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From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz)
Subject: sci.math FAQ: Formula for Compound Interest
Summary: Part 36 of many, New version,
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Archive-Name: sci-math-faq/compoundInterest
Last-modified: December 8, 1994
Version: 6.2
Formula to compute compound interest.
Here's a formula which can be used in 123, Excel, Wings and Dynaplan:
------- Input this data -------------------------------
principal amount = E9 ( in dollars )
Amortization Period = d10 ( in years ie 6 mon = .5 )
Payments / year = D11 ( 12 = monthly, 52 = weekly )
Published Interest rate = D12 ( ie 9 % = 0.09 )
Times per year Int calculated = d13 ( CDN mortgage use 2
US mortgage use 12
all other loans use 12 )
----- Calculate the proper rate of interest -----------
e14 = Effective annual rate = EXP(D13*LN(1+(D12/D13)))-1
e15 = Interest rate per payment = (EXP(LN(E14+1)/(D10*D11))-1)*D10*D11
e17 = Payments = APMT(E9,E15/D11,D10*D11) ( both these functions are
= PMT (E9,E15/D11,D10*D11) ( identical,diff spreadsheet)
APMT( principal amount,interest rate per period,# periods )
( this is a standard function on any true commercial spreadsheet)
OR use the following if done using a calculator
= Payments = P*I/[1-(I+1)^-T]
= E9*(E15/D11)/(1-((E15/D11) +1)**(-1*D10*D11))
Total interest cost = E17*D10*D11-E9
-- Use these formulas if you wish to generate an amortization table --
always add up to 'Payments (e17)'
Interest per payment = current balance * ( E15 / D11 )
Principal per payment = current balance - Interest per payment
new current balance = current balance - Principal per payment -
(extra payment)
keep repeating until 'new current balance' = 0
Derivation of Compound Interest Rate Formula
Suppose you deposited a fixed payment into an interest bearing account
at regular intervals, say monthly, at the end of each month. How much
money would there be in the account at the end of the nth month (at
which point you've made n payments)?
Let i be the monthly interest rate as a fraction of principle.
Let x be the amount deposited each month.
Let n be the total number of months.
Let p[k] be the principle after k months.
So the recursive formula is:
p[n] = x + ((1 + i) p[n - 1]) eq 1
This yields the summation:
p[n] = sum_(k = 0)^(n - 1)x (1 + i)^k
The way to solve this is to multiply through by (1 + i) and subtract
the original equation from the resulting equation. Observe that all
terms in the summation cancel except the last term of the multiplied
equation and the first term of the original equation:
pi p[n] = x ((1 + i)^n - 1)
or
p[n] = x ((1 + i)^n - 1)/i
Now suppose you borrow p at constant interest rate i . You make
monthly payments of x . It turns out that this problem is identical to
taking out a balloon loan of p (that is it's all due at the end of
some term) and putting payments of x into a savings account. At the
end of the term you use the principle in the savings account to pay
off the balance of the loan. The loan and the savings account, of
course, must be at the same interest rate. So what we want to know is:
what monthly payment is needed so that the balance of the savings
account will be identical to the balance of the balloon loan after n
payments?
The formula for the principal of the balloon loan at the end of the n
th month is:
p[n] = p[0] (1 + i)^n
So we set this expression equal to the expression for the the savings
account, and we get:
p[0] (1 + i)^n = x ((1 + i)^n - 1)/i
or solving for x:
x = p[0] (1 + i)^n i/((1 + i)^n - 1)
If (1 + i)^n is large enough (say greater than 5), here is an
approximation for determining n from x , p , and i :
n ~= -ln(ln(x/(ip)))/ln(1 + i)
The above approximation is based upon the following approximation:
ln(y - 1) ~= ln y - 1/y
Which is within 2 For example, a $100000loan at 1%monthly, paying
$1028.61 per month should be paid in 360 months. The approximation
yields 358.9 payments.
If this were your 30 year mortgage and you were paying $1028.61 per
month and you wanted to see the effect of paying $1050per month, the
approximation tells you that it would be paid off in 303.5 months (25
years and 3.5 months). If you stick 304 months into the equation for y
>= 5 , you get $1051.04, so it is fairly close. This approximation
does not work, though, for very small interest rates or for a small
number of payments. The rule is to get a rough idea first of what x
is. If that is greater than 5, the approximation works pretty well. In
the examples given, (1 + i)^n is about 36.
Finding (1 + i)^n given i , n , and x is not as easy. If i is less
than 5%per payment period, the following equation approximately holds
for i:
p
There is no direct solution to this, but you can do it by
Newton-Raphson approximation. Begin with a guess, i[0]. Then apply:
i = -(1/n) ln(1 - ip/x)
You must start with i[k + 1] = i[k] - (x(1 - i[k]p/x) (ni[k] + ln(1 -
i[k]p/x)))/(xn(1 - i[k]p/x) - p) too big, because the equation for i
has a solution at i , and that's not the one you want to end up with.
Example: Let the loan be for i = 0 , p = $10000 per week for 5 years (
x = $50 ). Let n = 260 %per annum or 0.3846%per week. Since i[0] = 20
must be a fraction rather than a percent, i . Then, applying eq 11:
i[0] = 0.003846
The series is clearly beginning to converge here.
To get i[1] = 0.003077 ; i[2] = 0.002479 ; i[3] = 0.002185 ; i[4] =
0.002118 ; i[5] = 0.002115 ; as an annual percentage rate, multiply by
52 weeks in a year and then by 100%, so i[5] %per annum. Substituting
i[5] = 10.997 back into eq 7, we get i[5] , so it works pretty well.
References
The theory of interest. Stephen G. Kellison. Homewood, Ill., R. D.
Irwin, 1970.
_________________________________________________________________
alopez-o@barrow.uwaterloo.ca
Tue Apr 04 17:26:57 EDT 1995