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Lecture no31 depreciation methods

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Depreciation Methods
Lecture No. 31
Chapter 9
Contemporary Engineering Economics
Copyright © 2016

Contemporary Engineering Economics, 6 th edition
Park

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Book Depreciation Methods
• Purpose: Used in reporting net income to
stockholders/investors

• Types of Depreciation Methods
o Straight-line method
o Declining balance method
o Unit production method

Contemporary Engineering Economics, 6 th edition
Park

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Straight-Line (SL) Method
 Principle:


A fixed asset as providing its
service in a uniform fashion over
its life
 Formula
o Annual depreciation
Dn = (I − S) / N, and
for all n.
o 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

constant

Contemporary Engineering Economics, 6 th edition
Park

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Declining Balance (DB) Method

 Principle:
A fixed asset as
providing its service
in a decreasing
fashion
 Formula

 Example:
o
o
o
o

I = $10,000
S = $778
N = 5 years
α = 0.40

D n   Bn  1   (1   ) n  1

Bn  I(1   )n
where 0 < α < 2(1/N)

Contemporary Engineering Economics, 6 th edition
Park

Copyright © 2016 by Pearson Education, Inc.
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Example 9.5: DB Switching to SL
 Given:
o Depreciation Base =$10,000
o Salvage Value = 0
o Depreciation rate = 200%
DB
o Depreciable life = 5 years

 Without Switching

 Find:

When switching to SL

• SL Dep. Rate = 1/5
• α = (200%) (1/5) =
0.40

Contemporary Engineering Economics, 6 th edition
Park

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.

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Case 1: S < BN
• Switch from DB to SL
after n’

Contemporary Engineering Economics, 6 th edition
Park

• Example: S = 0

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Case 2: S > BN
• No further depreciation
after n”

• Example: S = $2,000

Note: Tax law does not permit a business
to depreciate assets below their salvage
value.
Contemporary Engineering Economics, 6 th edition
Park

Copyright © 2016 by Pearson Education, Inc.
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Units-of-Production Method

• Given: I = $55,000, S = $5,000, total

 Principle:
Service units will be
consumed in a non-timephased fashion.

service unit = 250,000 miles, service
units consumed = 30,000 miles

• Find: Dn
• Solution:

 Formula:
Service units consumed


during year n
Dn  (I _ S ) �
� Total service units










I = Initial investment

S = Salvage value
Contemporary Engineering Economics, 6 th edition
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Tax Depreciation
 Purpose: To determine the income taxes owed for the
IRS
• Assets placed in service prior to 1981
o Used the book depreciation methods (SL, DB, SOYD)

• Assets placed in service from 1981 to 1986
o Used the ACRS (Accelerated Cost Recovery System) Table

• Assets placed in service from 1986 to present
o Used the MACRS (Modified ACRS) Table

Contemporary Engineering Economics, 6 th edition
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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, 6 th edition
Park

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MACRS Depreciation Schedules for Personal
Property with Half-Year Convention

Contemporary Engineering Economics, 6 th edition
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Comparison Between DDB with Switching
to SL and MACRS Method
 Conventional DDB
Method
o
o
o
o

Cost basis: $10,000
Salvage value: $0
Depreciable life: 5 years
DB rate: 200%

 MACRS Method
o Property class: 5-year
o Salvage value: $0
o Half-year convention

Contemporary Engineering Economics, 6 th edition
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MACRS for Real Property

 Types
o 27.5-year (residential)
o 39-year (commercial)
 SL Method
 Mid-month convention
 Zero salvage value
 Example:

D1 = (9.5/12)(100%/27.5)
= 2.8788%

Placed a residential property in
service in March. Find the
depreciation allowance in year
1.
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Depletion
• Depletion is the physical reduction of natural
resources (not time-phased).
• Two types of depletion
o Cost depletion
o Percentage depletion

Contemporary Engineering Economics, 6 th edition

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Cost Depletion
 Concept: Units-ofproduction method
 Cost depletion formula

Example:
o Cost basis = $120,000,
o Total recoverable volume = 1.5MBF
o Amount sold this year = 0.5 MBF
o Allowed depletion this year?

Depletion allowance = 0.5 MBF 

$120,000
1.5 MBF

$40, 000

Contemporary Engineering Economics, 6 th edition
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Percentage Depletion
 Concept: Based on a prescribed
percentage of the gross income
from the property during the tax
year
 Example 9.11
• Given:

• Solution

 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

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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

Contemporary Engineering Economics, 6 th edition
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$4,500,000

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Summary
• 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, 6 th edition
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Copyright © 2016 by Pearson Education, Inc.
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Summary
• 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, 6 th edition
Park

Copyright © 2016 by Pearson Education, Inc.
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