AMORTIZATION, SINKING FUND

AND DEPRECIATION

PROBLEM

AMORTIZATION PROBLEM:

A 400,000 loan at 10% compounded semi-annually is to be amortized every 6 months for 8 years. Find the semi annual payment and construct the amortization schedule table.

GIVEN:A= 400,000m=2 r=

10%8

years

A. Find the periodic paymentR= Ai

1-(1+i)^-ni = r/m n= t x m

i= 10% / 2 = 0.05

n= 8 x 2 = 16

R= 400,000 (0.05) 1-(1+0.05)^-16

R = 36,907.96

B. Amortization ScheduleInterest = Unpaid Balance x i

Principal Repaid = Periodic payment – Interest

Unpaid Balance = Previous Unpaid Balance – Principal RepaidPERIOD Unpaid

BalanceInterest Periodic

PaymentPrincipal Repaid

1 400,000 20,000 36,907.96 16,907.96

2 383,092.04

19,154.60 36,907.96 17,753.36

3 365,338.68

18,266.93 36,907.96 18,641.03

4 346,697.65

17,334.88 36,907.96 19,573.08

5 327,124.57

16,356.23 36,907.96 20,551.73

6 306,572.84

15,328.64 36,907.96 21,579.32

7 284,993.52

14,249.68 36,907.96 22,658.29

Period Unpaid Balance

Interest Periodic Payment

Principal Repaid

8 262,335.23

13,116.76 36,907.96 23,791.20

9 238,544.03

11,927.20 36,907.96 24,980.76

10 213,563.27

10,678.16 36,907.96 26,229.80

11 187,333.47

9,366.67 36,907.96 27,541.29

12 159,792.18

7,989.61 36,907.96 28,918.35

13 130,873.83

6,543.69 36,907.96 30,354.27

14 100,509.56

5,025.48 36,907.96 31,882.49

15 68,627.07 3,431.35 36,907.96 33,476.61

16 35,150.46 1,757.52 36,907.96 35,150.44

TOTAL 190,527.40

590,527.40

400,000

Sinking Fund Problem:The sum of 40,000 will be needed at the end of 8 years. If money can be invested at 10% converted annually . Find the annual deposit and create a sinking fund schedule.GIVE

N:S= 40,000

J= 10%

m= 1

8 YEARS

A. FIND THE ANNUAL DEPOSIT

R= J x S m((1+J/m)^n-1)

n= t x m

n= 8 x 1 = 8R= 0.10 x 40,000

1((1+0.10/1)^8-1)

R = 3,497.76

B. Sinking Fund ScheduleInterest = Amount in Fund x i

Increased = Deposit + Interest

Amount in Fund = Previous Amt. in Fund + IncreasedPERIOD DEPOSIT INTEREST INCREASE

D AMT. IN FUND

1 3,497.76 0 3,497.76 3,497.76

2 3,497.76 349.76 3,847.54 7,345.30

3 3,497.76 734.53 4,232.29 11,577.59

4 3,497.76 1,157.76 4,655.52 16,233.11

5 3,497.76 1,623.31 5,121.07 21,354.18

6 3,497.76 2,135.42 5,633.18 26,987.36

7 3,497.76 2,698.74 6,196.50 33,183.86

8 3,497.76 3,318.39 6,816.15 40,000.00

TOTAL 27,982.32 12,017.91 40,000

DEPRECIATION PROBLEM:An equipment worth 10,000 has a scrap value of 2000 and has a useful life of 8 years. Find the book value after 6 years and construct a depreciation schedule.GIVEN:O = 10,000S =

2000n= 8 years

WEARING VALUEW = 0 - S

W = 10,000 - 2,000W = 8,000

PERIODIC DEPRECIATION

R = W/nR =

8,000/8R = 1,000

BOOK VALUE AFTER 6 YEARS:BV = O – (R x

n)BV = 10,000 – (1,000 X 6)BV = 4,000

Depreciation Schedule:Accumulated Dep. = Previous Accum. Dep. + Annual Dep.Book Value = Previous Book Value – Annual Dep.YEAR Annual

DepreciationAccumulated Depreciation

BOOK VALUE

0 0 0 10,000

1 1,000 1,000 9,000

2 1,000 2,000 8,000

3 1,000 3,000 7,000

4 1,000 4,000 6,000

5 1,000 5,000 5,000

6 1,000 6,000 4,000

7 1,000 7,000 3,000

8 1,000 8,000 2,000

THE END

A-131AGUILON, CLARE ANNCARBUNGCO, KRISSIECASTRO, RIO KATRINA

FERMIN, JAN JONAHGARCIA, BIEN

MRS. MA. ESPERANZA S. MALANG