Sections 5.3, 5.4

1. Sections 5.3, 5.4 When a loan is being repaid with the amortization method, each payment is partially a repayment of principal and partially a payment of interest. Determining the amount of each for a payment can be important (for income tax purposes, for example). Table 5.1 on page 157 of the textbook displays the format for an amortization schedule. The entries are for a loan of at interest rate i per period being repaid with payments of 1 at the end of each period for n periods. Observe each of the following from this table: 1. 2. a – n| At the end of period 1, the interest paid is i = 1 – vn, the principal repaid is vn, and the outstanding loan balance is . a – n| a ––– n–1| At the end of period k, the interest paid is i = 1 – vn–k+1, the principal repaid is vn–k+1, and the outstanding loan balance is = Bkp . a ––––– n–k+1| a ––– n–k|

2. 3. 4. Sum of Principal Repayments = Original Amount of Loan Sum of Interest Payments = Sum of Total Payments – Sum of Principal Repayments For convenience, we shall often let It = Pt = the amount of interest in the tth payment, the amount of principal in the tth payment,

3. A $5000 loan is being repaid by payments of $X at the end of each half year for as long as necessary until a smaller final payment is made. The nominal rate of interest convertible semiannually is 14%. (a) If X = 400, find the principal and the interest in the sixth payment. (Note: it is not necessary to calculate any of the formulas in the amortization table.) The outstanding balance at the beginning of the sixth half-year is B5r = 5000(1.07)5– 400 = 7012.76 – 400(5.75074) = $4712.46 s– 5| The interest in the sixth payment is I6 = (0.07)(4712.46) = $329.87 The principal in the sixth payment is P6 = 400 – 329.87= $70.13

4. (b) If X = 350, find the principal in the sixth payment, and interpret this. The outstanding balance at the beginning of the sixth half-year is B5r = 5000(1.07)5– 350 = 7012.76 – 350(5.75074) = $5000.00 s– 5| If X 350, then the loan will never be paid off, because every payment will count only toward interest and nothing toward principal.

5. Jones borrows $20,000 from Smith and agrees to repay the loan with equal quarterly installments of principal and interest at 10% convertible quarterly over eight years. At the end of three years, Smith sells the right to receive future payments to Collins at a price which produces a yield rate of 12% convertible quarterly for Collins. Find the total amount of interest received (a) by Collins, and (b) by Smith. 20000 ———– = 20000 ———– = 21.84918 $915.37 (a) Each quarterly payment by Jones is a –– 32 |0.025 Total payments by Jones to Collins over the last five years (i.e., the five years immediately after sale) are (20)915.37 = $18307.40 After three years, the price Collins pays to Smith is 915.37 = a –– 20 |0.03 915.37(14.87747) = $13618.39 Total amount of interest received by Collins is 18307.40 – 13618.39 = $4689.01 (b) Total payments by Jones to Smith over the first three years are (12)915.37 = $10984.44

6. Jones borrows $20,000 from Smith and agrees to repay the loan with equal quarterly installments of principal and interest at 10% convertible quarterly over eight years. At the end of three years, Smith sells the right to receive future payments to Collins at a price which produces a yield rate of 12% convertible quarterly for Collins. Find the total amount of interest received (a) by Collins, and (b) by Smith. (b) Total payments by Jones to Smith over the first three years are (12)915.37 = $10984.44 Recall that after three years, the price Collins pays to Smith is 915.37 = 915.37(14.87747) = $13618.39 a –– 20 |0.03 Total amount of interest received by Smith is 13618.42 + 10984.44 – 20000 = $4602.86 After three years, the outstanding loan balance is 915.37 = a –– 20 |0.025 915.37(15.58916) = $14269.85 (Compare this to the price Collins pays to Smith.)

7. An amount is invested at an annual effective rate of interest i which is exactly sufficient to pay 1 at the end of each year for n years. In the first year, the fund earns rate i and 1 is paid at the end of the year. However, in the second year, the fund earns rate j > i. If X is the revised payment which could be made at the end of years 2 to n, then find X assuming that (a) the rate reverts back to i again after this one year, From the amortization table, the balance after one year is , and therefore the balance after two years must be a ––– n–1| i (1 + j) – X . a ––– n–1| i However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j) – X = X a ––– n–1| i a ––– n–2| i X (1 + ) = (1 + j) a ––– n–2| i a ––– n–1| i X (1 + i) = (1 + j) 1 + j X = —— 1 + i a ––– n–1| i a ––– n–1| i

8. An amount is invested at an annual effective rate of interest i which is exactly sufficient to pay 1 at the end of each year for n years. In the first year, the fund earns rate i and 1 is paid at the end of the year. However, in the second year, the fund earns rate j > i. If X is the revised payment which could be made at the end of years 2 to n, then find X assuming that (b) the rate earned remains at j for the rest of the n-year period. From the amortization table, the balance after one year is , and therefore the balance after two years must be a ––– n–1| i (1 + j) – X . a ––– n–1| i However, after two years, the balance must be equal to the present value of all future payments. Consequently, we have that (1 + j) – X = X a ––– n–1| i a ––– n–2| j X (1 + ) = (1 + j) a ––– n–2| j a ––– n–1| i X (1 + j) = (1 + j) X = ———— a ––– n–1| j a ––– n–1| i a ––– n–1| i a ––– n–1| j

9. Instead of repaying a loan in installments by the amortization method, a borrower can accumulate a fund which will exactly repay the loan in one lump sum at the end of a specified period of time. This fund is called a sinking fund. It is generally required that the borrower periodically pay interest on the loan, sometimes referred to as a service. Consider a loan of amount 1 repaid over n periods. 1 — With the amortization method, the payment each period is . a – n| With the sinking fund method, the payment at the end of each period is the sum of the periodic sinking fund deposit necessary to accumulate the amount of the loan at the end of n periods and the amount of the interest paid on the loan each period: 1 — + i s – n| 1 — = 1 — + i Recall that , which implies that if the rate of interest paid on the loan equals the rate of interest earned on the sinking fund, then the sinking fund method and the amortization method are equivalent. a – n| s – n|

10. Table 5.2 on page 159 of the textbook and Table 5.3 on page 166 of the textbook are respectively an amortization schedule and a sinking fund schedule for a loan of $1000 repaid over 4 years at 8%. Suppose i = the rate of interest paid on the loan is not necessarily equal to j = the rate of interest earned on the sinking fund. (Usually j < i but not necessarily). We let = the present value of an annuity which pays 1 at the end of each period for n periods with i and j as previously defined. a – n| i&j

11. Consider a loan of 1 with level payments for n periods. Each payment must be , but each of these payments must pay interest rate i on the loan and provide for a sinking deposit which will accumulate at rate j to the amount of the loan at the end of n periods. Consequently, we have 1 ——— a – n| i&j 1 ——— 1 —— + i = a – n| i&j s – n| j 1 ——— 1 —— – j + i = a – n| i&j a – n | j 1 ————— = 1 —— + (i – j) a – n| i&j a – n | j a – n | j = —————— 1 + (i – j) a – n| i&j a – n | j

12. The sinking fund schedule in Table 5.3 on page 166 in the textbook is for i = j = 0.08. Consider how this schedule would change if i = 0.10.

13. Abernathy wishes to borrow $5000. Lender Barnaby offers a loan in which the principal is to be repaid at the end of six years. In the meantime, 12% effective is to be paid on the loan and Abernathy is to accumulate the amount necessary to repay the $5000 by means of annual deposits in a sinking fund earning 9% effective. Lender Cromwell offers a loan for six years in which Abernathy repays the loan by the amortization method. Find which offer would be better for Abernathy to take if Cromwell charges an effective rate of (a) 10% and (b) 12%. With the sinking fund offer from Barnaby, the annual payment is 5000 ———— 1 —— + 0.12 = 1 ——— + 0.12 = 7.52333 = 5000 5000 $1264.60 a – 6| 0.12&0.09 s– 6| 0.09 With the amortization offer from Cromwell, the annual payment at effective rate In either case, the better offer, the one involving smaller payments, is 5000 ——— = 5000 ——— = 4.35526 10% is $1148.04 a – 6| 0.10 the amortization offer from Cromwell. 5000 ——— = 5000 ——— = 4.11141 12% is $1216.13 a – 6| 0.12

14. How high can the interest rate on Cromwell’s amortization method offer be until it is not better than the sinking fund offer from Barnaby? With the amortization offer from Cromwell, the annual payment at effective rate The better offer is 5000 ——— = 5000 ——— = 4.99755 the amortization offer from Cromwell. 13% is $1250.77 a – 6| 0.13 With the amortization offer from Cromwell, the annual payment at effective rate The better offer is 5000 ——— = 5000 ——— = 3.88867 the sinking fund offer from Barnaby. 14% is $1285.79 a – 6| 0.14

15. With the sinking fund method, the borrower is not only paying i per unit borrowed but is investing in a sinking fund on which interest is being sacrificed at rate i– j per period. Roughly speaking, the average balance in the sinking fund is 1/2 per unit borrowed (i.e., half the original loan amount). Consequently, the average rate interest is being sacrificed per unit borrowed is approximately (i– j)/2. The total interest cost per unit borrowed, which would be the interest charged in the amortization method in order for payments to be equal to those in the sinking fund method, is approximately i + (i– j)/2. In the example just completed, we have i + (i– j)/2 = 0.12 + (0.03)/2 = 0.135 Consider Example 5.7 on page 170 of the textbook.