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 \( 5.25 \% \).
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