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1 Depreciation -definitions A decline in the value of a property due to general wear and tear or obsolescence Method of writing off wear and tear on assets that are used to produce income. Amount of value that a possession loses over time. Depreciation basis is that part of the asset’s purchase price that is spread over the depreciation period (service life). Depreciation The cost is spread out over its estimated useful life in the form of an expense given various depreciation methods.

Engineering Economics - Depreciation

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Page 1: Engineering Economics - Depreciation

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Depreciation -definitions

A decline in the value of a property due to

general wear and tear or obsolescence

Method of writing off wear and tear on assets

that are used to produce income.

Amount of value that a possession loses over

time.

Depreciation basis is that part of the asset’s

purchase price that is spread over the

depreciation period (service life).

Depreciation

The cost is spread out over its

estimated useful life in the form

of an expense given various

depreciation methods.

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Causes of Depreciation

Question: What are the causes of depreciation?

Answers:

Fixed assets are those assets bought by the company

for the intention to be used for a long period of time.

Fixed assets are said to depreciate over a period of

time due to the following factors:

Causes of Depreciation

1) Physical deterioration

i) Wear and tear – When a motor vehicle or machinery or

fixtures and fittings are used, they eventually wear out.

Some last many years, others last only a few year.

ii) Erosion, rust, rot and decay – Land may be eroded or

wasted away by the action of wind, rain, sun and other

elements of nature. Similarly, the metals in motor

vehicles or machinery will rust away.

Causes of Depreciation

2) Economic factors

i) Obsolescence

This is the process of becoming out of date.

For instance, replacing a computer with old operatingsystem with a new computer with XP system.

ii) Inadequacy

This arises when an asset is no longer used because of thegrowth and changes in the size of the firm. Forinstance, a small ferryboat that is operated by a firm ata coastal resort will become entirely inadequate whenthe resort becomes more popular, to be more efficientand economical, the firm may replace it with a largeferryboat.

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Causes of Depreciation

3) The time factor (the effluxion of time)

Some assets might have a legal life fixed in terms of

years. For example, the patents, and leasehold. You may

agree to rent some buildings for 10 years. This is

normally called a lease. When the years are finished, the

lease is worth nothing to you, as it has finished.

Whatever you paid for the lease is now of no value.

Causes of Depreciation

4) Depletion

Other assets are of wasting character, perhaps due to the

extraction of raw materials from them. These materials

are then either used by the firm to make something else, or

are sold in their raw state to other firms. Natural

resources such as mines, quarries and oil wells come

under this heading.

Factors that affect the calculation of Depreciation

Question: What are the factors that affect the calculation of depreciation?

Answers:

1) Cost of asset (include expenses and capital expenditure incurred eg. The installation fees, the legal fees)

2) Estimated useful life of asset

This is the number of years that the asset is expected to be used)

3) Residual or scrap value of the asset

This is the value of the asset at the end of its life.

4) Method of calculating depreciation

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4 factors required for depreciation

calculations

1. Cost Net purchase price.

All reasonable and necessary expenditures to get the asset in place and ready for use.

2. Residual Value (Salvage or Scrap Value)– Asset value at the end of its expected “useful life”.

3. Depreciable Cost Cost less residual value.

Depreciable cost is allocated over the useful life of an asset.

4. Life useful life period of the asset

Depreciation Methods

Straight-line Method

Accelerated Methods

– Declining balance method

– Sum-of-the-years digits

Production or Use Methods

Cost - Residual Value

Estimated Useful Life

Straight Line Method

This method spreads the depreciable costs

evenly over the asset’s estimated useful life.

Annual depreciation is computed as follows:

$10,000 - $1000

5 years= = $1800 / year

Check the help for SLN function in Excel for more details

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The straight line formula

N

SId

N

SIADA

N

SIAIS A

Where: -

N = life of the structure in years

I = the original cost

S = the values at the end of the life of structure

d = the annual cost of depreciation

DA = depreciation up to age A years

SA = the value of the end of A years

B C D E F

2 Depreciation Schedule3 Straight Line Method - SLN function

4

5 Scrap Value - 1000

6 Years - 5

7

8

9perio d yrs C o st

A nnual

D epreciat io n

A ccumulated

D epreciat io nC arrying Value

10 DOP 0 10000 0 0 10000

11 End of Yr 1 1 10000 1800 1800 8200

12 End of Yr 2 2 10000 1800 3600 6400

13 End of Yr 3 3 10000 1800 5400 4600

14 End of Yr 4 4 10000 1800 7200 2800

15 End of Yr 5 5 10000 1800 9000 1000

Same Increases Decreases Stop at

each Uniformly Uniformly Residual Value

year

Carrying Value(CV) = Cost – Accumulated Depreciation

=D10-F10

=SLN(D11,$D$5,$D$6)

Book Value

The book value of a plant asset is

the net cost of the asset after

accumulated depreciation (the

contra account) has been

subtracted.

$50,000 Beg. Book Value

-12,000 Accum. Depre.

$38,000 End. Book Value

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Example

An equal amount of depreciation

is recorded each fiscal year.

Straight-Line Depreciation Schedule

Straight-line method (Examples)

Example 1:

ABC Ltd. Bought a machine at a cost of £80,000. Themachine has an expected useful life of 5 years and at the endof the 5th year, it can be sold for £10,000. calculatedepreciation per annum?

Cost Annual

Depreciation

Provision for

Depreciation

NBV

Date of

purchase

80,000 80,000

End of 1st year 80,000 14,000 14,000 66,000

End of 2cd year 80,000 14,000 28,000 52,000

End of 3rd year 80,000 14,000 42,000 38,000

End of 4th year 80,000 14,000 56,000 24,000

End of 5th year 80,000 14,000 70,000 10,000

Depreciation for 5 years would be:

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Example 2:

ABC Ltd. Bought a machine at a cost of £80,000. The

depreciation is to be charged at a 20% per annum on cost.

Solution:-

Depreciation per annum = £80,000 x 20%

= £16,000 per year

Straight-line method (Examples)

Decling-Balance (DB) Method -Type

1. Fixed DB method

– computes depreciation at a fixed rate.

2. Double DB method

– computes depreciation at an accelerated rate.

3. Variable DB method

– Also computes depreciation at an accelerated rate.

– switches to straight-line depreciation when depreciation is greater than the declining balance calculation

Check the help for DB, DDB and VDB functions in Excel for more details

Accelerated Depreciation

Declining balance method

Depreciation is calculated on a fixed percentage on the

Diminishing Balance of the Asset (the NBV). This

results in a higher depreciation charge in the earlier years

of the asset’s estimated useful life.

Example 3: A machine costs £50,000 is to be

depreciated at 15% on Reducing Balance Method.

Accelerated Depreciation

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Declining balance method

Cost Annual Depreciation Provision for

Depreciation

NBV

Date of purchase 50,000 0 - 50,000

End of 1st year 50,000 50,000 x 15% = 7,500 7,500 42,500

End of 2cd year 50,000 42,500 x 15% = 6,375 13,875 36,125

End of 3rd year 50,000 36,125 x 15% = 5,419 19,294 30,706

End of 4th year 50,000 30,706 x 15% = 4,606 23,900 26,100

End of 5th year 50,000 26,100 x 15% = 3,915 27,815 22,185

Accelerated Depreciation

The ratio of the depreciation in any one year to the

salvage value at the beginning of that year is constant

throughout the life of the structure and is designated

by X.

X = the annual fixed ratio of depreciation

Deprecation during the first year (d1) = I X

Depreciation value (Book value) at the end of the

1st year = I – I X = I (1 – X)

Depreciation value (Book Value) at the end of n

years (S) = I (1 – X)n

S/I = (1 – X)n

X = 1 – (S/I)(1/n)

The Matheson formula or declining balance method

Accelerated Depreciation

Double-declining-balance method, it would

Multiply the straight-line rate by two, e.g. 2/1 x

¼ = 2/4

$25,000 = $50,000 x2

4

Double-declining-balance

depreciation expense= Book Value x

2 x straight-line

Expected useful

life

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A B C D E

(A xB) ($50k - D)

Year

2004 2/4 $50,000 $25,000 $25,000 $25,000

2005 2/4 25,000 12,500 37,500 12,500

2006 2/4 12,500 6,250 43,750 6,250

2007 2/4 6,250 4,250 48,000 2,000

Ending

Book

Value

Double-

Declining-

Balance

Rate

Beginning

Book

Value

Debit

Depreciation

Expense

Credit

Accumulated

Depreciation

Double-Declining Balance

Depreciation Schedule

Accelerated Depreciation

$6,250 – $2,000

(residual value)

Reasons for Using Accelerated Depreciation

1. An asset is more useful earlier in its life than

later, and the useful life may be difficult to

estimate.

2. Depreciation expense is deductible in computing

taxable income and income taxes.

The second reason is the most

common reason for using

accelerated depreciation.

Income before depreciation

and taxes $100,000 $100,000

Depreciation expense 12,000 25,000

Pretax income 88,000 75,000

Income taxes (35%) 30,800 26,250

Net income $ 57,200 $ 48,750

Straight-Line Accelerated

Comparison of Straight-Line and

Accelerated Depreciation Methods in 2004

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B C D E F

2 Depreciation Schedule3 Fixed Decling Method - DB function

4

5 Scrap Value - 1000

6 Years - 5

7

8

9perio d yrs C o st

A nnual

D epreciat io n

A ccumulated

D epreciat io nC arrying Value

10 DOP 0 10000 0 0 10000

11 End of Yr 1 1 10000 3690 3690 6310

12 End of Yr 2 2 10000 2328 6018 3982

13 End of Yr 3 3 10000 1469 7488 2512

14 End of Yr 4 4 10000 927 8415 1585

15 End of Yr 5 5 10000 585 9000 1000

Increases Decreases Can't go

Quickly Quickly below scrap

Value

Carrying Value(CV) = Cost – Accumulated Depreciation

=D10-F10

=DB(D11,$D$5,$D$6,C11)

B C D E F

2 Depreciation Schedule3 Accelerated Double Decling Method - DDB function

4

5 Scrap Value - 1000

6 Years - 5

7

8

9perio d yrs C o st

A nnual

D epreciat io n

A ccumulated

D epreciat io nC arrying Value

10 DOP 0 10000 0 0 10000

11 End of Yr 1 1 10000 4000 4000 6000

12 End of Yr 2 2 10000 2400 6400 3600

13 End of Yr 3 3 10000 1440 7840 2160

14 End of Yr 4 4 10000 864 8704 1296

15 End of Yr 5 5 10000 296 9000 1000

Increases Decreases Can't go

Quickly Quickly below scrap

Value

Carrying Value(CV) = Cost – Accumulated Depreciation

=D10-F10

=DDB(D11,$D$5,$D$6,C11)

Summary - depreciation

VDB Variable-decling balance

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B C D E F

2 Depreciation Schedule3 Formulae

4 Asset - Equipment C12 =SLN($D$5,$D$6,$D$7)

5 Cost 100,000 D12=DB($D$5,$D$6,$D$7,B12)

6 Scrap Value - 2000 E12=DDB($D$5,$D$6,$D$7,B12)

7 Years - 5 F12=VDB($D$5,$D$6,$D$7,B11,B12)

8 D20=D19-C11

9 Depreciation Amt.

10 Years SLN DB DDB VDB

11 1 19,600 54,300 40,000 40,000

12 2 19,600 24,815 24,000 24,000

13 3 19,600 11,341 14,400 14,400

14 4 19,600 5,183 8,640 9,800

15 5 19,600 2,368 5,184 9,8001617 Value of Asset

18 period yrs SLN D B D D B VD B

19 DOP 0 100,000 100,000 100,000 100,000

20 End of Yr 1 1 80,400 45,700 60,000 60,000

21 End of Yr 2 2 60,800 20,885 36,000 36,000

22 End of Yr 3 3 41,200 9,544 21,600 21,600

23 End of Yr 4 4 21,600 4,362 12,960 11,800

24 End of Yr 5 5 2,000 1,993 7,776 2,000

Asset value vs Life time

0

20,000

40,000

60,000

80,000

100,000

120,000

0 1 2 3 4 5 6

Period (years)

Asset V

alu

e

SLN

DB

DDB

VDB

Depreciation Vs Period

-

10,000

20,000

30,000

40,000

50,000

60,000

0 1 2 3 4 5 6

Period (years)

De

pre

cia

tio

n A

mt

SLN

DB

DDB

VDB

Comparison of Depreciation Methods

To obtain the depreciation charge in any year of

life by the sum-of-the year digits method

(commonly designated by SYD), the digits

corresponding to the number of each year of life

are listed in reverse order.

The sum of the digits is then determined.

The depreciation factor for any year is the reverse

digit for that year divided by the sum of the digits.

For example, for a property having a life of five

years,

Sum of digits plan

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year Number of the year in

reverse order (digits)

Depreciation

1 5 5/15

2 4 4/15

3 3 3/15

4 2 2/15

5 1 1/15

Sum of the digits 15

Sum of digits plan

The depreciation for any year is the product of the SYD

depreciation factor for that year and the depreciable

value, I – S.

The general expression for the annual cost of

depreciation for any year, A, when the total life is N, is

)()1(

)1(2SI

NN

AN

DA =

Sum of digits plan

Units-of-Production Depreciation

At the beginning of 2005, Mom’s Cookie Company

purchased a truck for $30,000. Management

expects the useful life of the truck to be 100,000

miles, at which time it will be sold for $10,000.

Cost – Residual Value

Expected Units

$30,000 – $10,000

100,000 miles

Activity Depreciation

$0.20 per mile

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If the truck were driven

12,000 miles in 2005,

Hydro would record

depreciation expense of

$2,400 (12,000 x $0.20).

Activity Depreciation

Units-of-Production Depreciation