numpy.fv¶

numpy.
fv
(rate, nper, pmt, pv, when='end')[source]¶ Compute the future value.
 Given:
 Return:
 the value at the end of the
nper
periods
Parameters:  rate : scalar or array_like of shape(M, )
Rate of interest as decimal (not per cent) per period
 nper : scalar or array_like of shape(M, )
Number of compounding periods
 pmt : scalar or array_like of shape(M, )
Payment
 pv : scalar or array_like of shape(M, )
Present value
 when : {{‘begin’, 1}, {‘end’, 0}}, {string, int}, optional
When payments are due (‘begin’ (1) or ‘end’ (0)). Defaults to {‘end’, 0}.
Returns:  out : ndarray
Future values. If all input is scalar, returns a scalar float. If any input is array_like, returns future values for each input element. If multiple inputs are array_like, they all must have the same shape.
Notes
The future value 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
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, PreDraft 12. Organization for the Advancement of Structured Information Standards (OASIS). Billerica, MA, USA. [ODT Document]. Available: http://www.oasisopen.org/committees/documents.php?wg_abbrev=officeformula OpenDocumentformula20090508.odt Examples
What is the future value after 10 years of saving $100 now, with an additional monthly savings of $100. Assume the interest rate is 5% (annually) compounded monthly?
>>> np.fv(0.05/12, 10*12, 100, 100) 15692.928894335748
By convention, the negative sign represents cash flow out (i.e. money not available today). Thus, saving $100 a month at 5% annual interest leads to $15,692.93 available to spend in 10 years.
If any input is array_like, returns an array of equal shape. Let’s compare different interest rates from the example above.
>>> a = np.array((0.05, 0.06, 0.07))/12 >>> np.fv(a, 10*12, 100, 100) array([ 15692.92889434, 16569.87435405, 17509.44688102]) # may vary