LOGO
MATH 2040
Introduction to
Mathematical Finance
Instructor: Miss Liu Youmei
1

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
2

• 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 lumpsum 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.
3

• 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.
4

• 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.
5

:
– (a) Prospective method
– (b) Retrospective method.
6

• The original loan at time 0,
L
0
, 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 , L t
, 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.
7

• 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.
8

• 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 < t < n . The value of the n payments at time t is:
• Therefore
• So “
Prospective method =
Retrospective method”
9

• 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.
10

• 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.
11

• 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.
12

• 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 halfyear. So it is
• Retrospectively, the outstanding balance is: to the nearest dollar.
• Thus the prospective and retrospective methods produce the same answer.
13

• 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.
14

• 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
15

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.
16

Solution:
17

• 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.
18

• 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 I t and P t be the amount of interest and principal included in the t -th payment.
19

20

•
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:
21

• 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.
22

If the installment payment at the end of each period is R , then we have the relationship
B t

1

B t
( 1
 i )

R which represents the recursion method.
B t
: the outstanding loan balance at the end of t th period
I t
: the amount of interest paid in the t th installment
P t
: the amount of principal repaid in the same installment
I t

1
 v n
 t

1
P t
 v n
 t

1
23

• 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:
24

• 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
25

The outstanding load balance at the beginning of the fourth quarter, i.e. the end of the third quarter, is
B
3

1000 ( 1 .
04 )
3 
100 s
3 |

1124 .
86

312 .
16

$ 812 .
7
The interest contained in the fourth payment is
I
4

.
04 ( 812 .
7 )

$ 32 .
51
The principle contained in the fourth payment is
P
4

100

32 .
51

$ 67 .
49
26

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:
27

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:
28

•
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:
(1) By C.
(2) By B.
29

• 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
8459.36-6902.31=1557.05
30

• (2) There are methods to calculate the total interest received by B.
• a. The outstanding loan balance on B’s original amortization schedule at the end of two years is
• The total principal repaid by A over the first two years is
• 10,000-7178.67=2821.33
• The total payments made by A over this period are
•
• (8)(528.71)=4229.68
• Thus, the total interest received by B apparently is
4229.68-2821.33=1408.35
31

• b. By lending out $10,000, B gets
(8)(528.71)+6902.31=$11131.99 in return.
•
• So the total interest received by B is
11131.99 -10,000=1131.99
• Note that a and b result in different answers, which one is more reasonable?
32

• 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 j>i . 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
33

•
• 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:
34

• And must equal the present value of the future payments. Thus, we have
• Which gives
35

• 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
36

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: a) The amount of interest paid in the 5th installment.
b) The amount of principal repaid in the 15th installment.
37

38

• 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 .
39

• 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.
40

• The accumulated value of the sinking fund at time t , denoted by SF t
, 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 .
41

• The net amount of loan can be calculated as follows:
Net Loan t
= Loan

SF t
• 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.
42

• 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

SF t -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.
43

•
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.
44

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:
45

• 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:
46

• 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?
47

• 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%
48

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
49

• 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 33 rd payment
• The principle repaid will be a geometric progression with common ration
( 1
 i )
1 / 12
• The interval from the 3 rd payment to the 33 rd payment is
(33-3)/12=2.5 years.
• Thus the principal in the 33 rd payment is
1000 ( 1 .
11 )
2 .
5 
$ 1298 .
1
50

• 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:
51

• 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
52

• The interest payments on the loan are 200 at the end of each year. Let the sinking fund deposit be D. Then s
8 .| 02
D s
2 .| 02

2000
• or
D

2000 / s
8 .| 02 s
2 .| 02

2 .
02
2000
8 .
5830

470 .
7
53

• Assume that the varying payments by the borrower are i
 j 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
L

( R
1
 iL )( 1
 j ) n

1 
( R
2
 iL )( 1
 j ) n

2 
...

( R n
 iL )( 1
 j )

• Or t n 

1
R t
( 1
 j ) n

1  iLs n | j
L
 t n 

1
R t
( 1

1
 is n | j j ) n
 t

1
 t n 

1
( i
R t
 v j t j ) a n | j
54

• 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
10 th 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
L

100 a
10 |

10 ( Da )
10 |

100 ( 7 .
72 )

10
10

7 .
72
.
05

1227 .
83
55

• We have
R
5

160 ; B
4 p 
100 a
6 |

10 ( Da )
6 |
I
5
 iB
4 p 
100 ( 1
 v
6
)

10 ( 6
 a
6 |
)

100 ( 1

.
746 )

10 ( 6

5 .
0757 )

$ 34 .
62
P
5

R
5

I
5

160

34 .
62

$ 125 .
38
56

• 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.
57

• Each year A pays $1000 principal plus interest on the unpaid balance at 3%. The price to C at the end of the 10 th year is the present value of the remaining payments, i.e.
1000 a
10 .| 05

30 ( Da )
10 .| 05

100 ( 7 .
72 )

30
10
.

7 .
72

9089
05
• The answer must be less than the outstanding loan balance of
$10,000, since C has a yield rate in excess of 3%.
58

• 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.
59

Solution:
60

• 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.
61

• Solution
62