Title: MATH 2040 Introduction to Mathematical Finance
1MATH 2040 Introduction to Mathematical Finance
Instructor Miss Liu Youmei
2Chapter 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
3Example 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.
4Introduction
- 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.
5Purpose 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.
6Finding 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.
7Prospective 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.
8Retrospective 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.
9Basic 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
10Comparing 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.
11Example 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.
12Example 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.
13Example 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.
14Example 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.
15Example 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
16Example 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.
17 Example 5.3
18Amortization 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.
19Amortization 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.
20An Amortization Schedule
21Remarks (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
22Remarks (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.
23Remarks (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
24Example 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
25Example 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
26Example 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
27Example 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
28Example 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
29Example 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.
30Example 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
31Example 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
32Example 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?
33Example 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
34Example 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 -
35Example 5.9(1)
- And must equal the present value of the
future payments. Thus, we have - Which gives
36Example 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
37Example 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.
38Example 5.10
39Sinking 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.
40Sinking 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.
41Net 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.
42Net 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.
43Net 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.
44Sinking 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.
45Sinking 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
46Sinking 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 -
47Example 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?
48Example 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
49Differing 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
50Example 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
51Example 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
52Example 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
53Example 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
54Varying 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
55Example 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
56Example 5.15
57Example 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.
58Example 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.
59Example 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.
60Example 5.17
61Example 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.
62Example 5.18