# Mortgage Formula with Examples

 

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 $$5.25 \%$$.