xirr function returns None

[cmodi369]

Details

import datetime
from pyxirr import xirr

dates = [
    datetime.date(2020, 5, 27),
    datetime.date(2020, 5, 27),
    datetime.date(2020, 5, 27),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28),
    datetime.date(2020, 5, 28)
]

values = [187.5, -30, 187.5, 187.5, 187.5, -188, -188, -188, -188, -188]

# This returns `None`
xirr(dates, values)

# Update value from -30 to -25 
values[1] = -25

# This returns `8.181769983749503e+75`
xirr(dates, values)

It would be good if instead of None exception can be raised. I am not able to understand why xirr function is returning None in the first case. Let me know how I can fix this.

Let me know your thoughts.

Thanks.

[Anexen]

Hi @cmodi369. The xirr function uses an iterative root‑finding algorithm to solve the following equation:

$$\sum_{i=1}^n \frac{P_i}{(1 + rate)^{(d_i - d_0)/365}} = 0$$

as noted in the docs, if it can't find a solution it simply returns None. If you prefer an exception instead of None, you can simply wrap the call in your code:

rate = xirr(dates, values)

if rate is None:
    raise ValueError

In your example, the solver fails to converge - most likely because the cash flow sequence doesn't produce a well-defined IRR. For the IRR to be meaningful, there must be an exactly one sign change. Multiple sign changes can result in multiple mathematical roots that the solver is able to find (or none in some cases).

In your example, the algorithm is working with very small exponents (1/365), which makes it highly sensitive to even minor changes. Some data preprocessing techniques may help you, e.g. aggregating cash flows that occur on the same date can reduce the problem's complexity and improve numerical stability. In your case, everything is aggregated into two payments, for which XIRR can be solved analytically (unfortunately, for three or more payments, it's impossible to solve the equation analytically).