Solution to Homework 5

ACTS 4308
Instructor: Natalia A. Humphreys
SOLUTION TO HOMEWORK 5
Section 9: Amortization of a loan.
Section 10: The sinking fund method of loan repayment.
Problem 1
Suppose that a $15,000 loan for a new car is to be repaid by level payments at the end of each month
for 5 years at a 3% effective annual rate of interest. Fill in the blanks corresponding to the first 6
months of the amortization schedule.
Month
0
1
2
3
4
5
6
Payment
Interest Paid Principal Repaid
Balance
15,000
Solution. Use calculator to do this problem. n = 5 · 12 = 60, j = 3% ⇒ i(12) = 2.9595%:
2nd ICONV ↓ EFF=3 Enter ↓ C/Y=12 Enter ↓ NOM appears CPT 2.959523727,
(12)
j = i 12 = 0.00246627 ≈ 0.0025
Ka60 j = 15,000 ⇒ K = 269.2606: 60 N Enter , 0.2466 I/Y Enter , 15,000 PV Enter , 0 FV
Enter , CPT PMT ⇒ −269.2606395
Foe each period 2nd AMORT to look at amortization table
The display reads P1=. Key in 1 ENTER ↓ The display reads P2=. Key in 1 ENTER
↓ BAL=14,767.73. This is OB1
↓ PRN=-232.27. This is −P R1
↓ INT=-36.99. This is −I1 .
Repeat with periods 2, 3, . . . , 6.
Month Payment Interest Paid Principal Repaid
0
1
269.26
36.99
232.27
2
269.26
36.42
232.84
3
269.26
35.85
233.41
4
269.26
35.27
233.99
5
269.26
34.69
234.57
6
269.26
34.12
235.14
Balance
15,000.00
14,767.73
14,534.89
14,301.48
14,067.49
13,832.92
13,597.78
Problem 2
A loan is amortized over n years with annual payments of 180 at the end of each year. After the n2 th
payment, the outstanding balance is 75% of the amount of the loan. (Assume n is an even integer.)
Calculate the amount of principal repaid in the first payment.
ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 5.
Solution.
P R1 = Kv n = 180v, OB n2 = Ka n = 0.75L, Kan i = L ⇒
2
Ka n = 0.75Kan i ⇔ 1 − v
2
n
2
i
i
n
= 0.75(1 − v n ) ⇔ 0.75v n − v 2 + 0.25 = 0
n
x = v 2 ⇒ 0.75x2 − x + 0.25 = 0 ⇔ x2 − 1.3333x + 0.3333 = 0
D = 1.33332 − 4 · 0.3333 = 0.4444 = 0.66672 ⇒
1.3333 − 0.6667
x=
= 0.3333 or x = 1 − reject since i 6= 0
2
n
v 2 = 0.3333 ⇒ v n = 0.33332 = 0.1111 ⇒ P R1 = 180 · 0.1111 = 20 Problem 3
A $8,000 loan is being repaid by payments of $100 each at the end of each month for as long as
necessary, plus a smaller final payment. If the nominal rate of interest convertible monthly is 12%,
find the amount of interest in the 50th payment.
A) $66.30
B) $66.78
C) $67.10
D) $67.43
E) $67.76
Solution. First, n > 50. Indeed, if payments are 100, then
100an 1% = 8000 ⇒ n = 161.75
I50 = OB49 j, j =
i(12)
= 0.01
12
Using retrospective form for the OB:
OB49 = L(1 + j)49 − Ks49 j = 8000 · 1.0149 − 100s49 1% =
= 13,026.79 − 6289.48 = 6743.30 ⇒
I50 = 6743.30 · 0.01 = 67.433 ⇒ D Problem 4
A loan is amortized over five years with monthly payments at a nominal interest rate of 6% compounded monthly. The first payment is 1,000 and is to be paid one month from the date of the loan.
Each succeeding monthly payment will be 3% lower than the prior payment.
Calculate the outstanding loan balance immediately after the 50th payment is made.
A) 1702
B) 1859
C) 1917
Solution.
j=
0
1000
1
D) 2011
E) 2074
0.06
= 0.005 = 0.5%, n = 5 · 12 = 60
12
1000 · 0.97
2
1000 · 0.972
3
...
...
1000 · 0.9759
60
This is the annuity whose payments change in a geometric progression with 1 + r = 0.97.
r = 0.97 − 1 = −0.03 6= j = 0.005 ⇒ use
n
0.97 60
1 − 1+r
1 − 1.005
1+j
L=K·
= 1000 ·
= 25,615.21
j−r
0.005 + 0.03
1.00550 − 0.9750
= 32,292.65 − 30,433.16 = 1859.49 ⇒ B
OB50 = 25,615.21 · 1.00550 − 1000 ·
0.005 + 0.03
Copyright ©Natalia A. Humphreys, 2014
Page 2 of ??
ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 5.
Or, using prospective approach, the 51st payment is:
K51 = 1000 · 0.9750 = 218.07
0.97 10
1 − 1.005
OB50 = 218.07 ·
= 1859.49 ⇒ B 0.005 + 0.03
Problem 5
A loan is repaid with level annual payments based on an annual effective interest rate of 6%. The
10th payment consists of $908.28 of interest and $91.72 of principal. Calculate the amount of interest
paid in the 24th payment.
A) 767
B) 780
C) 793
D) 804
E) 815
Solution.
P Rt = Kv n−t+1 , It = K(1 − v n−t+1 ), K = P Rt + It , K = 908.28 + 91.72 = 1000
1000v n−9 = 91.72
1000(1 − v n−9 ) = 908.28
91.72
x
v n−9
=
= 0.101, x = v n−9 ⇒
= 0.101 ⇔ 0.101 − 0.101x = x ⇔
n−9
1−v
908.28
1−x
1.101x = 0.101 ⇒ x = 0.09172 ⇔ v n−9 = 0.09172
I24 = K(1 − v n−23 ) = K(1 − v n−9 v −14 ) = 1000(1 − 0.09172 · 1.0614 ) = 792.63 ≈ 793 ⇒ C Problem 6
Complete the sinking fund schedule for a loan of $10,000 at 3% per annum for which the borrower
repays the interest on the loan at the end of each year and deposits a level amount (also at the end
of each year) necessary to repay the principal at the end of 4 years into a sinking fund which earns
4% per annum.
Year
Interest
Sinking
Interest on
Amount in
Fund Deposit Sinking Fund Sinking Fund
1
2
3
4
Solution. The interest: I = Li = 10,000 · 0.03 = 300
Sinking fund deposit:
L
10,000
10,000
D=
=
=
= 2354.90
sn j
s4 4%
4.246464
Interest on a sinking fund:
Dst−1 j · j = D (1 + j)t−1 − 1
1st year : 0
2nd year : D · j = 2354.90 · 0.04 = 94.20
1.042 − 1
3rd year : D · s2 j · j = 2354.90 ·
· 0.04 = 192.16
0.04
1.043 − 1
4th year : D · s3 j · j = 2354.90 ·
· 0.04 = 294.04
0.04
Copyright ©Natalia A. Humphreys, 2014
Page 3 of ??
ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 5.
Amount in the sinking fund Dst j :
1st year : D = 2354.90
2nd year : Ds2 j = 4804
3rd year : Ds3 j = 7351.06
4th year : Ds4 j = 10,000
Sinking
Interest on
Amount in
Year Interest Fund Deposit Sinking Fund Sinking Fund
1
2
3
4
300
300
300
300
2354.90
2354.90
2354.90
2354.90
0
94.20
192.16
294.04
2354.90
4804.00
7351.06
10,000
Problem 7
Helen borrows 1,000 for 5 years at a nominal annual rate of 6% convertible monthly. At the end of
each month, she pays the interest on the loan and deposits the level amount necessary to repay the
principal to a sinking fund earning a nominal annual rate of j% convertible monthly. The total of all
interest payments and sinking fund deposits made by Helen over the 5-year period is 1200. Determine
j, rounded to at least three decimal places.
Solution. n = 5 · 12 = 60, i = 0.5%


Payment per period: K = L i +
1
s
n

1000 0.005 +
 . Total payments: 60K ⇒
j%
12

1
s
60
 · 60 = 1200 ⇒ s
60
j%
12
j%
12
= 66.67 ⇒
j%
= 0.3515 ⇒ j = 4.2182 12
Problem 8
Smith borrows 1,000 for 5 years at an annual effective rate of interest of 8%. At the end of each
month, he pays the interest on the loan and deposits the level amount necessary to repay the principal to a sinking fund earning an annual effective rate of 6%. The total of all interest payments and
sinking fund deposits made by Smith over the 5-year period is X. Calculate X.
A) 1,152
B) 1,187
C) 1,250
D) 1,287
E) 1,302
Solution. We are given: L = 1000, n = 5, i = 8%, j = 6%. Since payments are monthly periodic,
i(12)
7.7208
j (12)
5.8411
=
= 0.6434,
=
= 0.4868
12
12
12
12
Therefore,
"
#
0.6434
1
X = 60 · 1000
+
=
100
s60 0.4868%
1
= 60,000 0.006434 +
= 1249.53 ≈ 1250 ⇒ C 69.4858
Problem 9
Copyright ©Natalia A. Humphreys, 2014
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ACTS 4308. AU 2014. SOLUTION TO HOMEWORK 5.
A 10-year loan may be repaid under the following two methods:
(1) amortization method with equal annual payments of X at the end of each year at an annual
effective rate of 6%
(2) sinking fund method in which the lender receives interest payments I at the end of each year
at an annual effective rate of 7%. Also, at the end of each year, level deposits of D are made
into a sinking fund which accumulates at an annual effective rate of j.
If X = I + D, calculate j.
A) 7.0%
B) 7.5%
C) 8.0%
D) 8.5%
E) 9.0%
Solution.
L = Xa10 6%
"
K = L 0.07 +
1
s10 j
"
#
= I + D = Xa10 6% 0.07 +
1
s10 j
#
=X⇔
a10 6% = 7.36 ⇒ s10 j = 15.1819 ⇒ j = 8.9848% ≈ 9% ⇒ E Problem 10
Howard repays a loan of $20,000 by establishing a sinking fund and making 30 equal payments at
the end of each year. The sinking fund earns 5% effective annually. Immediately after the seventh
payment, the yield on the sinking fund increases to 7% effective annually. At that time Howard
adjusts his sinking fund payment to Z so that the sinking fund will accumulate to $20,000 30 years
after the original loan date. Determine Z.
Solution. The original sinking fund payment is
Xs30 5% = 20,000 ⇒ X = 301.0287
Balance in the account after the 7th payment: Xs7 5% = 301.0287 · s7 5% = 2450.98. Thus, Joe still
needs to collect:
20,000 − 2450.98 = 17,549.02
With the new interest rate of 7%:
Zs23 7% + 2450.98 · 1.0723 = 20,000
Zs23 7% = 8381.064 ⇒ Z = 156.8426 ≈ 156.84 Copyright ©Natalia A. Humphreys, 2014
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