Contemporary Engineering Economics, 4th edition, © 2007
Book Depreciation
Lecture No. 33Chapter 9Contemporary Engineering EconomicsCopyright © 2006
Contemporary Engineering Economics, 4th edition, © 2007
Book Depreciation Methods Purpose: Used to report net income to
stockholders/investors Types of Depreciation Methods:
Straight-Line Method Declining Balance Method Unit Production Method
Contemporary Engineering Economics, 4th edition, © 2007
Straight – Line (SL) Method
• Principle A fixed asset as providing its service in a uniform fashion over its life
• Formula•Annual Depreciation
Dn = (I – S) / N, and constant for all n.•Book Value
Bn = I – n (D)where I = cost basis
S = Salvage valueN = depreciable life
Contemporary Engineering Economics, 4th edition, © 2007
Example 9.3 – Straight-Line Method
n Dn Bn 1 1,600 8,400 2 1,600 6,800 3 1,600 5,200 4 1,600 3,600 5 1,600 2,000
I = $10,000N = 5 YearsS = $2,000D = (I - S)/N
n
Contemporary Engineering Economics, 4th edition, © 2007
Declining Balance Method• Principle: A fixed asset as providing its service in a decreasing fashion• Formula
• Annual Depreciation
• Book Value
1 nn BD 1)1( nI
nn IB )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 (DDB) method.
Contemporary Engineering Economics, 4th edition, © 2007
Example 9.4 – Declining Balance Method
n012345
Dn
$4,0002,4001,440
864518
Bn$10,000
6,0003,6002,1601,296
778
I
N
S
D B
I
B I
n n
n
nn
= $10,
= years
= $778
=
= ( -
000
5
1
1
1
1
( )
Contemporary Engineering Economics, 4th edition, © 2007
• SL Dep. Rate = 1/5• (DDB rate) = (200%) (SL rate)
= 0.40
Asset: Invoice Price $9,000Freight 500Installation 500
Depreciation Base $10,000Salvage Value 0Depreciation 200% DBDepreciable life 5 years
Example 9.5 DB Switching to SL
Contemporary Engineering Economics, 4th edition, © 2007
Adjustments to the DB Method
Switch from DB to SL after n’
No further depreciation allowances are availableafter n”
Contemporary Engineering Economics, 4th edition, © 2007
n Depreciation
Book
Value
12345
10,000(0.4) = 4,000 6,000(0.4) = 2,400 3,600(0.4) = 1,440 2,160(0.4) = 864 1,296(0.4) = 518
$6,0003,6002,1601,296
778
n
Book
Depreciation Value
12345
4,000 $6,0006,000/4 = 1,500 < 2,400 3,6003,600/3 = 1,200 < 1,440 2,1602,160/2 = 1,080 > 864 1,0801,080/1 = 1,080 > 518 0
(a) Without switching (b) With switching to SL
Note: Without switching, we have not depreciated the entirecost of the asset and thus have not taken full advantage of depreciation’s tax deferring benefits.
Case 1: S = 0
Contemporary Engineering Economics, 4th edition, © 2007
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.
Case 2: S = $2,000
Contemporary Engineering Economics, 4th edition, © 2007
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)