loan analysis program (Proof)
leung at imsvax.UUCP
leung at imsvax.UUCP
Wed Feb 8 04:42:49 AEST 1984
Posted Feb 2 stevens at hsi.UUCP in net.source has a program for
loan analysis. There is no analytic proof of the formula he used.
The following is a proof of the formula.
The following lemma will be useful.
Lemma.
n
_
\ i n+1
1 + a / (1 + a) = (1 + a) (*)
-
i=0
Proof:
For i=0, lhs = 1 + a = rhs
Suppose for i=n, (*) holds
For i=n+1,
n+2
lhs = (1 + a)
n+1
= (1 + a) (1 + a)
n
_
\ i
= (1 + a) (1 + a / (1 + a) )
-
i=0
n+1
_
\ i
= 1 + a + a / (1 + a)
-
i=1
n+1
_
\
= 1 + a / (1 + a) = rhs
-
i=0 Q.E.D.
Back to the Loan Equation:
Theorem:
n
AI (1 + I)
P = ------------
n
(1 + I) - 1
where P = Monthly Payment
A = Loan Amount
I = Monthly Interest Rate
(Annual Interest Rate / 12)
n = Number of Payments
Proof:
For n=1, the final balance is
A(1 + I) - P = 0
For n=2, the final balance is
(A(1 + I) - P)(1 + I) - P = 0
2
A(1 + I) - P(1 + (1 + I)) = 0
For n=3, the final balance is
2
(A(1 + I) - P(1 + (1 + I)))(1 + I) - P = 0
3 2
A(1 + I) - P(1 + (1 + I) + (1 + I) ) = 0
For n=n, the final balance is
n-1
_
n \ j
A(1 + I) - p / (1 + I) = 0
-
j=0
then,
n-1
_
n \ j
AI(1 + I) - pI / (1 + I) = 0
-
j=0
Using the above Lemma,
n n
AI(1 + I) - p((1 + I) - 1) = 0
This completes the proof.
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