Rahul Patel - 11 months ago 50

iOS Question

I am developing a financial app and require

`IRR (in-built functionality of Excel)`

`C`

`C#`

I implemented code of the C language above, but it gives a perfect result when IRR is in positive. It is not returning a negative value when it should be. Whereas in Excel

`=IRR(values,guessrate)`

I have referred to code in above C# link too, and it seems that it follows good procedures and returns errors and also hope that it returns negative IRR too, the same as Excel. But I am not familiar with C#, so I am not able to implement the same code in Objective-C or C.

I am writing C code from the above link which I have implemented for helping you guys.

`#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;

}

I have attached the result for some value which are different in Excel and the above C language code.

`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

Answer

Since low_rate and high_rate are both positive, you're not able to get a negative score. You have to change:

```
#define LOW_RATE 0.01
```

to, for example,

```
#define LOW_RATE -0.5
```