amortization

1. 2.2  Amortization2.2  Amortization
Amortization

2. 2.2  Amortization2.2  Amortization
• Amortization – a debt-repayment scheme
wherein the original amount borrowed is
repaid by making equal payments periodically
• In amortization problems, we usually want to
find the following values:
– Periodic payment
– Outstanding principal at the end of any period
– Interest payment for any period
– Principal repayment for any period
– Final irregular payment, if there is any

3. 2.2  Amortization2.2  Amortization
• Outstanding principal – refers to the amount
of debt still unpaid
• Amortization schedule – a table which shows
how a debt is completely repaid through
periodic payments, parts of which go to
interest payments and principal repayments

4. 2.2  Amortization2.2  Amortization

5. 2.2  Amortization2.2  Amortization
Formulas:
• Periodic payment
• Outstanding principal
R =
Ai
1−(1+i)−n
OBk = R
1−(1+i)−(n−k)
i
⎡
⎣
⎢
⎤
⎦
⎥
OBk = A(1+i)k
− R
(1+ i)k
−1
i
⎡
⎣
⎢
⎤
⎦
⎥

6. 2.2  Amortization2.2  Amortization
Formulas:
• Interest payment
• Principal repayment
• Total interest (if all payments are regular)
IPk = OBk−1( )i
PRk = R − IPk
IPT = nR − A

7. 2.2  Amortization2.2  Amortization
3. Find R given A = Php2.75M, j = 8%, m = 1,
and t = 12 years.
R =
Ai
1−(1+i)−n
=
(2,750,000)(.08)
1−1.08−12
= Πηπ364,911.30

8. 2.2  Amortization2.2  Amortization
5. Find OB115 given R = Php17,315, i = 1%, and n
= 180.
OBk = R
1−(1+i)−(n−k)
i
⎡
⎣
⎢
⎤
⎦
⎥
OB115 =17,315
1−1.01−(180−115)
.01
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ824,654.71

9. 2.2  Amortization2.2  Amortization
9. Find OB42 given A = Php1.9M, R = Php21,525,
j = 10%, and m = 4.
OBk = A(1+ i)k
− R
(1+i)k
−1
i
⎡
⎣
⎢
⎤
⎦
⎥
OB42 =1,900,000(1.02542
) −21,525
1.02542
−1
.025
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ3,792,014.01

10. 2.2  Amortization2.2  Amortization
17. A loan is to be amortized via equal payments
of Php119,764.71 each at the end of six
months for 9 years. If the interest is based on
10% compounded semi-annually, find
a) the original amount of the loan
b) outstanding principal after the 8th
payment
c) outstanding principal after the 8th
year.

11. 2.2  Amortization2.2  Amortization
17. A loan is to be amortized via equal payments
of Php119,764.71 each at the end of six
months for 9 years. If the interest is based on
10% compounded semi-annually, find
a) the original amount of the loan
A = R
1−(1+i)−n
i
⎡
⎣
⎢
⎤
⎦
⎥ =119,764.71
1−1.05−18
.05
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ1.4M

12. 2.2  Amortization2.2  Amortization
17. A loan is to be amortized via equal payments
of Php119,764.71 each at the end of six
months for 9 years. If the interest is based on
10% compounded semi-annually, find
b) outstanding principal after the 8th
payment
OB8 =119,764.71
1−1.05−(18−8)
.05
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ924,791.34

13. 2.2  Amortization2.2  Amortization
17. A loan is to be amortized via equal payments
of Php119,764.71 each at the end of six
months for 9 years. If the interest is based on
10% compounded semi-annually, find
c) outstanding principal after the 8th
year.
OB16 =119,764.71
1−1.05−(18−16)
.05
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ222,691.75

14. 2.2  Amortization2.2  Amortization
19. Goriotik obtains a Php13M bank loan at 12%
interest compounded semi-annually to
construct another studio. The company repays
the loan by paying Php0.5M every 6 months.
What is the outstanding principal after the 10th
payment?
OB10 =13,000,000(1.0610
) −500,000
1.0610
−1
.06
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ16,690,622.58

15. 2.2  Amortization2.2  Amortization
21. A Php1.75M loan is to be repaid through
annual payments of Php360,000. Construct an
amortization schedule up to the end of the
third period if effective interest rate is 9%.

16. 2.2  Amortization2.2  Amortization
Period Regular
payment
Interest
payment
Principal
repayment
Outstanding
balance

17. 2.2  Amortization2.2  Amortization
Period Regular
payment
Interest
payment
Principal
repayment
Outstanding
balance
0 1,750,000
1
2
3

18. 2.2  Amortization2.2  Amortization
Period Regular
payment
Interest
payment
Principal
repayment
Outstanding
balance
0 1,750,000
1 360,000 157,500 202,500 1,547,500
2
3

19. 2.2  Amortization2.2  Amortization
Period Regular
payment
Interest
payment
Principal
repayment
Outstanding
balance
0 1,750,000
1 360,000 157,500 202,500 1,547,500
2 360,000 139,275 220,725 1,326,775
3

20. 2.2  Amortization2.2  Amortization
Period Regular
payment
Interest
payment
Principal
repayment
Outstanding
balance
0 1,750,000
1 360,000 157,500 202,500 1,547,500
2 360,000 139,275 220,725 1,326,775
3 360,000 119,409.75 240,590.25 1,086,184.75

21. 2.2  Amortization2.2  Amortization
25. A loan of Php400,000 with interest at 8%
payable semi-annually is to be amortized
through equal semi-annual payments for 5
years.
a) Find the amount of each semi-annual
payment.
b) How much of the 6th
payment goes to interest
payment? How much is allocated for
repayment of principal?
c) How much is the total interest paid?

22. 2.2  Amortization2.2  Amortization
25. A loan of Php400,000 with interest at 8%
payable semi-annually is to be amortized
through equal semi-annual payments for 5
years.
a) Find the amount of each semi-annual
payment.
R =
(400,000)(.04)
1−1.04−10
= Πηπ49,316.38

23. 2.2  Amortization2.2  Amortization
25. A loan of Php400,000 with interest at 8%
payable semi-annually is to be amortized
through equal semi-annual payments for 5
years.
b) How much of the 6th
payment goes to interest
payment? How much is allocated for
repayment of principal?
OB5 = 49,316.38
1−1.04−(10−5)
.04
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ219,547.76
IP6 = OB5i
= Πηπ8,781.91
PR6 = R − IP6
= Πηπ40,534.47

24. 2.2  Amortization2.2  Amortization
25. A loan of Php400,000 with interest at 8%
payable semi-annually is to be amortized
through equal semi-annual payments for 5
years.
c) How much is the total interest paid?
IPT = nR − A
= (10)(49,316.38) − 400,000
= Πηπ93,163.80

25. 2.2  Amortization2.2  Amortization
27. How much will be the quarterly amortization
for a Php1.34M loan with interest at 10%
converted quarterly for a term of 6 years?
How much interest will be paid on the 4th
payment? What is the outstanding principal in
5 years?

26. 2.2  Amortization2.2  Amortization
27. How much will be the quarterly amortization
for a Php1.34M loan with interest at 10%
converted quarterly for a term of 6 years?
How much interest will be paid on the 4th
payment? What is the outstanding principal in
5 years?
R =
(1,340,000)(.025)
1−1.025−24
= Πηπ74,923.18

27. 2.2  Amortization2.2  Amortization
27. How much will be the quarterly amortization
for a Php1.34M loan with interest at 10%
converted quarterly for a term of 6 years?
How much interest will be paid on the 4th
payment? What is the outstanding principal in
5 years?
OB3 = 74,923.18
1−1.025−(24 −3)
.025
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ1,212,597.85
IP4 = OB3i
= Πηπ30,314.95

28. 2.2  Amortization2.2  Amortization
27. How much will be the quarterly amortization
for a Php1.34M loan with interest at 10%
converted quarterly for a term of 6 years?
How much interest will be paid on the 4th
payment? What is the outstanding principal in
5 years?
OB20 = 74,923.18
1−1.025−(24−20)
.025
⎡
⎣
⎢
⎤
⎦
⎥
= Πηπ281,859.07

29. 2.2  Amortization2.2  Amortization
29. To restructure a loan payment supposedly
due now, a debtor agrees to Php111,500
payment at the end of each 6 months for 4
years including interest payments at 7%,
m = 2.
a) Determine the outstanding principal after the
4th
payment.
b) What part of the 7th
payment is interest
payment?
c) What part of the 7th
payment is allotted for
principal repayment?

30. 2.2  Amortization2.2  Amortization
29. To restructure a loan payment supposedly
due now, a debtor agrees to Php111,500
payment at the end of each 6 months for 4
years including interest payments at 7%,
m = 2.
a) Determine the outstanding principal after the
4th
payment.
OB4 =111,500
1−1.035−(8−4)
.035
⎡
⎣
⎢
⎤
⎦
⎥ = Πηπ409,548.33

31. 2.2  Amortization2.2  Amortization
29. To restructure a loan payment supposedly
due now, a debtor agrees to Php111,500
payment at the end of each 6 months for 4
years including interest payments at 7%,
m = 2.
b) What part of the 7th
payment is interest
payment?
IP7 =111,500
1−1.035−(8−6)
.035
⎡
⎣
⎢
⎤
⎦
⎥(.035)
= Πηπ7,413.56

32. 2.2  Amortization2.2  Amortization
29. To restructure a loan payment supposedly
due now, a debtor agrees to Php111,500
payment at the end of each 6 months for 4
years including interest payments at 7%,
m = 2.
c) What part of the 7th
payment is allotted for
principal repayment?
PR7 =111,500 − 7,413.56 = Πηπ104,086.44

33. 2.2  Amortization2.2  Amortization
• Final irregular payment
– When n is not an integer, a smaller final payment
is needed to completely settle a debt
– Unless specified, this final payment is made one
period after the last regular payment
• Formulas:
n =
log 1− Ai
R( )
−log(1+i)
x = OBLR (1+i) IPT = n⎣ ⎦R + x − A

34. 2.2  Amortization2.2  Amortization
1. A debt of Php450,000 will be amortized by
semi-annual payments of Php58,000 for as
long as necessary. If interest is paid at 5 ½%
compounded semi-annually, find
a) the number of full payments
b) the final or concluding payment

35. 2.2  Amortization2.2  Amortization
1. A debt of Php450,000 will be amortized by
semi-annual payments of Php58,000 for as
long as necessary. If interest is paid at 5 ½%
compounded semi-annually, find
a) the number of full payments
n =
log 1− Ai
R( )
−log(1+i)
=
log 1− (450,000)(.0275)
58,000( )
−log(1.0275)
= 8.85
Τηερε αρε 8 φυλλ παψµεντσ.

36. 2.2  Amortization2.2  Amortization
1. A debt of Php450,000 will be amortized by
semi-annual payments of Php58,000 for as
long as necessary. If interest is paid at 5 ½%
compounded semi-annually, find
b) the final or concluding payment
x = OBLR (1+i) = OB8(1+i)
OB8 = 450,000(1.02758
) −58,000 1.02758
−1
.0275[ ]= Πηπ47,868.63
x = (47,868.63)(1.0275) = Πηπ49,185.02

37. 2.2  Amortization2.2  Amortization
3. A Php25,000 office equipment is bought with
a downpayment of Php5,000 and monthly
installments of Php2,000. If the buyer pays
18% interest compounded monthly,
a) how much will be the outstanding principal
after the 5th
installment?
b) how much interest is paid on the 9th
installment?
c) how much of the principal has been reduced
by the 7th
installment?
d) what is the final irregular installment?

38. 2.2  Amortization2.2  Amortization
3. A Php25,000 office equipment is bought with
a downpayment of Php5,000 and monthly
installments of Php2,000. If the buyer pays
18% interest compounded monthly,
a) how much will be the outstanding principal
after the 5th
installment?
OB5 = 20,000(1.0155
) −2,000 1.0155
−1
.015[ ] = Πηπ11,241.15

39. 2.2  Amortization2.2  Amortization
3. A Php25,000 office equipment is bought with
a downpayment of Php5,000 and monthly
installments of Php2,000. If the buyer pays
18% interest compounded monthly,
b) how much interest is paid on the 9th
installment?
IP9 = OB8i
OB8 = 20,000(1.0158
) −2,000 1.0158
−1
.015[ ] = Πηπ5,664.17
IP9 = (5,664.17)(.015) = Πηπ84.96

40. 2.2  Amortization2.2  Amortization
3. A Php25,000 office equipment is bought with
a downpayment of Php5,000 and monthly
installments of Php2,000. If the buyer pays
18% interest compounded monthly,
c) how much of the principal has been reduced
by the 7th
installment?
PR7 = R −OB6i
OB6 = 20,000(1.0156
) −2,000 1.0156
−1
.015[ ] = Πηπ9,409.76
PR7 = 2,000 −(9,409.76)(.015) = Πηπ1,858.85

41. 2.2  Amortization2.2  Amortization
3. A Php25,000 office equipment is bought with
a downpayment of Php5,000 and monthly
installments of Php2,000. If the buyer pays
18% interest compounded monthly,
d) what is the final irregular installment?
n =
log 1− (20,000)(.015)
2,000( )
−log(1.015)
=10.92
OB10 = 20,000(1.01510
) −2,000 1.01510
−1
.015[ ] = Πηπ1,805.37
x = (1,805.37)(1.015)= Πηπ1,832.45