numpy.ipmt(rate, per, nper, pv, fv=0, when='end')
[source]
Compute the interest portion of a payment.
Parameters: 


Returns: 

The total payment is made up of payment against principal plus interest.
pmt = ppmt + ipmt
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
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