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

Question

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.

more information from the comments:

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.

Answer 1

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