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30

Prospective Method = Retrospective Method.

We will use the notation developed in Chapter 5 and denote the outstanding loan balance at time / by the symbol . As an aid in using this symbol, we can denote the prospective form by and the retrospective form by B,. However, the usage of the superscripts p and r should be considered optional. The original loan balance Bq is often denoted by L.

In specific cases it is possible to prove that the prospective method is equal to the retrospective method algebraically. For example, consider a loan of at interest rate / per period being repaid with payments of 1 at die end of each period for n periods, for which the outstanding loan balance t periods after the inception date of the loan is desired where 0<t<n . The outstanding loan balance at time / is computed after making the rth payment.

The prospective method gives

= a.

(6.1)

The retrospective method gives

B; = a (l+i)-s . (6.2)

We can show that the retrospective form is equal to the prospective form: m -St 1 -v"n 1 (1 +0- 1

(1 f o-v"-(l + 0+ 1

1 - V

In any given problem the prospective method or the retrospective method may be more efficient depending upon the nature of the problem. If the size and number of payments are known, then the prospective method is usually more efficient. If the number of payments or the size of a final irregular payment is not known, then the retrospective method is usually more efficient.

Example 6.1 A loan is being repaid with 10 payments of \$2000followed by 10 payments of \$1000 at the end of each half-year. If the nominal rate of interest convertible semiannually is 10%, find the outstanding loan balance

y. 6060.70

6.3 AMORTIZATION SCHEDULES

If a loan is being repaid by the amortization method, each payment is partially repayment of principal and partially payment of interest. Section 6.3 is concerned with determining how each payment can be divided between principal and interest.

Determining the amount of principal and interest contained in each payment is important for both the borrower and lender. For example, principal and interest are generally treated quite differently for income tax 08 8.

An amortization schedule is a table which shows the division of each payment into principal and interest, together with the outstanding loan balance

immediately after five payments have been made by both the prospective method and the retrospective method.

1. The rate of interest is 5% per half-year. Prospecdvely, die outstanding loan balance is

= 1000(aj +«5]) = 1000(10.3797 -1- 4.3295) = \$14,709 to the nearest dollar.

2. The original loan was

L = 1000(a5g, +«Tol) = 1000(12.4622 -I- 7.7217) = \$20,184

to the nearest dollar. Retrospectively, the outstanding loan balance is

5 = 20,184(1.05)5-200055, = 20,184(1.27628) - 2000(5.5256)

= \$14,709

to the nearest dollar. Thus, the prospective and retrospective methods produce the same answer.

Example 6.2 A loan is being repaid with 20 annual payments of \$1000 each. At the time of the fifth payment, the borrower wishes to pay an extra \$2000 and then repay the balance over 12 years with a revised annual payment. If the effective rate of interest is 9%, find the amount of the revised annual payment.

The balance after five years, prospectively, is

= = 1000(8.0607) = \$8060.70.

If die borrower pays an additional \$2000, die balance becomes \$6060.70. An equation of value for die revised payment, denoted by X, is

= 6060.70

Present Value of Future Payments = Accumulated Value of Loan -Accumulated Value of Past Payments,

 Payment Interest Principal Outstanding Period amount paid repaid loan balance v" ",7TTTn=l-v"-* a„ /+il-v"-* = a;rr71 n - 1 = 1 - «31- = "11 = 1 - > Total

Consider the first period of the loan. At the end of the first period the interest due on the balance at the beginning of the period is ia = 1 - v". The rest of the total payment of 1, i.e. v", must be principal repaid. The outstanding loan balance at the end of the period equals the outstanding loan balance at the beginning of the period less the principal repaid, i.e. -v" = a. The same reasoning applies for each successive line of the schedule.

Several additional observations are possible. First, it should be noted that the outstanding loan balance agrees with that obtained by the prospective method in formula (6.1.). Second, the sum of the principal repayments equals the original amount of the loan. Third, the sum of the interest payments is equal to the difference between the sum of the total payments and the principal repayments. Fourth, the principal repayments form a geometric progression with common ratio 1 + i. Thus, it is a simple matter to find any one principal repayment knowing any other principal repayment and the rate of interest.

Further insight into the nature of the amortization schedule can be gained by the following argument. The original loan balance will accumulate to {1 + i) = at die end of the first period. However, = 1 + aj ,

P. = v"-+l

(6.3) (6.4)

The reader should note that the notation /, was also used in Chapter 1 in a different context. However, in both cases the symbol refers to an amount of interest earned between times r - 1 and t, so there should be little risk of ambiguity. Finally, we will let the installment payment be denoted by R.

It should be noted that Table 6.1 is based on an original loan of - . If the original loan were some other amount, then all the values in the schedule would be proportional. For example, if the original loan were \$1( , then each number in the last four columns of the schedule would be multiplied

by imia.

For a specific problem, the amortization schedule can be constructed from basic prmciples. For example, consider the construction of an amortization schedule for a \$10( loan repaid in four annual payments if the annual effective rate of interest is 8%. Then we have

= \$301.92.

3.3121 Rable 6.2 is the amortization schedule for this example.

Table 6.2 Amortization Schedule for a Loan of \$1000 Repaid Over Four Years at 8%

 Payment Interest Principal Outstanding Year amount paid repaid loan balance 1000.00 301.92 80.00 221.92 778.08 301.92 62.25 239.67 538.41 301.92 43.07 258.85 279.56 301.92 22.36 279.56

i.e. dj is sufficient to make the annuity payment of 1 and leave an outstanding balance of a- at the end of the first period. The same reasoning applies for each successive line of the schedule.

For convenience, we will denote the amount of interest paid in the rth installment by /, and the amount of principal repaid in the same installment by P[- Thus, for the amortization schedule given in Table 6.1 we have

, n-t+l

after each payment is made. Consider a loan of at interest rate i per period being repaid with payments of 1 at the end of each period for n periods. Table 6.1 is an amortization schedule for this case.

Table 6.1 Amortization Schedule for a Loan of Repaid Over n Periods at Rate U

irregular payment will be considered in Example 6.3 and in the exercises. The general case of a varying series of payments will be considered in Section 6.6.

Certain special types of loan transactions are discussed in more detail in Chapter 8. In particular, real estate mortgages are considered in Section 8.3. One important new type of mortgage considered in Section 8.3 is the adjustable rate mortgage. This type of mortgage involves features not considered in Chapter 6.

Example 6.3 A \$1000 loan is being repaid by payments of \$100 at the end of each quarter for as long as necessary, plus a smaller final payment. If the nominal rate of interest convertible quarterly is 16%, find the amount of principal and interest in the fourth payment.

The outstanding loan balance at die beginning of die fourth quarter, i.e. the end of the diird quarter, is

b4 = 1000(1.04)-1005,, = 1124.86-312.16 = \$812.70.

The interest contained in die fourth payment is

/4 = .04(812.70) = \$32.51.

The principal contained in die fourth payment is

p4 = 100.00-32.51 = \$67.49.

Note diat it was not necessary to find die duration and amount of die smaller final payment in order to solve this example.

Example 6.4 A borrows \$10,000from and agrees to repay it with equal Muarterly installments of principal and interest at 8% convertible quarterly over years. At the end of two years sells the right to receive future payments * at a price which produces a yield rate of 10% convertible quarterly. Find total amount of interest received: (1) by C, and (2) by B. The quarterly installment paid by A is

10,000 10,000 J528 71 «541.02 18.9139 The price pays is dien

528.7la jg, = (528.71)(13.0550) = \$6902.31 . The total payments made by A over die last four years are

(16)(528.71) = \$8459.36. Thus, die total interest received by is

8459.36 - 6902.31 = \$1557.05.

In the first line of Table 6.2 we have the following calculations. The interest contained in the first payment is

/j = i.Bq = .08(1000) = \$80.00. The principal contained in the first payment is

= R-Iy = 301.92 - 80.00 = \$221.92. The outstanding loan balance at the end of the first year is Bj = - = 1000.00 - 221.92 = \$778.08.

Each successive line of the schedule is calculated in similar fashion.

In this particular example the last line exactly balances. However, more typically some roundoff error is likely to accumulate. Standard practice is to adjust the last payment so that it is exactly equal to the amount of interest for the final period plus the outstanding loan balance at the beginning of the final period (i.e. at the end of the next-to-final period). This adjustment will bring the outstanding loan balance exactly to zero at the end of the term of the loan.

It is also possible to construct the amortization schedule by alternate methods utilizing the various relationships in the table. As one example, the values of the outstanding loan balance can be calculated as in Section 6.2, and then the rest of the schedule can be deduced from these values. As a second example, the principal repaid column can be calculated using the fact that the successive values are in geometric progression, and then the rest of the schedule can be deduced from these values.

It should be noted that if it is desired to fmd the amount of principal and interest in one particular payment, it is not necessary to construct the entire amortization schedule. The outstanding loan balance at the beginning of the period in question can be determined by the methods of Section 6.2, and then that one line of the amortization schedule can be calculated.

The reader might wonder about the amortization schedule for a 1 11 . It is clear that for a pe etuity the entire payment represents interest and die outstanding loan balance remains unchanged. Thus, amortization of a pe etuity is really a contradiction in terms.

Several assumptions have been implicit in the preceding discussion of amortization schedules. First, we have assumed a constant rate of interest. An amortization schedule at a fixed, but varying rate of interest will be considered in the exercises. Second, we have assumed that the annuity payment period and the interest conversion period are equal. The cases in which they are not equal will be considered in Section 6.5. Third, we have assumed that aimuity payments are level. The situation in which they are level except for a final

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