Loan Amortization
May 12, 2009
Kate was struggling with her task, so I decided to help by showing her a complete version of the program. I wrote in awk; I didn’t want to give away the whole answer by writing in Java, but awk is similar enough to be genuinely helpful. Here is my solution:
awk -v bal=10000 -v rate=7 -v term=36 ' BEGIN {
rate = rate / 12 / 100
pmt = bal * rate / (1 - (1+rate) ^ -term)
printf " N PRIN INT BALANCE\n"
printf "-- ------ ------ --------\n"
printf "%2d %8.2f\n",0,bal
for (n=1; n<term; n++) {
intr = round(rate * bal, 2)
prin = pmt - intr
bal = bal - prin
printf "%2d %6.2f %6.2f %8.2f\n", n, prin, intr, bal }
printf "%2d %6.2f %6.2f %8.2f\n", term, bal, pmt - bal, 0 }
function round(x, n) { return int(x * 10^n + 0.5) / 10^n } '
The output looks like this:
N PRIN INT BALANCE
-- ------ ------ --------
0 10000.00
1 250.44 58.33 9749.56
2 251.90 56.87 9497.66
3 253.37 55.40 9244.29
4 254.84 53.93 8989.45
5 256.33 52.44 8733.12
6 257.83 50.94 8475.28
7 259.33 49.44 8215.95
8 260.84 47.93 7955.11
9 262.37 46.40 7692.74
10 263.90 44.87 7428.84
11 265.44 43.33 7163.40
12 266.98 41.79 6896.42
13 268.54 40.23 6627.88
14 270.11 38.66 6357.77
15 271.68 37.09 6086.09
16 273.27 35.50 5812.81
17 274.86 33.91 5537.95
18 276.47 32.30 5261.48
19 278.08 30.69 4983.40
20 279.70 29.07 4703.70
21 281.33 27.44 4422.37
22 282.97 25.80 4139.40
23 284.62 24.15 3854.78
24 286.28 22.49 3568.50
25 287.95 20.82 3280.55
26 289.63 19.14 2990.91
27 291.32 17.45 2699.59
28 293.02 15.75 2406.57
29 294.73 14.04 2111.84
30 296.45 12.32 1815.39
31 298.18 10.59 1517.21
32 299.92 8.85 1217.29
33 301.67 7.10 915.62
34 303.43 5.34 612.19
35 305.20 3.57 306.99
36 306.99 1.78 0.00
A similar loan calculator is available from Bret Whissel at http://www.bretwhissel.net/cgi-bin/amortize.
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[…] Praxis – Loan Amortization By Remco Niemeijer Today’s Programming Praxis problem is about loan amortization. For those of you who, like me, had no idea […]
My Haskell solution (see http://bonsaicode.wordpress.com/2009/05/12/programming-praxis-loan-amortization/ for a version with comments):
import Text.Printf amortize :: Float -> Float -> Int -> [(Int, Float, Float, Float)] amortize _ _ 0 = [] amortize b r l = (l - 1, prn, int, b') : amortize b' r (l - 1) where int = b * r prn = int / (1 - (1 + r) ^^ (-l)) - int b' = b - prn amortization :: Float -> Float -> Int -> IO () amortization b r l = do putStrLn "Left Principal Interest Balance" mapM_ table $ (l, 0, 0, b) : amortize b (r / 12 / 100) l where table (m, p, i, t) = printf "%4d %9.2f %8.2f %9.2f\n" m p i t main :: IO () main = amortization 10000 7 36A Solution in PLT Scheme.
This is the first Scheme program I had ever wrote; and its present form is the result of many kind folks helping me out :)
#lang scheme (define mamort (lambda (bal pcnt pay) (format "Payment #,Payment Amount,Interest Paid,Principle Paid,Balance,Total Interest~%") (amort bal pcnt pay 12))) (define amort (lambda (bal pcnt pay freq) (let loop ((bal bal) (tot-int 0) (pmt 1)) (let* ((int (* pcnt bal (/ freq))) (prin (min (- pay int) bal)) (new-bal (- bal prin)) (new-int (+ int tot-int)) (done (< (+ int prin) pay))) (if done (printf "~s,~s,~s,~s,~s,~s~n" pmt (+ int prin) int prin new-bal new-int) (begin (printf "~s,~s,~s,~s,~s,~s~n" pmt pay int prin new-bal new-int) (loop new-bal new-int (+ 1 pmt)))))))) (mamort 10000 .12 350)This is my favorite posting yet, Dad. :)
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