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MATH 2040 Introduction to Mathematical Finance

Instructor Miss Liu Youmei

Chapter 5 Amortization Schedules and Sinking

Funds

Introduction

Finding the outstanding loan balance

Amortization schedules

Sinking funds

Differing payment periods and interest conversion

periods

Varying series of payments

Example 3.3

- Compare the total amount of interest that would

be paid on a 1000 loan over a 10-year period,

if the effective rate of interest is 9 per

annum, under the following three repayment

methods - (1) The entire loan plus interest is paid in one

lump-sum at the end of 10 years. - (2) Interest is paid each year and the principal

is repaid at the end of 10 years. - (3) The loan was repaid by level payments over

the 10 year period.

Introduction

- There are two methods of paying off a loan
- (a) Amortization Method Borrower makes

installment payments at periodic intervals. - (b) Sinking Fund Method Borrower makes

installment payments - (i) As the annual interest comes due and pay

back the original loan as a lump-sum at the end, - (ii) The lump-sum is built up with periodic

payments going into a fund called sinking fund.

Purpose of this chapter

- Besides discussing the two methods of paying off

a loan, this chapter also discuss how to

calculate - (a) the outstanding balance once the repayment

schedule has begun, and - (b) what portion of annual payment is made up of

the interest payment and the principal repayment.

Finding the Outstanding Balance

- There are two methods for determining the

outstanding loan balance once the re-payment

processes begin - (a) Prospective method
- (b) Retrospective method.

Prospective method (see the future)

- The original loan at time 0, L0, represents

the present value of future repayments. If the

repayments, P, are to be level and payable at

the end of each year, then the original loan can

be represented as follows - The outstanding loan at time t, Lt, represents

the present value of the remaining future

repayments - Note that this assumes that the installments

prior to time t has been paid on time as

scheduled.

Retrospective method (see the past)

- If the repayments, P, are to be level and

payable at the end of each year, then the

outstanding loan at time t is equal to the

accumulated value of the loan at time t, less

the accumulated value of the repayments made to

date - This also assumes that the installments prior to

time t has been paid on time as scheduled.

Otherwise, the accumulated value of past payments

will need to be adjusted accordingly.

Basic relationship

- A basic relationship is
- Prospective method Retrospective method
- Suppose a loan L is to be repaid with

end-of-year payments of 1 over the next n

years. - Let t be an integer with 0 lt t lt n. The

value of the n payments at time t is - Therefore
- So Prospective method Retrospective method

Comparing two methods

- The prospective method is preferable when the

size of each level payment and the number of

remaining payments is known. - The retrospective method is preferable when the

number of remaining payments or a final irregular

payment is unknown.

Example 5.1

- A loan is being repaid by 10 payments of 2000

followed 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 immediately after 5

payments are made by using both the prospective

method and the retrospective method.

Example 5.1 Prospective method

- Immediately after 5 payments are made, there

will be 5 more payments of 2000 followed by

10 more payments of 1000 at the end of each

year. - These payments may be viewed as 15 payments of

1000 plus 5 payments of 1000 at the end of

each half-year starting the end of this

half-year. - So the present value of these future payments

are - To the nearest dollar.

Example 5.1 Retrospective method

- The original loan amount is the present value of

20 payments of 1000 plus 10 payments of

1000 at the end of each half-year starting the

end of this half-year. So it is - Retrospectively, the outstanding balance is
- to the nearest dollar.
- Thus the prospective and retrospective methods

produce the same answer.

Example 5.2

- A loan is being repaid by 20 payments of 1000

each. At the time of the 5-th 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 this revised annual

payment.

Example 5.2

- The balance after five years, prospectively is
- If the borrow makes an additional payment of

2000, then the loan balance becomes 6060.70. - So to repay this balance by 12 more payments, the

equation of value is - Solving for X, we get

Example 5.3

- A 20,000 mortgage is being repaid with 20

annual installments at the end of each year. The

borrower makes five payments and then is

temporarily unable to make payments for the next

two years. Find an expression for the revised

payment to start at the end of the 8th year if

the loan is still to be repaid at the end of the

original 20 years.

Example 5.3

- Solution

Amortization Methods

- If a loan is repaid by the amortization method,

each payment consists of interest and principal. - Determining the amount of interest and principal

is important for both the lender and the

borrower. - For example, interest and principal are generally

treated differently for income tax purposes.

Amortization Schedules

- An amortization schedule is a table which shows

the division of each payment into principal and

interest, - Together with the outstanding balance after each

payment. - Suppose a loan requires repayment by n payments

of 1 at the end of year. Then the initial loan

is - Let It and Pt be the amount of interest and

principal included in the t-th payment.

An Amortization Schedule

Remarks (1)

- The total of all interest payments is represented

by the total of all amortization payments less

the original loan - The total of all the principal payments must

equal to the original loan

Remarks (2)

- Note that the outstanding loan at t n is

equal to 0. - The whole point of amortizing is to reducing the

loan to 0 within n years. - Principal repayments increase by a factor of (1

i) in each period. This is because in each

period, the outstanding balance is decreasing,

and as a result the interest charged is also

decreasing. So more principal is paid in each

subsequent payment.

Remarks (3)

- If the installment payment at the end of each

period is R, then we have the relationship - which represents the recursion method.
- the outstanding loan balance at the end of

tth period - the amount of interest paid in the tth

installment - the amount of principal repaid in the same

installment

Example 5.4

- Ron is repaying a loan with payments of 1 at the

end of each year for n years. The amount of

interest paid in period t plus the amount of

principal repaid in period t 1 equals X. - Calculate X.
- Solution

Example 5.5

- 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

Example 5.5

- The outstanding load balance at the beginning of

the fourth quarter, i.e. the end of the third

quarter, is - The interest contained in the fourth payment is
- The principle contained in the fourth payment is

Example 5.6

- A loan is being repaid with quarterly

installments of 1000 at the end of each quarter

for five years at 12 convertible quarterly. Find

the amount of principal and interest in the sixth

installment. - Solution

Example 5.7

- A loan is being repaid with a series of payments

at the end of each quarter for five years. If the

amount of principal in the third payment is 100,

find the amount of principal in the last five

payments. Interest is at the rate of 10

convertible quarterly. - Solution

Example 5.8

- A borrows 10,000 from B and agrees to repay it

with equal quarterly installments of principal

and interest at 8 convertible quarterly over six

years. At the end of two years B sells the right

to receive future payments to C at a price that

will yield C 10 convertible quarterly. Find the

total amount of interest received - By C.
- By B.

Example 5.8 (1)

- The quarterly installment paid by A is
- (1)The price C pays is the present value of the

remaining payments at a rate of interest equal to

2.5 per quarter, i.e. - The total payments made by A over the last four

years is - (16)(528.71)8459.36
- The total interest received by C is
- 8459.36-6902.311557.05

Example 5.8 (2)

- (2) There are methods to calculate the total

interest received by B. - a. The outstanding loan balance on Bs original

amortization schedule at the end of two years is - The total principal repaid by A over the first

two years is - 10,000-7178.672821.33
- The total payments made by A over this period are
- (8)(528.71)4229.68
- Thus, the total interest received by B apparently

is - 4229.68-2821.331408.35

Example 5.8 (2)

- b. By lending out 10,000, B gets

(8)(528.71)6902.3111131.99 in return. - So the total interest received by B is
- 11131.99 -10,0001131.99
- Note that a and b result in different answers,

which one is more reasonable?

Example 5.9

- An amount is invested at an annual effective rate

of interest i which is just sufficient to pay 1

at the end of each year for n years. In the first

year the fund actually earns rate i and 1 is paid

at the end of the year. However, in the second

year the fund earns rate j where jgti. Find the

revised payment which could be made at the ends

of years 2 through n - (1) Assuming the rate earned reverts back to i

again after this one year. - (2)Assuming the rate earned remains at j for the

rest of the n-year period

Example 5.9(1)

- The initial investment is and the

account balance at the end of the first year is

- .Let X be the revised payment. We can get the

followings

Example 5.9(1)

- And must equal the present value of the

future payments. Thus, we have - Which gives

Example 5.9 (2)

- 2. The development is identical to case 1 above,

except that the present value of the future

payments, which equals , is computed at rate

j instead of i. Thus, we have - Which gives

Example 5.10

- A loan is being repaid with installments of

1 at the end of each year for 20 years. Interest

is at effective rate i for the first 10 years and

effective rate j for the second 10 years. Find

expressions for - The amount of interest paid in the 5th

installment. - The amount of principal repaid in the 15th

installment.

Example 5.10

- Solution

Sinking Funds

- Suppose of a loan of is repaid with single

lump-sum at t n. If annual end-of-year

interest payment of are being met

each year, then that lump-sum required is . - Suppose the lump-sum required at time n is to

be built up in a sinking fund, and this fund is

credited with effective interest rate i.

Sinking Fund Payments

- If the lump-sum is to be built up with annual

end-of-year payments for the next n years, then

the sinking fund payment is - Then the total annual payment made by the

borrower is the annual interest due on the loan

plus the sinking fund payment, i.e.

Net amount of loan (1)

- The accumulated value of the sinking fund at time

t, denoted by SFt, is the accumulated value

of the sinking fund payments made to date and is

calculated as follows - The loan itself will not grow if the annual

interest is paid at the end of each year. - We shall call the amount of the loan over the

accumulated amount of sinking fund the net amount

of loan.

Net amount of loan (2)

- The net amount of loan can be calculated as

follows - Net Loant Loan ? SFt
- In other words, the net amount of the loan under

the sinking fund method is the same as the

outstanding loan under the amortization method.

Net amount of interest

- Each year, the amount of interest the borrower

pays interest to the lender is . - Each year the borrower also earns interest from

the sinking fund to the amount of i ? SFt-1. - So the actual interest paid by the borrower in

year t, called the net amount of interest, is - So we see that the net amount of interest paid

under the sinking fund method is the same as the

interest payment under the amortization method.

Sinking Fund Increase

- The sinking fund grows each year by the amount of

interest it earns and by the end-of-year

contribution that it receives. - In other words, the annual increase in sinking

fund is the same as the principal repayment under

the amortization method. - Both methods are aiming at paying back the

principal. In amortization method, that was done

every year. In the sinking fund method that was

done at the very end, and at the same time, an

amount was set aside to accumulate to the final

payment.

Sinking fund with different interest rate (1)

- Usually, the interest rate on borrowing, i, is

greater than the interest rate offered by

investing in a fund, j. - The total payment under sinking fund approach is

then - We wish now to determine the interest rate i',

for which the amortization method would provide

for the same level of payment

Sinking with different interest rate (2)

- Therefore, the amortization payment, using this

mixed interest rate, will cover the smaller

amortization payment at rate j and the interest

rate shortfall, i ? j, that the smaller payment

does not recognize. - The mixed interest rate can be approximated by

the formula

Example 5.11

- John wants to borrow 1000.
- HSBC offers a loan in which the principal is to

be repaid at the end of four years. In the

meantime, 10 effective is to be paid on the loan

and John is to accumulate the amount necessary to

repay the loan by means of annual deposits in a

sinking fund earning 8 effective. - HS Bank offers a loan for four years in which

John repays the loan by amortization method. - What is the rate charged by HS Bank if the two

offers make no difference to John?

Example 5.11

- Under both method, John has to make 4 annual

payments. So if the two offers make no difference

to John, the annual payments must be equal. - Annual payment under HSBC plan is
- Suppose the amortization offer is i effective,

then - or
- By iteration method, we can determine i

10.94. - Note that the approximation method gives

10 0.5(10?8) 11, - Which is close to the above result of 10.94

Differing payment periods and interest conversion

periods

- Find the rate of interest, convertible at the

same frequency as payments are made, that is

equivalent to the given rate of interest - Using this new rate of interest, construct the

amortization schedule

Example 5.12

- A debt is being amortized by means of monthly

payments at an annual effective rate of interest

of 11. If the amount of principal in the third

payment is 1000, find the amount of principal in

the 33rd payment - The principle repaid will be a geometric

progression with common ration - The interval from the 3rd payment to the 33rd

payment is - (33-3)/122.5 years.
- Thus the principal in the 33rd payment is

Example 5.13

- A borrows 10,000 for five years at 12

convertible semiannually. A replaces the

principal by means of deposits at the end of

every year for five years into a sinking fund

which earns 8 effective. Find the total dollar

amount which A must pay over the five year period

to completely repay the loan. - Solution

Example 5.14

- A borrower takes out a loan of 2000 for two

years. Find the sinking fund deposit if the

lender receives 10 effective on the loan and if

the borrower replaces the amount of the loan with

semiannual deposits in a sinking fund earning 8

convertible quarterly. - All three frequencies differ
- Interest payments on the loan are made annually
- Sinking fund deposits are made semiannually
- Interest on the sinking fund is convertible

quarterly

Example 5.14

- The interest payments on the loan are 200 at the

end of each year. Let the sinking fund deposit be

D. Then - or

Varying Series of Payments

- Assume that the varying payments by the borrower

are R1,R2., Rn and that . Let the

amount of the loan be denoted by L. Then the

sinking fund deposit for the t-th period is

Rt-iL. - The accumulated value of the sinking fund at the

end of n periods must be L, we have - Or

Example 5.15

- A borrower is repaying a loan at 5 effective

with payments at the end of each year for 10

years, such that the payment the first year is

200, the second year 190, and so forth, until

the 10th year it is 110. Find (1) the amount of

the loan (2)the principal and interest in the

fifth payment. - The amount of the loan is

Example 5.15

- We have

Example 5.16

- A borrows 20,000 from B and agree to repay it

with 20 equal annual installments of principal

plus interest on the unpaid balance at 3

effective. After 10 years B sells the right to

future payments to C, at a price that yields C 5

effective over the remaining 10 years. Find the

price which C should pay to the nearest dollar.

Example 5.16

- Each year A pays 1000 principal plus interest on

the unpaid balance at 3. The price to C at the

end of the 10th year is the present value of the

remaining payments, i.e. - The answer must be less than the outstanding loan

balance of 10,000, since C has a yield rate in

excess of 3.

Example 5.17

- A loan is amortized over five years with monthly

payments at a nominal interest rate of 9

compounded monthly. The first payment is 1000 and

is to be paid one month from the date of the

loan. Each succeeding monthly payment will be 2

lower than the prior payment. Calculate the

outstanding loan balance immediately after the

40th payment is made.

Example 5.17

- Solution

Example 5.18

- A loan is repaid with payments which start at

200 the first year and increase by 50 per year

until a payment of 1000 is made, at which time

payments cease. If interest is 4 effective, find

the amount of principal in the fourth payment.

Example 5.18

- Solution