根据XIRR函数 openoffice文档（公式与excel中的相同），您需要在以下f（xirr）方程中求解XIRR变量：
您可以通过以下方式计算xirr值：
- 计算上述函数的导数 – > f’（xirr）
- 在有了
f(xirr)和f'(xirr)你可以用迭代牛顿法解决xirr值 – 着名的公式 – >
编辑
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); } }
顺便说一句，请记住，由于公式和/或牛顿法的限制，并非所有付款都会产生有效的XIRR！
干杯！
我从0x69的解决方案开始，但最终一些新的场景导致牛顿的方法失败。 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。 参考XIRR，是吗？ 有趣的讨论。
使用系统;
使用System.Collections.Generic;
使用System.Linq;
//请参阅以下文章：
// 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
//基于Excel doc的默认值
// http://office.microsoft.com/en-us/excel-help/xirr-function-HP010062387.aspx
命名空间Xirr
{
公共课程
{
private const DoublePersYear = 365.0;
private const int MaxIterations = 100;
private const double DefaultTolerance = 1E-6;
private const double DefaultGuess = 0.1;
private static readonly Func, Double> NewthonsMethod = cf => NewtonsMethodImplementation(cf, Xnpv, XnpvPrime); private static readonly Func, 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 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 cashFlow, Func, 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 cashFlow, Func, Double, Double> f, Func, 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 cashFlow, Func, 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 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 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, Double, Double> f, IEnumerable 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; } } }
}
