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Transcript of ALI SALMAN1 LECTURE - 12 ASST PROF. ENGR ALI SALMAN ceme.nust.edu.pk DEPARTMENT OF ENGINEERING...
![Page 1: ALI SALMAN1 LECTURE - 12 ASST PROF. ENGR ALI SALMAN ceme.nust.edu.pk DEPARTMENT OF ENGINEERING MANAGEMENT COLLEGE OF E & ME, NUST DEPARTMENT.](https://reader031.fdocuments.in/reader031/viewer/2022020802/5a4d1ad57f8b9ab059972af0/html5/thumbnails/1.jpg)
ALI SALMAN 1
LECTURE - 12 ASST PROF. ENGR
ALI SALMANalisalman@
ceme.nust.edu.pkDEPARTMENT OF ENGINEERING MANAGEMENTCOLLEGE OF E & ME, NUST
ENGINEERING ECONOMICS
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DepreciationDepreciation may be defined as the decrease in the value of physical assets with the passage of time as a result of wear, deterioration and technological obsolescence.
It is used in the books of accounts for preparing a balance sheet of assets.Depreciation is viewed as a part of business expenses that reduce taxable income.
Why Do We Consider Depreciation?
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Depreciation ExampleYou purchased a car worth $15,000 at the beginning of year 2004.
Depreciatio
nEnd of Year
Market Value
Loss of Value
012345
$15,00010,000
8,0006,0005,0004,000
$5,0002,0002,0001,0001,000p
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Factors to Consider in Asset Depreciation
Depreciable life (how long?)
Salvage value (disposal value)
Cost basis (depreciation basis)
Method of depreciation (how?)
Salvage value is the price of an equipment
that can be obtained after it has been used.
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What Can Be Depreciated?
Assets used in business or held for production of income
Assets having a definite useful life and a life longer than one year
Assets that must wear out, become obsolete or lose value
A qualifying asset for depreciation must satisfy all of the three conditions above.
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Cost BasisCost of a new hole-punching machine (Invoice price) $62,500+ Freight 725
+ Installation labor 2,150
+ Site preparation 3,500
Cost basis to use in depreciation calculation
$68,875
Depreciation Methods
Straight-Line MethodDeclining Balance Method
Unit Production Method
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Straight Line (SL) MethodThis method assumes a uniform decrease in the value of asset with the passage of time.
• Formula•Annual Depreciation
Dn = (I – S) / N, and constant for all n.
•Book ValueBn = I – n (D)
where I = cost basis/valueS = Salvage valueN = depreciable life
Book value is the worth of an asset as shown on the accounting record of a company.
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Example – Straight Line Method
D1
D2
D3
D4
D5
B1
B2B3
B4
B5
$10,000
$8,000
$6,000
$4,000
$2,000
0 1 2 3 4 5
Total depreciation at end of life
n Dn Bn
1 1,600 8,4002 1,600 6,8003 1,600 5,2004 1,600 3,6005 1,600 2,000
I = $10,000N = 5 YearsS = $2,000D = (I - S)/N
Annual Depreciation
Book Value
n
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Declining Balance MethodIn this method the depreciation cost is highest in the first year and reduces year after year.
Formula• Annual Depreciation
• Book Value
1 nn BD 1)1( nI
nIB )1( where 0 << 2(1/N)Note: if is chosen to be the upper bound, = 2(1/N),we call it a 200% DB or double declining balance method.
n
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Example – Declining Balance Method
D1
D2
D3
D4D5
B1
B2
B3
B4 B5
$10,000
$8,000
$6,000
$4,000
$2,000
0 1 2 3 4 5
Total depreciation at end of life
$778
Annual Depreciation
Book Value
n012345
Dn
$4,0002,4001,440
864518
Bn
$10,0006,0003,6002,1601,296
778
INS
D B
I
B I
n n
n
nn
= $10, = years = $778 =
= ( -
0005
1
1
1
1
( )
n
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Example – Declining Balance Method (if B<salvage value)
D1
D2
D3 D4
B1
B2
B3
B4 B5
$10,000
$8,000
$6,000
$4,000
$2,000
0 1 2 3 4 5
Total depreciation at end of life
$778
Annual Depreciation
Book Value
n012345
Dn
$4,0002,4001,440
1600
Bn
$10,0006,0003,6002,16020002000
INS
D B
I
B I
n n
n
nn
= $10, = years = $778 =
= ( -
0005
1
1
1
1
( )
n
2000
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When S = $2,000
End of Year
Depreciation Book Value
1 0.4($10,000) = $4,000 $10,000 - $4,000 = $6,000
2 0.4(6,000) = 2,400 6,000 – 2,400 = 3,600
3 0.4(3,600) = 1,440 3,600 –1,440 = 2,160
4 0.4(2,160) = 864 > 160 2,60 – 160 = 2,000
5 0 2,000 – 0 = 2,000
Note: Tax law does not permit us to depreciate assets belowtheir salvage values.
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Units-of-Production Method
• PrincipleService units will be consumed in a non
time-phased fashion
• Formula•Annual Depreciation
Dn = Service units consumed for yeartotal service units
(I - S)
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Example
Given: I = $55,000, S = $5,000, Total service units = 250,000 miles, usage for this year = 30,000 miles
Solution:
30,000 ($55,000 $5,000)250,000
3 ($50,000)25
$6,000
Dep
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15