# Mortgage Formula with Examples

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The mortgage formula for repayment of a mortgage is presented along with exmaples and their solutions.

Also if needed, a mortgage calculator is included.

## Mortgage Payment Process

A mortgage is a type of loan that helps people or businesses buy a property like a house or a commercial building. When you take out
a mortgage, the bank or lender gives you the money you need to pay for the property right away. You then pay back the loan, along with interest,
over a number of years. This can be a long time, usually around 15 to 30 years.

Let \( R \) be the annual interest and \( N \) the number of years to repay the loan.

Since the payment of the loan is on a monthly basis, let the monthly rate of interest be:

\( \qquad r = \dfrac{R}{12} \)

and the number of months in \( N \) years be

\( \qquad n = 12 \times N \)

let \( P \) be the amount you need to buy a house for example. So the bank will agree to lend you the money on the condition that you pay back the loan plus interest.

Let \( M \) be the monthly payments for the mortgage.

At the start, the amount owed is: \( \; A_0 = P \)

The amount owed at the end of the first month is

\( \qquad A_1 = A_0(1 + r) - M \\ \qquad \qquad = P(1 + r) - M \)

Explanantion: within a month, the amount owed \( p \) has inreased by \( r P \) which gives \( P(1+r) \) but you pay back \( M \) and hence the \( - M \) in the above expression.

The amount owed at the end of the second month is

\( \qquad A_2 = A_1 (1+r) - M \\ \qquad \qquad = \left( P(1 + r) - M \right)(1+r) - M \\ \qquad \qquad \qquad = P(1+r)^2 - M(1+(1+r)) \)

Explanantion: within a month, the last amount owed \( A_1 \) has inresred by \( r A_1 \) which gives \( A_1(1+r) \) but you pay back \( M \) and hence the \( - M \) in the above expression.

The amount owed at the end of the third month is

\( \qquad A_3 = A_2(1 + r) - M \\ \qquad \qquad = \left( P(1+r)^2 - M(1+(1+r)) \right) (1+r) - M \\ \qquad \qquad \qquad = P(1+r)^3 - M (1 + (1+r) + (1+r)^2 ) \)

Explanantion: within a month, the last amount owed \( A_2 \) has inresred by \( r A_2 \) which gives \( A_2(1+r) \) but you pay back \( M \) and hence the \( - M \) in the above expression.

.

.
and so on

Examining \( A_1 \), \( A_2 \) and \( A_3 \), we can write that the amount owed at the end of the \( n^{th} \) month is

\( \qquad A_n = P(1+r)^n - M (1 + (1+r) + (1+r)^2 + ... + (1+r)^{n-1} ) \qquad (I) \)

The sum

\( \qquad 1 + (1+r) + (1+r)^2 + ... + (1+r)^{n-1} \qquad (II)\)

is a geometric sequence sum and we need to simplify it.

## Review Geometric Sequence Sum

The geometric sequence sum \( S_n \) defined by

\( \qquad S_n = a_1 + R \times a_1 + R^2 \times a_1 .... R^n \times a_1 \)

is given by the formula

\[ \qquad S = a_1 \dfrac{R^{n+1} - 1}{R - 1} \]
\( \qquad a_1 \) is the first term and \( R \) is the common factor.

## Formula for Payment

Use the above formula to simplify the geometric sequence sum in \( II \) above

\( \qquad 1 + (1+r) + (1+r)^2 + ... + (1+r)^{n-1} \\ \qquad \qquad = \dfrac{(1+r)^{n}-1}{(1+r) - 1} \\ \qquad \qquad \qquad = \dfrac{(1+r)^{n} - 1}{r} \)

Hence

The amount \( A_n \) owed after n months is given by

\( \qquad A_n = P(1+r)^n - M (1 + (1+r) + (1+r)^2 + ... + (1+r)^{n-1} ) \\ \qquad\qquad = P(1+r)^n - M \dfrac{(1+r)^{n}-1}{r} \)

Payment is made until the amount \( A_n \) owed is equal to zero. Hence the equation

\( \qquad P(1+r)^n - M \dfrac{(1+r)^{n} - 1}{r} = 0 \)

Solve the above to find the monthly payment

\[ \qquad M = \dfrac{r \; P\; (1+r)^n}{(1+r)^{n} - 1} \]

The formula may also be written as

\[ \qquad M = \dfrac{ \left(\frac{R}{12}\right) \; P\; (1+\frac{R}{12})^{12\times N}}{(1+\frac{R}{12})^{12 \times N} - 1} \qquad (III) \]

where \( R \) is the annual interest, \( N \) is the number of years to repay the loan and \( P \) is the initial loan.

## Examples

Example 1

A loan of \( $ 250,000 \) was obtained at a bank with a fixed rate of \( 4.5 \% \). What should be the monthly payment so that the loan and the interest are paid in \( 25 \) years?

Solutions

Given

\( R = 5.25 \% = \dfrac{5.25}{100} = 0.0525 \)

\( N = 25 \)

and \( P = 250,000 \)

Substitute the known quantities in formula \( III \) above:

\( \qquad M = \dfrac{ \frac{0.045}{12} \times \; 250000\; (1+\frac{0.045}{12})^{300}}{(1+\frac{0.045}{12})^{300} - 1} \approx $ 1390\)

Example 2

You need a loan of \( $ 300,000 \) and the bank offered a fixed rate of \( 5.25 \% \). You can afford \( $ 2100 \) as a monthly payment for the loan. What should be the term of payments if you want to pay the loan in the shortest period of time?

Solutions

Given

\( P = 300,000 \)

\( M = 2100 \)

and \( R = 5.25 \% = \dfrac{5.25}{100} = 0.025 \)

Substitute the known quantities in formula \( III \) above to obtain the equation

\( \qquad \dfrac{ \frac{0.0525}{12} \times \; 300000\; (1+\frac{0.0525}{12})^{12 \times N}}{(1+\frac{0.0525}{12})^{12 \times N} - 1} = 2100 \)

Let \( x = (1+\frac{0.0525}{12})^{12 \times N} \)

and rewrite the equation as

\( \qquad \dfrac{ 1312.5 \; x }{x - 1} = 2100 \)

Rewrite the above equation as

\( 1312.5 \; x = 2100 x - 2100 \)

and solve for \( x \)

\( x = 2.66666 \)

Substitute \( x \) by its expression

\( (1+\frac{0.0525}{12})^{12 \times N} = 2.66666 \)

Take \( \ln \) of both sides

\( 12 \times N \; \ln (1+\frac{0.0525}{12}) = \ln 2.66666 \)

Solve for \( N \approx 18.72325\)

A term of \( 20 \) years is suitable to repay the loan of \( $ 300,000 \) at the fixed rate of \( 5.25 \% \).

## Links and References