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Scott,
I read with interest your article,
"Amortized
Loans." At the end of the article you indicate the
explanation of how to derive the formulae as being covered in
another article.
Can you please tell me where I can
find this?
Thank you very much,
Luke
Luke,
Here is the derivation of the loan
formula from scratch:
F --> Future Value
P --> Principal
A --> Annuity
n --> period number
i --> interest
#1) F0=P
Simply states that the future value at time 0 is equal to the
principal.
#2) F1=P+Pi-A=P(1+i)-A
The future value, after one period, is equal to the principal (P)
with the addition of the interest in dollars (Pi) and then you
subtract the periodic payment (A). The second part of the equation
factors P from P+Pi. This step is repeated for all periods.
#3)
F2=F1(1+i)-A=P((1+i)^2)-A(1+i)-A
The future value, after the second period is equal to the new
balance (F1) with the addition of the interest in dollars (F1i) and
then you subtract the periodic payment (A). The second part of the
equation substitutes P(1+i)-A for F1 (from equation #2).
#4)
F3=F2(1+i)-A=P((1+i)^3)-A((1+i)^2)-A(1+i)-A
Applying the same procedure in steps #2 and #3 we get the future
value at period #3.
#5)
F4=F3(1+i)-A=P((1+i)^4)-A((1+i)^3)-A((1+i)^2)-A(1+i)-A
Applying the same procedure in steps #2 and #3 we get the future
value at period #4.
#6)
Fn=P((1+i)^n)-A[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]
This is the future value at period n, or any period, based on the
general equation being expanded. Additionally, A is factored from
all terms.
 The
first part of #6, P((1+i)^n), is easy to work with because it is a
closed form, however, we need to find the closed form for the
remaining part of that formula, therefore, for the general case of
[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]:
#7)
y=1+a^1+a^2+a^3+a^4+a^5+a^6+....a^k
This formula is very similar to the case we are trying to solve
because it's the addition of a to all powers from 0 to k.
#8)
y-1=a^1+a^2+a^3+a^4+a^5+a^6+....a^k
Subtracting 1 from both sides.
#9)
(y-1)/a=1+a^1+a^2+a^3+a^4+a^5+....a^(k-1)
Dividing all terms by a.
#10) (y-1)/a+a^k=1+a^1+a^3+a^3+a^4+a^4+a^5+....a^k
Adding a^k to both sides because...
#11)
1+a^1+a^2+a^3+a^4+a^5+....a^k=y
we get the very same equation we started with (#7).
Therefore,
#12) (y-1)/a+a^k=y
Substituting #7 into #10.
#13) y-1+a^(k+1)=ya
Multiplying all terms by a.
#14) y-ya=1-a^(k+1)
Putting like terms on opposite sides of the equation.
#15) y(1-a)=1-a^(k+1)
Factoring y out from left side.
#16) y=(1-a^(k+1))/(1-a)
Solving for y.
Going back to the general case for
the loan formula in #6:
#17) [((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]
#18)
y=[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]
Set y equal to #17.
#19) y=(1-a^(k+1))/(1-a)
This is the solution to #18 based on #16 where k=n-1 (the highest
power in #18) and a=1+i.
#20) y=(1-((1+i)^n))/(-i)
Substitute k=n-l and a=1+i in #19.
#21) y=(((1+i)^n)-1)/I
Multiplying numerator and denominator by -1.
Back to #6:
#22)
Fn=P((1+i)^n)-A[((1+i)^(n-1))+((1+i)^(n-2))+((1+i)^(n-3))+....+1]
#23) Fn=P((1+i)^n)-A[y]
Substituting y from equation defined in #18.
#24) Fn=P((1+i)^n)-A[(1-((1+i)^n))/i]
Substituting again for y in equation #21.
Note: Fn=0, the last future value of
the loan, when the loan is repaid, therefore,
#25) 0=P((1+i)^n)-A[(((1+i)^n)-1)/i]
(at period n)
Substituting 0 for Fn.
#26) A[(((1+i)^n)-1)/i]=P((1+i)^n)
Separating terms with A and P on both sides of the equation.
#27) A=[P((1+i)^n)][i/(((1+i)^n)-1)]
Solving for A.
#27) A=Pi/(1-((1+i)^(-n)))
Simplifying terms by multiplying numerator and denominator by
((1+i)^(-n)).
We have it! Equation #27 is the
formula used to calculate the periodic payment (A) for a loan of
principal (P) at the periodic rate (i) for n periods.
Here's a quick example:
What is the monthly payment for a $10,000 loan with an APR of 24%
for 5 years?
Solution:
P=10000
i=0.24/12=0.02 (period is monthly so 2% is the monthly rate)
n=5(12)=60 months or 60 periods
Formula is:
A=Pi/(1-((1+i)^(-n)))
A=10000(0.02)/(1-((1+0.02)^(-60)))
A=200/(1-0.30478)=200/0.69522
A=287.68
Therefore, the monthly payment is
$287.68.
I hope you enjoyed this derivation!
Regards,
Scott
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