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# numpy.nper

`numpy.``nper`(rate, pmt, pv, fv=0, when='end')

  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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