Emm 3604 r My Depreciation

8/2/2019 Emm 3604 r My Depreciation

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EMM3604 Cost Accounting and Engineering EconomyAssoc.Prof Dr.Rosnah Mohd.Yusuff

Depreciation

Is the decrease in value of physical

properties with use and time.

It is an accounting conceptan

annual deduction against before tax

income such that the effect of time

can be reflected in a firms financial

statement.

A non-cash cost that affects income

taxes.

Can involve deterioration or

obsolescence

Depreciation accounting is to

account for the cost of fixed assets in

a pattern that matches their decline

in value over time.


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Asset Depreciation

Fixed Assetsresources that are

acquired to provide future cash

flows.

Depreciation

Economic depreciation

The gradual decrease inutility in an asset with useand time.

Accounting depreciationThe systematic allocation ofan assets value in portions

over its depreciable life

Physical dep

Functional dep

Book dep

Tax dep


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Physical depa reduction in an assets capacity to

perform its intended service due tophysical impairment. due tointeraction with environmentcorrosion, rotting and due to wear andtear of use.

Functional depas a result of changes in the

organization or in technology obsolescence, declining need of theasset, or inability to meet increasedquantity and/or quality demands.


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Economic dep

Economic dep =purchase pricemarket

valueAccounting dep

use to assess financial position of

organization.

Asset dep, cost of fixed assets arecapitalizedtheir costs are distributed by

subtracting them as expenses from gross

income.

A fraction of the cost of the asset ischargeable as an expense in each

accounting period in which the asset

provide service to the firm.


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Depreciation Concepts &

Terminology

Depreciating an asset requires:1. What is the cost of the asset

2. What is the assets value at the end of

its useful life3. What is depreciable life of an asset

4. What method of dep do we choose?


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Cont

Depreciable Propertya. It must be used in business or held toproduce income

b. It must have a definite service life,

and that life must be longer than ayear.

c. It must be something that wears out,

decays, get used up, obsolete, or

loses value from natural causes.

d. It is not inventory, stock in trade, or

investment property.


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Definitions Adjusted (Cost) basis

original cost basis of the asset, adjustedby allowable increases or decreases. Ex.cost of any improvement to a capital

asset with a useful life greater than oneyear, the original cost basis. Casualtyor theft loss.

Basis, or cost basis initial cost of acquiring an asset

(purchase price + any sales taxes),including transportation expenses andother expenses making the assetserviceable.


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Cont

Book Value (BV)the worth of a depreciable property shownon the accounting records of a company.

Original cost basis include anyadjustments, less all allowabledepreciation or depletion deductions.

(Book Value) = adjusted cost basis - (dep deduction)

Market Value (MV)the amount that will be paid by a willing

buyer to a willing seller. The MVapproximates the present value of whatwill be received through ownership of theproperty, including the time value ofmoney.

J=1


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Definitioncont

Recovery periodThe number of years over which the

basis of a property is recovered

through the accounting process.

Recovery rateA percentage (in decimal form) for

each year of the MACRS recovery

period that is utilized to compute anannual depreciation deduction.


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Definitioncont

Recovery periodThe number of years over which the

basis of a property is recovered

through the accounting process.

Recovery rateA percentage (in decimal form) for

each year of the MACRS recovery

period that is utilized to compute anannual depreciation deduction.


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Definitioncont

Salvage Value (SV) the estimated value of a property at the end of its

useful life. It is the expected value or selling priceof a property when the asset can no longer be usedproductively by its owner. The term net salvagevalue is used when expenses are incur in disposingproperty.

a cash outflow deducted from cash inflow to obtainfinal net SV.

Useful life the expected (estimated) period of time that a

property will be used to produce income/owners

expects to productively use it. Referred asDepreciable Life sometimes. Actual Useful Life ofan asset, maybe different than its depreciable life.Determine service life of an asset IRS guidelines.


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Example:

A company purchased an automatic hole-punching

machine priced at RM62, 500. The vendors invoiceincluded a sales tax of RM3263. The company alsopaid the inbound transportation charges of RM725as well as labour cost of RM2150 to install themachine in the factory. The site was also preparedbefore installation at a cost of RM3500. Determinethe cost basis for the new machine for depreciationpurpose.

Cost of new machine RM62,500

Freight 725

Installation Labour 2,150

Site Preparation 3,500

Cost of machine (cost basis) RM68,875


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Depreciation Methods

1. Book dep - for financial reports,balance sheet, income statement.2. Tax dep - for calculating taxes.

depreciating assets more quickly

allows firms to defer paying income

taxes.

Permit a higher dep in earlier years

than book dep, tax benefit is enjoyed

earlier. Pay lower taxes in earlier years,

better cash position because of timevalue of the funds.


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Book Depreciation Methods

EE -dep in the context of income

tax computation. firms use book dep methods for

financial reporting to stockholders and

outside parties to reflect the actualloss in value of asset

still use for income tax purposes.

1. straight line method

2. accelerated methodDB, SOYD.

3. unit of prod method


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Straight line method (SL)

Assumes fixed assets provides its

services in a uniform fashion.

A constant amount is depreciatedeach year over the useful life of

the asset.


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SLcont

dk = (B - SVn) / N

d* = dfor 1 k N

BV = B - d

N =depreciable life of the assets in years

B =cost basis d =annual dep deduction in year (1

N)

BV=book value at the end of year

SV=estimated salvage value at end year N d* =cumulative dep through year


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SLexample

A new electric saw for cuttingsmall pieces of lumber in a

furniture manuf. Plant has a cost

basis of RM4000 and a 10-year

depreciable life. The estimated SV

of the saw is zero at the end of 10

yrs. Determine the annual dep

using SL and the BV at the end ofeach year.


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SLexample (cont)

d = (40000) / 10

at year 5,

d5 = RM 400

d*5 = 5 (40000) =RM 2000

10

BV = 40005(4000-0)

10

= RM 2000


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SLexample (cont)

EOY, dk BVR

0 - RM 40001 RM400 3600

2 400 3200

3 400 2800

4 400 24005 400 2000

6 400 1600

7 400 1200

8 400 800

9 400 400

10 400 0

The final SV is the final book value


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Declining Balance (DB) method sometimes called constant percentage

method or the Matheson Formula. Assumes the annual cost of dep is a fixed

percentage of BV at the beginning of the

year.

The fixed fraction, is = (1/N) (multiplier)

Multiplier commonly used 1.5 (150% DB)

and 2.0 (200%, or DDB).

As N , dep is highest in thefirst year and decreases over the assets

depreciable life.


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DBcont

D1 = Ii

D2 = (I D1) = I (1-)

D3 = (ID1- D2) = I (1)2

Dn = I (1-) n-1

Bn = I (1-) n


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DBExample

The following information for acomputer system:

Cost Basis of the asset, I =

RM10,000

Useful Life, N = 5 years.

Estimated SV = RM778.


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DBExample (cont)

Use same example as SL:-a. = 2/N = 0.2

d6 = 4000 (1-0.2)5 (0.2) = RM262.14

d*6 = 4000 [ 1-(1-0.2)6 ] = RM2951.42

BV6= 4000 (1-0.2)6 = RM1,048.58

b. = 1.5/N = 0.15

d6 = 4000 ( 1- 0.15)5 (0.15) = RM266.22

d*6 = 4000 [ 1(1 - 0.15)6] = RM2491.40BV6 = 4000 (10.15)

6 = RM1508.60

Using = 0.2


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DBExample (cont)

EOY, d BV0 - 4000

1 800 3200

2 640 2560

3 512 20484 409.60 1638.40

5 327.68 1310.72

6 262.14 1048.58

7 209.72 838.86

8 167.77 671.09

9 134.22 536.87

10 107.37 429.50


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Sum of the yearsdigit (SOYD)

If N = depreciable life of the asset,SOYD = n (n+1)

2

d = (B - SVN) . [ 2 (N + 1) ]

N ( N + 1 )

Book Value

BV= B[ 2 ( B- SVN) ] + [ (B - SVN) ]

( + 1)N N ( N + 1 )

d* = B - BV


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SOYDexample

Using same ex:d4 = 400 [ 7 / 25 ], N = 10

= 509.09

SOYD = 10 (11) = 55

2


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SOYDexample (cont)

EOY, d BV

0 - 4000

1 727.27 3272.73

2 645.55 2618.183 581.82 2036.36

4 509.09 1527.27

5 436.36 109.91

10 72.73 0

d1 = 4000 (10 / 55)

d2 = 4000 (9 / 55)

d3 = 4000 (8 / 55)


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DB switchover to SL

DB - BV never reaches

zero can switch to SL

Switchover occur when SLDep >

DBdep


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DB switchover to SLex

d6 = RM262.14,

BV6 = RM1048.58

BV10 = RM429.50

With switchover, BV10 = 0, since

in year 6

Dbdep = SLdep= RM262.14

switch to SL

thru year 7-10, SLdep>DBdep


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Example (cont)

BOYear, DB SL BV

6 262.14 = 262.14 1310.72

7 209.72< 262.14 1048.58

8 167.77< 262.14 786.44

9 134.22< 262.14 524.3010 107.37< 262.14 262.16

Total depreciation :

DDB = RM3,570.50SL = RM4000.


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Units of Production Method

SL, DB, SOYD -decrease in valueas a function of time.

When decrease in value is mostly afunction of use, the cost basis (minus

final SV) is allocated equally over theestimated number of units producedduring the useful life of the asset.The depreciation rate:

Depreciation per unit prod= B -SVN

estimated lifetime prod in units


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Example

A piece of equipment has a basis ofRM50,000 and SV is RM10,000, whenreplaced after 30,000 hrs. of use. Find thedep rate per hour of use, and its BV after10,000 hrs of operation.

Dep per unit ofprod

=RM50,00010,000 = RM1.33

30, 000

After 10, 000 hrs,BV = 50, 000 RM1.33 (10,000)

hr

Or BV = RM36, 700 #


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MODIFIED ACCELERATED COST

RECOVERY SYSTEM (MACRS)

Practice in US

1981-ACRS 1986 - MACRS

asset dep law

dictate depreciation rates for allpersonal and real property

take advantage of accelerated methods

of capital recovery


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MACRS (cont)

MACRS

Dt = dtB

dt = dep rate provided by

the Govt.Book Value,

BVt = BVt-1 - Dt

BVt = first costsum ofacc. dep

= B - Dj

j=t

j=1


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MACRS (cont)

First cost B is always completely depreciated

as MACRS assumes that the estimated SV=0. Recovery periods are standardized to the

values of 3,5,7,10,15 and 20 years forpersonal property. For real property

commonly 39 years, possible to justify 27.5year recovery period.

For Annual Dep: First Cost (unadjusted)

X MACRS rate (from table).

the dep rate incorporate the DDB method (d= 2/n) and switch to SL dep during recoveryperiod for property starts with DDB (dt = 2/n) SL rate (dt = 1/n) when SL method offers afaster write-off.


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MACRS (cont)

Real property -SL method for n = 39.

Annual percentage dep rate =1/39 =0.02564. the MACRS forces partial

recovery in years 1 to 40.

Year 1 100 d1 =1.391%

Year 239 100 dt =2.564%

Year 40 100 d40 = 1.177%.

MARCS dep rates are presented for 1 yearlonger than stated recovery period n.

- is built in half year convention imposed bythe MACRS system.


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Replacement AnalysisWhy???a. Reduced Performance -

deterioration of parts, expected levelof reliability/productivity is notachieved, increased cost of operation,higher scrap and rework costs, lostsales, and larger maintenanceexpenses.

b. Altered requirements - Newrequirements accuracy, speed etc.

c. Obsolescence - competition, rapidlychanging technology of automation,computers, communication.


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Replacement Analysis (cont)

Basic concepts -compare

alternatives.Defenderasset currently owned

(or in place).

Challenger(s)other alternative(s)Perspective - of analysis

consultant or outsider.

- neither own or use the asset

To acquire the defender, we must

invest the going market value of the

defender alternative.


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Estimated market or trade-in-value

the First Cost of the defender

alternative.

- economic life, AOC & SV for defender

past cost are irrelevant inreplacement analysis

- includes sunk cost.

Sunk cost should not be included in

the economic analysis. It represents

capital loss.

Sunk cost = present BV-present MV


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Time Value of Money

Interest- manifestation of the time value ofmoney

- the increase between an original sum of

money borrowed and final amount owed,original amount owned (investment) andfinal amount accrued.

Principal - original investment orloan amount.

Ifve lose money and no interest.

Interest=total amount now - original principal


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If you borrowed money

Interest rate :Percent interest rate

= interest accrued per time unit x

100%

original amount

unit of time used - interest period

Interest =amount owed now-original principal


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Examples1) Investment RM100, 000 and

withdrew a total of RM106, 000

exactly 1 year later.

a. Interest gained:

RM106, 000100, 000 = RM60, 000

b. Interest rate:

6000 / yr x 100% = 6% per year.

100,000


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2) You borrowed RM20,000 from a

bank for a year at 9% interest.

a. interest : RM20,000 (0.09) =

RM1800

b. Total due at end of year :

= sum of principal + interest

= RM20, 000 + RM1800

= RM 21,800


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3) How much money has to be

deposited one year ago to have

RM1000 now at an interest rate of 5%

per year. What is the interest earned?Let X be the original deposit

Total = original + original

(interest rate)

RM1000= + ( 0.05) = (1+0.05)

= 1000 / (1.05) = RM952.38.

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Emm 3604 r My Depreciation