**FORMULA AND DERIVATION**

Mortgages are generally amortisation loans which include repayments consisting of both principal and interest payments. The lender loses out on the time value of the money that has been lent. So, the amount lent will be equivalent to the amount that will be repaid by the borrower in the future.

Let us first assume a simple case, where a bank provides an amount of \(\$100\) as a mortgage loan to a private builder. Let us assume (for simplicity) that the loan will be repaid in one instalment. The bank is losing \(10\%\) interest that it could have earned on that amount if it hadn’t lent it to the builder. So, by the end of the year, \(\$100\) would have accumulated to \(\$110\) at \(10\%\) interest. The builder, therefore, has to make a payment of \(\$110\) to the bank at the end of the year. In simple terms, the amount lent by the bank was the present value of what it would receive from the borrower at the end of the year.

The same logic is applied in calculating the amount of each instalment that the borrower has to pay. If \(L\) is the amount of loan granted for \(n\) years at effective interest rate \(i\), then the annual instalment amount \(X\) can be calculated from the following equation:

\[L=X{\left(1+i\right)}^{-1}+X{\left(1+i\right)}^{-2}\dots \dots \dots X{\left(1+i\right)}^{-n}\]
\[\ \ \ =Xv+{Xv}^2+{Xv}^3\dots \dots \dots {Xv}^n\]
\[\ \ \ =Xa_{\overline{n|}}\]

where,

\(a_{\overline{n|}}=\frac{1-v^n}{i}\)
(This is the annuity function used to calculate the present value of a series of level cash flows)

The above equation can be rearranged to give:

\[X=\frac{L}{a_{\overline{n|}}}\]

or

\[X=\ \frac{L\times i}{1-v^n}\]

The above equation considers that \(X\) is the annual instalment amount. But in most cases mortgage loans include instalments that are paid more frequently than once per year (half yearly, quarterly, monthly, etc.). Now let us consider a loan of amount \(L\) for \(n\) years at effective interest rate \(i\) p.a. and \(p\) instalments per year. If \(X\) is the amount of each instalment paid \(p\) times a year, we can write the loan equation as follows:

\[L=X{\left(1+i\right)}^{{-1}/{p}}+X{\left(1+i\right)}^{{-2}/{p}}\dots \dots \dots X{\left(1+i\right)}^{{-np}/{p}}\]
\[\ \ \ =pXa^{\left(p\right)}_{\overline{n|}}\]

where,

\(a^{(p)}_{\overline{n|}}=\frac{1-v^n}{i^{(p)}}\)
(This is the annuity function used to calculate the present value of a series of level cash flows, made p times a year)

\[i^{\left(p\right)}=p\left[{\left(1+i\right)}^{{1}/{p}}-1\right]\]

The above equation can be rearranged to give:

\[X=\frac{L}{pa^{(p)}_{\overline{n|}}}\]

or

\[X=\frac{{L\times i}^{\left(p\right)}}{p\left(1-v^n\right)}\]

**EXAMPLES**

Example 1

A 3 year mortgage loan for \(\$1000\) is given by a building society at an effective interest rate of \(10\%\) p.a. The annual instalment needs to be determined.

Here,

\[L=\$1000\]
\[n=3\]
\[i=10\%\]

\[X=\frac{L}{a_{\overline{n|}}}\]
\[\ \ \ =\frac{1000}{a_{\overline{3|}}}\]
\[\ \ \ =\frac{1000}{\frac{1-v^3}{0.1}}\]
\[\ \ \ =\frac{1000\times 0.1}{1-{1.1}^{-3}}\]
\[\ \ \ =402.11\]

The borrower has to pay three instalments of \(402.11\). Now, let us consider the amount of principal and interest in each of the instalments paid.

Initially, the total principal, due to be repaid, is \(\$1000\). Interest of \(10\%\) on \(\$1000\) is \(\$100\) So, in the first instalment, the amount of interest paid would be \(\$100\) and the principal repaid would be \(402.11-100=\$302.11\).

Now that \(\$302.11\) has been repaid, the remaining principal would be \(1000-302.11=\$697.89\). So, for the second year the interest will be \(10\%\) of \($697.89\), which is equal to \(\$69.79\).

The second instalment consists of \(\$69.79\) interest, and the remaining portion will be the principal repayment \((402.11-69.79=\$332.32)\) .

The remaining principal amount would now be \(1000-302.11-332.32=365.57\) . The interest for the third year will be \(10\%\) of \(365.57\) that is \(\$36.56\).

In the third instalment, \(\$36.56\) would go towards interest payment and the remaining amount will be the principal payment \(402.11-36.56=365.57\) .

The amount of principal remaining now would be zero as the balance of \(\$365.57\) was paid off in the last instalment.

Example 2

A bank lends \(\textrm{₹}100000\) as a mortgage loan for 10 years, to be repaid through monthly instalments, with interest charged at \(10\%\) p.a. effective. The monthly instalment amount needs to be determined.

Here,

\[L=100000\]
\[n=10\]
\[p=12\]
\[i=10\%\]

i

^{(p)} = i

^{(12)} = 12[(1.1)

^{1/2} – 1] = 9.5689%

\[X=\frac{L}{pa^{(p)}_{\overline{n|}}}=\frac{{L\times i}^{\left(p\right)}}{p\left(1-v^n\right)}\]
\[\ \ \ =\frac{100000\times 0.095689}{12\left(1-{1.1}^{-10}\right)}\]
\[\ \ \ =1297.75\]