CAT 2025 Lesson : Interest & Growth - Present Value & EMI

3.3 Present Value of Multiple Payments

To get the present value of future payments, we discount each of these payments at the given interest rate for the time periods they correspond to. This is an extension of the formula given in the above Section 3.2.

Let $p$ be the principal or present value of a loan lent at $r \%$ interest rate per annum. If $A$, $B$ and $C$ are the repayments made at the end of $a$, $b$ and $c$ years, wherein the loan is fully repaid, then

$p = \dfrac{A}{\left(1 + \dfrac{r}{100} \right)^{a}} + \dfrac{B}{\left(1 + \dfrac{r}{100} \right)^{b}} + \dfrac{C}{\left(1 + \dfrac{r}{100} \right)^{c}}$

3.4 Equated Monthly Instalment (EMI)

For most retail loans (such as home loans, auto loans and personal loans), the borrowers repay by paying a fixed amount every month over the tenure of the loan. This fixed amount is called Equated Monthly Instalment or EMI. EMI includes interest and principal.

For instance, where Rs. $25,000$ was lent on $1^{\text{st}}$t January $2019$ and the interest is $12 \%$ per annum, which is $1 \%$ per month, and the EMI is Rs. $1250$, the following table provides the impact on the principal outstanding, i.e. the principal remaining.

Date	EMI	Interest	Principal Repaid	Principal Outstanding
$1^{\text{st}}$ January				$25,000$
$31^{\text{st}}$ January	$1,250$	$250$	$1,000$	$24,000$
$28^{\text{th}}$ February	$1,250$	$240$	$1,010$	$22,990$
$31^{\text{st}}$ March	$1,250$	$229.90$	$1,020.10$	$21,969.90$

As a certain portion of the EMI is principal repayment, the principal outstanding decreases over time. As a result, interest decreases over time and principal repayment portion of the EMI increases with time.

While the above are general observations, we need to understand how EMI is calculated. The present value of all future payments should equal the loan amount. Compound interest applies for these loans.

Let the loan amount be $\bm{p}$, the EMI be $\bm{e}$, interest rate per time period be $\bm{r \%}$ and number of time periods be $\bm{n}$.

$p = \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{1}} + \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{2}} + ... + \dfrac{e}{\left(1 + \dfrac{r}{100} \right)^{n}}$

$p$ is the sum of $n$ terms in Geometric Progression. We are not going create another formula here. You simply need to remember the sum to $n$ terms of GP formula and apply here.

$S_{n} = \dfrac{a(1 - r^{n})}{(1 - r)}$, where $\bm{a}$ is the first term and $\bm{r}$ is the common ratio.
[Note: $\bm{r}$ in the above formula is common ratio and not rate of interest.]