numpy.pmt¶
- numpy.pmt(rate, nper, pv, fv=0, when='end')[source]¶
Compute the payment against loan principal plus interest.
- Given:
- Return:
- 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: out : 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.
Notes
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.
References
[WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May). Open Document Format for Office Applications (OpenDocument)v1.2, Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version, Pre-Draft 12. Organization for the Advancement of Structured Information Standards (OASIS). Billerica, MA, USA. [ODT Document]. Available: http://www.oasis-open.org/committees/documents.php ?wg_abbrev=office-formulaOpenDocument-formula-20090508.odt 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%?
>>> np.pmt(0.075/12, 12*15, 200000) -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.