numpy.ipmt()

Compute the interest portion of a payment.

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