Chapter 6 - Amortization Schedules and
Sinking Funds
William Odefey
January 5, 2006
1
1 FINDING THE OUTSTANDING LOAN BALANCE
1
2
Finding the Outstanding Loan Balance
The outstanding loan balance goes by many names
• Outstanding principal
• Unpaid balance
• Remaining loan indebtedness
Let’s analanyze setp-by-step what happens. Suppose we have a loan L
with payments P at the end of each year for n years at an annual effective
rate of interest i. Then our equation of value is
L0 = P an
(1)
(2)
The value of the loan at time t = 1 is L0 (1 + i). After the end of year, we
have the value of the loan L1 equal to
L1 = L0 (1 + i) − P
= L0 − (P − L0 · i)
= L0 − (P − P an · i)
µ
¶
1 − νn
= L0 − P − P ·
·i
i
= L0 − (P − P + P ν n )
= L0 − P ν n
¡
¢
= P + P ν + P ν2 + · · · + P νn − P νn
= P + P ν + P ν 2 + · · · + P ν n−1
= P an−1
(3)
(4)
(5)
(9)
(10)
(11)
L2 = L1 (1 + 1) − P
= (L0 (1 + i) − P ) (1 + i) − P
= L0 (1 + i)2 − (P (1 + i) + P )
= L0 (1 + i)2 − P s2
(12)
(13)
(14)
(15)
(6)
(7)
(8)
At time t = 2
1 FINDING THE OUTSTANDING LOAN BALANCE
3
Also,
L2 = L1 (1 + i) − P
= L1 − (P − L1 · i)
= L1 − (P − P an−1 · i)
µ
¶
1 − ν n−1
= L1 − P − P ·
·i
i
¡
¢
= L1 − P − P + P ν n−1
= L1 − P ν n−1
(16)
(17)
(18)
L2 = L1 − P ν n−1
¡
¢
= P + P ν + P ν 2 + · · · + P ν n−1 − P ν n−1
= P + P ν + P ν 2 + · · · + P ν n−2
= P an−2
(23)
(19)
(20)
(21)
(22)
And finally,
(24)
(25)
(26)
In general, to determine the loan balance at time t = k
L = P an
L (1 + i)k = P an (1 + i)k
1 − νn
· (1 + i)k
=P·
i
(1 + i)k − ν n−k
=P·
i
(1 + i)k − 1 + 1 − ν n−k
=P·
i
µ
¶
(1 + i)k − 1 1 − ν n−k
=P
+
i
i
= P sk + P an−k
L(1 + i)k − P sk = P an−k
(27)
(28)
(29)
(30)
(31)
(32)
(33)
(34)
Therefore, we can use any of the following three methods to determine the
otstanding balance of the loan at time t = k
1. Prospective Method
Lk = P an−k
2 AMORTIZATION SCHEDULE
4
2. RetrospectiveMethod
Lk = L0 (1 + i)k − P sk
3. The previous balance
Lk = Lk−1 − P ν n−k+1
2
Amortization Schedule
An amortization schedule is a table which shows the division of each payment between principal and interest, as well as the otstanding loan balance.
2.1
Construction of the Amortization Schedule
Payment
amount
Interest
paid
Principal
repaid
1
1
..
.
ian = 1 − ν n
ian−1 = 1 − ν n−1
..
.
νn
ν n−1
..
.
Outstanding
loan balance
an
an − ν n = an−1
an−1 − ν n−1 = an−2
..
.
t
..
.
1
..
.
ian−t+1 = 1 − ν n−t+1
..
.
ν n−t+1
..
.
an−t+1 − ν n−t+1 = an−t
..
.
n-1
n
Total
1
1
n
ia2 = 1 − ν 2
ia1 = 1 − ν
n − an
ν2
ν
an
a2 − ν 2 = a1
a1 − ν = 0
Period
0
1
2
..
.
Clearly there is no amortization table for a perpetuity
• The original outstanding balance is
• The payment amount is 1.
• The interest pais is
1
· i = 1.
i
• The principal repaid is 1 − 1 = 0.
1
.
i
2 AMORTIZATION SCHEDULE
2.2
5
Observations
2.2.1
The outstanding balance is the balance according to the
prospective method
2.2.2
The amount of principal repaid forms a geometric progression with ratio (1 + i).
May 2001 Exam Problem 31 Seth borrows X for four years at an annual
effective interest rate of 8%, to be repaid with equal payments at the end
of each year. The outstanding loan balance at the end of the second year is
1076.82 and at the end of the third year is 559.12 . Calculate the principal
repaid in the first payment.
Solution: The principal reapid during the third year is
1076.82 − 559.12 = 517.70
(35)
Since the principal repaid forms a geometric progression, the amount repaid
during the first year is
517.70
= 443.84
1.082
(36)
May 2005 Exam Problem 25 A bank customer takes out a loan of
500 with a 16% nominal interest rate convertible quarterly. The customer
makes payments of 20 at the end of each quarter. Calculate the amount of
principal in the fourth payment.
Solution: The amount of interest paid during time period 1 is
500 · 0.04 = 20.0
(37)
Therefore, there is no interest repaid. Thus we also have that there is no
interest repaid in any succeeding payment either. Notice that the value of a
perpetity that pays 20 is
20 ·
1
= 500
0.04
(38)
November 2005 Exam Problem 18 A loan is repaid with level
annual payments based on an annual effective interest rate of 7%. The 8th
payment consists of 789 of interest and 211 of principal. Calculate the amount
of interest paid in the 18th payment.
2 AMORTIZATION SCHEDULE
6
Solution:
P = 789 + 211
= 1000
(39)
(40)
The principal repaid in the 18th payment is
211 · 1.0710 = 415.07
(41)
Therefore, the amount of principal repaid is
1000 − 415.07 = 584.93
2.2.3
(42)
The sum of the principal payments is the original loan balance
Problem The aount of principal repaid in the 3rd payment of a 5-year
loan at 9% is 300. What is the original loan value?
Solution:
L0 = P ν 5 + P ν 4 + P ν 3 + P ν 2 + P ν
= P ν 5 + P ν 4 + 300 + P ν 2 + P ν
300
300
=
+
+ 300 + 300(1.09) + 300(1.09)2
2
1.09
1.09
300
=
· s5
1.092
= 252.5040 · 5.9847
= 1511.16
2.2.4
(43)
(44)
(45)
(46)
(47)
(48)
The sum of the interest payments is equal to the difference
between the sum of the total payments and the principal
payments k − ak
November 2000 Exam Problem 12 Kevin takes out a 10-year loan
of L, which he repays by the amortization method at an annual effective
interest rate of i. Kevin makes payments of 1000 at the end of each year .
The total amount of interest repaid during the life of the loan is also equal to
L. Calculate the amount of interest repaid during the first year of the loan.
3 SINKING FUNDS
7
Solution:
L = 1000a10
L = 1000 (10 − a10 )
1000a10 = 10000 − 1000a10
a10 = 5
i = 15.0984%
I1 = 1000 · a10 · i
= 1000 · 5 · .150984
= 755
3
(49)
(50)
(51)
(52)
(53)
(54)
(55)
(56)
Sinking Funds
Rather than repay a loan in installments by the amortization method, a
borrower may choose to repay it by paying the interest as it accrues on the
loan, and a lump sum payment at the end
1. The interest payment is L0 · i. Using our recursion relationship, we
have
L1 = L0 (1 + i) − P
= L0 (1 + i) − L0 · i
= L0
(57)
(58)
(59)
Therefore, the lump sum amount that needs to be paid of at the end
of the loan
2. Since we need to accumulate L0 in the side-fund at t = n, so the interest
L
rate credited to the side fund is j, the requred deposit is .
sn
For insatnce, consider a loan for 1000 at an annual effective rate of 5% for
10 years.
1. The interest payment would be 1000 · .05 = 50
2. The deposit to the side fund would be
1000
1000
=
= 79.50
s10
12.5779
3 SINKING FUNDS
8
3. The total annual payment would be 50.00 + 79.50 = $129.50
4. If the loan would be repaid by the normal amortization method, the
1000
1000
annual payment would be
=
= 129.50
a10
7.7217
This will be always be the case if the interest rate on the loan and the
side fund are the same. This can be show as follows:
µ
¶
i
L
=L i+
(60)
L·i+
sn
(1 + i)n − 1
µ
¶
(1 + i)n − 1 + 1
= Li
(61)
(1 + i)n − 1
µ
¶
i
=L
(62)
1 − νn
L
=
(63)
an
If the two rates are not the same, then
1
an|i&j
an|i&j
=
1
+i
sn|j
¶
µ
1
−j +i
=
an|j
1 + (i − j)an|j
=
an|j
an|j
=
1 + (i − j)an|j
(64)
(65)
(66)
(67)
May 2005 Exam Problem 2 Lori borrows 10,000 for 10 years at an
annual effective interest rate of 9%. At the end of each year, she pays the
interest on the loan and deposits the level amount necessary to repay the
principal to a sinking fund earning an annual effective interest rate of 8%.
The total payments made by Lori over the 10-year period is X. Calculate X.
3 SINKING FUNDS
9
Solution:
·
P = 10000 ·
1
¸
+ 0.09
s10|0.08
·
¸
1
= 10000 ·
+ 0.09
14.4866
= 10000 · .159029
= 1590.23
X = 10P
= 15902.93
May 2001 Exam Problem 4
under the following two methods:
(68)
(69)
(70)
(71)
(72)
(73)
A 20-year loan of 20,000 may be repaid
1. The amortization method with equal annual payments at an annual
effective rate of 6.5%.
2. The sinking fund method in which the lender receives an annual effective rate of 8% and the sinking fund earns an annual effective rate of
j.
Both methods require a payment of X to be made at the end of each year
for 20 years. Calculate j.
Solution:
20000
a20|0.65
20000
=
11.0185
= 1815.13
20000
+ 20000 · 0.08
1815.13 =
s20|j
20000
215.13 =
s20|j
X=
s20|j = 92.9670
j = 14.179%
(74)
(75)
(76)
(77)
(78)
(79)
(80)
3 SINKING FUNDS
10
We can determine the amortization schedule for the sinking-fund method
where the interest on the side fund is the same as the loan as follows:
L1 = L0 (1 + i) − P
µ
¶
L0
= L0 + L0 · i − L0 · i +
sn
L0
= L0 −
sn
L0
= L0 −
· s1
sn
L2 = L1 (1 + i) − P
µ
¶
µ
¶
L0
L0
= L0 −
(1 + i) − L0 · i +
sn
sn
L0
= L0 −
[(1 + i) + 1]
sn
L0
= L0 −
· s2
sn
·
¸
s2
= L0 1 −
sn
sn − s2
= L0 ·
sn
(1 + i)2 sn−2
= L0 ·
sn
(1 + i)n an−2
= L0 ·
(1 + i)n an
L0
· an−2
=
an
= P an−2
Therfore, the outstanding balance at time t = k
1. Under the sinfing-fund method is L0 −
2. Under the amortization method is
3. The two are equivalent.
L0
· sk
sn
L0
· an−k
an
(81)
(82)
(83)
(84)
(85)
(86)
(87)
(88)
(89)
(90)
(91)
(92)
(93)
(94)
3 SINKING FUNDS
11
Problem A loan of 100,000 is to be repaid by 20 level annual payments.
The lender wishes to earn 12% on the full loan amount and wil deposit the
remainder of the annual payment to a sinking fund earning 8% annually.
1. Find the amount of the loan.
100, 000
100, 000
100, 000 · 0.12 +
= 12, 000 +
s20|.08
45.7620
(95)
= 12, 000.00 + 2, 185.22 = 14, 185.22 (96)
2. Just after receiving the 10th payment, the lender sells the remaining 10
payments. The purchaser considers two ways of valuing the remaining
payments:
(a) amortization at 10%, or
(b) earning an annual return of 12% on his investment while recovering
his principal in a sinking fund earning 8%.
Find the amount of the original lender’s sinking fund at the time the
remainder of the loan is sold, and in each case find the amount paid by
the investor to the original lender.
Solution:
The amount in the sinking fund is
2, 185.22s10 = 2, 185.22 · 14.4866
= 31, 656.41
(97)
(98)
The remaining 10 payments have a value using the amortization method
and 10% of
14, 185.22a10|.10 = 14, 185.22 · 6.1446
= 87, 162.50
(99)
(100)
Based on the alternative metod
14, 185.22 = .12 · L0 +
L0
s10|.08
L0
= .12 · L0 +
14.4866
= .12 · L0 + .0690 · L0
= .1890 · L0
L0 = 75, 054.07
(101)
(102)
(103)
(104)
(105)
4 VARYING SERIES OF PAYMENTS
4
12
Varying Series of Payments
Consider a loan L to be repaid with n periodic payments R1 , R2 , . . . , Rn . The
equation of value is
L=
n
X
ν t Rt
(106)
t=1
At this point we have considered level payments with the additional possibility of a balloon or drop payment at the end. We now consider other
possible scenarios.
May 1984 CAS Exam Problem 8 A loan is repaid with 20 increasing
annual installments of 1,2,3,. . . ,20. The payments begin one year after the
loan is made. Find the principal contained in the 10t h payment, if the annual
interest rate is 4%.
Solution: the outstanding balance at t = 9 by the prospective method
is
L9 = 10a11 + (Ia)11
ä11 − 11ν 11
= 9 · 8.7605 +
.04
= 78.84 + 49.14
= 127.98
(107)
(108)
(109)
(110)
The principal repaid will be the payment minus the interest on the outstanding balance.
P R = 10 − 127.98 · 0.04
= 4.88
(111)
(112)
Problem A loan of 3000 at an effective quarterly interest rate of j = .02
is amortized by means of 12 quarterly payments , beginning one quarter after
the loan is made. Each payment consists of a principal repayment of 250 plus
interest due on the previous quarter’s outstanding balance. Construct the
amortization schedule.
Solution:
4 VARYING SERIES OF PAYMENTS
13
t
0
1
2
3
..
.
Payment
Interest
Due
Principal
Repaid
310
305
300
..
.
60
55
50
..
.
250
250
250
..
.
Outstanding
Balance
3000
2750
2500
2250
..
.
11
12
260
255
10
5
250
250
250
0
We see from the amortization table that this loan is equivalent to
L = 250a12|.02 + 5(Da)12|.02
12 − 10.5753
= 250 · 10.5753 + 5
.02
= 2643.83 + 356.17
= 3000.00
(113)
(114)
(115)
(116)
Problem In order to repay a school loan, a payment schedule of 200
at the end of the year for the first 5 years, 1200 at the end of the year for the
next 5 years, and 2200 at the end of the year for the final 5 years is agreed
upon.
1. If interest is at the annual effective rate of i = 6%, what is the loan
value?
L = 2200a15 − 1000a10 − 1000a5
= 21366.95 − 7360.09 − 4212.36
= 9794.50
(117)
(118)
(119)
2. Construct the amortization scedule for the loan for the first 7 years.
4 VARYING SERIES OF PAYMENTS
t
0
1
2
3
4
5
6
7
Interest
Payment
Due
200.00
200.00
200.00
200.00
200.00
1200.00
1200.00
587.67
610.93
635.59
661.72
689.42
718.79
689.92
14
Principal
Repaid
-387.67
-410.93
-435.59
-461.72
-489.42
481.21
510.08
Outstanding
Balance
9794.50
10182.17
10593.10
11028.69
11490.41
11979.83
11498.62
10988.54
Notice that the during the first 5 years of the loan repayment, the
principal repaid is negative (yet, it still forms a geometric progression).
3. Verify the outstanding balance of the loan at the end of the 7th year by
the prospective method.
OB = 2200a8 − 1000a3
= 13661.55 − 2673.01
= 10988.54
(120)
(121)
(122)