numpy.nper()

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

Compute the number of periodic payments.

Parameters:

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))

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:

1	`fv` `+` `pv(1+rate)nper` `+` `pmt(1+ratewhen)/rate((1+rate)**nper-1)` `=` `0`

but if rate = 0 then:

1	`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]]])

Links:

https://docs.scipy.org/doc/numpy-1.11.0/reference/generated/numpy.nper.html

doc_NumPy

2025-01-10 15:47:30

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