Transcript Book Depreciation

Lecture No. 31
Chapter 9
Contemporary Engineering Economics
Copyright © 2010
Contemporary Engineering Economics, 5th edition, © 2010
Purpose: Used to report net income to
stockholders/investors
Types of Depreciation Methods:
Straight-Line Method
Declining Balance Method
Units-of-Production Method
Contemporary Engineering Economics, 5th edition, © 2010
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
 Example:
I = $10,000
 S = $2,000
 N = 5 years
Contemporary Engineering Economics, 5th edition, © 2010
Declining Balance (DB) Method
Principle:
A fixed asset as providing
its service in a decreasing
fashion
 Formula:
Dn  Bn1  I (1   )n1
Bn  I(1   )n
where 0 < α < 2(1/N)
 Example:
I = $10,000
S = $778
N = 5 years
α = 0.40
Contemporary Engineering Economics, 5th edition, © 2010
Example 9.5 DB Switching to SL
 Without Switching
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
• a (DDB rate)
= (200%) (SL rate) = 0.40
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
Book
Value
$6,000
3,600
2,160
1,296
778
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, 5th edition, © 2010
Adjustments to the DB Method
 Switch from DB to SL after n’
 Case 1: S = 0
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
Contemporary Engineering Economics, 5th edition, © 2010
Book
Value
$6,000
3,600
2,160
1,080
0
Adjustments to the DB Method
 No further depreciation
after n”
 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 value.
Contemporary Engineering Economics, 5th edition, © 2010
Units-of-Production Method
Principle:
 Given: I = $55,000, S = $5,000, total
Service units will be
consumed in a nontime-phased fashion
service unit = 250,000 miles, service
units consumed = 30,000 miles
 Find: Dn
 Solution:
 Formula:
I = Initial investment
S = Salvage value
30,000
Dn 
($55,000  $5,000)
250,000
 3 
   ($50,000)
 25 
 $6,000
Contemporary Engineering Economics, 5th edition, © 2010
Purpose: To determine the income taxes owed for the IRS
Assets placed in service prior to 1981
 Used the book depreciation methods (SL, DB, SOYD)
Assets placed in service from 1981 to 1986
 Used the ACRS (Accelerated Cost Recovery System)
Table
Assets placed in service after 1986

Use the MACRS (Modified ACRS) Table
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Modified Accelerated Cost Recovery
Systems (MACRS)
Personal Property
Depreciation
schedule based
on the DB method
switching to SL
Half-year
convention
Zero salvage
value
 Real Property
SL Method
Mid-month
convention
Zero salvage
value
Contemporary Engineering Economics, 5th edition, © 2010
MACRS Depreciation Schedules for Personal
Property with Half-Year Convention
 Property Class
 3, 5, 7, 10-Year with 200%
DB
 15 and 20-Year with 150%
DB
 Sample Calculation – 5-Year
MACRS:
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Example 9.8 MACRS Depreciation
Asset cost = $10,000
 Annual Depreciation
Property class = 5-year
MACRS
DB method = Half-year
convention, zero salvage
value, 200% DB switching
to SL
20%
32% 19.20% 11.52% 11.52% 5.76%
$2000 $3200 $1920 $1152 $1152
Full
1
2
Full
3
Full
4
Half-year Convention
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$576
Full
5
6
Comparison between DDB with Switching to SL and
MACRS Method
 Conventional DDB
 Depreciation Rates
Method:
 Cost basis: $10,000
 Salvage value: $0
 Depreciable life: 5
years
 DB rate: 200%
 MACRS Method:
 Property class: 5-
year
 Salvage value: $0
 Half-year convention
Contemporary Engineering Economics, 5th edition, © 2010
MACRS for Real Property
 Types:
 Depreciation Allowances for a 10-year
 27.5-Year (Residential) Ownership of the Property
 39-Year (Commercial)
Year (n)
Calculation
Allowed Depreciation (%)
1
(9.5/12)(100%/27.5)
2.8788%
 Mid-month convention
2
100%/27.5
3.6364%
 Zero salvage value
3
100%/27.5
3.6364%
4
100%/27.5
3.6364%
5
100%/27.5
3.6364%
 Example:
6
100%/27.5
3.6364%
Placed a residential property in
service in March. Find the
depreciation allowance in year
1.
D1 = (9.5/12)(100%/27.5)
= 2.8788%
7
100%/27.5
3.6364%
8
100%/27.5
3.6364%
9
100%/27.5
3.6364%
10
(11.5/12)(100%/27.5)
3.4848%
 SL Method
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Depletion
 Unlike depreciation and amortization, which mainly
describe the deduction of expenses due to the aging of
equipment and property, depletion is the physical
reduction of natural resources.
 Two types of depletion:
 Cost depletion
 Percentage depletion
Contemporary Engineering Economics, 5th edition, © 2010
Cost Depletion
Concept: Units-of-production
method
Cost depletion formula:
Depletion allowance
Adjusted Basis


= 

 Total number of recoverable units 
 (Number of units sold)
Example 9.10: Cost basis = $120,000,
total recoverable volume = 1.5MBF,
amount sold this year = 0.5 MBF
$120,000
Depletion allowance = 0.5 MBF 
1.5 MBF
 $40,000
Contemporary Engineering Economics, 5th edition, © 2010
Percentage Depletion
Concept: Based on a prescribed
percentage of the gross income
from the property during the tax
year
Example 9.11:
Given:
Basis = $30 million,
 Total recoverable volume
= 120,000 ounces of gold,
Amount sold this year =
18,000 ounces,
 Gross income =
$16,425,000,
 Depletion expenses =
$12,250,000
Find: Maximum depletion
allowance
 Solution:
Calculation
Gross income from sale of
18,000 ounces
Depletion percentage
Computed percentage
depletion
$16,425,000
× 15%
$2,463,750
Gross income from sale of
18,000 ounces
Less mining expenses
Taxable income from mine
Deduction limitation
$16,425,000
12,250,000
4,175,000
× 50%
Maximum depletion
deduction
$2,087,500
Contemporary Engineering Economics, 5th edition, © 2010
Calculating the Allowable Depletion
Deduction for Federal Tax
$2,463,750
$2,088,000
$2,088,000
 $30, 000, 000 
Cost depletion = 
 (45, 000)
 300, 000 
 $4,500, 000
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$4,500,000
Summary 1
 Because it employs accelerated methods of
depreciation and shorter-than-actual depreciable
lives, the MACRS (Modified Accelerated Cost
Recovery System) gives taxpayers a break: It allows
them to take earlier and faster advantage of the taxdeferring benefits of depreciation.
 The total amount of taxes to pay remains unchanged
regardless of depreciation methods adopted. It only
changes the timing of the payment.
Contemporary Engineering Economics, 5th edition, © 2010
Summary 2
• Many firms select straight-line depreciation for
book depreciation because of its relative ease of
calculation.
 Given the frequently changing nature of
depreciation and tax law, we must use whatever
percentages, depreciable lives, and salvage
values mandated at the time an asset is
acquired.
Contemporary Engineering Economics, 5th edition, © 2010