How the formula for EMI is derived

I was looking for a formula to calculate EMI (Equated Monthly Installments). I have some fixed known parameters like, Principal Amount, Rate of Interest and No. Of Installments. By googling, I came across the formula,

$$Installment Amount = \frac {P*i*(1 + i)^n}{(1 + i)^n - 1}$$

      where i  =  interest rate per installment payment period,
n =  number of Installments,
P  = principal amount of the loan


This formula does my job, but I actually want to understand the formula in detail, that how it derived. I have done googling to decode it but no luck.

Can anybody help me to understand the formula? Like, what each operation in the formula stands for?

• You need parentheses in the denominator as the $-1$ term needs to be part of it. Jan 16, 2013 at 5:36

Let $I$ be the installment payment and $B(j)$ the balance remaining after $j$ payments. We want to choose $I$ so that $B(n)=0$. We are given $B(0)=P$. Each month, the interest is applied and the payment deducted to get the new balance, so $B(j)=(1+i)\cdot B(j-1)-I$ If we write this out we get:

$B(0)=P \\ B(1)=(1+i)P-I \\ B(2)=(1+i)((1+i)P-I)-I=(1+i)^2P-(1+i)I-I \\ B(j)=(1+i)^jP-(1+i)^{(j-1)}I-(1+i)^{j-2}I-\ldots I=(1+i)^jP-\frac {(1+i)^j-1}{i}I \\ 0=(1+i)^nP-\frac{(1+i)^n-1}{i}I$

where the second equals sign in $B(j)$ comes from summing the geometric series

• In geometric series, denominator will be i instead of (1+i). Jan 16, 2013 at 9:00
• @RumitParakhiya: You are right. Thanks. Jan 16, 2013 at 14:42
• Hi Ross, I'm having trouble understanding the first step. Shouldn't the balance after one payment be the principal minus the installment? Namely, shouldn't we have $B(1)=P-I$? Why do we have the $(1+i)$ term? Jul 1 at 10:24
• @AlexD: because you owe interest for the time between you took out the load and the time you make the first payment. Each factor of $(1+i)$ accounts for the interest on one term of the loan. This is described in the first paragraph. Jul 1 at 13:39

Hi while googling I found the below detailed answer posted by someone, how EMI is computed.

"Thanks to the person who posted it".

Below is the link to reach that post

http://rmathew.com/2006/calculating-emis.html

• Welcome to Math.SE. While your link may provide a good answer, it is always good to include the essence of the answer in-line here and provide the link, since some links get broken eventually. Apr 28, 2016 at 7:45

Let's say, the amount $$P$$ is borrowed on which interest is payable at the rate of $$i$$ per installment period. This amount is to be paid back along with interest in $$n$$ equal installments.

Let these installments be $$I_1,I_2,I_3,\ldots,I_n$$ respectively. Since they are all equal, we can say:$$I_1=I_2=I_3=\dots=I_n=I\label{a}\tag{1}$$

Each installment comprises some part of the principal $$P$$ that is paid back, the rest is interest on the amount that was outstanding for that installment period. Let the part of the principal that is paid back with $$I_1,I_2,I_3,\ldots,I_n$$ respectively be $$P_1,P_2,P_3,\ldots,P_n$$. These principal payments must add up to the original amount $$P$$ that was borrowed:$$P_1+P_2+P_3+\cdots+P_n=P\label{b}\tag{2}$$ Now let's talk about interest. At the end of the first installment period, the entire principal $$P$$ will have been outstanding for the duration of one installment period. The interest payable in this regard is $$P \times i$$ which will be paid along with $$P_1$$ when $$I_1$$ is paid. Therefore, $$I_1=P_1+P\times i$$. At the end of the second installment period, only the amount $$P-P_1$$ will have been outstanding for the same period. Therefore, only $$[P-P_1]\times i$$ is payable as interest for the second installment period which will be paid along with $$P_2$$ when $$I_2$$ is paid. Similarly, at the end of the third installment period, interest will be paid only on $$[P-(P_1+P_2)]$$ for one installment period along with $$P_3$$ when $$I_3$$ is paid.

Making the above argument for all installments, we get:

$$I_1=P_1+P\times i\label{c}\tag{3}$$ $$I_2=P_2+[P-P_1]\times i$$ $$I_3=P_3+[P-(P_1+P_2)]\times i$$ $$\vdots$$ $$I_n=P_n+[P-(P_1+P_2+\dots+P_{n-1})]\times i$$

If we add up the above equations, we get:

$$I_1+I_2+I_3+\cdots+I_n=[P_1+P\times i]+[P_2+[P-P_1]\times i]+[P_3+[P-(P_1+P_2)]\times i]+\cdots+[P_n+[P-(P_1+P_2+\dots+P_{n-1})]\times i]$$ $$I_1+I_2+I_3+\cdots+I_n=[P_1+P_2+P_3+\cdots+P_n]+P\times i+[P-P_1]\times i+[P-(P_1+P_2)]\times i+\cdots+[P-(P_1+P_2+\cdots+P_n)]\times i$$ $$I_1+I_2+I_3+\cdots+I_n=[P_1+P_2+P_3+\cdots+P_n]+P\times i+P\times i-P_1\times i+P\times i-(P_1+P_2)\times i+\cdots+P\times i-(P_1+P_2+\cdots+P_n)\times i$$ $$I_1+I_2+I_3+\cdots+I_n=[P_1+P_2+P_3+\cdots+P_n]+n\times P\times i-[P_1+(P_1+P_2)+\cdots+(P_1+P_2+\cdots+P_{n-1})]\times i$$

Remember that $$I_1=I_2=I_3=\dots=I_n=I$$ (from \ref{a}) and $$P_1+P_2+P_3+\cdots+P_n=P$$ (from \ref{b}):

$$n\times I=P+n\times P\times i-[P_1+(P_1+P_2)+\cdots+(P_1+P_2+\cdots+P_{n-1})]\times i\label{d}\tag{4}$$

Now, let $$j$$ be a natural number such that $$j \in [2,n-1]$$. $$I_j$$ and $$I_{j+1}$$ are given as follows:

$$I_j=P_j+[P-\sum_{k=1}^{j-1} P_k]\times i$$ $$I_{j+1}=P_{j+1}+[P-\sum_{k=1}^{j} P_k]\times i$$

Consider $$I_{j+1}-I_j$$:

$$I_{j+1}-I_j=P_{j+1}+[P-\sum_{k=1}^{j} P_k]\times i-(P_j+[P-\sum_{k=1}^{j-1} P_k]\times i)$$ $$I_{j+1}-I_j=P_{j+1}+P\times i-\sum_{k=1}^{j} P_k\times i-P_j-[P-\sum_{k=1}^{j-1} P_k]\times i$$ $$I_{j+1}-I_j=P_{j+1}+\require{cancel} \cancel{P\times i}-\sum_{k=1}^{j} P_k\times i-P_j-\require{cancel} \cancel{P\times i}+\sum_{k=1}^{j-1} P_k\times i$$ $$I_{j+1}-I_j=P_{j+1}-(\sum_{k=1}^{j-1} P_k\times i+P_j\times i)-P_j+\sum_{k=1}^{j-1} P_k\times i$$ $$I_{j+1}-I_j=P_{j+1}-\require{cancel}\cancel{\sum_{k=1}^{j-1} P_k\times i}-P_j\times i-P_j+\require{cancel}\cancel{\sum_{k=1}^{j-1} P_k\times i}$$ $$I_{j+1}-I_j=P_{j+1}-P_j\times i-P_j$$

Remember that $$I_j=I_{j+1}=I$$ (from \ref{a}):

$$\require{cancel}\cancel{I_{j+1}}-\require{cancel}\cancel{I_j}=P_{j+1}-P_j\times i-P_j$$ $$0=P_{j+1}-P_j\times i-P_j$$ $$P_{j+1}=P_j\times(i+1);\ j \in [2,n-1]\label{e}\tag{5}$$

Now, consider $$I_2-I_1$$:

$$I_2-I_1=P_2+[P-P_1]\times i-(P_1+P\times i)$$ $$I_2-I_1=P_2+\require{cancel}\cancel{P\times i}-P_1\times i-P_1-\require{cancel}\cancel{P\times i}$$ $$I_2-I_1=P_2-P_1\times i-P_1$$

Again, since $$I_1=I_2=I$$ (from \ref{a}):

$$\require{cancel}\cancel{I_2}-\require{cancel}\cancel{I_1}=P_2-P_1\times i-P_1$$ $$0=P_2-P_1\times i-P_1$$ $$P_2=P_1\times(i+1)$$

If we expand out $$\ P_{j+1}=P_j\times(i+1);\ j \in [2,n-1]$$ (from \ref{e}) and add $$P_2=P_1\times(i+1)$$ to the resulting set of equations, we get:

$$P_1=P_1$$ $$P_2=P_1\times(i+1)$$ $$P_3=P_2\times(i+1)$$ $$\vdots$$ $$P_j=P_{j-1}\times(i+1)$$ $$\vdots$$ $$P_{n-1}=P_{n-2}\times(i+1)$$ $$P_n=P_{n-1}\times(i+1)$$

Multiplying the above equations gives us: $$P_1\times P_2\times P_3\times\cdots\times P_{n-1}\times P_n=P_1\times [P_1\times(i+1)]\times [P_2\times(i+1)]\times\cdots\times [P_{n-2}\times(i+1)]\times [P_{n-1}\times(i+1)]$$ $$\require{cancel}\cancel{P_1}\times \require{cancel}\cancel{P_2}\times \require{cancel}\cancel{P_3}\times\cdots\times \require{cancel}\cancel{P_{n-1}}\times P_n=P_1\times [\require{cancel}\cancel{P_1}\times(i+1)]\times [\require{cancel}\cancel{P_2}\times(i+1)]\times\cdots\times [\require{cancel}\cancel{P_{n-2}}\times(i+1)]\times [\require{cancel}\cancel{P_{n-1}}\times(i+1)]$$ $$P_n=P_1\times [(i+1)\times (i+1)\times\cdots\text{(n-1) times}]$$ $$P_n=P_1\times(i+1)^{n-1}\label{f}\tag{6}$$

Now we would like to have an expression for $$\sum_{k=1}^j P_k;\ j \in [1,n]$$. The same set of equations that helped us figure out \ref{f} will be useful in this regard:

$$\sum_{k=1}^{j} P_k=P_1+P_2+P_3+\cdots+P_j$$ $$\sum_{k=1}^{j} P_k=P_1+P_1\times(i+1)+P_2\times(i+1)+\cdots+P_{j-1}\times(i+1)$$ $$\sum_{k=1}^{j} P_k=P_1+[P_1+P_2+\cdots+P_{j-1}]\times (i+1)$$ $$\sum_{k=1}^{j} P_k=P_1+[P_1+P_2+\cdots+P_{j-1}+P_j-P_j]\times (i+1)$$ $$\sum_{k=1}^{j} P_k=P_1+[\sum_{k=1}^{j} P_k-P_j]\times (i+1)$$ $$\sum_{k=1}^{j} P_k=P_1+\left(\sum_{k=1}^{j} P_k\right)\times (i+1)-P_j\times (i+1)$$ $$\require{cancel}\cancel{\sum_{k=1}^{j} P_k}=P_1+i\times \sum_{k=1}^{j}P_k+\require{cancel}\cancel{\sum_{k=1}^{j} P_k}-P_j\times (i+1)$$ $$0=P_1+i\times \sum_{k=1}^{j}P_k-P_j\times (i+1)$$ $$i\times \sum_{k=1}^{j}P_k=P_j\times (i+1)-P_1$$ $$\sum_{k=1}^{j}P_k=\frac{P_j\times (i+1)-P_1}{i}; \ j \in [1,n]\label{g}\tag{7}$$

We get the following equations by using \ref{g}:

$$P_1=\frac{P_1\times (i+1)-P_1}{i}$$ $$P_1+P_2=\frac{P_2\times (i+1)-P_1}{i}$$ $$P_1+P_2+P_3=\frac{P_3\times (i+1)-P_1}{i}$$ $$\vdots$$ $$P_1+P_2+P_3+\cdots+P_{n-1}=\frac{P_{n-1}\times (i+1)-P_1}{i}$$

Add these up to get the following:

$$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{P_1\times (i+1)-P_1}{i}+\frac{P_2\times (i+1)-P_1}{i}+\frac{P_3\times (i+1)-P_1}{i}+\cdots+\frac{P_{n-1}\times (i+1)-P_1}{i}$$ $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{(P_1\times (i+1)-P_1)+(P_2\times (i+1)-P_1)+(P_3\times (i+1)-P_1)+\cdots+(P_{n-1}\times (i+1)-P_1)}{i}$$ $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{(P_1+P_2+P_3+\cdots+P_{n-1})\times (i+1)-P_1\times (n-1)}{i}$$ $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{(P_1+P_2+P_3+\cdots+P_{n-1}+P_n-P_n)\times (i+1)-P_1\times (n-1)}{i}$$

Substitute the value of $$P_1+P_2+P_3+\cdots+P_{n-1}+P_n$$ from \ref{b}:

$$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{(P-P_n)\times (i+1)-P_1\times (n-1)}{i}$$

Substitute the value of $$P_n$$ from \ref{f}:

$$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{(P-P_1\times(i+1)^{n-1})\times (i+1)-P_1\times (n-1)}{i}$$ $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{P\times (i+1)-P_1\times(i+1)^n-P_1\times (n-1)}{i}$$ $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})=\frac{P\times (i+1)-P_1\times\Bigl((i+1)^n+(n-1)\Bigr)}{i}\label{h}\tag{8}$$

Substitute the value of $$P_1+(P_1+P_2)+(P_1+P_2+P_3)+\cdots+(P_1+P_2+P_3+\cdots+P_{n-1})$$ from \ref{h} in \ref{d}:

$$n\times I=P+n\times P\times i-\left(\frac{P\times (i+1)-P_1\times\Bigl((i+1)^n+(n-1)\Bigr)}{\require{cancel}\cancel{i}}\right)\times \require{cancel}\cancel{i}$$ $$n\times I=P+n\times P\times i-\biggl(P\times (i+1)-P_1\times\Bigl((i+1)^n+(n-1)\Bigr)\biggr)$$ $$n\times I=P+n\times P\times i-P\times (i+1)+P_1\times\Bigl((i+1)^n+(n-1)\Bigr)$$ $$n\times I=\require{cancel}\cancel{P}+n\times P\times i-P\times i-\require{cancel}\cancel{P}+P_1\times\Bigl((i+1)^n+(n-1)\Bigr)$$ $$n\times I=n\times P\times i-P\times i+P_1\times\Bigl((i+1)^n+(n-1)\Bigr)$$ $$P_1\times\Bigl((i+1)^n+(n-1)\Bigr)=n\times I+P\times i-n\times P\times i$$ $$P_1=\frac{n\times I+P\times i-n\times P\times i}{(i+1)^n+(n-1)}\label{i}\tag{9}$$

Substitute the value of $$P_1$$ from \ref{i} in \ref{c}:

$$I_1=\frac{n\times I+P\times i-n\times P\times i}{(i+1)^n+(n-1)}+P\times i$$

We know that $$I_1=I$$ from \ref{a}:

$$I=\frac{n\times I+P\times i-n\times P\times i}{(i+1)^n+(n-1)}+P\times i$$ $$I\times \Bigl((i+1)^n+(n-1)\Bigr)=\left(\frac{n\times I+P\times i-n\times P\times i}{\require{cancel}\cancel{(i+1)^n+(n-1)}}\right)\times \require{cancel}\cancel{\Bigl((i+1)^n+(n-1)\Bigr)}+P\times i\times \Bigl((i+1)^n+(n-1)\Bigr)$$ $$I\times \Bigl((i+1)^n+(n-1)\Bigr)-n\times I=P\times i-n\times P\times i+P\times i\times \Bigl((i+1)^n+(n-1)\Bigr)$$ $$I\times \Bigl((i+1)^n+(\require{cancel}\cancel{n}-1)-\require{cancel}\cancel{n}\Bigr)=P\times i\times(1-n)+P\times i\times (i+1)^n+P\times i\times(n-1)$$ $$I\times \Bigl((i+1)^n-1\Bigr)=-\require{cancel}\cancel{P\times i\times(n-1)}+P\times i\times (i+1)^n+\require{cancel}\cancel{P\times i\times(n-1)}$$ $$I\times \Bigl((i+1)^n-1\Bigr)=P\times i\times (i+1)^n$$ $$I=\frac{P\times i\times (i+1)^n}{(i+1)^n-1}$$

• wow so detatailed derivation Jul 24, 2021 at 16:53

This question is old, but I'd like to provide an alternative answer that provides more intuition. The geometric series is good for rigor.

In an installment account, the goal is to have equal payments $$A$$ (installment amount) such that the loan (initial amount $$P$$) is paid off after $$n$$ periods. If there were no interest, this would be simple and we could define $$A = \frac{P}{n}$$. However, if we include interest and it compounds, things become more complicated. Payments made earlier on prevent interest from compounding later and we need to take this into account.

If this loan were a pure compound interest loan, the amount due at the end would be $$P(1+i)^n$$. Note the $$(1+i)^n$$ term appears in the top and bottom of the installment formula.

The Original Formula is $$EMI = \frac{(1+i)^n}{[(1+i)^n -1]} \times (P \times i)$$

I'm going to rewrite the formula a bit so we can break it up. We have $$EMI = \frac{P \times (1+i)^n}{P \times [(1+i)^n -1]} \times (P \times i)$$

Notice I have multiplied by $$\frac{P}{P}$$. Now the top of the fraction $$P \times (1+i)^n$$ is the same as in the compound interest formula. It is the total amount due with interest. The bottom part of the fraction is similar, except it takes away $$P$$. We have \begin{align*} P \times [(1+i)^n -1] &= P(1+i)^n - P \end{align*} So the denominator is the total amount of interest due (because we subtracted the loan amount). So the fraction gives us how many total dollars are needed per interest dollar.

Lastly, we have $$(P \times i)$$ which is the amount of interest that the original loan amount gains every month. When we multiple that by the fraction we get the amount of total dollars that need to be paid every month. In words $$\text{Monthly Payment} = \frac{\text{Total Amount with Interest}}{\text{Total Amount of Interest}} \times \text{Principle Interest Per Month}$$

• very nice reasoning Jul 24, 2021 at 16:51

I prefer to think of the EMI formula in the following way.

First, I have the remuneration of the capital:

$$\text{Future Value} = VF = P(1+i)^n$$

where P = principal amount.

Then, I have a series of installments to be paid:

$$\text{Installments} = I = I_1 + I_2 + ... + I_n$$

Each installment is corrected retroactively by the number of months until the final payment is due, using the same interest rate:

$$I_1 = f(1+i)^{n-1}$$
$$I_2 = f(1+i)^{n-2}$$
...
$$I_n = f$$

where $$f = \text{fixed payment}$$.

So:

$$I = f(1+i)^{n-1} +f(1+i)^{n-2} ... + f = f\frac{(1+i)^n - 1}{i}$$

(If you have trouble understanding the above part, just look at the sum of geometric series with finite terms and substitute $$a = f$$, $$r = (1+i)$$)

Basically, the formula is the point where those two curves intersect:

$$P(1+i)^n = f\frac{(1+i)^n - 1}{i}$$

After separating the terms we have:

$$f = P(1+i)^n\frac{i}{(1+i)^n-1}$$

First thing you should realise is that gp is at heart of CI

$$\text I*(1+r)^n+I*(1+r)^{n-1}+......+I*(r)=P*(1+r)^n$$

Here each Instalment's future value is being caculated i.e. if the interest is charged on intalment what would be the resultant. In the end the sum of resultant would to equal if it is applied to Principal itself p.s. like conservation of momentum

Just apply sum of gp to lhs and reaarange and voila