XIRR计算

Question

如何使用C＃计算Excel的XIRR函数？

Answer 1

根据XIRR function openoffice文档（公式与excel中的公式相同），您需要在以下 f（xirr）等式中求解XIRR变量：

您可以通过以下方式计算xirr值：

1. 计算上述函数的导数 - ＆gt; f'（xirr）
2. 拥有f(xirr)和f'(xirr)后，您可以使用迭代Newton's method - 着名的公式 - ＆gt;来求解xirr值。
   

修改
我有一点时间，这里是 - 用于XIRR计算的完整C＃代码：

class xirr
    {
        public const double tol = 0.001;
        public delegate double fx(double x);

        public static fx composeFunctions(fx f1, fx f2) {
            return (double x) => f1(x) + f2(x);
        }

        public static fx f_xirr(double p, double dt, double dt0) {
            return (double x) => p*Math.Pow((1.0+x),((dt0-dt)/365.0));
        }

        public static fx df_xirr(double p, double dt, double dt0) {
            return (double x) => (1.0/365.0)*(dt0-dt)*p*Math.Pow((x+1.0),(((dt0-dt)/365.0)-1.0));
        }

        public static fx total_f_xirr(double[] payments, double[] days) {
            fx resf = (double x) => 0.0;

            for (int i = 0; i < payments.Length; i++) {
                resf = composeFunctions(resf,f_xirr(payments[i],days[i],days[0]));
            }

            return resf;
        }

        public static fx total_df_xirr(double[] payments, double[] days) {
            fx resf = (double x) => 0.0;

            for (int i = 0; i < payments.Length; i++) {
                resf = composeFunctions(resf,df_xirr(payments[i],days[i],days[0]));
            }

            return resf;
        }

        public static double Newtons_method(double guess, fx f, fx df) {
            double x0 = guess;
            double x1 = 0.0;
            double err = 1e+100;

            while (err > tol) {
                x1 = x0 - f(x0)/df(x0);
                err = Math.Abs(x1-x0);
                x0 = x1;
            }

            return x0;
        }

        public static void Main (string[] args)
        {
            double[] payments = {-6800,1000,2000,4000}; // payments
            double[] days = {01,08,16,25}; // days of payment (as day of year)
            double xirr = Newtons_method(0.1,
                                         total_f_xirr(payments,days),
                                         total_df_xirr(payments,days));

            Console.WriteLine("XIRR value is {0}", xirr);
        }
    }

顺便说一下，请记住，由于公式和/或牛顿方法的限制，并非所有付款都会产生有效的XIRR！

喝彩！

Answer 2

我从0x69的解决方案开始，但最终一些新方案导致牛顿方法失败。我创建了一个“智能”版本，当牛顿失败时，它使用Bisection Method（较慢）。

请注意我用于此解决方案的多个来源的内联引用。

最后，您无法在Excel中重现某些场景，因为Excel本身使用牛顿的方法。有关此问题的详细讨论，请参阅XIRR, eh?。

using System;
using System.Collections.Generic;
using System.Linq;


// See the following articles:
//   http://blogs.msdn.com/b/lucabol/archive/2007/12/17/bisection-based-xirr-implementation-in-c.aspx
//   http://www.codeproject.com/Articles/79541/Three-Methods-for-Root-finding-in-C
//   http://www.financialwebring.org/forum/viewtopic.php?t=105243&highlight=xirr
// Default values based on Excel doc 
//   http://office.microsoft.com/en-us/excel-help/xirr-function-HP010062387.aspx

namespace Xirr
{
    public class Program
    {
        private const Double DaysPerYear = 365.0;
        private const int MaxIterations = 100;
        private const double DefaultTolerance = 1E-6;
        private const double DefaultGuess = 0.1;

    private static readonly Func<IEnumerable<CashItem>, Double> NewthonsMethod =
        cf => NewtonsMethodImplementation(cf, Xnpv, XnpvPrime);

    private static readonly Func<IEnumerable<CashItem>, Double> BisectionMethod =
        cf => BisectionMethodImplementation(cf, Xnpv);

    public static void Main(string[] args)
    {
        RunScenario(new[]
            {
                // this scenario fails with Newton's but succeeds with slower Bisection
                new CashItem(new DateTime(2012, 6, 1), 0.01),
                new CashItem(new DateTime(2012, 7, 23), 3042626.18),
                new CashItem(new DateTime(2012, 11, 7), -491356.62),
                new CashItem(new DateTime(2012, 11, 30), 631579.92),
                new CashItem(new DateTime(2012, 12, 1), 19769.5),
                new CashItem(new DateTime(2013, 1, 16), 1551771.47),
                new CashItem(new DateTime(2013, 2, 8), -304595),
                new CashItem(new DateTime(2013, 3, 26), 3880609.64),
                new CashItem(new DateTime(2013, 3, 31), -4331949.61)
            });
        RunScenario(new[]
            {
                new CashItem(new DateTime(2001, 5, 1), 10000),
                new CashItem(new DateTime(2002, 3, 1), 2000),
                new CashItem(new DateTime(2002, 5, 1), -5500),
                new CashItem(new DateTime(2002, 9, 1), 3000),
                new CashItem(new DateTime(2003, 2, 1), 3500),
                new CashItem(new DateTime(2003, 5, 1), -15000)
            });
    }

    private static void RunScenario(IEnumerable<CashItem> cashFlow)
    {
        try
        {
            try
            {
                var result = CalcXirr(cashFlow, NewthonsMethod);
                Console.WriteLine("XIRR [Newton's] value is {0}", result);
            }
            catch (InvalidOperationException)
            {
                // Failed: try another algorithm
                var result = CalcXirr(cashFlow, BisectionMethod);
                Console.WriteLine("XIRR [Bisection] (Newton's failed) value is {0}", result);
            }
        }
        catch (ArgumentException e)
        {
            Console.WriteLine(e.Message);
        }
        catch (InvalidOperationException exception)
        {
            Console.WriteLine(exception.Message);
        }
    }

    private static double CalcXirr(IEnumerable<CashItem> cashFlow, Func<IEnumerable<CashItem>, double> method)
    {
        if (cashFlow.Count(cf => cf.Amount > 0) == 0)
            throw new ArgumentException("Add at least one positive item");

        if (cashFlow.Count(c => c.Amount < 0) == 0)
            throw new ArgumentException("Add at least one negative item");

        var result = method(cashFlow);

        if (Double.IsInfinity(result))
            throw new InvalidOperationException("Could not calculate: Infinity");

        if (Double.IsNaN(result))
            throw new InvalidOperationException("Could not calculate: Not a number");

        return result;
    }

    private static Double NewtonsMethodImplementation(IEnumerable<CashItem> cashFlow,
                                                      Func<IEnumerable<CashItem>, Double, Double> f,
                                                      Func<IEnumerable<CashItem>, Double, Double> df,
                                                      Double guess = DefaultGuess,
                                                      Double tolerance = DefaultTolerance,
                                                      int maxIterations = MaxIterations)
    {
        var x0 = guess;
        var i = 0;
        Double error;
        do
        {
            var dfx0 = df(cashFlow, x0);
            if (Math.Abs(dfx0 - 0) < Double.Epsilon)
                throw new InvalidOperationException("Could not calculate: No solution found. df(x) = 0");

            var fx0 = f(cashFlow, x0);
            var x1 = x0 - fx0/dfx0;
            error = Math.Abs(x1 - x0);

            x0 = x1;
        } while (error > tolerance && ++i < maxIterations);
        if (i == maxIterations)
            throw new InvalidOperationException("Could not calculate: No solution found. Max iterations reached.");

        return x0;
    }

    internal static Double BisectionMethodImplementation(IEnumerable<CashItem> cashFlow,
                                                         Func<IEnumerable<CashItem>, Double, Double> f,
                                                         Double tolerance = DefaultTolerance,
                                                         int maxIterations = MaxIterations)
    {
        // From "Applied Numerical Analysis" by Gerald
        var brackets = Brackets.Find(Xnpv, cashFlow);
        if (Math.Abs(brackets.First - brackets.Second) < Double.Epsilon)
            throw new ArgumentException("Could not calculate: bracket failed");

        Double f3;
        Double result;
        var x1 = brackets.First;
        var x2 = brackets.Second;

        var i = 0;
        do
        {
            var f1 = f(cashFlow, x1);
            var f2 = f(cashFlow, x2);

            if (Math.Abs(f1) < Double.Epsilon && Math.Abs(f2) < Double.Epsilon)
                throw new InvalidOperationException("Could not calculate: No solution found");

            if (f1*f2 > 0)
                throw new ArgumentException("Could not calculate: bracket failed for x1, x2");

            result = (x1 + x2)/2;
            f3 = f(cashFlow, result);

            if (f3*f1 < 0)
                x2 = result;
            else
                x1 = result;
        } while (Math.Abs(x1 - x2)/2 > tolerance && Math.Abs(f3) > Double.Epsilon && ++i < maxIterations);

        if (i == maxIterations)
            throw new InvalidOperationException("Could not calculate: No solution found");

        return result;
    }

    private static Double Xnpv(IEnumerable<CashItem> cashFlow, Double rate)
    {
        if (rate <= -1)
            rate = -1 + 1E-10; // Very funky ... Better check what an IRR <= -100% means

        var startDate = cashFlow.OrderBy(i => i.Date).First().Date;
        return
            (from item in cashFlow
             let days = -(item.Date - startDate).Days
             select item.Amount*Math.Pow(1 + rate, days/DaysPerYear)).Sum();
    }

    private static Double XnpvPrime(IEnumerable<CashItem> cashFlow, Double rate)
    {
        var startDate = cashFlow.OrderBy(i => i.Date).First().Date;
        return (from item in cashFlow
                let daysRatio = -(item.Date - startDate).Days/DaysPerYear
                select item.Amount*daysRatio*Math.Pow(1.0 + rate, daysRatio - 1)).Sum();
    }

    public struct Brackets
    {
        public readonly Double First;
        public readonly Double Second;

        public Brackets(Double first, Double second)
        {
            First = first;
            Second = second;
        }

        internal static Brackets Find(Func<IEnumerable<CashItem>, Double, Double> f,
                                      IEnumerable<CashItem> cashFlow,
                                      Double guess = DefaultGuess,
                                      int maxIterations = MaxIterations)
        {
            const Double bracketStep = 0.5;
            var leftBracket = guess - bracketStep;
            var rightBracket = guess + bracketStep;
            var i = 0;
            while (f(cashFlow, leftBracket)*f(cashFlow, rightBracket) > 0 && i++ < maxIterations)
            {
                leftBracket -= bracketStep;
                rightBracket += bracketStep;
            }

            return i >= maxIterations
                       ? new Brackets(0, 0)
                       : new Brackets(leftBracket, rightBracket);
        }
    }

    public struct CashItem
    {
        public DateTime Date;
        public Double Amount;

        public CashItem(DateTime date, Double amount)
        {
            Date = date;
            Amount = amount;
        }
    }
}

}

Answer 3

其他答案显示如何在C＃中实施 XIRR，但如果只需要计算结果，您可以直接在下面调用Excel的XIRR函数方式：

首先添加对Microsoft.Office.Interop.Excel的引用，然后使用以下方法：

    public static double Xirr(IList<double> values, IList<DateTime> dates)
    {
        var xlApp = new Application();

        var datesAsDoubles = new List<double>();
        foreach (var date in dates)
        {
            var totalDays = (date - DateTime.MinValue).TotalDays;
            datesAsDoubles.Add(totalDays);
        }

        var valuesArray = values.ToArray();
        var datesArray = datesAsDoubles.ToArray();

        return xlApp.WorksheetFunction.Xirr(valuesArray, datesArray);
    }

Answer 4

感谢位于Excel Financial Functions的nuget包的贡献者。它支持许多金融方法 - AccrInt，Irr，Npv，Pv，XIrr，XNpv等，

1. 安装并导入包。
2. 由于所有方法在Financial类中都是静态的，因此直接使用必需参数调用特定方法Financial.<method_name>。

示例：

using Excel.FinancialFunctions;

namespace ExcelXirr
{
    class Program
    {
        static void Main(string[] args)
        {
            List<double> valList =new List<double>();
            valList.Add(4166.67);
            valList.Add(-4166.67);
            valList.Add(-4166.67);
            valList.Add(-4166.67);
            List<DateTime> dtList = new List<DateTime>();
            dtList.Add(new DateTime(2014, 9, 1));
            dtList.Add(new DateTime(2014, 10, 1));
            dtList.Add(new DateTime(2014, 11, 1));
            dtList.Add(new DateTime(2014, 12, 1));
            double result = Financial.XIrr(valList, dtList);
            Console.WriteLine(result);
            Console.ReadLine();
        }
    }
}

结果与Excel相同。