2 2 amortization

2.2 Amortization2.2 Amortization

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


• 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


Formulas:• Periodic payment

• Outstanding principal

€

R =Ai

1− (1+ i)−n

€

OBk = R1− (1+ i)−(n−k )

i

⎡

⎣ ⎢

⎤

⎦ ⎥

€

OBk = A(1+ i)k − R(1+ i)k −1

i

⎡

⎣ ⎢

⎤

⎦ ⎥


Formulas:• Interest payment

• Principal repayment

• Total interest (if all payments are regular)€

IPk = OBk−1( )i

€

PRk = R − IPk

€

IPT = nR − A


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

€

=Php364,911.30


5. Find OB115 given R = Php17,315, i = 1%, and n = 180.

€

OBk = R1− (1+ i)−(n−k )

i

⎡

⎣ ⎢

⎤

⎦ ⎥

€

OB115 =17,3151−1.01−(180−115)

.01

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php824,654.71


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) − 21,5251.02542 −1

.025

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php3,792,014.01


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 loanb) outstanding principal after the 8th paymentc) outstanding principal after the 8th year.



a) the original amount of the loan

€

A = R1− (1+ i)−n

i

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=119,764.711 −1.05−18

.05

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php1.4M



b) outstanding principal after the 8th payment

€

OB8 =119,764.711−1.05−(18−8)

.05

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php924,791.34



c) outstanding principal after the 8th year.

€

OB16 =119,764.711−1.05−(18−16)

.05

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php222,691.75


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) − 500,0001.0610 −1

.06

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php16,690,622.58


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%.


Period Regular payment

Interest payment

Principal repayment

Outstanding balance



Interest payment

Principal repayment

Outstanding balance

0 1,750,000

1

2

3



Interest payment

Principal repayment

Outstanding balance

0 1,750,000

1 360,000 157,500 202,500 1,547,500

2

3



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



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


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?



a) Find the amount of each semi-annual payment.

€

R =(400,000)(.04)

1 −1.04−10

€

=Php49,316.38



b) How much of the 6th payment goes to interest payment? How much is allocated for repayment of principal?

€

OB5 = 49,316.381−1.04−(10−5)

.04

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php219,547.76 €

IP6 =OB5i

€

=Php8,781.91

€

PR6 = R − IP6

€

=Php40,534.47



c) How much is the total interest paid?

€

IPT = nR − A

€

=(10)(49,316.38) − 400,000

€

=Php93,163.80


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

€

=Php74,923.18



€

OB3 = 74,923.181 −1.025−(24 −3)

.025

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php1,212,597.85

€

IP4 =OB3i

€

=Php30,314.95



€

OB20 = 74,923.181−1.025−(24 −20)

.025

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php281,859.07


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?



a) Determine the outstanding principal after the 4th payment.

€

OB4 =111,5001 −1.035−(8−4 )

.035

⎡

⎣ ⎢

⎤

⎦ ⎥

€

=Php409,548.33



b) What part of the 7th payment is interest payment?

€

IP7 =111,5001−1.035−(8−6)

.035

⎡

⎣ ⎢

⎤

⎦ ⎥(.035)

€

=Php7,413.56



c) What part of the 7th payment is allotted for principal repayment?

€

PR7 =111,500 − 7,413.56

€

=Php104,086.44


• 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


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 paymentsb) the final or concluding payment



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

€

8 .There are full payments



b) the final or concluding payment

€

x =OBLR (1+ i) =OB8(1+ i)

€

OB8 = 450,000(1.02758) − 58,000 1.02758 −1.0275[ ]

€

=Php47,868.63

€

x = (47,868.63)(1.0275)

€

=Php49,185.02


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?



a) how much will be the outstanding principal after the 5th installment?

€

OB5 = 20,000(1.0155) − 2,000 1.0155 −1.015[ ]

€

=Php11,241.15



b) how much interest is paid on the 9th installment?

€

IP9 =OB8i

€

OB8 = 20,000(1.0158) − 2,000 1.0158 −1.015[ ]

€

=Php5,664.17

€

IP9 = (5,664.17)(.015)

€

=Php84.96



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[ ]

€

=Php9,409.76

€

PR7 = 2,000 − (9,409.76)(.015)

€

=Php1,858.85



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[ ]

€

=Php1,805.37

€

x = (1,805.37)(1.015)

€

=Php1,832.45

2 2 amortization

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