### EMI computations

I am recently working on setting up my farmland. To compute the cost of paying back my own investments to myself from the farm income, I was faced with an EMI like situation where the infrastructure investment is amortized over (say) 10 years and I expect some constant amounts to be payed back towards that investment at regular intervals (say once a year or once in 3 months or whatever). Obviously I will also charge myself an interest rate tied to the interest rates of some pessimistic instruments like fixed deposits at a bank (8.5% in India now).

This is a formula I have derived from first principles 2-3 times over in my life so far and each time I seem to lose it (easy as it is to re derive). So as blogging is the right idiom to log that (as long as my blog website is around , at least) here goes.

Some symbols I need to use:

Now We define a handy term called

This represents the rate at which the residual amount inflates due to interest burden. The residual amount is taken to inflate in bursts or as a single step once in every

So a principal amount

But every

After m units of time, whats remaining to be payed back is

During the second m units the remaining amount grows again by a factor

And during the third unit the same procedure gives

After

Or with some basic regrouping (where ^ is again exponentiation and 1 is chosen to be written as R^0):

Now using the summation formula for a geometric series the same equation can be written as:

Therefore

The above closed form formula for computation of EMI is very amenable for entry into a good calculator. In fact on my Casio calculator I have now stored the above as a function and I simply assign values to P, r, m, y and n and I can compute E very easily using that function.

A small example computation.

Say you have borrowed 1 lakh rupees. You want to pay back that 1 lakh rupees in proper EMIs, every month over ten years. Say the rate of interest charged to you is 10% (more then the housing loans in India but less than the rate for other loans).

So

Its good to calculate

Now therefore the EMI is

.

So after 120 payments you would have payed back

If you wanted to do the same repayment in annual instalments.

E = 100000 * 2.59374 * (1.1 - 1) /(2.59374 - 1) = 16274.55 or 16275

.

Now you would have payed back 162750 rupees.

This is a formula I have derived from first principles 2-3 times over in my life so far and each time I seem to lose it (easy as it is to re derive). So as blogging is the right idiom to log that (as long as my blog website is around , at least) here goes.

Some symbols I need to use:

- r
- the rate of interest as a fraction of 1. 8% is 0.08 and so on. rate of interest in percentage divided by 100, if you will.
- m
- Number of units of time elapsed between payments.(See y below). Number of months, or number of weeks or whatever - as long as it is in the same level as y below.
- y
- number of units that go into making up an year. If weeks - you can take it to be 52. If months you can take it to be 12. If days you can take to be 365.
- P
- Initial Investment - P for 'principal'.
- E
- Equated Monthly Instalment - when payed monthly. For this more generalized computation its more like Equated Periodical Instalment
- n
- Number of payments (once in each m units of time) over which the principal would be completely payed back , along with compound interest.

Now We define a handy term called

*R*which is basically*(1+ (r*(m/y)))*This represents the rate at which the residual amount inflates due to interest burden. The residual amount is taken to inflate in bursts or as a single step once in every

*m*units of time. [[ Obiter Dictum: rather than in a continuous manner - though even that could be done : using calculus.]]So a principal amount

*P*grows to be*P*R*after m units of time. Left to itself, without any payments, it grows exponentially. After n units of times it would amount to*R*R^n*, where*^*means exponentiation, "raised to the power".But every

*m*units of time a sum*E*is payed back.[[Obiter Dictum: part of which goes to pay the interest and the remaining part goes to pay the principal. The part towards principal in negligible in the beginning and most of the payment is towards interest. the situation is reversed by the end of n payments.]]. This changes the situation.After m units of time, whats remaining to be payed back is

*P*R - E*During the second m units the remaining amount grows again by a factor

*R*to*(P*R - E)*R*And during the third unit the same procedure gives

*((P*R - E)*R - E)*R*After

*n*paybacks the remaining amount should be 0(zero, cipher, nill, shunya).After all, that's the definition of EMIs. So we have*(...((P*R - E)*R - E)*R ...)*R - E = 0*Or with some basic regrouping (where ^ is again exponentiation and 1 is chosen to be written as R^0):

*P*R^n - E(R^(n-1) + R^(n-2) + ... + R^0) = 0*Now using the summation formula for a geometric series the same equation can be written as:

*P*R^n - E*((R^n - 1)/(R-1)) = 0*Therefore

*E*can be found to be:*E=P*R^n*(R-1)/(R^n - 1)*The above closed form formula for computation of EMI is very amenable for entry into a good calculator. In fact on my Casio calculator I have now stored the above as a function and I simply assign values to P, r, m, y and n and I can compute E very easily using that function.

A small example computation.

Say you have borrowed 1 lakh rupees. You want to pay back that 1 lakh rupees in proper EMIs, every month over ten years. Say the rate of interest charged to you is 10% (more then the housing loans in India but less than the rate for other loans).

So

*R*for this situation is*(1 + 0.1 * (1/12))*.*m*is 1 as it is payed every one unit of time (a month in this case). and*y*is 12 as there are 12 months in a year. Notice that if we artifically chose weeks for this case*m*would be 4 and*y*would 52 making it the same R. So in short R is 1.00833.Its good to calculate

*R^n*and keep it handy (in case you dont have functions - like on this windows calculator I am using right now).*R^n*is 2.70704 as n is 120 (12*10).Now therefore the EMI is

*E = 100000 * 2.70704 * (1.00833 - 1) / (2.70704 - 1) = 1320.97 or 1321*.

So after 120 payments you would have payed back

*1321*120 = 158520 rupees*.If you wanted to do the same repayment in annual instalments.

*R*is*(1 + 0.1 * (1/1))*or 1.1.*n*is 10.*R^n*is 2.59374. And E isE = 100000 * 2.59374 * (1.1 - 1) /(2.59374 - 1) = 16274.55 or 16275

.

Now you would have payed back 162750 rupees.

## 0 Comments:

Post a Comment

Subscribe to Post Comments [Atom]

## Links to this post:

Create a Link

<< Home