numpy.ipmt()

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

Compute the interest portion of a payment.

Parameters:

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.

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.

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?

1	`>>> 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

Links:

https://docs.scipy.org/doc/numpy-1.11.0/reference/generated/numpy.ipmt.html

doc_NumPy

2025-01-10 15:47:30

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