NumPy Financial functions: ipmt() function

Last update on August 19 2022 21:51:10 (UTC/GMT +8 hours)

numpy.ipmt() function

The ipmt() function is used to compute the interest portion of a payment.

Syntax:

numpy.ipmt(rate, per, nper, pv, fv=0.0, when='end')

Version: 1.15.0

Parameter:

Name	Description	Required / Optional
rate	Rate of interest as decimal (not per cent) per period scalar or array_like of shape(M, )	Required
per	Interest paid against the loan changes during the life or the loan. The per is the payment period to calculate the interest amount. scalar or array_like of shape(M, )	Required
nper	Number of compounding periods scalar or array_like of shape(M, )	Required
pv	Present value scalar or array_like of shape(M, )	Required
fv	Future value scalar or array_like of shape(M, )	Optional
when	When payments are due ('begin' (1) or 'end' (0)). Defaults to {'end', 0}. {{'begin', 1}, {'end', 0}}, {string, int}	Optional

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.

Notes:

The total payment is made up of payment against principal plus interest.

pmt = ppmt + ipmt

NumPy.ipmt() method Example - 1:

>>> import numpy as np
>>> np.allclose(ipmt + ppmt, pmt)

Output:

True

NumPy.ipmt() method Example - 2:

What is the amortization schedule for a 1 year loan of $10000 at 8.5% interest per year compounded monthly?

>>> import numpy as np
>>> principal = 10000.00
>>> per = np.arange(1*12) + 1 # The 'per' variable represents the periods of the loan. Remember that financial equations start the period count at 1!
>>> ipmt = np.ipmt(0.0850/12, per, 1*12, principal)
>>> ppmt = np.ppmt(0.0850/12, per, 1*12, principal)
>>> pmt = np.pmt(0.0850/12, 1*12, principal) # Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal 'pmt'.
>>> 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))

Output:

 1  -801.36   -70.83  9198.64
 2  -807.04   -65.16  8391.59
 3  -812.76   -59.44  7578.84
 4  -818.51   -53.68  6760.32
 5  -824.31   -47.89  5936.01
 6  -830.15   -42.05  5105.86
 7  -836.03   -36.17  4269.83
 8  -841.95   -30.24  3427.88
 9  -847.92   -24.28  2579.96
10  -853.92   -18.27  1726.03
11  -859.97   -12.23   866.06
12  -866.06    -6.13    -0.00

NumPy.ipmt() method Example - 3:

What is the amortization schedule for a 1 year loan of $10000 at 8.5% interest per year compounded monthly?

>>> import numpy as np
>>> principal = 10000.00
>>> per = np.arange(1*12) + 1 # The 'per' variable represents the periods of the loan. Remember that financial equations start the period count at 1!
>>> ipmt = np.ipmt(0.0850/12, per, 1*12, principal)
>>> ppmt = np.ppmt(0.0850/12, per, 1*12, principal)
>>> pmt = np.pmt(0.0850/12, 1*12, principal) # Each element of the sum of the 'ipmt' and 'ppmt' arrays should equal 'pmt'.
>>> interestpd = np.sum(ipmt)
>>> np.round(interestpd, 2)

Output:

-466.37

NumPy.ipmt() method Example - 4:

What is the amortization schedule for a 1 year loan of $10000 at 8.5% interest per year compounded monthly?

>>> import numpy as np
>>> principal = 10000.00
>>> per = np.arange(1*12) + 1 # The 'per' variable represents the periods of the loan. Remember that financial equations start the period count at 1!
>>> ipmt = np.ipmt(0.0850/12, per, 1*12, principal)
>>> interestpd = np.sum(ipmt)
>>> np.round(interestpd, 3)

Output:

-466.374

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