The fixed monthly payment for a fixed rate mortgage is the amount paid by the borrower every month that ensures that the loan is paid off in full with interest at the end of its term. This monthly payment c depends upon the monthly interest rate r (expressed as a fraction, not a percentage, i.e., divide the quoted yearly nominal percentage rate by 100 and by 12 to obtain the monthly interest rate), the number of monthly payments N called the loan's term, and the amount borrowed P0 known as the loan's principal; rearranging the formula for the present value of an ordinary annuity we get the formula for c:

c = (r / (1 − (1 + r) − N))P0

For example, for a home loan for $200,000 with a fixed yearly nominal interest rate of 6.5% for 30 years, the principal is P0 = 200000, the monthly interest rate is r = 6.5 / 100 / 12, the number of monthly payments is N = 30 * 12 = 360, the fixed monthly payment equals $1264.14. This formula is provided using the financial function PMT in a spreadsheet such as Excel. In the example, the monthly payment is obtained by entering either of the these formulas:

=PMT(6.5/100/12,30*12,200000)

=((6.5/100/12)/(1-(1+6.5/100/12)^(-30*12)))*200000

= 1264.14

This monthly payment formula is easy to derive, and the derivation illustrates how fixed-rate mortgage loans work. The amount owed on the loan at the end of every month equals the amount owed from the previous month, plus the interest on this amount, minus the fixed amount paid every month.

Amount owed at month 0:

P0

Amount owed at month 1:

P1 = P0 + P0 * r − c ( principle + interest - payment)

P1 = P0(1 + r) − c (equation 1)

Amount owed at month 2:

P2 = P1(1 + r) − c

Using equation 1 for P1

P2 = (P0(1 + r) − c)(1 + r) − c

P2 = P0(1 + r)2 − c(1 + r) − c (equation 2)

Amount owed at month 3:

P3 = P2(1 + r) − c

Using equation 2 for P2

P3 = (P0(1 + r)2 − c(1 + r) − c)(1 + r) − c

P3 = P0(1 + r)3 − c(1 + r)2 − c(1 + r) − c

Amount owed at month N:

PN = PN − 1(1 + r) − c

PN = P0(1 + r)N − c(1 + r)N − 1 − c(1 + r)N − 2.... − c

PN = P0(1 + r)N − c((1 + r)N − 1 + (1 + r)N − 2.... + 1)

PN = P0(1 + r)N − c(S) (equation 3)

Where S = (1 + r)N − 1 + (1 + r)N − 2.... + 1 (equation 4) (see geometric progression)

S(1 + r) = (1 + r)N + (1 + r)N − 1.... + (1 + r) (equation 5)

With the exception of two terms the S and S(1 + r) series are the same so when you subtract all but two terms cancel:

Using equation 4 and 5

S(1 + r) − S = (1 + r)N − 1

S((1 + r) − 1) = (1 + r)N − 1

S(r) = (1 + r)N − 1

S = ((1 + r)N − 1) / r (equation 6)

Putting equation 6 back into 3:

PN = P0(1 + r)N − c(((1 + r)N − 1) / r)

PN will be zero because we have paid the loan off.

0 = P0(1 + r)N − c(((1 + r)N − 1) / r)

We want to know c

c = (r(1 + r)N / ((1 + r)N − 1))P0

Divide top and bottom with (1 + r)N

c = (r / (1 − (1 + r) − N))P0

This derivation illustrates three key components of fixed-rate loans: (1) the fixed monthly payment depends upon the amount borrowed, the interest rate, and the length of time over which the loan is repaid; (2) the amount owed every month equals the amount owed from the previous month plus interest on that amount, minus the fixed monthly payment; (3) the fixed monthly payment is chosen so that the loan is paid off in full with interest at the end of its term and no more money is owed.

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