# Difference between revisions of "Haskell Quiz/Internal Rate of Return/Solution Dolio"

From HaskellWiki

m (category) |
(use until instead of iterate) |
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Line 9: | Line 9: | ||

secant :: (Double -> Double) -> Double -> Double |
secant :: (Double -> Double) -> Double -> Double |
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− | secant f delta = fst |
+ | secant f delta = fst $ until err update (0,1) |

where |
where |
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update (x,y) = (x - (x - y)*(f x)/(f x - f y), x) |
update (x,y) = (x - (x - y)*(f x)/(f x - f y), x) |
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− | err (x,y) = abs (x - y) |
+ | err (x,y) = abs (x - y) < delta |

npv :: Double -> [Double] -> Double |
npv :: Double -> [Double] -> Double |

## Latest revision as of 13:41, 9 February 2008

My solution for this quiz uses the secant method, which is quite easy to implement.

```
import Data.Function
import Numeric
import System.Environment
secant :: (Double -> Double) -> Double -> Double
secant f delta = fst $ until err update (0,1)
where
update (x,y) = (x - (x - y)*(f x)/(f x - f y), x)
err (x,y) = abs (x - y) < delta
npv :: Double -> [Double] -> Double
npv i = sum . zipWith (\t c -> c / (1 + i)**t) [0..]
main = do (s:t) <- getArgs
let sig = read s
cs = map read t
putStrLn . ($"") . showFFloat (Just sig) $ secant (flip npv cs) (0.1^sig)
```

The resulting program expects the first argument to be the number of digits to be displayed after the decimal point, while the rest are the yearly income. For instance:

./IRR 4 -100 30 35 40 45 0.1709