Hi I can compute a Basic Loan Amortization Schedule, But now I'm trying to Calculate a Seasonal Loan Amortization Schedule . So for example A loan starts at January And Will go on for 3 years
, with Monthly Payments
. Interest is at 3% Annually and is compounded Annually
. During the months of September, October, November, December
It'll only accumulate Interest
But no payments and principal. Principal is 100000
. How Do I calculate Payments and interest.
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Is this homework? What country are you in? I don't claim to know everything, but I've never heard of such a loan.– JTP - Apologise to Monica ♦Feb 24, 2014 at 19:31
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Not my Homework, A question I made off the top of my head. And It's in Canada. Some loan companies do give off seasonal loans. <hemafinance.com/en/calculators/loan/15/seasonal-loan-calculator> A calculator is already Made for a type of question like this, so It's not common but it's also not uncommon. Dont mind the double negatives. Just trying to portray a point.– user3276954Feb 24, 2014 at 19:36
1 Answer
You can calculate the payments as shown below. But first a basic monthly payment calculation to show the method :-
s = 100000
t = 3
n = 12
i = 0.03
The basic case using the formula here :-
This formula needs the periodic rate, r
r = (1 + i)^(1/n) - 1
monthly payment = (r*s)/(1 - (1 + r)^(-(n*t)))
2906.34
Now the same calculation using a more complicated formula given here :-
In the formula each a(1 + i)^(-k/n)
term is a payment. Solving gives the monthly payment ,a
, as demonstrated earlier.
By omitting payments for September, October, November & December, (since a
for these dates is zero), solving finds the payment amount for the remaining 24 months.
So to repay the loan in 24 equal payments each payment should be 4338.17.
The interest remains 3% annually.
(This method uses EU APR, so you may need to convert your rate first.)
Edit
Taking another go to match the output of the Hema calculator.
Here is the output to be matched :-
First a simple longhand calculation, solved using Mathematica :-
r = 0.03/12;
p[0] = 100000;
p[1] = p[0] - (a - r p[0]);
p[2] = p[1] - (a - r p[1]);
p[3] = p[2] - (a - r p[2]);
p[4] = p[3] - (a - r p[3]);
p[5] = p[4] - (a - r p[4]);
p[6] = p[5] - (a - r p[5]);
p[7] = p[6] - (a - r p[6]);
p[8] = p[7] - (a - r p[7]);
p[13] = p[8] - (a - r p[8]);
p[14] = p[13] - (a - r p[13]);
p[15] = p[14] - (a - r p[14]);
p[16] = p[15] - (a - r p[15]);
p[17] = p[16] - (a - r p[16]);
p[18] = p[17] - (a - r p[17]);
p[19] = p[18] - (a - r p[18]);
p[20] = p[19] - (a - r p[19]);
p[25] = p[20] - (a - r p[20]);
p[26] = p[25] - (a - r p[25]);
p[27] = p[26] - (a - r p[26]);
p[28] = p[27] - (a - r p[27]);
p[29] = p[28] - (a - r p[28]);
p[30] = p[29] - (a - r p[29]);
p[31] = p[30] - (a - r p[30]);
p[32] = p[31] - (a - r p[31]);
sol = Solve[p[32] == 0, a] ;
a = sol[[1, 1, 2]]
4298.12
Which matches the Hema calculation exactly. The seasonal interest charges match too :-
r p[8]
168.326
r p[20]
85.0034
The main payment amount can also be calculated in one step using the loan formula given at the beginning of this answer like so :-
r = 0.03/12
s = 100000
n = 8
t = 3
monthly payment = (r*s)/(1 - (1 + r)^(-(n*t)))
4298.12
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EU APR is the "effective annual rate" according to Mortgage Interest Calculations in Canada. Feb 24, 2014 at 22:05
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hi Chris if I may bother you again, What if the compounding interest wasn't annually and changed to Monthly. Feb 25, 2014 at 14:48
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Hi. Once you have your interest as effective annual rate (EAR) any compounding interval produces the same result. That's why it is used. In the mortgage example
EAR = 0.0609; monthly = (1 + EAR)^(1/12) - 1 = 0.00493862
. E.g. $100 for 2 years =100*(1 + EAR)*(1 + EAR) = $112.551
, which is the same as $100 for 24 months =100*(1 + monthly)^24 = $112.551
. Feb 25, 2014 at 15:11 -
Sweet! Another Question, Does this formula take into account interest that still occurs during the months that have no principal? Feb 25, 2014 at 17:19