AMORTIZATION SCHEDULES AND SINKING FUNDS
INTRODUCTION TO AMORTIZATION SCHEDULES AND SINKING FUNDS Amortization method : • Borrower repays the lender by means of installment payments at periodic intervals. This process is called AMORTIZATION of the loan.
Sinking fund method : • Borrower repays the lender by means of one lump-sum payment at the end of the term of the loan.
FINDING THE OUTSTANDING LOAN BALANCE
Outstanding loan balance is also called : • Outstanding principal • Unpaid balance • Remaining loan indebtedness
FINDING THE OUTSTANDING LOAN BALANCE Two approaches used to find the amount of the outstanding loan balance are : • Prospective method - Calculate the outstanding loan balance looking into the p future.
B a t
n t |
• Retrospective method - Calculate the outstanding loan balance looking into the past. r
B
t
an| (1 i)t st|
FINDING THE OUTSTANDING LOAN BALANCE Original loan balance is often denoted by L . In specific cases, possible to prove that the prospective method is equal to retrospective method algebraically. t ( 1 i ) 1 t t v 1 i i ( 1 ) ( 1 ) an| st | i i n t (1 i)t v (1 i)t 1 i n
1v i
a n t |
n t
EXAMPLE 1
A loan of $10,000 is being repaid by installments of $2000 at the end of each year, and a smaller final payments made one year after the last regular payments. Interest is at the effective rate of 12%. Find the amount of outstanding loan balance remaining when the borrower has made payments equal to the amount of the loan. Answer to the nearest dollar. 10000 2000 an|
a
n|
5
t i 10000 ( 1 ) 2000 s5| B5 r
(1.12)5 1 10000(1.12) 2000 0 . 12 5
17623.41683 12705.69472 4918
EXAMPLE 2
A loan of $1000 is being repaid with quarterly payments at the end of each quarter for 5 years at 6% convertible quarterly. Find the outstanding loan balance at the end of second year. 1000 a12| p 0.06 B8 R a12|
j
4 0.015
R
1000
a
20|
a
20|
1000(10.90751) 17.16864
635.32
EXAMPLE 3
A loan made at an annual rate of 6.5% has 7 remaining payments of 950. What is the loan balance?
B
p
n 7
950 an( n7 )| 950 a7| 1 (1.065) 7 950 0.065
5210.293783
Example 4
A loan for 40,000 at a 7% annual rate has an annual payments of 9,755.63. Find the balance after the fourth payments. 4 40000 ( 1 . 07 ) 9755.63 s4| B4 r
(1.07) 4 52431.8404 9755.63 0 . 07
52431.8404 43314.44113
9117.399271
AMORTIZATION SCHEDULES
If a loan is being repaid by the amortization method, each payment is partially repayment of principal and partially payments of interest. An amortization schedules is a table which shows the division of each payment into principal and interest, together with the outstanding loan balance after each payment is made.
Consider a loan of at interest rate i per period being repaid with payments of 1 at the end of each period for n periods.
AMORTIZATION SCHEDULES Period 0 1 2 .. .. t .. ..
n 1
Payment amount 1 1
Interest paid
ian| 1 v
Principal repaid OutstandingaLoan n| Balance
n
n 1
ian1| 1 v
v v
n
a v a n
n|
n 1
n 1
an1| v
n 1|
an2|
n t 1
1
n
1 1
Total
n
n t 1
n t 1|
1 v
ia
1 v
ia
2|
2
ia 1 v na 1|
n|
ant 1| v
v
an t
n t 1
v v
a
2
2
a v a a v 0 2|
1|
1|
n|
AMORTIZATION SCHEDULES
ian| 1v
v
n
:
The interest due on the balance at the beginning of the period at the end of the first period.
: Principal repaid
n
an| v an1| n
: The outstanding loan balance at the end of the period equals the outstanding loan balance at the beginning of the period less the principal repaid.
AMORTIZATION SCHEDULES
I1 i.Bo
: Interest paid
P1 R I1
: Principal paid
B1 Bo P1
: Balance
EXAMPLE 5
a) A borrower would like to borrow 30,000 at 8% for five years, but would like to pay only 5,000 for the first two years and then catch up with a higher payment for the final three years. What is the payment for the final three years? Year
Payment
Interest
Principal
0
Balance 30,000
1
5,000
2,400
2,600
27,400
2
5,000
2,192
2,808
24,592
3
9,542.520177
1,967.36
7,575.160177
17,016.83982
4
9,542.520177
1,361.347186
8,181.172991
8,835.666829
5
9,542.520177
706.8533463
8,835.666829
0
24592 Ra3|
R
24592
a
3|
24592 1 (1.08) 3 0.08
9542.520177
EXAMPLE 5
b) What would the payment for the final 3 years if the borrower in the above loan wanted to pay only 4000 in each of the first 2 years? 26,672 Ra3| Year
Payment
Interest
Principal
0
Balance 30,000
1
4000
2400
1600
28,400
2
4000
2272
1728
26,672
3
10349.62989
4
10349.62989
5
10349.62989
26,672 R 1 v3 0.08 26,672 1 (1.08) 3 0.08 10349.62989
EXAMPLE 6
A loan for 50,000 has level payments to be made at the end of each year for 10 years at an annual rate of 9%. Find a)The balance at the end of 3 years b)The principal and interest paid in the 3rd payment
SOLUTION
L Ran|
50,000 Ra10|
R
50,000 1 (1.09) 10 0.09
7791.004495
EXAMPLE 6
Year
Payment
Interest
Principal
0
Balance 50,000
1
7791.004495
4500
3291.004495
46,708.99551
2
7791.004495
4203.809595
3587.1949
43,121.80061
3
7791.004495
3880.962055
3910.04244
39,211.75817
a) Balance = 39211.75817
b) Principal paid = 3910.04244 Interest = 3880.962055
EXAMPLE 7
A borrower is repaying a $1000 loan with 10 equal payments of principal. Interest at 6% convertible semiannually is paid on the outstanding balance each year. Find the price to yield an investor 10% convertible semiannually. Year
Payment
Interest
Principal
0
Balance 1000
1
130
30
100
900
2
127
27
100
800
3
124
24
100
700
4
121
21
100
600
5
118
18
100
500
6
115
15
100
400
7
112
12
100
300
8
109
9
100
200
9
106
6
100
100
10
103
3
100
0
EXAMPLE 7
i ( 2) 6%
J1
0.06 0.03 2
i ( 2) 10% 0.10 J2 0.05 2
PV 100 a10| 3( Da)10| 1 (1.05) 10 10 10 0.05 1 (1.05) 100 3 0 . 05 0.05
772.1734929 136.6959042 908.8693971