Transcript MathFinLec5

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
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 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
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.
4
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.
5
Finding the Outstanding Balance
• There are two methods for determining
the outstanding loan balance once the repayment processes begin:
– (a) Prospective method
– (b) Retrospective method.
6
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.
7
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.
8
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 < t < n. The value of the
n payments at time t is:
• Therefore
• So “Prospective method = Retrospective method”
9
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.
10
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.
11
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.
12
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 halfyear. So it is
• Retrospectively, the outstanding balance is:
to the nearest dollar.
• Thus the prospective and retrospective methods
produce the same answer.
13
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.
14
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
15
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.
16
Example 5.3
Solution:
17
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.
18
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.
19
An Amortization Schedule
20
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:
21
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.
22
Remarks (3)
If the installment payment at the end of each period is R,
then we have the relationship
Bt 1  Bt (1  i)  R
which represents the recursion method.
Bt: the outstanding loan balance at the end of tth period
I t : the amount of interest paid in the tth installment
Pt : the amount of principal repaid in the same installment
I t  1  v n t 1
Pt  v n t 1
23
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:
24
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
25
Example 5.5
The outstanding load balance at the beginning of the
fourth quarter, i.e. the end of the third quarter, is
B3  1000(1.04)3  100s3|  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
P4  100 32.51  $67.49
26
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:
27
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:
28
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:
(1) By C.
(2) By B.
29
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
•
8459.36-6902.31=1557.05
30
Example 5.8 (2)
• (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
Example 5.8 (2)
• 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
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 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
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:
•
34
Example 5.9(1)
• And must equal the present value of the future
payments. Thus, we have
• Which gives
35
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
36
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:
a) The amount of interest paid in the 5th installment.
b) The amount of principal repaid in the 15th
installment.
37
Example 5.10
Solution
38
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.
39
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.
40
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.
41
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.
42
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.
43
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.
•
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
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:
46
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?
47
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%
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
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
(1  i)1/12
• The interval from the 3rd payment to the 33rd payment is
(33-3)/12=2.5 years.
• Thus the principal in the 33rd payment is
1000(1.11) 2.5  $1298.1
50
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:
51
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
52
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
s8|.02
D
 2000
s2|.02
• or
D  2000/
s8|.02
s2|.02
2.02
 2000
 470.7
8.5830
53
Varying Series of Payments
• Assume that the varying payments by the borrower are
R1,R2…., Rn and that i  j . 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
L  ( R1  iL)(1  j ) n1  ( R2  iL)(1  j ) n2  ...  ( Rn  iL)(1  j )
n
  Rt (1  j ) n1  iLsn| j
n
t 1
• Or
L
 Rt (1  j)
t 1
1  isn| j
n
n t

 Rt v j
t
t 1
1  (i  j )an| j
54
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
L  100 a10|  10( Da )10|
10  7.72
 100 (7.72)  10
 1227 .83
.05
55
Example 5.15
• We have
R5  160; B4  100a6|  10( Da) 6|
p
I 5  iB4  100(1  v 6 )  10(6  a6| )  100(1  .746)  10(6  5.0757)
p
 $34.62
P5  R5  I 5  160 34.62  $125.38
56
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.
57
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.
10  7.72
1000 a10|.05  30( Da )10|.05  100 (7.72)  30
 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
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.
59
Example 5.17
Solution:
60
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.
61
Example 5.18
• Solution
62