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.

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