我正在开发一个金融应用程序,需要IRR (in-built functionality of Excel)
计算,在这里找到了这么好的教程,在这里C
找到了这样的答案。C#
我实现了上面 C 语言的代码,但是当 IRR 为正时它给出了完美的结果。它不应该返回负值。而在 Excel=IRR(values,guessrate)
中,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,也希望它也返回负IRR,与Excel相同。但是我不熟悉 C#,所以我无法在 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,我已经实现了这些代码来帮助你们。
#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;
}
我附上了一些在 Excel 和上述 C 语言代码中不同的值的结果。
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