# How can I calculate amortized payment when the payments are made in advance?

I'm trying to calculate the payment amount on an amortized loan with interest where the payments are made in advance.

If the payments are made in arrears there's a simple formula, but since the first payment of my loan is made before the first interest accrual it's throwing everything out.

For some reason I also can't find any calculators online that allow advance instalment plans on amortized loans.

Just to clarify by an example: You borrow, say, \$10,00.00 at 5% compounded annually, on Jan 1, 2023. Then you immediately make a payment of x dollars, and then pay x more at the start of each following year for 9 more years to completely pay off the loan with the 10 payments. You want to calculate the value of x, using the appropriate formula.

Well, imagine for a moment that you take out the loan on Jan 1, 2022 (one year early), with two other differences: You do not make an immediate payment, and the loan is for a reduced amount. Specifically, you borrow A, where A = 10000/(1.05) = 9523.81. This amount is chosen so that, in the imaginary loan, by the time of the first payment of x, on Jan 1, 2023, your principal plus interest will be ... \$10,00.00! You then imagine make all ten payments as scheduled in the real loan.

Your real loan and the imaginary loan now match in all important details: the amount you owe on Jan 1, 2023, and the amount and timing of all 10 payments. But the imaginary loan is just a simple or ordinary annuity paying off a loan of 9523.81 with 10 equal payments at the end of each year at 5% interest.

By discounting the the principal amount one payment period earlier, you have converted the annuity due to a simple annuity, and all the simple annuity formulae can be used with the revised numbers.

BTW, this answer does by hand-waving what is done with more rigor in the correct answer of @Chris Degnen

• I added an example using this method to my answer. Commented Jul 9, 2022 at 7:40

With

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

This would be your arrears formula

$s=\sum_{k=1}^{n}\frac{d}{(1+r)^k}=\frac{d-d(r+1)^{-n}}{r}$

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

Example figures

``````  s = 800
r = 0.1
n = 6

∴ d = 183.69
``````

Stepping through amortisation

``````0.   s = 800
1.   s = s (1 + r) - d
2.   s = s (1 + r) - d
3.   s = s (1 + r) - d
4.   s = s (1 + r) - d
5.   s = s (1 + r) - d
6.   s = s (1 + r) - d = 0
``````

$\dpi{110}s=\sum_{k=0}^{n-1}\frac{d}{(1+r)^{k}}=\frac{d(1+r)^{1-n}((1+r)^{n}-1)}{r}$

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

Same example figures

``````  s = 800
r = 0.1
n = 6

∴ d = 166.99
``````

Stepping through amortisation

``````0.   s = 800 - d
1.   s = s (1 + r) - d
2.   s = s (1 + r) - d
3.   s = s (1 + r) - d
4.   s = s (1 + r) - d
5.   s = s (1 + r) - d = 0
``````

Alternatively using DJohnM's method to obtain the same

``````s = 800
r = 0.1
n = 6

s = s/(1 + r) = 727.27
``````

Using the arrears payment formula with the discounted principal

``````d = r s (1 + 1/((1 + r)^n - 1)) = 166.99
``````

Amortisation with payments in advance as before

``````0.   s = 800 - d
1.   s = s (1 + r) - d
2.   s = s (1 + r) - d
3.   s = s (1 + r) - d
4.   s = s (1 + r) - d
5.   s = s (1 + r) - d = 0
``````