# 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, 2023 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, 2023 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, 2023 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[1] = 31; d[2] = 29; d[3] = 31; d[4] = 30; d[5] = 31; d[6] = 30;
d[7] = 31; d[8] = 31; d[9] = 30; d[10] = 31; d[11] = 30; d[12] = 31;
p = 1000;
r = 0.05/365;
p = p (1 + r d[9]) - c;
p = p (1 + r d[10]) - c;
p = p (1 + r d[11]) - c;
p = p (1 + r d[12]) - (c + x);
p = p (1 + r d[1]) - c;
p = p (1 + r d[2]) - c;
p = p (1 + r d[3]) - c;
p = p (1 + r d[4]) - c;
p = p (1 + r d[5]) - c;
p = p (1 + r d[6]) - c;
p = p (1 + r d[7]) - c;
p = p (1 + r d[8]) - 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
``````