Chapter 11 Depreciation
EGN 3615
ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL
IMPLICATIONS
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Chapter Outline
 Basic Aspects of Depreciation
 Straight-Line Depreciation
 Declining Balancing
 Modified Accelerated Cost Recovery
System (MACRS)
 Unit of Production
 Unit of Operating Time
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Basic Aspects of Depreciation
 Depreciation is the reduction in the value of an asset
due to usage, passage of time, wear and tear,
technological outdating or obsolescence, depletion, rot,
rust, decay or other such factors.
 Business costs are generally either expensed or
depreciated.
 Used for planning purposes and taxation
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Basic Aspects of Depreciation
Depreciable property (by 3 requirements):
1. Used for the production of income
2. Determinable useful life > 1 year
3. Something that wears out, decays, gets used up, or
loses value from natural causes.
Land is NOT depreciable property. In fact,
land increases in value over time, in general.
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Amortization of Intangible Assets
 Intangible assets: nonphysical and long lived; useful
life is greater than one year.
 Copyrights- legal rights to written or other creative
works
 Trademarks- legal rights to names and logos.
 Patents- legal rights to inventions, designs, and
processes.
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Amortization of Intangible Assets
 Goodwill—economic value of the reputation and
profitability of a business.
 Franchise—contractual …
 Leasehold improvements—made by the tenant
Rental property can not be depreciated by tenant.
Only property owner can claim depreciation of a property.
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Examples of Depreciation
Example: Consider the costs that are incurred by
a local pizza business. Identify each cost as either
expensed or depreciated and describe why that
term applies.
Cost for pizza dough and toppings
Expensed, life<1 year; lose value immediately
Cost to pay wages for janitor
Expensed, life<1 year; lose value immediately
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Example of Depreciation
Cost of a new baking oven
Depreciated
Cost of new delivery van
Depreciated
Cost of furnishings in dining room
Depreciated
Utility costs for soda refrigerator
Expensed , life<1 year; lose value immediately
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Depreciation Methods
Prior to 1981
 Three basic methods to choose from. Much
flexibility.
•
Straight-line – Uniform write-off (simplest)
•
Sum-of-years Digits (SOYD)
•
Declining Balance & Double Declining
Balance (DDB)
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Depreciation Methods
 Owner’s choice of method, recovery period
and salvage values
Accelerated Cost Recovery System
1981 -- 1986
(ACRS)
 Development of recovery property classes;
zero salvage value.
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Depreciation Methods
Modified Accelerated Cost Recovery System
(MACRS)
1987 -- Present
 Uses modified property classes, and the
general depreciation system (GDS)
 May elect the alternative depreciation system
(ADS) when appropriate
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Straight-Line Depreciation
Straight – Line
Uniform write-off (still used today in other
countries, but not for US taxes)
Depreciation per year
B  SV
Dt 
n
Book value (unrecovered investment, EOY t)
(B  SV)
BVt  B 
t
n
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Straight-Line Depreciation
Example:
B = $10,500
B  SV
Dt 
n
n = 6 years SV6 = $500
10,500 - 500
Dep/yr 
 $1666.67
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Straight-Line (SL) Depreciation
End of Year t
Depreciation (Dept)
0
Book value (BVt )
10500.00
1
1666.67
8833.33
2
1666.67
7166.67
3
1666.67
5500.00
4
1666.67
3833.33
5
1666.67
2166.67
6
1666.67
500.00
Book valuet = Book valuet-1 - Depreciationt, t = 1, 2, …, n and BV0 = B
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Declining Balance Depreciation
Declining Balance
Accelerated write – off
Depreciate a fixed %-age (f) of remaining
book value each year
Dt = f*BVt-1
=> Dt = f*B*(1 – f)t-1
=> BVt = B*(1 – f)t
Typically f is a multiple of the
straight-line (SL) percent. Most
commonly, the multiple is 1.5 or 2
(times the SL depreciation value).
2 times is called Double Declining
Balance (DDB).
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DDB Setup
 Initial cost:
$100
 Salvage value:
$10
 Recovery period: 5 years
 The SL rate would be 1/5 or 20% of the original basis.
 The DDB would thus be 2 * SL: 2/5 or 40%.
However this percent is applied to the current book
value, while the SL is applied to the original value.
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 Initial cost:
DDB Example
$100
 Salvage value:
$10
 Recovery period: 5 years
 SL depreciation %age: 1/5 (20%)
BV (*1000)
DDB
100.00
1
(2/5)(100)
40.00
60.00
2
(2/5)(60)
24.00
36.00
3
(2/5)(36)
14.40
28.80
4
(2/5)(21.6)
8.64
12.96
5
(2/5)(12.96)
5.184
7.78
Note that the final value does not match the salvage value.
In fact, it never goes to zero.
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MACRS Depreciation
Modified Accelerated Cost Recovery System
(MACRS)
 General Depreciation System (GDS)
 Alternative Depreciation System (ADS) –rarely used
1. Determine if a property is eligible for depreciation
2. Determine the asset’s cost basis (B)
 Cost to obtain and place the asset in service fit for use
 For real property, the basis may include certain fees and
charges, such as legal and recording fees, abstract fees,
survey charges, transfer taxes, title insurance, …
3. Determine placed-in-service date
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MACRS Depreciation
4. Determine the property class and recovery period
• Use property class given in problem
• Match asset name with MACRS-GDS
property classes definition (Table 11-2, p. 385)
• Use IRS publication, such as Table 11-1
• Use ADR class life to determine property class
5. Use Table 11-3 MACRS Depreciation…half-year
convention
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MACRS Setup
 Initial cost:
$100
 Salvage value:
$10
 Recovery period: 5 years (we have been told this)
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MACRS CALC.
 Initial cost:
$100
 Salvage value:
$10
 Recovery period: 5 years
BV (*1000)
Why the “1/2”?
MACRS
100.00
1
½(2/5)(100-0)
20.00
80.00
2
(2/5)(100- 20.00)
32.00
48.00
3
(2/5)(100- 52)
19.20
28.80
4
(2/5)(100- 71.20) = [SL] 28.8/2.5
Why the “1/2”?
11.52
17.28
5
[SL] 11.52 = [17.28/1.5]
11.52
5.76
6
[SL] ½ (11.52)= ½[17.28/1.5]
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5.76
0
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Example 11-6
Office equipment
Purchase price:
Salvage value:
$150,000
$30,000 (at end of depreciable life)
Find yearly depreciations and book values
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Example 11-6
Solution
1. The assets qualify as depreciable property
2. The cost base B = $150,000
3. Property is placed in use in yr 1 of our analysis
4. MACRS GDS applies (Tables 11-1 & 11-2) with
a 7-year depreciable life
5. Salvage value is not used with MACRS, and
dt = B*rt, t = 1, 2, … , 8,
(11-5)
where rt = MACRS depreciation rate in year t,
given in Tables 11-3 and 11-4.
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Example 11-6
yr t
rt
dt
0
1 .1429 $21435
2 .2449
36735
3 .1749
26235
4 .1249
18375
5 .0893
13395
6 .0892
13380
7 .0893
13395
8 .0446
6690
sum 1.0000 $150000
∑dt
$21435
58170
84405
103140
116535
129915
143310
150000
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BVt
$150,000
$128,565
91,830
65,595
46,860
33,465
29,085
6,690
0
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Use of EXCEL
Straight line depreciation
sln(B, S, n) - returns the constant annual depreciation
Double declining balance depreciation
ddb(B, S, n, t, factor) – returns yr-t depreciation
Sum-of-years’-digits depreciation
syd(B, S, n, t) – returns year-t depreciation
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Use of EXCEL
MACRS depreciation
vdb(B, S, n, start t1, end t2, factor, no-switch) - returns
the MACRS depreciation from t1 to t2.
Remark
1. S must be zero.
2. n = 3, 5, 7, 10, 15 or 20 years.
3. Yr 1 is from t1 = 0 to t2 = 0.5, yr 2: t1 = 0.5 to t2 = 1.5,
……, yr n: t1 = n-0.5 to t2 = n.
4. Factor = 2 for n = 3, 5, 7, 10; and 1.5 for n = 15 & 20.
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Other Methods of Depreciation
These methods are used for depreciating equipment
used in exploring natural resources, such as mines,
wells, etc…
1. Units of Production
Dept = (B - SV) Ut / U
where: Ut= # units produced in year t
U = total units produced during useful life.
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Other Methods of Depreciation
2. Units of Operating Time (usage depreciation)
Dept = (B - SV) Qt / Q
where: Qt = # hours (days) used in year t
Q = total # hours (days) used during
useful life
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Other Methods of Depreciation
 Example:
A welding machine costs $50,000 and has a
useful life of 12,000 hours and a zero salvage
value at that time. Based upon estimated usage,
determine the depreciation for each year.
Dept = (B – SV) Qt / Q
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Usage Depreciation
Yr
Usage (hrs)
1
5,000
2
5,000
3
2,000
Dept = (P – SV) Qt / Q
Depreciation Schedule
($50,000 – 0) x (5,000 /
12,000) = $20,833
(50,000 – 0) x (5,000 /
12,000) = $20,833
(50,000 – 0) x (2,000 /
12,000) = $8,334
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Other Methods of Depreciation
3. Units of Depletion
Dept = (B - SV) Ut / U
where: Ut= quantity produced in year t
U = total quantity which is expected to be
produced during useful life.
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Init. Capacity: 12 Mbbls
Init. Value: $960 M
Dept = (P – SV) Qt / Q
Depletion Dep Example
Yr
Usage (Mbbls) Depreciation Schedule
1
5,000
2
5,000
3
2,000
($960 M – 0) x (5,000 /
12,000) = $400 M
($960 M – 0) x (5,000 /
12,000) = $400 M
($960 M – 0) x (2,000 /
12,000) = $160 M
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Problem 11-30
Dept = (P – SV) Qt / Q
B = $200,000
S = $20,000
n = 10 years
At r = 5%, which depreciation is preferred?
(a)
(b)
(c)
(d)
Straight-line depreciation
Sum-of-years’-digits depreciation
MACRS depreciation
Double declining balance depreciation
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Problem 11-30 - Solution
Assume equipment is in the 5-yr MACRS property class
yr
1
2
3
4
5
6
7
8
9
10
sum
SL
18000
18000
18000
18000
18000
18000
18000
18000
18000
18000
180000
SOYD
32727
29455
26182
22909
19636
16364
13091
9818
6545
3273
180000
MACRS
$40,000
$64,000
$38,400
$23,040
$23,040
$11,520
0
0
0
0
200000
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DDB
$40,000
$32,000
$25,600
$20,480
$16,384
$13,107
$10,486
$8,389
$6,711
$5,369
$178,525
DDB w SL
$40,000
$32,000
$25,600
$20,480
$16,384
$13,107
$10,486
$8,389
$6,777
$6,777
$180,000
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Problem 11-30 - Solution
Present worth (PW) at interest rate of 5%
year
1
2
3
4
5
6
7
8
9
10
PW
SL
$17,143
$16,327
$15,549
$14,809
$14,103
$13,432
$12,792
$12,183
$11,603
$11,050
$138,991
SOYD
$31,169
$26,716
$22,617
$18,847
$15,386
$12,211
$9,303
$6,645
$4,219
$2,009
$149,123
MACRS
$38,095
$58,050
$33,171
$18,955
$18,052
$8,596
$0
$0
$0
$0
$174,920
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DDB
$38,095
$29,025
$22,114
$16,849
$12,837
$9,781
$7,452
$5,678
$4,326
$3,296
$149,453
DDB w SL
$38,095
$29,025
$22,114
$16,849
$12,837
$9,781
$7,452
$5,678
$4,369
$4,160
$150,361
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End of Chapter 11
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