来,如果我觉得它太慢,我总是可以改变它或代码我自己的实现... – 2012-07-16 22:05:27