You cannot select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
831 lines
25 KiB
Python
831 lines
25 KiB
Python
"""Some simple financial calculations
|
|
|
|
patterned after spreadsheet computations.
|
|
|
|
There is some complexity in each function
|
|
so that the functions behave like ufuncs with
|
|
broadcasting and being able to be called with scalars
|
|
or arrays (or other sequences).
|
|
|
|
Functions support the :class:`decimal.Decimal` type unless
|
|
otherwise stated.
|
|
"""
|
|
from __future__ import division, absolute_import, print_function
|
|
|
|
from decimal import Decimal
|
|
import functools
|
|
|
|
import numpy as np
|
|
from numpy.core import overrides
|
|
|
|
|
|
array_function_dispatch = functools.partial(
|
|
overrides.array_function_dispatch, module='numpy')
|
|
|
|
|
|
__all__ = ['fv', 'pmt', 'nper', 'ipmt', 'ppmt', 'pv', 'rate',
|
|
'irr', 'npv', 'mirr']
|
|
|
|
_when_to_num = {'end':0, 'begin':1,
|
|
'e':0, 'b':1,
|
|
0:0, 1:1,
|
|
'beginning':1,
|
|
'start':1,
|
|
'finish':0}
|
|
|
|
def _convert_when(when):
|
|
#Test to see if when has already been converted to ndarray
|
|
#This will happen if one function calls another, for example ppmt
|
|
if isinstance(when, np.ndarray):
|
|
return when
|
|
try:
|
|
return _when_to_num[when]
|
|
except (KeyError, TypeError):
|
|
return [_when_to_num[x] for x in when]
|
|
|
|
|
|
def _fv_dispatcher(rate, nper, pmt, pv, when=None):
|
|
return (rate, nper, pmt, pv)
|
|
|
|
|
|
@array_function_dispatch(_fv_dispatcher)
|
|
def fv(rate, nper, pmt, pv, when='end'):
|
|
"""
|
|
Compute the future value.
|
|
|
|
Given:
|
|
* a present value, `pv`
|
|
* 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 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,
|
|
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-formula
|
|
OpenDocument-formula-20090508.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])
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
(rate, nper, pmt, pv, when) = map(np.asarray, [rate, nper, pmt, pv, when])
|
|
temp = (1+rate)**nper
|
|
fact = np.where(rate == 0, nper,
|
|
(1 + rate*when)*(temp - 1)/rate)
|
|
return -(pv*temp + pmt*fact)
|
|
|
|
|
|
def _pmt_dispatcher(rate, nper, pv, fv=None, when=None):
|
|
return (rate, nper, pv, fv)
|
|
|
|
|
|
@array_function_dispatch(_pmt_dispatcher)
|
|
def 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
|
|
|
|
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.
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
(rate, nper, pv, fv, when) = map(np.array, [rate, nper, pv, fv, when])
|
|
temp = (1 + rate)**nper
|
|
mask = (rate == 0)
|
|
masked_rate = np.where(mask, 1, rate)
|
|
fact = np.where(mask != 0, nper,
|
|
(1 + masked_rate*when)*(temp - 1)/masked_rate)
|
|
return -(fv + pv*temp) / fact
|
|
|
|
|
|
def _nper_dispatcher(rate, pmt, pv, fv=None, when=None):
|
|
return (rate, pmt, pv, fv)
|
|
|
|
|
|
@array_function_dispatch(_nper_dispatcher)
|
|
def nper(rate, pmt, pv, fv=0, when='end'):
|
|
"""
|
|
Compute the number of periodic payments.
|
|
|
|
:class:`decimal.Decimal` type is not supported.
|
|
|
|
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]]])
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
(rate, pmt, pv, fv, when) = map(np.asarray, [rate, pmt, pv, fv, when])
|
|
|
|
use_zero_rate = False
|
|
with np.errstate(divide="raise"):
|
|
try:
|
|
z = pmt*(1+rate*when)/rate
|
|
except FloatingPointError:
|
|
use_zero_rate = True
|
|
|
|
if use_zero_rate:
|
|
return (-fv + pv) / pmt
|
|
else:
|
|
A = -(fv + pv)/(pmt+0)
|
|
B = np.log((-fv+z) / (pv+z))/np.log(1+rate)
|
|
return np.where(rate == 0, A, B)
|
|
|
|
|
|
def _ipmt_dispatcher(rate, per, nper, pv, fv=None, when=None):
|
|
return (rate, per, nper, pv, fv)
|
|
|
|
|
|
@array_function_dispatch(_ipmt_dispatcher)
|
|
def ipmt(rate, per, nper, pv, fv=0, when='end'):
|
|
"""
|
|
Compute the interest portion of a payment.
|
|
|
|
Parameters
|
|
----------
|
|
rate : scalar or array_like of shape(M, )
|
|
Rate of interest as decimal (not per cent) per period
|
|
per : scalar or array_like of shape(M, )
|
|
Interest paid against the loan changes during the life or the loan.
|
|
The `per` is the payment period to calculate the interest amount.
|
|
nper : scalar or array_like of shape(M, )
|
|
Number of compounding periods
|
|
pv : scalar or array_like of shape(M, )
|
|
Present value
|
|
fv : scalar or array_like of shape(M, ), optional
|
|
Future 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
|
|
Interest portion of payment. If all input is scalar, returns a scalar
|
|
float. If any input is array_like, returns interest payment for each
|
|
input element. If multiple inputs are array_like, they all must have
|
|
the same shape.
|
|
|
|
See Also
|
|
--------
|
|
ppmt, pmt, pv
|
|
|
|
Notes
|
|
-----
|
|
The total payment is made up of payment against principal plus interest.
|
|
|
|
``pmt = ppmt + ipmt``
|
|
|
|
Examples
|
|
--------
|
|
What is the amortization schedule for a 1 year loan of $2500 at
|
|
8.24% interest per year compounded monthly?
|
|
|
|
>>> principal = 2500.00
|
|
|
|
The 'per' variable represents the periods of the loan. Remember that
|
|
financial equations start the period count at 1!
|
|
|
|
>>> per = np.arange(1*12) + 1
|
|
>>> ipmt = np.ipmt(0.0824/12, per, 1*12, principal)
|
|
>>> ppmt = np.ppmt(0.0824/12, per, 1*12, principal)
|
|
|
|
Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal
|
|
'pmt'.
|
|
|
|
>>> pmt = np.pmt(0.0824/12, 1*12, principal)
|
|
>>> np.allclose(ipmt + ppmt, pmt)
|
|
True
|
|
|
|
>>> fmt = '{0:2d} {1:8.2f} {2:8.2f} {3:8.2f}'
|
|
>>> for payment in per:
|
|
... index = payment - 1
|
|
... principal = principal + ppmt[index]
|
|
... print(fmt.format(payment, ppmt[index], ipmt[index], principal))
|
|
1 -200.58 -17.17 2299.42
|
|
2 -201.96 -15.79 2097.46
|
|
3 -203.35 -14.40 1894.11
|
|
4 -204.74 -13.01 1689.37
|
|
5 -206.15 -11.60 1483.22
|
|
6 -207.56 -10.18 1275.66
|
|
7 -208.99 -8.76 1066.67
|
|
8 -210.42 -7.32 856.25
|
|
9 -211.87 -5.88 644.38
|
|
10 -213.32 -4.42 431.05
|
|
11 -214.79 -2.96 216.26
|
|
12 -216.26 -1.49 -0.00
|
|
|
|
>>> interestpd = np.sum(ipmt)
|
|
>>> np.round(interestpd, 2)
|
|
-112.98
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
rate, per, nper, pv, fv, when = np.broadcast_arrays(rate, per, nper,
|
|
pv, fv, when)
|
|
total_pmt = pmt(rate, nper, pv, fv, when)
|
|
ipmt = _rbl(rate, per, total_pmt, pv, when)*rate
|
|
try:
|
|
ipmt = np.where(when == 1, ipmt/(1 + rate), ipmt)
|
|
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,
|
|
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-formula
|
|
OpenDocument-formula-20090508.odt
|
|
|
|
Examples
|
|
--------
|
|
What is the present value (e.g., the initial investment)
|
|
of an investment that needs to total $15692.93
|
|
after 10 years of saving $100 every month? Assume the
|
|
interest rate is 5% (annually) compounded monthly.
|
|
|
|
>>> np.pv(0.05/12, 10*12, -100, 15692.93)
|
|
-100.00067131625819
|
|
|
|
By convention, the negative sign represents cash flow out
|
|
(i.e., money not available today). Thus, to end up with
|
|
$15,692.93 in 10 years saving $100 a month at 5% annual
|
|
interest, one's initial deposit should also be $100.
|
|
|
|
If any input is array_like, ``pv`` returns an array of equal shape.
|
|
Let's compare different interest rates in the example above:
|
|
|
|
>>> a = np.array((0.05, 0.04, 0.03))/12
|
|
>>> np.pv(a, 10*12, -100, 15692.93)
|
|
array([ -100.00067132, -649.26771385, -1273.78633713])
|
|
|
|
So, to end up with the same $15692.93 under the same $100 per month
|
|
"savings plan," for annual interest rates of 4% and 3%, one would
|
|
need initial investments of $649.27 and $1273.79, respectively.
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
(rate, nper, pmt, fv, when) = map(np.asarray, [rate, nper, pmt, fv, when])
|
|
temp = (1+rate)**nper
|
|
fact = np.where(rate == 0, nper, (1+rate*when)*(temp-1)/rate)
|
|
return -(fv + pmt*fact)/temp
|
|
|
|
# Computed with Sage
|
|
# (y + (r + 1)^n*x + p*((r + 1)^n - 1)*(r*w + 1)/r)/(n*(r + 1)^(n - 1)*x -
|
|
# p*((r + 1)^n - 1)*(r*w + 1)/r^2 + n*p*(r + 1)^(n - 1)*(r*w + 1)/r +
|
|
# p*((r + 1)^n - 1)*w/r)
|
|
|
|
def _g_div_gp(r, n, p, x, y, w):
|
|
t1 = (r+1)**n
|
|
t2 = (r+1)**(n-1)
|
|
return ((y + t1*x + p*(t1 - 1)*(r*w + 1)/r) /
|
|
(n*t2*x - p*(t1 - 1)*(r*w + 1)/(r**2) + n*p*t2*(r*w + 1)/r +
|
|
p*(t1 - 1)*w/r))
|
|
|
|
|
|
def _rate_dispatcher(nper, pmt, pv, fv, when=None, guess=None, tol=None,
|
|
maxiter=None):
|
|
return (nper, pmt, pv, fv)
|
|
|
|
|
|
# Use Newton's iteration until the change is less than 1e-6
|
|
# for all values or a maximum of 100 iterations is reached.
|
|
# Newton's rule is
|
|
# r_{n+1} = r_{n} - g(r_n)/g'(r_n)
|
|
# where
|
|
# g(r) is the formula
|
|
# g'(r) is the derivative with respect to r.
|
|
@array_function_dispatch(_rate_dispatcher)
|
|
def rate(nper, pmt, pv, fv, when='end', guess=None, tol=None, maxiter=100):
|
|
"""
|
|
Compute the rate of interest per period.
|
|
|
|
Parameters
|
|
----------
|
|
nper : array_like
|
|
Number of compounding periods
|
|
pmt : array_like
|
|
Payment
|
|
pv : array_like
|
|
Present value
|
|
fv : array_like
|
|
Future value
|
|
when : {{'begin', 1}, {'end', 0}}, {string, int}, optional
|
|
When payments are due ('begin' (1) or 'end' (0))
|
|
guess : Number, optional
|
|
Starting guess for solving the rate of interest, default 0.1
|
|
tol : Number, optional
|
|
Required tolerance for the solution, default 1e-6
|
|
maxiter : int, optional
|
|
Maximum iterations in finding the solution
|
|
|
|
Notes
|
|
-----
|
|
The rate of interest is computed by iteratively solving the
|
|
(non-linear) equation::
|
|
|
|
fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1) = 0
|
|
|
|
for ``rate``.
|
|
|
|
References
|
|
----------
|
|
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-formula
|
|
OpenDocument-formula-20090508.odt
|
|
|
|
"""
|
|
when = _convert_when(when)
|
|
default_type = Decimal if isinstance(pmt, Decimal) else float
|
|
|
|
# Handle casting defaults to Decimal if/when pmt is a Decimal and
|
|
# guess and/or tol are not given default values
|
|
if guess is None:
|
|
guess = default_type('0.1')
|
|
|
|
if tol is None:
|
|
tol = default_type('1e-6')
|
|
|
|
(nper, pmt, pv, fv, when) = map(np.asarray, [nper, pmt, pv, fv, when])
|
|
|
|
rn = guess
|
|
iterator = 0
|
|
close = False
|
|
while (iterator < maxiter) and not close:
|
|
rnp1 = rn - _g_div_gp(rn, nper, pmt, pv, fv, when)
|
|
diff = abs(rnp1-rn)
|
|
close = np.all(diff < tol)
|
|
iterator += 1
|
|
rn = rnp1
|
|
if not close:
|
|
# Return nan's in array of the same shape as rn
|
|
return np.nan + rn
|
|
else:
|
|
return rn
|
|
|
|
|
|
def _irr_dispatcher(values):
|
|
return (values,)
|
|
|
|
|
|
@array_function_dispatch(_irr_dispatcher)
|
|
def irr(values):
|
|
"""
|
|
Return the Internal Rate of Return (IRR).
|
|
|
|
This is the "average" periodically compounded rate of return
|
|
that gives a net present value of 0.0; for a more complete explanation,
|
|
see Notes below.
|
|
|
|
:class:`decimal.Decimal` type is not supported.
|
|
|
|
Parameters
|
|
----------
|
|
values : array_like, shape(N,)
|
|
Input cash flows per time period. By convention, net "deposits"
|
|
are negative and net "withdrawals" are positive. Thus, for
|
|
example, at least the first element of `values`, which represents
|
|
the initial investment, will typically be negative.
|
|
|
|
Returns
|
|
-------
|
|
out : float
|
|
Internal Rate of Return for periodic input values.
|
|
|
|
Notes
|
|
-----
|
|
The IRR is perhaps best understood through an example (illustrated
|
|
using np.irr in the Examples section below). Suppose one invests 100
|
|
units and then makes the following withdrawals at regular (fixed)
|
|
intervals: 39, 59, 55, 20. Assuming the ending value is 0, one's 100
|
|
unit investment yields 173 units; however, due to the combination of
|
|
compounding and the periodic withdrawals, the "average" rate of return
|
|
is neither simply 0.73/4 nor (1.73)^0.25-1. Rather, it is the solution
|
|
(for :math:`r`) of the equation:
|
|
|
|
.. math:: -100 + \\frac{39}{1+r} + \\frac{59}{(1+r)^2}
|
|
+ \\frac{55}{(1+r)^3} + \\frac{20}{(1+r)^4} = 0
|
|
|
|
In general, for `values` :math:`= [v_0, v_1, ... v_M]`,
|
|
irr is the solution of the equation: [G]_
|
|
|
|
.. math:: \\sum_{t=0}^M{\\frac{v_t}{(1+irr)^{t}}} = 0
|
|
|
|
References
|
|
----------
|
|
.. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
|
|
Addison-Wesley, 2003, pg. 348.
|
|
|
|
Examples
|
|
--------
|
|
>>> round(irr([-100, 39, 59, 55, 20]), 5)
|
|
0.28095
|
|
>>> round(irr([-100, 0, 0, 74]), 5)
|
|
-0.0955
|
|
>>> round(irr([-100, 100, 0, -7]), 5)
|
|
-0.0833
|
|
>>> round(irr([-100, 100, 0, 7]), 5)
|
|
0.06206
|
|
>>> round(irr([-5, 10.5, 1, -8, 1]), 5)
|
|
0.0886
|
|
|
|
(Compare with the Example given for numpy.lib.financial.npv)
|
|
|
|
"""
|
|
# `np.roots` call is why this function does not support Decimal type.
|
|
#
|
|
# Ultimately Decimal support needs to be added to np.roots, which has
|
|
# greater implications on the entire linear algebra module and how it does
|
|
# eigenvalue computations.
|
|
res = np.roots(values[::-1])
|
|
mask = (res.imag == 0) & (res.real > 0)
|
|
if not mask.any():
|
|
return np.nan
|
|
res = res[mask].real
|
|
# NPV(rate) = 0 can have more than one solution so we return
|
|
# only the solution closest to zero.
|
|
rate = 1/res - 1
|
|
rate = rate.item(np.argmin(np.abs(rate)))
|
|
return rate
|
|
|
|
|
|
def _npv_dispatcher(rate, values):
|
|
return (values,)
|
|
|
|
|
|
@array_function_dispatch(_npv_dispatcher)
|
|
def npv(rate, values):
|
|
"""
|
|
Returns the NPV (Net Present Value) of a cash flow series.
|
|
|
|
Parameters
|
|
----------
|
|
rate : scalar
|
|
The discount rate.
|
|
values : array_like, shape(M, )
|
|
The values of the time series of cash flows. The (fixed) time
|
|
interval between cash flow "events" must be the same as that for
|
|
which `rate` is given (i.e., if `rate` is per year, then precisely
|
|
a year is understood to elapse between each cash flow event). By
|
|
convention, investments or "deposits" are negative, income or
|
|
"withdrawals" are positive; `values` must begin with the initial
|
|
investment, thus `values[0]` will typically be negative.
|
|
|
|
Returns
|
|
-------
|
|
out : float
|
|
The NPV of the input cash flow series `values` at the discount
|
|
`rate`.
|
|
|
|
Notes
|
|
-----
|
|
Returns the result of: [G]_
|
|
|
|
.. math :: \\sum_{t=0}^{M-1}{\\frac{values_t}{(1+rate)^{t}}}
|
|
|
|
References
|
|
----------
|
|
.. [G] L. J. Gitman, "Principles of Managerial Finance, Brief," 3rd ed.,
|
|
Addison-Wesley, 2003, pg. 346.
|
|
|
|
Examples
|
|
--------
|
|
>>> np.npv(0.281,[-100, 39, 59, 55, 20])
|
|
-0.0084785916384548798
|
|
|
|
(Compare with the Example given for numpy.lib.financial.irr)
|
|
|
|
"""
|
|
values = np.asarray(values)
|
|
return (values / (1+rate)**np.arange(0, len(values))).sum(axis=0)
|
|
|
|
|
|
def _mirr_dispatcher(values, finance_rate, reinvest_rate):
|
|
return (values,)
|
|
|
|
|
|
@array_function_dispatch(_mirr_dispatcher)
|
|
def mirr(values, finance_rate, reinvest_rate):
|
|
"""
|
|
Modified internal rate of return.
|
|
|
|
Parameters
|
|
----------
|
|
values : array_like
|
|
Cash flows (must contain at least one positive and one negative
|
|
value) or nan is returned. The first value is considered a sunk
|
|
cost at time zero.
|
|
finance_rate : scalar
|
|
Interest rate paid on the cash flows
|
|
reinvest_rate : scalar
|
|
Interest rate received on the cash flows upon reinvestment
|
|
|
|
Returns
|
|
-------
|
|
out : float
|
|
Modified internal rate of return
|
|
|
|
"""
|
|
values = np.asarray(values)
|
|
n = values.size
|
|
|
|
# Without this explicit cast the 1/(n - 1) computation below
|
|
# becomes a float, which causes TypeError when using Decimal
|
|
# values.
|
|
if isinstance(finance_rate, Decimal):
|
|
n = Decimal(n)
|
|
|
|
pos = values > 0
|
|
neg = values < 0
|
|
if not (pos.any() and neg.any()):
|
|
return np.nan
|
|
numer = np.abs(npv(reinvest_rate, values*pos))
|
|
denom = np.abs(npv(finance_rate, values*neg))
|
|
return (numer/denom)**(1/(n - 1))*(1 + reinvest_rate) - 1
|