Book Depreciation
Lecture No. 33
Chapter 9
Contemporary Engineering Economics
Copyright © 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 value
N = depreciable life
Contemporary Engineering Economics, 4th
edition, © 2007
Example 9.3 – Straight-Line Method
I = $10,000
N = 5 Years
S = $2,000
D = (I - S)/N
n
Contemporary Engineering Economics, 4th
edition, © 2007
n
1
2
3
4
5
Dn
1,600
1,600
1,600
1,600
1,600
Bn
8,400
6,800
5,200
3,600
2,000
Declining Balance Method
• Principle:
A fixed asset as providing its service in a
decreasing fashion
• Formula
• Annual Depreciation
Dn  Bn1 I (1   )n1
• Book Value
Bn  I (1   )n 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
I = $10,000
N = 5 years
S = $778
Dn =  Bn 1
=  I (1 -  n 1
Bn  I (1   ) n
n
0
1
2
3
4
5
Contemporary Engineering Economics, 4th
edition, © 2007
Dn
$4,000
2,400
1,440
864
518
Bn
$10,000
6,000
3,600
2,160
1,296
778
Example 9.5 DB Switching to SL
Asset: Invoice Price
Freight
Installation
Depreciation Base
Salvage Value
Depreciation
Depreciable life
$9,000
500
500
$10,000
0
200% DB
5 years
• SL Dep. Rate = 1/5
•  (DDB rate) = (200%) (SL rate)
= 0.40
Contemporary Engineering Economics, 4th
edition, © 2007
Adjustments to the DB Method
Switch from DB to SL after n’
No further depreciation
allowances are available
after n”
Contemporary Engineering Economics, 4th
edition, © 2007
Case 1: S = 0
(a) Without switching
n
Depreciation
1
2
3
4
5
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
(b) With switching to SL
Book
Value
$6,000
3,600
2,160
1,296
778
n
1
2
3
4
5
Depreciation
4,000
6,000/4 = 1,500 < 2,400
3,600/3 = 1,200 < 1,440
2,160/2 = 1,080 > 864
1,080/1 = 1,080 > 518
Book
Value
$6,000
3,600
2,160
1,080
0
Note: Without switching, we have not depreciated the entire
cost of the asset and thus have not taken full advantage of
depreciation’s tax deferring benefits.
Contemporary Engineering Economics, 4th
edition, © 2007
Case 2: 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 below
their salvage values.
Contemporary Engineering Economics, 4th
edition, © 2007
Units-of-Production Method
• Principle
Service units will be consumed in a non
time-phased fashion
• Formula
•Annual Depreciation
Dn = Service units consumed for year (I - S)
total service units
Contemporary Engineering Economics, 4th
edition, © 2007
Example 9.7 Units-of-Production


Given: I = $55,000, S = $5,000, Total service
units = 250,000 miles, usage for this year =
30,000 miles
Solution:
30, 000
Dep 
($55, 000  $5, 000)
250, 000
 3 

 ($50, 000)
 25 
 $6, 000
Contemporary Engineering Economics, 4th
edition, © 2007