(6.17)

The progression of loan balances is illustrated in Figure 6.3.

Figure 6.3 Outstanding loan balances for a loan involving step-rate amounts of principal

Arrwrtization schedules and sinking funds 193 A formula for R can be developed by the following procedure:

1. Find RP.

2. Find B.

3. Equate the two and solve for R. The prospective approach gives

The retrospective approach gives

(6.18)

fi = (Z.-Z.)(l +jr + L+iLsj-Rsj.

Equating formulas (6.18) and (6.19) and solving for R produces Y

(6.19)

R = (-K +jy + L+iLsj On=a\i + Sa\j

(6.20)

Formula (6.20) is the desired formula for the level periodic payment. Although formula (6.20) is the desired formula for R, the crossover period a is unknown. A simple criterion can be developed to minimize the i frial-and-error involved in finding a.

If formulas (6.18) and (6.20) are substituted into formula (6.17), then a is the smallest integer such that

(L-Z.)(l +f)" + L +iLsj

< L .

\1 + a\j

[The above inequality is a criterion which could be used to determine a. However, it is possible to substantially simplify the criterion as follows:

is In

(L-L)(l +; , + iLsjOji < Lsj {L-LXl +j)"o,- < Lv-sj

L-L L

(6.21)

6.8 STEP-RATE AMOUNTS OF PRINCIPAL

A problem often encountered in practical applications involves the amortization of a loan in which the loan balance is subdivided and different rates of interest are charged on the subdivided portions. For example, a credit institution may charge 11/2% per month on the first $1000 of outstanding loan balance and 1 % per month on any excess. We call these subdivided portions of loan balance step-rate amounts of principal.

The determination of a level periodic payment which will repay a loan mvolving step-rate amounts of principal is surprisingly a non-trivial problem which involves trial-and-error.

In this section we give a solution to this type of problem for loans involving two step-rates. The approach can be generalized to more than two step-rates.

Consider a loan L which is being repaid with level payments of R over n periods. Interest is computed at rate / per period on the first L of outstanding loan balance, 0 < L < L, and at rate j per period on the excess over L. In practice / > j in most siUiations, although this is not required for the derivation. It is desired to fmd the level payment R.

Before R can be determined it is necessary to find the "crossover" period when the outstanding loan balance first becomes less than or equal to Z,. Denote this duration by a, so that a is the smallest integer such that

Formula (6.21) is a simple criterion which can be used to determine a.

Another computer-oriented approach to deal with this type of trial-and-error problem is to solve for R by iteration. The procedure would be as follows:

1. Determine an approximate starting value /?q . For example, 7?q could be based on a rate of interest between i and j.

2. Bn = L.

Table 6.7 Amortization Schedule for a Loan of $3000 Repaid Over 12 Months at 1 1/2% Per Mondi on the First $1000 and 1% on Any Excess

3. Use the recursion formulas:

B, = B, i-/? + /L+j(B, i-L), ifB, i > L B, = B, , - + , 1, if B, , < L\

B„ = 0.

(6.22a) (6.226)

Successive values of R, R2, ... can be generated with these formulas by standard numerical methods on a computer.

Example 6.16 Consider a loan of $3000 which is being repaid with levell monthly payments over 12 months. Interest is computed at 11/2% per month \ on the first $1000 of outstanding loan balance and at 1% per month on any \ excess over $1000. Find the level payment which will exacUy amortize this loan.

We have L = 3000, L = 1000, « = 12, / = .015, and j = .01, so diat formula (6.21) gives

2 V2! .

12=51.015

The smaUest integer a satisfying diis inequaUty is a = 9. The level monddy payment given by formula (6.20) is

2000(1.01)5-1-1000+ 15551 „, R =---bOi = 270.98545.

"51 .015 + *9l .01

The complete amortization schedule carrying five decimal places is given in Table! 6.7. The criterion is verified, since Bg > 1000 and Bg < 1000. The interest paid I column is computed as 1 1/2% on die first $1000 of outstanding loan balance and 1 % on j any excess.

ICISES

6.2 Finding the outstanding loan balance

A loan of $1 0 is being repaid with quarterly payments at the end of each quarter for five years at 6% convertible quarterly. Find the outstanding loan balance at the end of the second year.

A loan of $10,000 is being repaid by installments of $2000 at die end of each year, and a smaller final payment made one year after die last regular payment. Interest is at the effective rate of 12%. Find die amount of outstanding loan balance remaining when the borrower has made payments equal to the amount of the loan. Answer to the nearest dollar.

A loan is being repaid by quarterly installments of $1500 at die end of each quarter at 10% convertible quarterly. If the loan balance at the end of the first year is $12,000, find die original loan balance. Answer to die nearest dollar.

A loan is being repaid by 15 annual payments at the end of each year. The first 5 installments are $4000 each, die next 5 are $3000 each, and die final 5 are $2000 each. Find expressions for the outstanding loan balance immediately after the second $3000 installment:

a) Prospectively.

b) Retrospectively.

5. A loan is to be repaid witfi level installments payable at die end of each half-year for 3 1/2 years, at a nominal rate of interest of 8% convertible semiannually. After the fourdi payment, die outstanding loan balance is $5000. Find die initial amount of the loan. Answer to the nearest dollar.

6. A $20,000 loan is to be repaid widi annual payments at die end of each year for 12 years. If (1 + i) -2, find die outstanding balance immediately after die fourth payment. Answer to die nearest dollar.

7. A $20,000 mortgage is being repaid widi 20 annual installments at die end of each year. The borrower makes five payments and dien is temporarily unable to make payments for die next two years. Find an expression for die revised payment to start i at die end of die 8di year if die loan is still to be repaid at die end of die original 20 i years.

8. A loan of 1 was originally scheduled to be repaid by 25 equal annual payments at die end of each year. An extra payment widi each of die 6di tiirough die 10th i scheduled payments will be sufficient to repay die loan 5 years earlier dian under die i original schedule. Show that

"251 "51

9. A loan is being repaid witfi level payments. If B,, B,+ ,, B,2> and B,j are four j successive outstanding loan balances, show diat:

a) (B, -B,+ ,)(B,+2 - ,+ ) = (B,+ , - B,+2)-

b) B, + B,+3 < B,+ ,-t-B,+2. 6.3 Amortization schedules

10. A loan is being repaid witfi quarterly instalhnents of $ 1000 at die end of each quarter j for five years at 12% convertible quarterly. Find die amount of principal in tfiej sixth instalhnent.

11. Consider a loan which is being repaid witfi installments of 1 at die end of each] period for n periods. Find an expression at issue for tfie present value of tfie interest! which will be paid over the life of the loan.

12. A loan of $10,000 is being repaid witfi 20 installments at die end of each year at] 10% effective. Show diat tfie amount of interest in tfie lltfi installment is

1000

1 -t-V

13. A loan is being repaid witfi 20 instalhnents at tfie end of each year at 9% effective.. In what installment are tfie principal and interest portions most nearly equal to each odier?

1-7 lei

A loan is being repaid witfi a series of payments at tfie end of each quarter for five years. If the amount of principal in the third payment is $100, find the amount of principal in die last five payments. Interest is at tfie rate of 10% convertible quarterly.

A loan is being repaid witfi installments of 1 at die end of each year for 20 years. Interest is at effective rate / for tfie first 10 years and effective ratey for tfie second 10 years. Find expressions for:

a) The amount of interest paid in die 5tii installment.

b) The amount of principal repaid in the 15th installment.

A mortgage widi original principal A is being repaid witfi level payments of at die end of each year for as long as necessary plus a smaller fuial payment. The effective rate of interest is /.

a) Find the amount of principal in the rth instalhnent.

b) Is the principal repaid column in the amortization schedule in geometric progression (excluding the irregular final payment)?

A borrower has a mortgage which calls for level annual payments of 1 at die end of each year for 20 years. At die time of die sevendi regular payment an additional payment is made equal to the amount of principal that according to the original amortization schedule would have been repaid by die eighdi regular payment. If payments of 1 continue to be made at tfie end of tfie eightfi and succeeding years until the mortgage is fiilly repaid, show that the amount saved in interest payments over the fiill term of the mortgage is

1-v".

A loan of L is being amortized witii payments at die end of each year for 10 years. If V* = 2/3, find tfie foUowing:

a) The amount of principal repaid in die first 5 payments.

b) The amount due at die end of 10 years if die final 5 payments are not made as scheduled.

119. A 35-year loan is to be repaid witfi equal instaUments at die end of each year. The amount of interest paid in die 8tfi instalhnent is $135. The amount of interest paid Sf: in die 22nd installment is $108. Calculate die amount of interest paid in die 29di installment.

6.4 Sinking funds

M. A has borrowed $10,000 on which interest is charged at 10% effective. A is accumulating a sinking fiind at 8% effective to repay die loan. At die end of 10 years die balance in die sinking fiind is $5000. At die end of die 1 Itii year A makes a total payment of $1500.

a) How much of die $1500 pays interest currendy on die loan?

b) How much of tfie $1500 goes into die sinking fiind?

c) How much of tfie $1500 should be considered as interest?