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_financial.py
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"""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 decimal import Decimal
import numpy as np
from . import _cfinancial
__all__ = ['fv', 'pmt', 'nper', 'ipmt', 'ppmt', 'pv', 'rate',
'irr', 'npv', 'mirr',
'NoRealSolutionError', 'IterationsExceededError']
_when_to_num = {'end': 0, 'begin': 1,
'e': 0, 'b': 1,
0: 0, 1: 1,
'beginning': 1,
'start': 1,
'finish': 0}
class NoRealSolutionError(Exception):
"""No real solution to the problem."""
class IterationsExceededError(Exception):
"""Maximum number of iterations reached."""
def _get_output_array_shape(*arrays):
return tuple(array.shape[0] for array in arrays)
def _ufunc_like(array):
try:
# If size of array is one, return scalar
return array.item()
except ValueError:
# Otherwise, return entire array
return array.squeeze()
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
--------
>>> import numpy as np
>>> import numpy_financial as npf
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?
>>> npf.fv(0.05/12, 10*12, -100, -100)
15692.92889433575
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
>>> npf.fv(a, 10*12, -100, -100)
array([15692.92889434, 16569.87435405, 17509.44688102])
"""
when = _convert_when(when)
rate, nper, pmt, pv, when = np.broadcast_arrays(rate, nper, pmt, pv, when)
fv_array = np.empty_like(rate)
zero = rate == 0
nonzero = ~zero
fv_array[zero] = -(pv[zero] + pmt[zero] * nper[zero])
rate_nonzero = rate[nonzero]
temp = (1 + rate_nonzero) ** nper[nonzero]
fv_array[nonzero] = (
- pv[nonzero] * temp
- pmt[nonzero] * (1 + rate_nonzero * when[nonzero]) / rate_nonzero
* (temp - 1)
)
if np.ndim(fv_array) == 0:
# Follow the ufunc convention of returning scalars for scalar
# and 0d array inputs.
return fv_array.item(0)
return fv_array
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
--------
>>> import numpy_financial as npf
What is the monthly payment needed to pay off a $200,000 loan in 15
years at an annual interest rate of 7.5%?
>>> npf.pmt(0.075/12, 12*15, 200000)
np.float64(-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(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
--------
>>> import numpy as np
>>> import numpy_financial as npf
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(np.round(npf.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.
>>> rates = [0.05, 0.06, 0.07]
>>> payments = [100, 200, 300]
>>> amounts = [7_000, 8_000, 9_000]
>>> npf.nper(rates, payments, amounts).round(3)
array([[[-30.827, -32.987, -34.94 ],
[-20.734, -22.517, -24.158],
[-15.847, -17.366, -18.78 ]],
<BLANKLINE>
[[-28.294, -30.168, -31.857],
[-19.417, -21.002, -22.453],
[-15.025, -16.398, -17.67 ]],
<BLANKLINE>
[[-26.234, -27.891, -29.381],
[-18.303, -19.731, -21.034],
[-14.311, -15.566, -16.722]]])
"""
when = _convert_when(when)
rates = np.atleast_1d(rate).astype(np.float64)
pmts = np.atleast_1d(pmt).astype(np.float64)
pvs = np.atleast_1d(pv).astype(np.float64)
fvs = np.atleast_1d(fv).astype(np.float64)
whens = np.atleast_1d(when).astype(np.float64)
out_shape = _get_output_array_shape(rates, pmts, pvs, fvs, whens)
out = np.empty(out_shape)
_cfinancial.nper(rates, pmts, pvs, fvs, whens, out)
return _ufunc_like(out)
def _value_like(arr, value):
entry = arr.item(0)
if isinstance(entry, Decimal):
return Decimal(value)
return np.array(value, dtype=arr.dtype).item(0)
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
--------
>>> import numpy as np
>>> import numpy_financial as npf
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 = npf.ipmt(0.0824/12, per, 1*12, principal)
>>> ppmt = npf.ppmt(0.0824/12, per, 1*12, principal)
Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal
'pmt'.
>>> pmt = npf.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)
np.float64(-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_array = np.array(_rbl(rate, per, total_pmt, pv, when) * rate)
# Payments start at the first period, so payments before that
# don't make any sense.
ipmt_array[per < 1] = _value_like(ipmt_array, np.nan)
# If payments occur at the beginning of a period and this is the
# first period, then no interest has accrued.
per1_and_begin = (when == 1) & (per == 1)
ipmt_array[per1_and_begin] = _value_like(ipmt_array, 0)
# If paying at the beginning we need to discount by one period.
per_gt_1_and_begin = (when == 1) & (per > 1)
ipmt_array[per_gt_1_and_begin] = (
ipmt_array[per_gt_1_and_begin] / (1 + rate[per_gt_1_and_begin])
)
if np.ndim(ipmt_array) == 0:
# Follow the ufunc convention of returning scalars for scalar
# and 0d array inputs.
return ipmt_array.item(0)
return ipmt_array
def _rbl(rate, per, pmt, pv, when):
"""Remaining balance on loan.
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, 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, 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
--------
>>> import numpy as np
>>> import numpy_financial as npf
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.
>>> npf.pv(0.05/12, 10*12, -100, 15692.93)
np.float64(-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
>>> npf.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):
# Evaluate g(r_n)/g'(r_n), where g =
# fv + pv*(1+rate)**nper + pmt*(1+rate*when)/rate * ((1+rate)**nper - 1)
t1 = (r + 1) ** n
t2 = (r + 1) ** (n - 1)
g = y + t1 * x + p * (t1 - 1) * (r * w + 1) / r
gp = (n * t2 * x
- p * (t1 - 1) * (r * w + 1) / (r ** 2)
+ n * p * t2 * (r * w + 1) / r
+ p * (t1 - 1) * w / r)
return g / gp
# 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=None,
tol=None,
maxiter=100,
*,
raise_exceptions=False):
"""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
raise_exceptions: bool, optional
Flag to raise an exception when at least one of the rates
cannot be computed due to having reached the maximum number of
iterations (IterationsExceededException). Set to False as default,
thus returning NaNs for those rates.
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 np.all(close):
rnp1 = rn - _g_div_gp(rn, nper, pmt, pv, fv, when)
diff = abs(rnp1 - rn)
close = diff < tol
iterator += 1
rn = rnp1
if not np.all(close):
if np.isscalar(rn):
if raise_exceptions:
raise IterationsExceededError('Maximum number of iterations exceeded.')
return default_type(np.nan)
else:
# Return nan's in array of the same shape as rn
# where the solution is not close to tol.
if raise_exceptions:
raise IterationsExceededError(f'Maximum iterations exceeded in '
f'{len(close) - close.sum()} rate(s).')
rn[~close] = np.nan
return rn
def _irr_default_selection(eirr):
""" default selection logic for IRR function when there are > 1 real solutions """
# check sign of all IRR solutions
same_sign = np.all(eirr > 0) if eirr[0] > 0 else np.all(eirr < 0)
# if the signs of IRR solutions are not the same, first filter potential IRR
# by comparing the total positive and negative cash flows.
if not same_sign:
pos = sum(eirr[eirr > 0])
neg = sum(eirr[eirr < 0])
if pos >= neg:
eirr = eirr[eirr >= 0]
else:
eirr = eirr[eirr < 0]
# pick the smallest one in magnitude and return
abs_eirr = np.abs(eirr)
return eirr[np.argmin(abs_eirr)]
def irr(values, *, raise_exceptions=False, selection_logic=_irr_default_selection):
r"""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.
raise_exceptions: bool, optional
Flag to raise an exception when the irr cannot be computed due to
either having all cashflows of the same sign (NoRealSolutionException) or
having reached the maximum number of iterations (IterationsExceededException).
Set to False as default, thus returning NaNs in the two previous
cases.
selection_logic: function, optional
Function for selection logic when more than 1 real solutions is found.
User may insert their own customised function for selection
of IRR values.The function should accept a one-dimensional array
of numbers and return a number.
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
--------
>>> import numpy_financial as npf
>>> round(npf.irr([-100, 39, 59, 55, 20]), 5)
0.28095
>>> round(npf.irr([-100, 0, 0, 74]), 5)
-0.0955
>>> round(npf.irr([-100, 100, 0, -7]), 5)
-0.0833
>>> round(npf.irr([-100, 100, 0, 7]), 5)
0.06206
>>> round(npf.irr([-5, 10.5, 1, -8, 1]), 5)
0.0886
>>> npf.irr([[-100, 0, 0, 74], [-100, 100, 0, 7]]).round(5)
array([-0.0955 , 0.06206])
"""
values = np.atleast_2d(values)
if values.ndim != 2:
raise ValueError("Cashflows must be a 2D array")
irr_results = np.empty(values.shape[0])
for i, row in enumerate(values):
# If all values are of the same sign, no solution exists
# We don't perform any further calculations and exit early
same_sign = np.all(row > 0) if row[0] > 0 else np.all(row < 0)
if same_sign:
if raise_exceptions:
raise NoRealSolutionError('No real solution exists for IRR since all '
'cashflows are of the same sign.')
irr_results[i] = np.nan
# We aim to solve eirr such that NPV is exactly zero. This can be framed as
# simply finding the closest root of a polynomial to a given initial guess
# as follows:
# V0 V1 V2 V3
# NPV = ---------- + ---------- + ---------- + ---------- + ... = 0
# (1+eirr)^0 (1+eirr)^1 (1+eirr)^2 (1+eirr)^3
#
# by letting g = (1+eirr), we substitute to get
#
# NPV = V0 * 1/g^0 + V1 * 1/g^1 + V2 * 1/x^2 + V3 * 1/g^3 + ... = 0
#
# Multiplying by g^N this becomes
#
# V0 * g^N + V1 * g^{N-1} + V2 * g^{N-2} + V3 * g^{N-3} + ... = 0
#
# which we solve using Newton-Raphson and then reverse out the solution
# as eirr = g - 1 (if we are close enough to a solution)
else:
g = np.roots(row)
eirr = np.real(g[np.isreal(g)]) - 1
# Realistic IRR
eirr = eirr[eirr >= -1]
# If no real solution
if len(eirr) == 0:
if raise_exceptions:
raise NoRealSolutionError("No real solution is found for IRR.")
irr_results[i] = np.nan
# If only one real solution
elif len(eirr) == 1:
irr_results[i] = eirr[0]
else:
irr_results[i] = selection_logic(eirr)
return _ufunc_like(irr_results)
def npv(rate, values):
r"""Return the NPV (Net Present Value) of a cash flow series.
Parameters
----------
rate : scalar or array_like shape(K, )
The discount rate.
values : array_like, shape(M, ) or shape(M, N)
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 or array shape(K, M)
The NPV of the input cash flow series `values` at the discount
`rate`. `out` follows the ufunc convention of returning scalars
instead of single element arrays.
Warnings
--------
``npv`` considers a series of cashflows starting in the present (t = 0).
NPV can also be defined with a series of future cashflows, paid at the
end, rather than the start, of each period. If future cashflows are used,
the first cashflow `values[0]` must be zeroed and added to the net
present value of the future cashflows. This is demonstrated in the
examples.
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
--------
>>> import numpy as np
>>> import numpy_financial as npf
Consider a potential project with an initial investment of $40 000 and
projected cashflows of $5 000, $8 000, $12 000 and $30 000 at the end of
each period discounted at a rate of 8% per period. To find the project's
net present value:
>>> rate, cashflows = 0.08, [-40_000, 5_000, 8_000, 12_000, 30_000]
>>> np.round(npf.npv(rate, cashflows), 5)
np.float64(3065.22267)
It may be preferable to split the projected cashflow into an initial
investment and expected future cashflows. In this case, the value of
the initial cashflow is zero and the initial investment is later added
to the future cashflows net present value:
>>> initial_cashflow = cashflows[0]
>>> cashflows[0] = 0
>>> np.round(npf.npv(rate, cashflows) + initial_cashflow, 5)
np.float64(3065.22267)
The NPV calculation may be applied to several ``rates`` and ``cashflows``
simulatneously. This produces an array of shape ``(len(rates), len(cashflows))``.
>>> rates = [0.00, 0.05, 0.10]
>>> cashflows = [[-4_000, 500, 800], [-5_000, 600, 900]]
>>> npf.npv(rates, cashflows).round(2)
array([[-2700. , -3500. ],
[-2798.19, -3612.24],
[-2884.3 , -3710.74]])
"""
values_inner = np.atleast_2d(values).astype(np.float64)
rate_inner = np.atleast_1d(rate).astype(np.float64)
if rate_inner.ndim != 1:
msg = "invalid shape for rates. Rate must be either a scalar or 1d array"
raise ValueError(msg)
if values_inner.ndim != 2:
msg = "invalid shape for values. Values must be either a 1d or 2d array"
raise ValueError(msg)
output_shape = _get_output_array_shape(rate_inner, values_inner)
out = np.empty(output_shape)
_cfinancial.npv(rate_inner, values_inner, out)
return _ufunc_like(out)
def mirr(values, finance_rate, reinvest_rate, *, raise_exceptions=False):
r"""
Return the Modified Internal Rate of Return (MIRR).
MIRR is a financial metric that takes into account both the cost of
the investment and the return on reinvested cash flows. It is useful
for evaluating the profitability of an investment with multiple cash
inflows and outflows.
Parameters
----------
values : array_like, 1D or 2D
Cash flows, where the first value is considered a sunk cost at time zero.
It must contain at least one positive and one negative value.
finance_rate : scalar or 1D array
Interest rate paid on the cash flows.
reinvest_rate : scalar or D array
Interest rate received on the cash flows upon reinvestment.
raise_exceptions: bool, optional
Flag to raise an exception when the MIRR cannot be computed due to
having all cash flows of the same sign (NoRealSolutionException).
Set to False as default,thus returning NaNs in the previous case.
Returns
-------
out : float or 2D array
Modified internal rate of return
Notes
-----
The MIRR formula is as follows:
.. math::
MIRR =
\\left( \\frac{{FV_{positive}}}{{PV_{negative}}} \\right)^{\\frac{{1}}{{n-1}}}
* (1+r) - 1
where:
- \(FV_{positive}\) is the future value of positive cash flows,
- \(PV_{negative}\) is the present value of negative cash flows,
- \(n\) is the number of periods.
- \(r\) is the reinvestment rate.
Examples
--------