Current File : //opt/alt/python37/lib64/python3.7/site-packages/numpy/lib/financial.py
"""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).
"""
from __future__ import division, absolute_import, print_function
import numpy as np
__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(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
miter = np.broadcast(rate, nper, pmt, pv, when)
zer = np.zeros(miter.shape)
fact = np.where(rate == zer, nper + zer,
(1 + rate*when)*(temp - 1)/rate + zer)
return -(pv*temp + pmt*fact)
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 == 0.0)
masked_rate = np.where(mask, 1.0, rate)
z = np.zeros(np.broadcast(masked_rate, nper, pv, fv, when).shape)
fact = np.where(mask != z, nper + z,
(1 + masked_rate*when)*(temp - 1)/masked_rate + z)
return -(fv + pv*temp) / fact
def nper(rate, pmt, pv, fv=0, when='end'):
"""
Compute the number of periodic payments.
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*(1.0+rate*when)/rate
except FloatingPointError:
use_zero_rate = True
if use_zero_rate:
return (-fv + pv) / (pmt + 0.0)
else:
A = -(fv + pv)/(pmt+0.0)
B = np.log((-fv+z) / (pv+z))/np.log(1.0+rate)
miter = np.broadcast(rate, pmt, pv, fv, when)
zer = np.zeros(miter.shape)
return np.where(rate == zer, A + zer, B + zer) + 0.0
def ipmt(rate, per, nper, pv, fv=0.0, 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), 0.0, 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(rate, per, nper, pv, fv=0.0, 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(rate, nper, pmt, fv=0.0, 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
miter = np.broadcast(rate, nper, pmt, fv, when)
zer = np.zeros(miter.shape)
fact = np.where(rate == zer, nper+zer, (1+rate*when)*(temp-1)/rate+zer)
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))
# 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.
def rate(nper, pmt, pv, fv, when='end', guess=0.10, tol=1e-6, 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 : float, optional
Starting guess for solving the rate of interest
tol : float, optional
Required tolerance for the solution
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)
(nper, pmt, pv, fv, when) = map(np.asarray, [nper, pmt, pv, fv, when])
rn = guess
iter = 0
close = False
while (iter < maxiter) and not close:
rnp1 = rn - _g_div_gp(rn, nper, pmt, pv, fv, when)
diff = abs(rnp1-rn)
close = np.all(diff < tol)
iter += 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(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.
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)
"""
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 = 1.0/res - 1
rate = rate.item(np.argmin(np.abs(rate)))
return rate
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(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, dtype=np.double)
n = values.size
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)**(1.0/(n - 1))*(1 + reinvest_rate) - 1
Mini Shell