在 Objective-C 中以编程方式计算 IRR（内部利率回报）和 NPV

我正在开发一个金融应用程序并需要IRR (in-built functionality of Excel)计算并发现了如此好的教程C here http://www.codeproject.com/Tips/461049/Internal-Rate-of-Return-IRR-Calculation?msg=4616844#xx4616844xx以及这样的答案C# here https://stackoverflow.com/questions/5179866/xirr-calculation/6348722#6348722.

我实现了上面的C语言代码，但是当IRR为正时，它给出了完美的结果。它没有在应该返回负值的时候返回负值。而在 Excel 中=IRR(values,guessrate)对于某些值也会返回负 IRR。

我也参考了上面 C# 链接中的代码，看起来它遵循良好的程序并返回错误，并且也希望它也返回负 IRR，与 Excel 相同。但我对 C# 不熟悉，所以无法在 Objective-C 或 C 中实现相同的代码。

我正在从上面的链接编写 C 代码，我已经实现了它来帮助你们。

#define LOW_RATE 0.01
#define HIGH_RATE 0.5
#define MAX_ITERATION 1000
#define PRECISION_REQ 0.00000001
double computeIRR(double cf[], int numOfFlows)
{
    int i = 0, j = 0;
    double m = 0.0;
    double old = 0.00;
    double new = 0.00;
    double oldguessRate = LOW_RATE;
    double newguessRate = LOW_RATE;
    double guessRate = LOW_RATE;
    double lowGuessRate = LOW_RATE;
    double highGuessRate = HIGH_RATE;
    double npv = 0.0;
    double denom = 0.0;
    for (i=0; i<MAX_ITERATION; i++)
    {
        npv = 0.00;
        for (j=0; j<numOfFlows; j++)
        {
            denom = pow((1 + guessRate),j);
            npv = npv + (cf[j]/denom);
        }

        /* Stop checking once the required precision is achieved */
        if ((npv > 0) && (npv < PRECISION_REQ))
            break;
        if (old == 0)
            old = npv;
        else
            old = new;
        new = npv;
        if (i > 0)
        {
            if (old < new)
            {
                if (old < 0 && new < 0)
                    highGuessRate = newguessRate;
                else
                    lowGuessRate = newguessRate;
            }
            else
            {
                if (old > 0 && new > 0)
                    lowGuessRate = newguessRate;
                else
                    highGuessRate = newguessRate;
                }
        }
        oldguessRate = guessRate;
        guessRate = (lowGuessRate + highGuessRate) / 2;
        newguessRate = guessRate;
    }
    return guessRate;
}

我附上了一些值的结果，这些值在 Excel 和上面的 C 语言代码中是不同的。

 Values:             Output of Excel: -33.5%
 1 = -18.5,          Output of C code: 0.010 or say (1.0%)
 2 =  -18.5,
 3 = -18.5,
 4 = -18.5,
 5 = -18.5,
 6 =  32.0

Guess rate: 0.1

由于 low_rate 和 high_rate 均为正数，因此您无法获得负分。你必须改变：

#define LOW_RATE 0.01

例如，

#define LOW_RATE -0.5