SciPy

numpy.nper

numpy.nper(rate, pmt, pv, fv=0, when='end')[source]

Compute the number of periodic payments.

Parameters :

rate : array_like

Rate of interest (per period)

pmt : array_like

Payment

pv : array_like

Present value

fv : array_like, optional

Future value

when : {{‘begin’, 1}, {‘end’, 0}}, {string, int}, optional

When payments are due (‘begin’ (1) or ‘end’ (0))

Notes

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

but if rate = 0 then:

fv + pv + pmt*nper = 0

Examples

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?

>>> print round(np.nper(0.07/12, -150, 8000), 5)
64.07335

So, over 64 months would be required to pay off the loan.

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[0.07/12: 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]]])

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