Calculate Loan Balance and Future Balance with Extra Repayments

Question

So, here's the current situation.

I took out a £7000 loan over 42 months at the insane interest rate of 47.4%.

I needed the loan, couldn't borrow it off family, and the interest rate is due to some averse credit history from when I was a teenager. I'm in a much more financially stable place now.

The monthly repayments are £309.59. I've paid 5 of these so far. I have also overpaid an additional £1000 which led to a further adjustment of £1,430.07 due to the reduction in interest.

Which brings the loan balance (according to the company) to £9,024.76.

I'm planning on overpaying by £1000/month until the balance is gone, as obviously I don't want to get reemed on interest.

How do I calculate how much I would actually be paying with this overpayment amount? Every calculator I've found so far only works if you've been regularly overpaying, not if you start overpaying part way through or over-pay sporadically. The lender will also only give me a settlement figure, they're unwilling to answer this for me.

Ideally, the answer would be an actual mathematical formula/process that I can apply to my numbers should they change. Presumably, the same formula that lenders use to calculate these adjustments when someone overpays.

Answer 1

Notwithstanding the stated £9024.76 balance.

Using the standard loan formula, where the initial loan principal is set equal to the sum of the payments discounted to present value, i.e. divided by (1 + r)^k where k is the month number.

s = principal
r = periodic rate
n = number of payments
d = payment amount

  s = (d - d (r + 1)^-n)/r
∴ d = r s (1 + 1/((1 + r)^n - 1))
& n = -(log(1 - (r s)/d)/log(1 + r))

Applying the loan figures. (APR in the UK is quoted as an effective rate.)

s = 7000
r = (1 + 0.474)^(1/12) - 1
n = 42

d = r s (1 + 1/((1 + r)^n - 1)) = 309.663

The balance b after payment in month x is given by

x = 5
b = (d + (1 + r)^x (r s - d))/r = 6574.74

Without changing the payment amount, if this was a new loan starting from month 5

s = (d + (1 + r)^x (r s - d))/r = 6574.74
n = 42 - 5 = 37

d = r s (1 + 1/((1 + r)^n - 1)) = 309.663
n = -(log(1 - (r s)/d)/log(1 + r)) = 37

So that checks out: payment is the same and loan finishes on time.

If instead £1000 is added to the month 5 payment and to every subsequent payment, the new loan would be

s = (d + (1 + r)^x (r s - d))/r - 1000 = 5574.74
d = 309.663 + 1000 = 1309.663

n = -(log(1 - (r s)/d)/log(1 + r)) = 4.66029

The balance in month 4 of the new loan (month 9 overall) is

x = 4
b = (d + (1 + r)^x (r s - d))/r = 841.83

requiring a final payment of (1 + r) b = 869.492

Total interest is

9*309.663 + 5*1000 + 869.492 - 7000 = 1656.46

Total interest of the original loan is

42*309.663 - 7000 = 6005.85