Depreciation

DEPRECIATION By: Aldinette Esto Business Math

Depreciation  Is

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

 Assets

depreciate through age, obso le sce nc e, d ec ay, de cr ea se in ef fici enc y a nd ina deq uac y.

Basic Principle of Business 

Capital intact Assets no longer profitable



Replacement or depreciation fund

Basic Notation in Depreciation    

   

O S W n

= original cost of the asset = Scrap or salvage value = wearing value = 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 remember  Scra

p valu e/Salva ge/Tra de-inval ue An

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

 We ar in g

va lue/To tal de pre cia tion

The

difference between the original cost and the scrap value; W = O - S

 Book

v al ue

Methods of determining the depreciation of various assets  Straight

Line Method  Service Hour Method  Production Units Method  Annuity Method

A. Straight-Line-Method 

It spr ea ds depreci atio n evenl y over the usef ul l ife of the asset. It assumes tha t depreci atio n is in proporti on to ti me.



Form ula fo r p eriodic depre ciatio n a) W = O – S b) R = W n



Example 1 A s teel ca binet which cos t P 8, 500 is expect ed to l ast 12 yrs ., an d at t hat time will have a t rad e-in-va lue of P500. F ind t he ann ua l depreciat ion ch arg e.

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 it em which co st P 5,200 an d af ter 5 yrs h as a scr ap value of P 400. W hat is the annu al depreciat io n? W hat per cent o f t he c ost is t he yearly depre cia tion?

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

• Example 3 A mach in e which co st s P 12 ,0 00 and after 6 yrs has a s alvag e va lue o f P 900. Det er min e t he yearly deprec iat ion and pr ep are a depreciat ion s che dule.

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 Book Year Depreciation Depreciation Value 0 0 SUB TRA CT O to 0AD P12,000 1ADD A P1,850 P1,850 10,150 D & RE 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

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?

So lut ion:  A) BV = O – (Rt ) = 20 0, 000 – [ ( 200,0 00 -10 ,000) x 3 ] 

8 = P1 28,75 0 

B) BV = O – (Rt ) = 20 0, 000 – [ ( 200,0 00 -10 ,000) x 5 ] 8

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

An nua l D epr eciati on E xpens e: 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 SUB TRA CT O to 0AD P4,500 D & RE 1ADD A P1,260 P1,260 3,240 2 1,148 2,408 2,092 3 980 3,388 1,112 4 812 4,200 300

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 3rd yr: 3,800 SH 5th yr: 3,000 SH 7th yr: 2,000 SH

f)

Find the depreciation charge per hour. Find the book value of the car after 2 yrs. Find the book value of the car after 4 yrs. Find the book value of the car after 5 yrs. Find the book value of the car after 7 yrs.

g) h) i) j)

2nd yr: 4,000 SH 4th yr: 3,200 SH 6th yr: 2,500 SH 8th yr: 1,500 SH

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 machi ne cost s P 7,5 00 has a salvag e value of P60 0. It is e st ima ted t hat t he mach in e can produ ce 2 5,00 0 unit s. This mach in e has been run as f ollows : 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 Schedule

Yr 0 5 6 7 8

Annu al Accu mulate d Boo k Depr ec iation D epr eci ati on Val ue 0 0 P7,500 P 772.80 P 772.80 6,727.20 883.20 1,656.00 5,844.00 1,131.60 2,787.60 4,712.40 1,518.00 4,305.60 3,194.40

Example 2 A cer tain machine co st s P 12 ,8 00 depre cia tes t o P 1, 200. T his t ype o f mach ine h as an e st im ated oper ating lif e of 3 5,5 00 h rs, a nd run as f ollows : 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 Schedule Yr

Annual Depreciation

Total Depreciation

Book Value

0 1 2 3 4 5 6 7

0 0 P12,800 P1,110.99P1,110.99 11,689.01 1,241.69 2,352.68 10,447.32 1,421.41 3,774.09 9,025.91 1,686.08 5,460.17 7,339.83 1,632.17 7,092.34 5,707.66 2,352.68 9,445.02 3,354.98 490.14 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” 