comparison DEPENDENCIES/mingw32/Python27/Lib/site-packages/numpy/lib/financial.py @ 87:2a2c65a20a8b

Add Python libs and headers

+"""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.asarray, [rate, nper, pv, fv, when])
+    temp = (1+rate)**nper
+    miter = np.broadcast(rate, nper, pv, fv, when)
+    zer = np.zeros(miter.shape)
+    fact = np.where(rate == zer, nper + zer,
+                    (1 + rate*when)*(temp - 1)/rate + zer)
+    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 res.size == 0:
+        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