diff options
Diffstat (limited to 'numpy/lib/financial.py')
| -rw-r--r-- | numpy/lib/financial.py | 24 |
1 files changed, 13 insertions, 11 deletions
diff --git a/numpy/lib/financial.py b/numpy/lib/financial.py index 7f62a81f9..8b77eb3fc 100644 --- a/numpy/lib/financial.py +++ b/numpy/lib/financial.py @@ -206,7 +206,7 @@ def pmt(rate, nper, pv, fv=0, when='end'): def nper(rate, pmt, pv, fv=0, when='end'): """ - Compute the number of periods. + Compute the number of periodic payments. Parameters ---------- @@ -225,16 +225,16 @@ def nper(rate, pmt, pv, fv=0, when='end'): ----- The number of periods ``nper`` is computed by solving the equation:: - fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1) == 0 + fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate*((1+rate)**nper-1) = 0 - or, when ``rate == 0``:: + but if ``rate = 0`` then:: - fv + pv + pmt * nper == 0 + fv + pv + pmt*nper = 0 Examples -------- - If you only had $150 to spend as payment, how long would it take to pay-off - a loan of $8,000 at 7% annual interest? + If you only had $150/month to pay towards the loan, how long would it take + to pay-off a loan of $8,000 at 7% annual interest? >>> np.nper(0.07/12, -150, 8000) 64.073348770661852 @@ -244,11 +244,13 @@ def nper(rate, pmt, pv, fv=0, when='end'): The same analysis could be done with several different interest rates and/or payments and/or total amounts to produce an entire table. - >>> np.nper(*(np.ogrid[0.06/12:0.071/12:0.01/12, -200:-99:100, 6000:7001:1000])) - array([[[ 32.58497782, 38.57048452], - [ 71.51317802, 86.37179563]], - [[ 33.07413144, 39.26244268], - [ 74.06368256, 90.22989997]]]) + >>> np.nper(*(np.ogrid[0.07/12: 0.08/12: 0.01/12, + ... -150 : -99 : 50 , + ... 8000 : 9001 : 1000])) + array([[[ 64.07334877, 74.06368256], + [ 108.07548412, 127.99022654]], + [[ 66.12443902, 76.87897353], + [ 114.70165583, 137.90124779]]]) """ when = _convert_when(when) |