DEPRECIATIONDEPRECIATIONBy: Aldinette Esto

Business Math

DepreciationDepreciation

Is the loss or decline in value of an asset through use and passage of time.

Assets depreciate through age, obsolescence, decay, decrease in efficiency and inadequacy.

Basic Principle Basic Principle of Businessof Business

Capital intact

Assets no longer profitable

Replacement or depreciation fund

Basic Notation in Depreciation

O = original cost of the asset S = Scrap or salvage value W = wearing value n = useful or economic life of an asset in yrs, months & the like R = depreciation charge per year t = given year or date BV = Book value r = rate of depreciation

Terms to rememberTerms to remember Scrap value/Salvage/Trade-in-value

An asset that can no longer be used for the purpose for which it was purchased

Wearing value/Total depreciationThe difference between the original

cost and the scrap value; W = O - S Book value

The difference between the original cost and amount of depreciation on a given date; BV = O – (Rt)

Methods of determining the depreciation of various assets

Straight Line Method Service Hour Method Production Units Method Annuity Method

A. Straight-Line-Method It spreads depreciation evenly

over the useful life of the asset. It assumes that depreciation is in proportion to time.

Formula for periodic depreciation a) W = O – S b) R = W n

Example 1

A steel cabinet which cost P8,500 is expected to last 12 yrs., and at that time will have a trade-in-value of P500. Find the annual depreciation charge.

Given: O = P8,500 S = P500 n = 12

Sol: W = O – S

= P8,500 – P500 = P8,000

R = W ÷ n

= P8,000 ÷ 12 = P666.67

Example 2

An item which cost P5,200 and after 5 yrs has a scrap value of P400. What is the annual depreciation? What percent of the cost is the yearly depreciation?

Given: O = P5,200 S = P400 n = 5

Sol:

W = O – S R = W/n r = R/W

= 5,200–400 = 4,800/5 = 960/4,800

= P4,800 = P960 = 20%

• Example 3

A machine which costs P12,000 and after 6 yrs has a salvage value of P900. Determine the yearly depreciation and prepare a depreciation schedule.

Given: O = P12,000 S = P900 n = 6

Sol: W = O – S R = W ÷ n

= 12,000 - 900 = 11,100 ÷ 6

= P11,100 = P1,850

DEPRECIATION SCHEDULE – STRAIGHT LINE METHOD

Annual Accumulated BookYear Depreciation Depreciation

Value

0 0 0 P12,000

1 P1,850 P1,850 10,150

2 1,850 3,700 8,300

3 1,850 5,550 6,450

4 1,850 7,400 4,600

5 1,850 9,250 2,750

6 1,850 11,100 900

SUBTRACT O to AD

ADD AD & RE

Try it yourself!!!• A piece of equipment which was

purchased for P200,000.00 has an estimated useful life of 8 years and a scrap of P10,000.00.

a) What is the book value of the equipment after 3 years?

b) What is the book value of the equipment after 5 years?

Solution: A) BV = O – (Rt) = 200,000 – [ (200,000 -

10,000) x 3 ] 8 = P128,750

B) BV = O – (Rt) = 200,000 – [ (200,000 -

10,000) x 5 ] 8 = P81,250

B. Service Hours Method Similar to the Straight-Line Method

except that it calculates “book value” based on the hours an object has been used. It assumes that the number of productive hours decreases as the property becomes older.

Formula: a) R = (O - S) ÷ n

b) RE = SH x R

Example 1

A sewing machine purchased for P4500 has an estimated life of four years and a scrap value of P300. Use the straight line method to find the depreciation. Assume that the useful life of the machine is estimated to be 15,000 service hours and the actual number of hours spent in production each year is as follows:

1st yr: 4,500 hrs 2nd yr: 4,100 hrs

3rd yr: 3,500 hrs 4th yr: 2,900 hrs

• Solution to Ex 1

R = O – S = 4,500 = 0.28 per hour

n 15,000

Annual Depreciation Expense: RE = SH x R

1st yr RE = 4,500 x 0.28 = 1,260

2nd yr RE = 4,100 x 0.28 = 1,148

3rd yr RE = 3,500 x 0.28 = 980

4th yr RE = 2,900 x 0.28 = 812

DEPRECIATION SCHEDULE – SERVICE HOURS METHOD

Annual Accumulated Book

Year Depreciation Depreciation Value

0 0 0 P4,500

1 P1,260 P1,260 3,240

2 1,148 2,408 2,092

3 980 3,388 1,112

4 812 4,200 300

SUBTRACT O to AD

ADD AD & RE

Try It Yourself!!!

A car purchased for P134,000 has an estimated life of eight years and a scrap value of P6,425. Use the service hour method to find the depreciation. Assume that the useful life of the car is estimated to be 24,300 service hours and the actual no. of hours spent in using the car each yr is as follows:

1st yr: 4,300 SH 2nd yr: 4,000 SH3rd yr: 3,800 SH 4th yr: 3,200 SH

5th yr: 3,000 SH 6th yr: 2,500 SH7th yr: 2,000 SH 8th yr: 1,500 SH

a) Find the depreciation charge per hour.b) Find the book value of the car after 2 yrs.c) Find the book value of the car after 4 yrs.d) Find the book value of the car after 5 yrs.e) Find the book value of the car after 7 yrs.

C. Production Units Method It is based on the number of hours

an asset is used or the number of units it produces.

This method allows for more depreciation during a busy period of an asset

Example 1

A machine costs P7,500 has a salvage value of P600. It is estimated that the machine can produce 25,000 units. This machine has been run as follows:1st yr – 2,800 units 2nd yr – 3,200 units

3rd yr – 4,100 units 4th yr – 5,500 units

5th yr – 2,500 units

* Prepare a depreciation table.

Given: O= P7,500 S = P600 n = 2,500hrs

Sol: W = O – S

= 7,500 – 600 = P6,900

Annual Depreciation:

1st yr P6,900 (2,800 ÷ 25,000) = P 772.80

2nd yr 6,900 (3,200 ÷ 25,000) = 883.20

3rd yr 6,900 (4,100 ÷ 25,000) = 1,131.60

4th yr 6,900 (5,500 ÷ 25,000) = 1,518.00

5th yr 6,900 (2,500 ÷ 25,000) = 690.00

• Depreciation ScheduleAnnual Accumulated

BookYr Depreciation Depreciation

Value0 0 0 P7,500

1 P 772.80 P 772.80 6,727.20

2 883.20 1,656.00 5,844.00

3 1,131.60 2,787.60 4,712.40

4 1,518.00 4,305.60 3,194.40

5 690.00 4,995.60 2,504.40

Example 2

A certain machine costs P12,800 depreciates to P1,200. This type of machine has an estimated operating life of 35,500 hrs, and run as follows:1st yr - 3,400 hrs 2nd yr - 3,800 hrs

3rd yr - 4,350 hrs 4th yr - 5,160 hrs

5th yr - 4,995 hrs 6th yr - 7,200 hrs

7th yr - 1,500 hrs

*Prepare a depreciation table.

• Sol: W = O – S

= P12,800 – P1,200 = P11,600

Annual Depreciation:

1st yr P11,600 (3,400÷35,500) = P1,110.99

2nd yr 11,600 (3,800÷35,500) = 1,241.69

3rd yr 11,600 (4,350÷35,500) = 1,421.41

4th yr 11,600 (5,160÷35,500) = 1,686.08

5th yr 11,600 (4,995÷35,500) = 1,632.17

6th yr 11,600 (7,200÷35,500) = 2,352.68

7th yr 11,600 (1,500÷35,500) = 490.14

• Depreciation ScheduleAnnual Total Book

Yr Depreciation Depreciation Value

0 0 0 P12,800

1 P1,110.99 P1,110.99 11,689.01

2 1,241.69 2,352.68 10,447.32

3 1,421.41 3,774.09 9,025.91

4 1,686.08 5,460.17 7,339.83

5 1,632.17 7,092.34 5,707.66

6 2,352.68 9,445.02 3,354.98

7 490.14 9,935.16 2,864.84

Try It Yourself!!!

D. Annuity Method The depreciation charge is composed

of the amount credited to the fund and the interest on the book value of the asset.

The investment in the asset is regarded, first, as the amount of the salvage value which earns interest, and second, an investment in an annuity to be equal to the periodic payments.

“compound interest method of depreciation”