Payment (PMT)

pmt(rate, nper, pv[, fv, when])

Compute the payment against loan principal plus interest.
syse.pmt(rate, nper, pv, fv=0, when='end')

Compute the payment against loan principal plus interest.

Given:

  • a present value, pv (e.g., an amount borrowed)

  • a future value, fv (e.g., 0)

  • an interest rate compounded once per period, of which there are

  • nper total

  • and (optional) specification of whether payment is made at the beginning (when = {‘begin’, 1}) or the end (when = {‘end’, 0}) of each period

Returns:

the (fixed) periodic payment.

Parameters:

  • rate (array_like) – Rate of interest (per period)

  • nper (array_like) – Number of compounding periods

  • pv (array_like) – Present value

  • fv (array_like, optional) – Future value (default = 0)

  • when ({{'begin', 1}, {'end', 0}}, {string, int}) – When payments are due (‘begin’ (1) or ‘end’ (0))

Returns:

ndarray: Payment against loan plus interest. If all input is scalar, returns a scalar float. If any input is array_like, returns payment for each input element. If multiple inputs are array_like, they all must have the same shape.

Note

The payment is computed by solving the equation::

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

or, when rate == 0::

fv + pv + pmt * nper == 0

for pmt. Note that computing a monthly mortgage payment is only one use for this function. For example, pmt returns the periodic deposit one must make to achieve a specified future balance given an initial deposit, a fixed, periodically compounded interest rate, and the total number of periods.

Examples:

What is the monthly payment needed to pay off a $200,000 loan in 15 years at an annual interest rate of 7.5%?

import syse as syse
syse.pmt(0.075/12, 12*15, 200000)
Output = -1854.0247200054619

In order to pay-off (i.e., have a future-value of 0) the $200,000 obtained today, a monthly payment of $1,854.02 would be required. Note that this example illustrates usage of fv having a default value of 0.