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831 lines
25 KiB
Python
831 lines
25 KiB
Python
"""Some simple financial calculations
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patterned after spreadsheet computations.
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There is some complexity in each function
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so that the functions behave like ufuncs with
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broadcasting and being able to be called with scalars
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or arrays (or other sequences).
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Functions support the :class:`decimal.Decimal` type unless
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otherwise stated.
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"""
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from __future__ import division, absolute_import, print_function
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from decimal import Decimal
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import functools
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import numpy as np
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from numpy.core import overrides
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array_function_dispatch = functools.partial(
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overrides.array_function_dispatch, module='numpy')
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__all__ = ['fv', 'pmt', 'nper', 'ipmt', 'ppmt', 'pv', 'rate',
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'irr', 'npv', 'mirr']
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_when_to_num = {'end':0, 'begin':1,
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'e':0, 'b':1,
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0:0, 1:1,
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'beginning':1,
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'start':1,
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'finish':0}
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def _convert_when(when):
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#Test to see if when has already been converted to ndarray
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#This will happen if one function calls another, for example ppmt
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if isinstance(when, np.ndarray):
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return when
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try:
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return _when_to_num[when]
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except (KeyError, TypeError):
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return [_when_to_num[x] for x in when]
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def _fv_dispatcher(rate, nper, pmt, pv, when=None):
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return (rate, nper, pmt, pv)
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@array_function_dispatch(_fv_dispatcher)
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def fv(rate, nper, pmt, pv, when='end'):
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"""
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Compute the future value.
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Given:
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* a present value, `pv`
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* an interest `rate` compounded once per period, of which
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there are
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* `nper` total
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* a (fixed) payment, `pmt`, paid either
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* at the beginning (`when` = {'begin', 1}) or the end
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(`when` = {'end', 0}) of each period
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Return:
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the value at the end of the `nper` periods
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Parameters
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----------
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rate : scalar or array_like of shape(M, )
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Rate of interest as decimal (not per cent) per period
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nper : scalar or array_like of shape(M, )
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Number of compounding periods
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pmt : scalar or array_like of shape(M, )
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Payment
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pv : scalar or array_like of shape(M, )
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Present value
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when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
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When payments are due ('begin' (1) or 'end' (0)).
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Defaults to {'end', 0}.
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Returns
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-------
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out : ndarray
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Future values. If all input is scalar, returns a scalar float. If
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any input is array_like, returns future values for each input element.
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If multiple inputs are array_like, they all must have the same shape.
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Notes
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-----
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The future value is computed by solving the equation::
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fv +
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pv*(1+rate)**nper +
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pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0
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or, when ``rate == 0``::
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fv + pv + pmt * nper == 0
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References
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----------
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.. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
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Open Document Format for Office Applications (OpenDocument)v1.2,
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Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
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Pre-Draft 12. Organization for the Advancement of Structured Information
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Standards (OASIS). Billerica, MA, USA. [ODT Document].
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Available:
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http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
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OpenDocument-formula-20090508.odt
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Examples
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--------
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What is the future value after 10 years of saving $100 now, with
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an additional monthly savings of $100. Assume the interest rate is
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5% (annually) compounded monthly?
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>>> np.fv(0.05/12, 10*12, -100, -100)
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15692.928894335748
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By convention, the negative sign represents cash flow out (i.e. money not
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available today). Thus, saving $100 a month at 5% annual interest leads
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to $15,692.93 available to spend in 10 years.
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If any input is array_like, returns an array of equal shape. Let's
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compare different interest rates from the example above.
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>>> a = np.array((0.05, 0.06, 0.07))/12
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>>> np.fv(a, 10*12, -100, -100)
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array([ 15692.92889434, 16569.87435405, 17509.44688102]) # may vary
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"""
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when = _convert_when(when)
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(rate, nper, pmt, pv, when) = map(np.asarray, [rate, nper, pmt, pv, when])
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temp = (1+rate)**nper
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fact = np.where(rate == 0, nper,
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(1 + rate*when)*(temp - 1)/rate)
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return -(pv*temp + pmt*fact)
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def _pmt_dispatcher(rate, nper, pv, fv=None, when=None):
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return (rate, nper, pv, fv)
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@array_function_dispatch(_pmt_dispatcher)
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def pmt(rate, nper, pv, fv=0, when='end'):
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"""
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Compute the payment against loan principal plus interest.
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Given:
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* a present value, `pv` (e.g., an amount borrowed)
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* a future value, `fv` (e.g., 0)
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* an interest `rate` compounded once per period, of which
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there are
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* `nper` total
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* and (optional) specification of whether payment is made
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at the beginning (`when` = {'begin', 1}) or the end
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(`when` = {'end', 0}) of each period
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Return:
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the (fixed) periodic payment.
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Parameters
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----------
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rate : array_like
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Rate of interest (per period)
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nper : array_like
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Number of compounding periods
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pv : array_like
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Present value
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fv : array_like, optional
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Future value (default = 0)
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when : {{'begin', 1}, {'end', 0}}, {string, int}
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When payments are due ('begin' (1) or 'end' (0))
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Returns
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-------
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out : ndarray
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Payment against loan plus interest. If all input is scalar, returns a
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scalar float. If any input is array_like, returns payment for each
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input element. If multiple inputs are array_like, they all must have
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the same shape.
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Notes
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-----
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The payment is computed by solving the equation::
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fv +
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pv*(1 + rate)**nper +
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pmt*(1 + rate*when)/rate*((1 + rate)**nper - 1) == 0
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or, when ``rate == 0``::
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fv + pv + pmt * nper == 0
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for ``pmt``.
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Note that computing a monthly mortgage payment is only
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one use for this function. For example, pmt returns the
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periodic deposit one must make to achieve a specified
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future balance given an initial deposit, a fixed,
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periodically compounded interest rate, and the total
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number of periods.
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References
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----------
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.. [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
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?wg_abbrev=office-formulaOpenDocument-formula-20090508.odt
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Examples
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--------
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What is the monthly payment needed to pay off a $200,000 loan in 15
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years at an annual interest rate of 7.5%?
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>>> np.pmt(0.075/12, 12*15, 200000)
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-1854.0247200054619
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In order to pay-off (i.e., have a future-value of 0) the $200,000 obtained
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today, a monthly payment of $1,854.02 would be required. Note that this
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example illustrates usage of `fv` having a default value of 0.
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"""
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when = _convert_when(when)
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(rate, nper, pv, fv, when) = map(np.array, [rate, nper, pv, fv, when])
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temp = (1 + rate)**nper
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mask = (rate == 0)
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masked_rate = np.where(mask, 1, rate)
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fact = np.where(mask != 0, nper,
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(1 + masked_rate*when)*(temp - 1)/masked_rate)
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return -(fv + pv*temp) / fact
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def _nper_dispatcher(rate, pmt, pv, fv=None, when=None):
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return (rate, pmt, pv, fv)
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@array_function_dispatch(_nper_dispatcher)
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def nper(rate, pmt, pv, fv=0, when='end'):
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"""
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Compute the number of periodic payments.
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:class:`decimal.Decimal` type is not supported.
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Parameters
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----------
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rate : array_like
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Rate of interest (per period)
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pmt : array_like
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Payment
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pv : array_like
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Present value
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fv : array_like, optional
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Future value
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when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
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When payments are due ('begin' (1) or 'end' (0))
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Notes
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-----
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The number of periods ``nper`` is computed by solving the equation::
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fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate*((1+rate)**nper-1) = 0
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but if ``rate = 0`` then::
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fv + pv + pmt*nper = 0
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Examples
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--------
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If you only had $150/month to pay towards the loan, how long would it take
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to pay-off a loan of $8,000 at 7% annual interest?
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>>> print(np.round(np.nper(0.07/12, -150, 8000), 5))
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64.07335
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So, over 64 months would be required to pay off the loan.
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The same analysis could be done with several different interest rates
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and/or payments and/or total amounts to produce an entire table.
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>>> np.nper(*(np.ogrid[0.07/12: 0.08/12: 0.01/12,
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... -150 : -99 : 50 ,
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... 8000 : 9001 : 1000]))
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array([[[ 64.07334877, 74.06368256],
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[108.07548412, 127.99022654]],
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[[ 66.12443902, 76.87897353],
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[114.70165583, 137.90124779]]])
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"""
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when = _convert_when(when)
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(rate, pmt, pv, fv, when) = map(np.asarray, [rate, pmt, pv, fv, when])
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use_zero_rate = False
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with np.errstate(divide="raise"):
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try:
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z = pmt*(1+rate*when)/rate
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except FloatingPointError:
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use_zero_rate = True
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if use_zero_rate:
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return (-fv + pv) / pmt
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else:
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A = -(fv + pv)/(pmt+0)
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B = np.log((-fv+z) / (pv+z))/np.log(1+rate)
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return np.where(rate == 0, A, B)
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def _ipmt_dispatcher(rate, per, nper, pv, fv=None, when=None):
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return (rate, per, nper, pv, fv)
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|
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@array_function_dispatch(_ipmt_dispatcher)
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def ipmt(rate, per, nper, pv, fv=0, when='end'):
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"""
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Compute the interest portion of a payment.
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Parameters
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----------
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rate : scalar or array_like of shape(M, )
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Rate of interest as decimal (not per cent) per period
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per : scalar or array_like of shape(M, )
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Interest paid against the loan changes during the life or the loan.
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The `per` is the payment period to calculate the interest amount.
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nper : scalar or array_like of shape(M, )
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Number of compounding periods
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pv : scalar or array_like of shape(M, )
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Present value
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fv : scalar or array_like of shape(M, ), optional
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Future value
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when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
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When payments are due ('begin' (1) or 'end' (0)).
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Defaults to {'end', 0}.
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|
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Returns
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-------
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out : ndarray
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Interest portion of payment. If all input is scalar, returns a scalar
|
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float. If any input is array_like, returns interest payment for each
|
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input element. If multiple inputs are array_like, they all must have
|
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the same shape.
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|
|
See Also
|
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--------
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ppmt, pmt, pv
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|
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Notes
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-----
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The total payment is made up of payment against principal plus interest.
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``pmt = ppmt + ipmt``
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|
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Examples
|
|
--------
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What is the amortization schedule for a 1 year loan of $2500 at
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8.24% interest per year compounded monthly?
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>>> principal = 2500.00
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The 'per' variable represents the periods of the loan. Remember that
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financial equations start the period count at 1!
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>>> per = np.arange(1*12) + 1
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>>> ipmt = np.ipmt(0.0824/12, per, 1*12, principal)
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>>> ppmt = np.ppmt(0.0824/12, per, 1*12, principal)
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Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal
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'pmt'.
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>>> pmt = np.pmt(0.0824/12, 1*12, principal)
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>>> np.allclose(ipmt + ppmt, pmt)
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True
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>>> fmt = '{0:2d} {1:8.2f} {2:8.2f} {3:8.2f}'
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>>> for payment in per:
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... index = payment - 1
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... principal = principal + ppmt[index]
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... print(fmt.format(payment, ppmt[index], ipmt[index], principal))
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1 -200.58 -17.17 2299.42
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2 -201.96 -15.79 2097.46
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3 -203.35 -14.40 1894.11
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4 -204.74 -13.01 1689.37
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5 -206.15 -11.60 1483.22
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6 -207.56 -10.18 1275.66
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7 -208.99 -8.76 1066.67
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8 -210.42 -7.32 856.25
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9 -211.87 -5.88 644.38
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10 -213.32 -4.42 431.05
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11 -214.79 -2.96 216.26
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12 -216.26 -1.49 -0.00
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|
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>>> interestpd = np.sum(ipmt)
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>>> np.round(interestpd, 2)
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-112.98
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|
|
"""
|
|
when = _convert_when(when)
|
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rate, per, nper, pv, fv, when = np.broadcast_arrays(rate, per, nper,
|
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pv, fv, when)
|
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total_pmt = pmt(rate, nper, pv, fv, when)
|
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ipmt = _rbl(rate, per, total_pmt, pv, when)*rate
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try:
|
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ipmt = np.where(when == 1, ipmt/(1 + rate), ipmt)
|
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ipmt = np.where(np.logical_and(when == 1, per == 1), 0, ipmt)
|
|
except IndexError:
|
|
pass
|
|
return ipmt
|
|
|
|
|
|
def _rbl(rate, per, pmt, pv, when):
|
|
"""
|
|
This function is here to simply have a different name for the 'fv'
|
|
function to not interfere with the 'fv' keyword argument within the 'ipmt'
|
|
function. It is the 'remaining balance on loan' which might be useful as
|
|
it's own function, but is easily calculated with the 'fv' function.
|
|
"""
|
|
return fv(rate, (per - 1), pmt, pv, when)
|
|
|
|
|
|
def _ppmt_dispatcher(rate, per, nper, pv, fv=None, when=None):
|
|
return (rate, per, nper, pv, fv)
|
|
|
|
|
|
@array_function_dispatch(_ppmt_dispatcher)
|
|
def ppmt(rate, per, nper, pv, fv=0, when='end'):
|
|
"""
|
|
Compute the payment against loan principal.
|
|
|
|
Parameters
|
|
----------
|
|
rate : array_like
|
|
Rate of interest (per period)
|
|
per : array_like, int
|
|
Amount paid against the loan changes. The `per` is the period of
|
|
interest.
|
|
nper : array_like
|
|
Number of compounding periods
|
|
pv : array_like
|
|
Present value
|
|
fv : array_like, optional
|
|
Future value
|
|
when : {{'begin', 1}, {'end', 0}}, {string, int}
|
|
When payments are due ('begin' (1) or 'end' (0))
|
|
|
|
See Also
|
|
--------
|
|
pmt, pv, ipmt
|
|
|
|
"""
|
|
total = pmt(rate, nper, pv, fv, when)
|
|
return total - ipmt(rate, per, nper, pv, fv, when)
|
|
|
|
|
|
def _pv_dispatcher(rate, nper, pmt, fv=None, when=None):
|
|
return (rate, nper, nper, pv, fv)
|
|
|
|
|
|
@array_function_dispatch(_pv_dispatcher)
|
|
def pv(rate, nper, pmt, fv=0, when='end'):
|
|
"""
|
|
Compute the present value.
|
|
|
|
Given:
|
|
* a future value, `fv`
|
|
* an interest `rate` compounded once per period, of which
|
|
there are
|
|
* `nper` total
|
|
* a (fixed) payment, `pmt`, paid either
|
|
* at the beginning (`when` = {'begin', 1}) or the end
|
|
(`when` = {'end', 0}) of each period
|
|
|
|
Return:
|
|
the value now
|
|
|
|
Parameters
|
|
----------
|
|
rate : array_like
|
|
Rate of interest (per period)
|
|
nper : array_like
|
|
Number of compounding periods
|
|
pmt : array_like
|
|
Payment
|
|
fv : array_like, optional
|
|
Future value
|
|
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
|
|
When payments are due ('begin' (1) or 'end' (0))
|
|
|
|
Returns
|
|
-------
|
|
out : ndarray, float
|
|
Present value of a series of payments or investments.
|
|
|
|
Notes
|
|
-----
|
|
The present 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
|
|
|
|
for `pv`, which is then returned.
|
|
|
|
References
|
|
----------
|
|
.. [WRW] Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May).
|
|
Open Document Format for Office Applications (OpenDocument)v1.2,
|
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Part 2: Recalculated Formula (OpenFormula) Format - Annotated Version,
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Pre-Draft 12. Organization for the Advancement of Structured Information
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Standards (OASIS). Billerica, MA, USA. [ODT Document].
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Available:
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http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
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OpenDocument-formula-20090508.odt
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Examples
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--------
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What is the present value (e.g., the initial investment)
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of an investment that needs to total $15692.93
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after 10 years of saving $100 every month? Assume the
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interest rate is 5% (annually) compounded monthly.
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>>> np.pv(0.05/12, 10*12, -100, 15692.93)
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-100.00067131625819
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By convention, the negative sign represents cash flow out
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(i.e., money not available today). Thus, to end up with
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$15,692.93 in 10 years saving $100 a month at 5% annual
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interest, one's initial deposit should also be $100.
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If any input is array_like, ``pv`` returns an array of equal shape.
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Let's compare different interest rates in the example above:
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>>> a = np.array((0.05, 0.04, 0.03))/12
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>>> np.pv(a, 10*12, -100, 15692.93)
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array([ -100.00067132, -649.26771385, -1273.78633713]) # may vary
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So, to end up with the same $15692.93 under the same $100 per month
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"savings plan," for annual interest rates of 4% and 3%, one would
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need initial investments of $649.27 and $1273.79, respectively.
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"""
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when = _convert_when(when)
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(rate, nper, pmt, fv, when) = map(np.asarray, [rate, nper, pmt, fv, when])
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temp = (1+rate)**nper
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fact = np.where(rate == 0, nper, (1+rate*when)*(temp-1)/rate)
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return -(fv + pmt*fact)/temp
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# Computed with Sage
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# (y + (r + 1)^n*x + p*((r + 1)^n - 1)*(r*w + 1)/r)/(n*(r + 1)^(n - 1)*x -
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# p*((r + 1)^n - 1)*(r*w + 1)/r^2 + n*p*(r + 1)^(n - 1)*(r*w + 1)/r +
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# p*((r + 1)^n - 1)*w/r)
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def _g_div_gp(r, n, p, x, y, w):
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t1 = (r+1)**n
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t2 = (r+1)**(n-1)
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return ((y + t1*x + p*(t1 - 1)*(r*w + 1)/r) /
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(n*t2*x - p*(t1 - 1)*(r*w + 1)/(r**2) + n*p*t2*(r*w + 1)/r +
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p*(t1 - 1)*w/r))
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def _rate_dispatcher(nper, pmt, pv, fv, when=None, guess=None, tol=None,
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maxiter=None):
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return (nper, pmt, pv, fv)
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# Use Newton's iteration until the change is less than 1e-6
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# for all values or a maximum of 100 iterations is reached.
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# Newton's rule is
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# r_{n+1} = r_{n} - g(r_n)/g'(r_n)
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# where
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# g(r) is the formula
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# g'(r) is the derivative with respect to r.
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@array_function_dispatch(_rate_dispatcher)
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def rate(nper, pmt, pv, fv, when='end', guess=None, tol=None, maxiter=100):
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"""
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Compute the rate of interest per period.
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Parameters
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----------
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nper : array_like
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Number of compounding periods
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pmt : array_like
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Payment
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pv : array_like
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Present value
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fv : array_like
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Future value
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when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
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When payments are due ('begin' (1) or 'end' (0))
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guess : Number, optional
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Starting guess for solving the rate of interest, default 0.1
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tol : Number, optional
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Required tolerance for the solution, default 1e-6
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maxiter : int, optional
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Maximum iterations in finding the solution
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Notes
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-----
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The rate of interest is computed by iteratively solving the
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(non-linear) equation::
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fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1) = 0
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for ``rate``.
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References
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----------
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Wheeler, D. A., E. Rathke, and R. Weir (Eds.) (2009, May). Open Document
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Format for Office Applications (OpenDocument)v1.2, Part 2: Recalculated
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Formula (OpenFormula) Format - Annotated Version, Pre-Draft 12.
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Organization for the Advancement of Structured Information Standards
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(OASIS). Billerica, MA, USA. [ODT Document]. Available:
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http://www.oasis-open.org/committees/documents.php?wg_abbrev=office-formula
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OpenDocument-formula-20090508.odt
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"""
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when = _convert_when(when)
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default_type = Decimal if isinstance(pmt, Decimal) else float
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# Handle casting defaults to Decimal if/when pmt is a Decimal and
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# guess and/or tol are not given default values
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if guess is None:
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guess = default_type('0.1')
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if tol is None:
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tol = default_type('1e-6')
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(nper, pmt, pv, fv, when) = map(np.asarray, [nper, pmt, pv, fv, when])
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rn = guess
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iterator = 0
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close = False
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while (iterator < maxiter) and not close:
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rnp1 = rn - _g_div_gp(rn, nper, pmt, pv, fv, when)
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diff = abs(rnp1-rn)
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close = np.all(diff < tol)
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iterator += 1
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rn = rnp1
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if not close:
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# Return nan's in array of the same shape as rn
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return np.nan + rn
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else:
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return rn
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def _irr_dispatcher(values):
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return (values,)
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@array_function_dispatch(_irr_dispatcher)
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def irr(values):
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"""
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Return the Internal Rate of Return (IRR).
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This is the "average" periodically compounded rate of return
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that gives a net present value of 0.0; for a more complete explanation,
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see Notes below.
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:class:`decimal.Decimal` type is not supported.
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Parameters
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----------
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values : array_like, shape(N,)
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Input cash flows per time period. By convention, net "deposits"
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are negative and net "withdrawals" are positive. Thus, for
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example, at least the first element of `values`, which represents
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the initial investment, will typically be negative.
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Returns
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-------
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out : float
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Internal Rate of Return for periodic input values.
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Notes
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-----
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The IRR is perhaps best understood through an example (illustrated
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using np.irr in the Examples section below). Suppose one invests 100
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units and then makes the following withdrawals at regular (fixed)
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intervals: 39, 59, 55, 20. Assuming the ending value is 0, one's 100
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unit investment yields 173 units; however, due to the combination of
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compounding and the periodic withdrawals, the "average" rate of return
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is neither simply 0.73/4 nor (1.73)^0.25-1. Rather, it is the solution
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(for :math:`r`) of the equation:
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.. math:: -100 + \\frac{39}{1+r} + \\frac{59}{(1+r)^2}
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+ \\frac{55}{(1+r)^3} + \\frac{20}{(1+r)^4} = 0
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In general, for `values` :math:`= [v_0, v_1, ... v_M]`,
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irr is the solution of the equation: [G]_
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.. math:: \\sum_{t=0}^M{\\frac{v_t}{(1+irr)^{t}}} = 0
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References
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----------
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.. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
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Addison-Wesley, 2003, pg. 348.
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Examples
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--------
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>>> round(np.irr([-100, 39, 59, 55, 20]), 5)
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0.28095
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>>> round(np.irr([-100, 0, 0, 74]), 5)
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-0.0955
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>>> round(np.irr([-100, 100, 0, -7]), 5)
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-0.0833
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>>> round(np.irr([-100, 100, 0, 7]), 5)
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0.06206
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>>> round(np.irr([-5, 10.5, 1, -8, 1]), 5)
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0.0886
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(Compare with the Example given for numpy.lib.financial.npv)
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"""
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# `np.roots` call is why this function does not support Decimal type.
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#
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# Ultimately Decimal support needs to be added to np.roots, which has
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# greater implications on the entire linear algebra module and how it does
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# eigenvalue computations.
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res = np.roots(values[::-1])
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mask = (res.imag == 0) & (res.real > 0)
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if not mask.any():
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return np.nan
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res = res[mask].real
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# NPV(rate) = 0 can have more than one solution so we return
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# only the solution closest to zero.
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rate = 1/res - 1
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rate = rate.item(np.argmin(np.abs(rate)))
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return rate
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def _npv_dispatcher(rate, values):
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return (values,)
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@array_function_dispatch(_npv_dispatcher)
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def npv(rate, values):
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"""
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Returns the NPV (Net Present Value) of a cash flow series.
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Parameters
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----------
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rate : scalar
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The discount rate.
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values : array_like, shape(M, )
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The values of the time series of cash flows. The (fixed) time
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interval between cash flow "events" must be the same as that for
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which `rate` is given (i.e., if `rate` is per year, then precisely
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a year is understood to elapse between each cash flow event). By
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convention, investments or "deposits" are negative, income or
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"withdrawals" are positive; `values` must begin with the initial
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investment, thus `values[0]` will typically be negative.
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Returns
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-------
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out : float
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The NPV of the input cash flow series `values` at the discount
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`rate`.
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Notes
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-----
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Returns the result of: [G]_
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.. math :: \\sum_{t=0}^{M-1}{\\frac{values_t}{(1+rate)^{t}}}
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References
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----------
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.. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
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Addison-Wesley, 2003, pg. 346.
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Examples
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--------
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>>> np.npv(0.281,[-100, 39, 59, 55, 20])
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-0.0084785916384548798 # may vary
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(Compare with the Example given for numpy.lib.financial.irr)
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"""
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values = np.asarray(values)
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return (values / (1+rate)**np.arange(0, len(values))).sum(axis=0)
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def _mirr_dispatcher(values, finance_rate, reinvest_rate):
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return (values,)
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@array_function_dispatch(_mirr_dispatcher)
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def mirr(values, finance_rate, reinvest_rate):
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"""
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Modified internal rate of return.
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Parameters
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----------
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values : array_like
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Cash flows (must contain at least one positive and one negative
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value) or nan is returned. The first value is considered a sunk
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cost at time zero.
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finance_rate : scalar
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Interest rate paid on the cash flows
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reinvest_rate : scalar
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Interest rate received on the cash flows upon reinvestment
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Returns
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-------
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out : float
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Modified internal rate of return
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"""
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values = np.asarray(values)
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n = values.size
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# Without this explicit cast the 1/(n - 1) computation below
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# becomes a float, which causes TypeError when using Decimal
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# values.
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if isinstance(finance_rate, Decimal):
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n = Decimal(n)
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pos = values > 0
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neg = values < 0
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if not (pos.any() and neg.any()):
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return np.nan
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numer = np.abs(npv(reinvest_rate, values*pos))
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denom = np.abs(npv(finance_rate, values*neg))
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return (numer/denom)**(1/(n - 1))*(1 + reinvest_rate) - 1
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