Mortgage Formula with Examples – Derivation and Applications

The mortgage formula for repayment of a mortgage is presented along with examples and detailed solutions. If needed, you can also use our mortgage calculator.

Mortgage Payment Process

A mortgage is a type of loan that helps individuals or businesses buy property like a house or a commercial building. When you take out a mortgage, the bank or lender provides the funds needed to pay for the property upfront, and you repay the loan plus interest over a set period of time (typically 15 to 30 years).

Let $ R $ be the annual interest rate and $ N $ the number of years to repay the loan. Since payments are made monthly, the monthly rate of interest is: \[ r = \frac{R}{12} \] and the total number of months in $ N $ years is: \[ n = 12 \times N \]

Let $ P $ be the initial principal amount of the loan. The lender agrees to lend this amount on the condition that you repay it with interest via monthly payments $ M $.

At the start ($ t = 0 $), the amount owed is: \[ A_0 = P \]
At the end of the first month ($ n = 1 $), the amount owed is: \[ A_1 = A_0(1 + r) - M = P(1 + r) - M \] Explanation: Within a month, the previous balance $ P $ increases by the interest charge $ rP $, resulting in $ P(1+r) $, minus your monthly payment $ M $.
At the end of the second month ($ n = 2 $), the amount owed is: \[ A_2 = A_1(1 + r) - M = \left( P(1 + r) - M \right)(1 + r) - M = P(1+r)^2 - M\left(1 + (1+r)\right) \]
At the end of the third month ($ n = 3 $), the amount owed is: \[ A_3 = A_2(1 + r) - M = \left( P(1+r)^2 - M(1 + (1+r)) \right)(1 + r) - M = P(1+r)^3 - M\left(1 + (1+r) + (1+r)^2\right) \]

By extension, the amount owed at the end of the $ n $-th month is: \[ A_n = P(1+r)^n - M \left( 1 + (1+r) + (1+r)^2 + \dots + (1+r)^{n-1} \right) \qquad (I) \]

The sum $ 1 + (1+r) + (1+r)^2 + \dots + (1+r)^{n-1} $ is a geometric sequence sum that can be simplified.

Review Geometric Sequence Sum

The geometric sequence sum $ S_n $ defined by: \[ S_n = a_1 + R \times a_1 + R^2 \times a_1 + \dots + R^n \times a_1 \] is given by the formula: \[ S = a_1 \frac{R^{n+1} - 1}{R - 1} \] where $ a_1 $ is the first term and $ R $ is the common ratio.

Formula for Payment

Using the geometric series sum formula to simplify the sum in expression (II): \[ 1 + (1+r) + (1+r)^2 + \dots + (1+r)^{n-1} = \frac{(1+r)^n - 1}{(1+r) - 1} = \frac{(1+r)^n - 1}{r} \]

Substituting this back into equation (I), the balance $ A_n $ owed after $ n $ months is: \[ A_n = P(1+r)^n - M \frac{(1+r)^n - 1}{r} \]

Payments continue until the loan balance $ A_n $ is fully paid off ($ A_n = 0 $): \[ P(1+r)^n - M \frac{(1+r)^n - 1}{r} = 0 \]

Solving for the monthly payment $ M $:

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

In terms of the annual interest rate $ R $ and loan term in years $ N $, the formula can be expressed as:

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

Examples and Detailed Solutions

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 interest are paid off in $ 25 $ years?

Solution:

Given parameters: $ P = 250,000 $, $ R = 4.5\% = 0.045 $, and $ N = 25 $ years.
Monthly interest rate: $ r = \frac{0.045}{12} = 0.00375 $
Total number of months: $ n = 12 \times 25 = 300 $
Substitute into formula (III): \[ M = \frac{ \frac{0.045}{12} \times 250000 \times \left(1+\frac{0.045}{12}\right)^{300} }{ \left(1+\frac{0.045}{12}\right)^{300} - 1 } \approx \$1390 \]

Example 2

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

Solution:

Given parameters: $ P = 300,000 $, $ M = 2100 $, $ R = 5.25\% = 0.0525 $.
Set up the equation using formula (III): \[ \frac{ \frac{0.0525}{12} \times 300000 \times \left(1+\frac{0.0525}{12}\right)^{12N} }{ \left(1+\frac{0.0525}{12}\right)^{12N} - 1 } = 2100 \]
Let $ x = \left(1+\frac{0.0525}{12}\right)^{12N} $, simplifying the numerator to $ 1312.5x $: \[ \frac{1312.5x}{x - 1} = 2100 \implies 1312.5x = 2100x - 2100 \]
Solving for $ x $: \[ 787.5x = 2100 \implies x \approx 2.66666 \]
Substitute back for $ N $: \[ \left(1+\frac{0.0525}{12}\right)^{12N} = 2.66666 \] Taking natural logarithms of both sides: \[ 12N \ln\left(1+\frac{0.0525}{12}\right) = \ln(2.66666) \implies N \approx 18.72 \text{ years} \]
A term of $ 20 $ years is suitable to comfortably repay the loan of $ \$300,000 $ at $ 5.25\% $.