我正在开发一个财务应用并需要IRR (in-built functionality of Excel)
计算,并在C
here中找到了这样精彩的教程,并在C#
here中找到了这样的答案。
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。它应该是没有返回负值。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,似乎它遵循了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。
我正在编写上述链接中的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;
}
我已将结果附加到Excel和上述C语言代码中不同的值。
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
答案 0 :(得分:4)
由于low_rate和high_rate均为正数,因此无法获得负分。你必须改变:
#define LOW_RATE 0.01
,例如,
#define LOW_RATE -0.5