Saturday, February 28, 2009

EMI Calculation formula -- proof

=== I'm confident that this proof is correct; but it is possible that there is a simpler proof existing for the same ==

P ==> Loan amount (principal)
E ==> EMI per month (to be derived)
n ==> tenure in months
R ==> rate of interest per month (to avoid cluttering of derivations, R is assumed to have been divided by 100. ie., for 12% p.a interest rate, it is 1% per month, so R is 0.01)

Let us derive the formula for outstanding-principal at the end of n EMIs (months):

outstanding-principal after month n = outstanding-principal after month n-1 minus principal component of the n'th EMI.

First month: P-(E-P*R) ==> P-E+PR ==> P(R+1)-E

Second month (n=2):
=> P-(E-PR)-(E-(P-E+PR)*R) => P-E+PR-(E+ER-(P+PR)*R) => -E-E-ER + P + PR + (P+PR)*R
=> -2E-ER + (P+PR)(1+R) => -E(R+2) + P(1+R)(R+1) = P(R+1)2 - E(R+2)

Third month (n=3):
=> P(R+1)2 - E(R+2) - E + PR(R+1)2 - ER(R+2)
=> P(R+1)2 + PR(R+1)2 -E -E(R+2) -ER(R+2) ==> P(R+1)3 - E(1+(R+2)(1+R))

Fourth month (n=4):
=> P(R+1)3 - E(1+(R+2)(R+1)) - E + PR(R+1)3 - ER(1+(R+2)(R+1))
=> P(R+1)4 -E -E(1+(R+2)(R+1)) -ER(1+(R+2)(R+1))
=> P(R+1)4 - E(1+(1+(R+2)(R+1))(R+1)) -------- (1)

Clearly the first portion of the outstanding principal has become a function of 'n'. I had to formulate the second portion ie., the co-efficient of E -> (1+(1+(R+2)(R+1))(R+1)) so I can come up with a outstanding-principal formula after n EMIs.

First approach:

1 ==> (R+1)0
R+2 ==> (R+1)0 + (R+1)1
1+(R+2)(R+1) == > 1 + R2 + R + 2R+2 ==> (R+1)0 + (R+1)1 + (R+1)2
1+(1+(R+2)(R+1))(R+1) ==> 1 + ((R+1)2+R+2)(R+1)) == > 1 + ((R+1)3+(R+2)(R+1))
==> (R+1)0 + (R+1)1 + (R+1)2+ (R+1)3

This translates to something like a0+a1+a2+a3...+an ; but I have no idea of any means to change this to a simple function of n.

So tried another approach to deduce:

R+2 => (R2+2R+1-1)/R ==> ((R+1)2-1)/R
1 + (R+2)(R+1) ==> 1 + ((R+1)2-1)*(R+1)/R ==> 1 + ((R+1)3-R-1)/R
==> 1 + ((R+1)3-1)/R - 1 == > ((R+1)3-1)/R
Similarly it can also be proved that for n=4 it is ((R+1)4-1)/R

Proof my mathematical induction:

Lets call this function P(k):

P(1) = ((R+1)1-1)/R => R+1-1/R => 1
Assuming P(k) is true, lets prove p(k+1).

based on the above recurrence, p(k+1) = 1+(R+1)P(k)
==> 1+(R+1) * ((R+1)k-1)/R ==> 1 + ((R+1)(k+1)-(R+1))/R
==> 1 + ((R+1)(k+1)-1)/R - 1 ==> ((R+1)(k+1)-1)/R ==> P(k+1)


Substituting the result in equation (1), the outstanding principal after n EMIs is: => P(R+1)n - E((R+1)n-1)/R

For the loan to end correctly, the outstanding principal at the end of n EMIs should be less than or equal to zero (if -ve, last month EMI will be smaller appropriately).

P(R+1)n - E((R+1)n - 1)/R <= 0 or E >= P(R+1)n / (((R+1)n - 1)/R) or
E >= PR(R+1)n / ((R+1)n -1)

EMI has to be as small as possible while completing the loan on time. So, computationally, for E to be a smallest possible round integer, >= can be converted to = by doing a ceil on the result.

So, E = ceil(PR(R+1)n / ((R+1)n -1))

where ceil(x) ==> smallest integer that is >= x e.g, ceil(1.5) = 2; ceil(2.1) = 3; ceil(2) = 2;

Hence proved :)

8 comments:

  1. Good. Till now I was blindly using that formula without knowing how it came but today I completely agree with ur proof.

    ReplyDelete
  2. It is true, If I have EMI how i can i calculate Principal amount. Please tell

    ReplyDelete
  3. let s = a^0+a^1+a^2+a^3+..+a^(n-1) ... Eqn(1)
    then s*a = a^1+a^2+a^3+..+a^(n-1)+a^n ... Eqn(2)

    Now Equation(2) - Equation(1)
    s*(a-1) = -a^0 + a^n
    s*(a-1) = a^n - 1 ..Now the value of a=R+1
    s*R = (R+1)^n - 1
    s = ((R+1)^n-1)/R

    ReplyDelete
  4. I can give you another way using simple technique of addition. The sum of numbers from say 1 to 100 = n(n+1)/2 where n = 100 sum = 100*101/2=101*50=5050.
    Like this every month principle paid in the EMI component will reduce its interest at the given rate.
    So if 1 l.20 lakh is the loan and 12 months is the period of repayment then principle paid per month is Rs 10000.
    Interest at the rate of say 12% is 10000*12/1200 = 100. So every month by paying Rs10000 interest will come down by Rs100. So Total interest will be 100*12*13/2= 7800. Total amount to be paid in 12 instalments = 120000+7800=127800. Permonth to be paid is 127800/12= 10650. So EMI = 10650 per month
    Sundar

    ReplyDelete
  5. Good try. I think you have over-simplified few things. Your assumption of '10K' principal repayment per month is incorrect -- realistically, interest/principal repayment is not uniform. For 1.2L, at 12% pa, for 1 year, the EMI should be Rs. 10662. Interest to-be-paid is: Rs. 7942 (against your Rs.7800). For bigger loans and tenure, your EMI value will be further away.

    ReplyDelete
  6. Read your blog its really informative and helpful keep updating with newer post on Property loan interest rate

    ReplyDelete
  7. Use the interactive home loan EMI Calculator to calculate your home loan EMI. Get all details on interest payable and tenure using the house loan calculator.

    ReplyDelete