The schedule based on monthly payment of $531.18 is:
YEAR |
INTEREST ($) |
PRINCIPAL ($) |
YEAR-END BALANCE ($) |
1 |
2317.41 |
4056.70 |
20943.30 |
2 |
1892.62 |
4481.49 |
16461.80 |
3 |
1423.35 |
4950.76 |
11551.04 |
4 |
904.94 |
5469.17 |
6041.87 |
5 |
332.25 |
6041.87 |
0 |
Monthly interest rate is 10/12%=⅚%=0.00833 approx. Period=5×12=60 months.
Growth of principal in 5 years=1.0083360=1.64531 approx. $25000×1.64531=$41132.72.
If m is the fixed monthly instalment then:
m=rP0(1+r)n/((1+r)n-1), where P0 is the initial principal, r is the monthly interest rate, n is the number of months over which the debt is paid. In this case rP0(1+r)n=$41132.72r=$342.77 approx.
m=$342.77/0.64531=$531.18 approx. This confirms the above amount of the monthly instalment.
Now let's see the month-by-month calculations.
At the end of the first month the new principal, P1=P0(1+r)-m. After n months, Pn=Pn-1(1+r)-m.
This recursive formula is used to derive m, because P60=0, and P60=P59(1+r)-m.
We can also see that the 1st year-end principal is P12=P11(1+r)-m.
P11=P10(1+r)-m and so on until we reach P0. We can similarly calculate P24 in terms of P12. This helps to construct the table above.
To calculate the first row of the table we calculate the principal P12 at the end of the first year:
P12=P0(1+r)12-m((1+r)12-1)/r=$20943.29701 or $20943.30 to the nearest cent. This is the year-end balance.
12 instalments have been paid=12×531.18=$6374.16. The principal has been reduced by 25000-20943.30=$4056.70 (PRINCIPAL column). So $4056.70 of the initial $25000 has been paid off.
Therefore $6374.16-$4056.70=$2317.41 interest has been paid (INTEREST column).
Other rows of the table have been similarly calculated.