# How to solve amortization schedule where 10% of the interest is paid back on 4th payment

I am trying to find an algorithm to solve this amortization schedule:

• P = 1000, term = 12 months, Interest=5%, calculation based on actual/365.
• 10% of the total interest of the loan is added to the 4th payment(in addition to regular payment).
• All payment amounts except the 4th payment must be same amounts.
• The principal must be paid down to 0 at the end of the 12th payment

The challenge is that the extra amount at 4th payment will reduce principal and it will in turn reduce the interest. I am looking for an analytical solution if possible. If not then an iterative algorithm.

Loan begins and the first payment is on 23-Oct-2023. 4th Payment is on 23-Jan-2024. 10% interest is added to 4th payment thus reducing the principal. The effective interest rate does not change. I have come up with an iterative algorithm to solve this problem so far.

• Actual/365 depends on using the actual number of days in a specific month e.g. 28 days for February, 31 for March, but you haven't said when the loan begins or what month the 4th payment is in. Nov 18 at 10:24
• Is the 10% interest in addition to the amortized interest (making the effective interest rate about 15%)? Any are you trying to find the payment amount that satisfies the criteria? Nov 18 at 21:41
• Loan begins and the first payment is on 23-Oct-2023. 4th Payment is on 23-Jan-2024. 10% interest is added to 4th payment thus reducing the principal. The effective interest rate does not change. I have come up whin an iterative algorithm to solve this problem so far. Nov 19 at 6:22

Here is a quick solution done in Mathematica, using its computer algebra. Since the first payment is in October I presume the loan commences September, so the first month gains 30 days interest. Should be fairly obvious what the symbols are.

``````Clear[p, r, d, c, i, x]
d = 31; d = 29; d = 31; d = 30; d = 31; d = 30;
d = 31; d = 31; d = 30; d = 31; d = 30; d = 31;
p = 1000;
r = 0.05/365;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - (c + x);
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c;
p = p (1 + r d) - c
`````` ``````p = Simplify[p]
``````

1051.31 - 12.2803 c - 1.03392 x

The above expression for `p` should equal 0, so solving:-

``````Reduce[{p == 0, i == (12 c + x) - 1000, 10 x == i}]
``````

x == 2.72788 && i == 27.2788 && c == 85.3792

Results (full precision)

``````x = 2.7278770882620518
i = 27.27877088262052
c = 85.37924114952988
``````