Question

set up an amortization schedule for a $25000 to be repaid in equal installment at the end of each of the next 5 year .the interest rate is 10%.

Answer 1

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, P_n=P_n-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.