4

I have an auto loan I wish to apply a very large amount to. I have a amortization table that shows me details for each payment I make. Here's where it gets complicated...

My lender does not allow applying payments directly to principal, so I'm forced to apply any additional payments towards the full balance. I asked them if there is an advantage to making such large, additional payments, and they said, ultimately, it can help reduce overall interest charged. Is this really true?

If so, I want to make a payment of roughly 1,800$. How would I go about recalculating my amortization table to account for that?

  • Can you edit and add country tag. Is your loan on reducing balance, or fixed repayment. – Dheer Dec 6 '15 at 14:56
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Just subtract your additional principal payment from your present principal. Then scan down the table until you find the entry where you have the same (reduced) principal. Basically you get to move down to a later entry in the amortization table. (This will be true assuming your payments stay the same each month). The total interest from that line on will be how much interest you'll now have to pay.

2

Assuming the term is fixed, so that additional payments reduce the future repayments.

Illustrating with a $10k loan over 3 years with monthly repayments at 5% APR compounded monthly.

pv is the present value of the loan
r is the periodic (monthly) interest rate
n is the number of periods
p is the periodic repayment amount

pv = 10000
r  = 0.05/12
n  = 36

Using formula 1

p = r*pv/(1 - (1 + r)^-n) = 299.71

total interest = n*p - pv = 789.56

Now suppose after one year an additional payment of $1800 is made.

Using formula 2 to find the loan balance after the 12th repayment.

q(0) = pv = 10000

n = 12

q(n) = (1 + r)^n*pv - ((1 + r)^n - 1)*p/r

∴ q(12) = 6831.52

Deducting the additional payment and continuing as if a new loan.

pv = 6831.52 - 1800 = 5031.52
r  = 0.05/12
n  = 24

p = r*pv/(1 - (1 + r)^-n) = 220.74

total interest = 12*299.71 + 24*220.74 + 1800 - 10000 = 694.28

The additional payment reduces the total interest by $95.28

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Formulae

Formula 1 - loan payment formula

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Formula 2 - inhomogeneous difference equation (Arne Jensen, Aalborg Uni.)

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