Amortization Formula

Then the loan amortization formula for monthly payment (P) is:

P = A * ( R - 1 ) / ( 1 - R^-N )

Similarly, other amortization formulas are as following:

Amount (A):
A = P * ( 1 - R^-N ) / ( R - 1 )

Term(N):
N = -ln [ 1 - A ( R - 1 ) / P ] / ln[ R ]

Payoff (or balloon) (P[n]):
P[n] = (A - P * ( 1 - R^-n ) / ( R - 1 ) ) Rⁿ

Amount owed after n payment (A[n]):
A[n] = ( A - P * ( 1 - R^-(n-1) ) / ( R - 1 ) ) R^(n-1) - P

If you are interested in derivative of Amortization formula, here it is:

	The Derivation

Let p be the monthly payment.
Let N be the number of monthly payments.
Let I be the annual interest rate.
So the monthly interest is i=I/12
Let P be the principal.
Let the amount borrowed be A, P[0].

At the moment the borrower gets the amount of the loan,
the principal owed is that amount borrowed.

A = P[0]

A month later, the principal has accrued interest.
The amount owed increases by the interest accrued.

P'[1] = P[0]*(1+i)
or
P'[n+1] = P[n]*(1+i) in general

But then the borrower pays the monthly payment.

P[1] = P'[1]-p = P[0]*(1+i) - p
or
P[n+1] = P'[n+1]-p = P[n]*(1+i) - p
 
Let R=1+i

P[1] = P[0]*R-p
P[2] = P[1]*R-p = P[0]*RR-pR-p
P[3] = P[2]*R-p = P[1]*RR-pr-p = P[0]*RRR-pRR-pR-p
.
.
.
P[N] = P[0]R^N-pR^(N-1)-pR^(N-2)...-pR-p

After N payments the loan is paid off, P[N]=0.
Remember that P[0]=A

0 = A*R^N-pR^(N-1)...-pR-p
A*R^N = p + pR + pRR + pR^3 + ... + pR^(N-1)
A = (1/R^N)*(p + pR + pRR + pR^3 + ... + pR^(N-1))
A = p( (1/R)^N + (1/R)^(N-1) + (1/R)^(N-2) + ... + 1/R )
A = (p/R)*( (1/R)^(N-1) + (1/R)^(N-2) + ... + (1) )

Note the finite sum of a geometric series.
S[n] = a(1-r^(n+1))/(1-r) =  a + ar + arr + ... + ar^n  <---Check

A = (p/R)*(1-(1/R)^(N))/(1-(1/R))
A = p*(1-(1/R)^(N))/(R-1)
p = A*(R-1)/(1-(1/R)^(N))

p = A*(R-1)/(1-R^-N)

A = p(1-R^-n)/(R-1)


(E.O.P.)
 n f r
 d   o
     o
     f


	Test:

  Let A=100,000.  i=6%/12=0.5%=0.005-->R=1.005
N=30years*12months/year*1pmt/month=360pmts

		p = 100000 * (1.005-1) / (1-1.005^(-360))
		p = 599.55

Survey says... 	p = 599.55  (Cheers!) 


	Check: The finite sum of a geometric series:

S[n] = a + ar + arr + arrr + ... + ar^n
     = a + r*(a + ar + arr + ... + ar^(n-1)) 
     = a + r*(S[n-1])

So...
S[n-1] = (S[n] - a)/r
Note also that...
S[n] = S[n-1] + ar^n

S[n] = (S[n] - a)/r + ar^n
S[n]r = S[n] - a + ar^(n+1)
S[n](r-1) = a(r^(n+1)-1)

S[n] = a(r^(n+1)-1)/(r-1)
or
S[n] = a(1-r^(n+1))/(1-r)

So, 1+r+r^2+r^3+...+r^n = (1-r^(n+1))/(1-r)

The term, N:
A = p(1-R^(-N))/(R-1)
A(R-1) = p(1-R^(-N))
A(R-1)/p = 1-R^(-N)
R^(-N) = 1-A(R-1)/p
ln[R^(-N)] = ln[1-A(R-1)/p] (you can use ln or log)
-N ln[R] = ln[1-A(R-1)/p]
N = -ln[1-A(R-1)/p]/ln[R]

Common Derivations, Schedules:

After you figure out the payment, p,
you can iterate through each month.
The initial amount borrowed, A, 
is P[0], the principal as it first starts out.
The principal grows to P[0]*(1+i) after one month (i=I/12).
The principal after the first monthly payment is

P[1]=P[0]*(1+i)-p

In general, the principal for the nth+1 payment is 
(base on the nth payment):

P[n+1] = P[n]*(1+i)-p

The interest portion is P[n]*i.
There are some rounding issues,
one resolution is to round the payment up to the nearest penny,
and have the last payment payoff the principal 
rather than it being exactly p.

If you don't want to iterate for the principal
you could use

P[n] = (A -p(1 - R^(-n))/(R - 1))R^n
or
P[n] = ( A - a[p,R,n] ) R^n
where
a[p,R,n] = p(1 - R^(-n))/(R-1)
and
R = (1+I/12)

Proof:
The principal starts out as the amount borrowed
A = P[0]
We'll denote the principle just before payment as P', so...
After the first month interest accrues
P'[1] = P[0] + i*P[0] = P[0] + i*P[0] = P[0]R
Then the fist payment comes
P[1] = P'[1] - p = P[0]R - p
In general
P[n] = P[n-1]R - p
P'[n] = P[n-1]R
P'[n] = (P'[n-1] - p)R
If we are primarily interested in P'[n] as a means at getting the final payment,
then we'll use
P'[n]	= (P'[n-1] - p)R

P'[n-1]	= {P'[n-2] - p}R
P'[n]	= {[(P'[n-2]-p)R] - p}R
	= {[P'[n-2]R-pR] - p}R
	= {P'[n-2]R - pR - p}R
	= {[(P'[n-3]-p)R]R - pR - p}R
	= {[(P'[n-3]R - pR)]R - pR - p}R
	= {P'[n-3]RR  - pRR - pR - p}R
	= {P'[n-3]R^2  - pR^2 - pR - p}R
	= P'[n-3]R^3  - pR^3 - pR^2 - pR
	= P'[n-m]R^m  - pR^m - pR^(m-1) ... - pR
Allow m ->n
	= P'[0]R^n  - pR^n - pR^(n-1) ... - pR
	= (P'[0]  - (pR^n - pR^(n-1) ... - pR)/R^n)R^n

Since A = P[0] = P'[0]-p (by definition)

	= (A - p) - (pR^n + pR^(n-1) ... + pR)/R^n)R^n
	= (A - (pR^n + pR^n + pR^(n-1) ... + pR)/R^n)R^n
	= (A - (pR^(n-1) ... + pR)/R^n)R^n
	= (A - pR^(-1) ... + pR^(1-n))R^n
	= (A - p(R^(-1) ... + R^(1-n))R^n
	= (A - p((1/R) ... + (1/R)^(n-1))R^n
	= (A - (p/R)(1 ... + (1/R)^n)R^n

remember that 1+r+r^2+r^3+...+r^n = (1-r^(n+1))/(1-r)
map r to 1/R to get (1-(1/R)^(n+1))/(1-(1/R))
so 1 ... + (1/R)^n = (1-(1/R)^(n+1))/(1-(1/R))
		= (1-R^(-n-1))/(1-(1/R))
		= (R-R^(-n))/(R-1)
		= R(1-R^(-n))/(R-1)

replacing this in...
	= (A - (p/R)(R(1-R^(-n))/(R-1))R^n
	= (A - p(1-R^(-n))/(R-1))R^n
	= (A - p(1-R^(-n))/(R-1))R^n
	= (A - a[p,R,n])R^n
E.O.P.

Common derivations, variable rates / split rate loans
you take the loan through each rate period separately.
The dynamic remains
Principal just before nth+1 payment
P'[n+1] = P[n]*(1+i)
Principal just after nth+1 payment
P[n+1] = P'[n+1]-p = P[n]*(1+i) - p
The particular case where
there are N payments, the first n of which are interest free.
Solve for the payment:
Consider this...
Let A be the initial amount.
The principal after n payments of amount p is (A-np).
Then the loan gets amortiziced over (N-n) payment at interest I.
Based on the formula above,
p = (A-np)(R-1)/(1-R^(n-N))
solve for p...
Let k= (R-1)/(1-R^(n-N))
p=(A-np)k
p=Ak-nkp
p+nkp=Ak
p(1+kn)=Ak

p=Ak/(1+kn)






 

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