1. ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS
Chapter 11 Depreciation
EGN 3615
ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS
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2. 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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3. 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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4. Basic Aspects of Depreciation
Depreciable property (by 3 requirements):
Used for the production of income
Determinable useful life > 1 year
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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5. 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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6. 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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7. 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
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8. Example of Depreciation
Cost of a new baking oven
Depreciated
Cost of new delivery van
Cost of furnishings in dining room
Utility costs for soda refrigerator
Expensed , life<1 year; lose value immediately
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9. 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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10. Accelerated Cost Recovery System
Depreciation Methods
Owner’s choice of method, recovery period and salvage values
Accelerated Cost Recovery System
(ACRS)
Development of recovery property classes; zero salvage value.
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11. Modified Accelerated Cost Recovery System (MACRS)
Depreciation Methods
Modified Accelerated Cost Recovery System (MACRS)
Present
Uses modified property classes, and the general depreciation system (GDS)
May elect the alternative depreciation system (ADS) when appropriate
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12. Straight-Line Depreciation
Uniform write-off (still used today in other countries, but not for US taxes)
Depreciation per year
Book value (unrecovered investment, EOY t)
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13. Straight-Line Depreciation
Example: B = $10,500 n = 6 years SV6 = $500
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14. Straight-Line (SL) Depreciation
End of Year t
Depreciation (Dept)
Book value (BVt ) 1 2 3 4 5 6
500.00
Book valuet = Book valuet Depreciationt, t = 1, 2, …, n and BV0 = B
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15. 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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16. 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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17. Initial cost: $100
Salvage value: $10
Recovery period: 5 years
SL depreciation %age: 1/5 (20%)
DDB Example
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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18. MACRS Depreciation Modified Accelerated Cost Recovery System (MACRS)
General Depreciation System (GDS)
Alternative Depreciation System (ADS) –rarely used
Determine if a property is eligible for depreciation
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, …
Determine placed-in-service date
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19. MACRS Depreciation Use property class given in problem
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
Use Table 11-3 MACRS Depreciation…half-year convention
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20. MACRS Setup Initial cost: $100 Salvage value: $10
Recovery period: 5 years (we have been told this)
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21. MACRS CALC. BV (*1000) MACRS 100.00 1 ½(2/5)(100-0) 20.00 80.00 2
Initial cost: $100
Salvage value: $10
Recovery period: 5 years
BV (*1000)
MACRS
100.00 1
½(2/5)(100-0)
20.00
80.00 2
(2/5)( )
32.00
48.00 3
(2/5)( )
19.20
28.80 4
(2/5)( ) = [SL] 28.8/2.5
11.52
17.28 5
[SL] = [17.28/1.5]
5.76 6
[SL] ½ (11.52)= ½[17.28/1.5]
Why the “1/2”?
Why the “1/2”?
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22. Example 11-6
Office equipment Purchase price: $150,000 Salvage value: $30,000 (at end of depreciable life) Find yearly depreciations and book values
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23. Example 11-6
Solution 1. The assets qualify as depreciable property 2. The cost base B = $150, 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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24. Example 11-6
yr t rt dt ∑dt BVt 0 $150, $21435 $21435 $128, , , , , , , sum $150000
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25. 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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26. 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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27. 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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28. 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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29. 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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30. Usage Depreciation Yr Usage (hrs) Depreciation Schedule
Dept = (P – SV) Qt / Q
Yr Usage (hrs) Depreciation Schedule
5, ($50,000 – 0) x (5,000 / ,000) = $20,833
5, (50,000 – 0) x (5,000 / ,000) = $20,833
2, (50,000 – 0) x (2,000 / ,000) = $8,334
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31. 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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32. Depletion Dep Example Yr Usage (Mbbls) Depreciation Schedule
Init. Capacity: 12 Mbbls
Init. Value: $960 M
Depletion Dep Example
Dept = (P – SV) Qt / Q
Yr Usage (Mbbls) Depreciation Schedule
5, ($960 M – 0) x (5,000 / ,000) = $400 M
5, ($960 M – 0) x (5,000 / ,000) = $400 M
2, ($960 M – 0) x (2,000 / ,000) = $160 M
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33. 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) Straight-line depreciation (b) Sum-of-years’-digits depreciation (c) MACRS depreciation (d) Double declining balance depreciation
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34. Problem Solution
Assume equipment is in the 5-yr MACRS property class yr SL
SOYD
MACRS
DDB
DDB w SL 1
18000
32727
$40,000 2
29455
$64,000
$32,000 3
26182
$38,400
$25,600 4
22909
$23,040
$20,480 5
19636
$16,384 6
16364
$11,520
$13,107 7
13091
$10,486 8
9818
$8,389 9
6545
$6,711
$6,777 10
3273
$5,369
sum
180000
200000
$178,525
$180,000
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35. Problem 11-30 - Solution Present worth (PW) at interest rate of 5%
year SL
SOYD
MACRS
DDB
DDB w SL 1
$17,143
$31,169
$38,095 2
$16,327
$26,716
$58,050
$29,025 3
$15,549
$22,617
$33,171
$22,114 4
$14,809
$18,847
$18,955
$16,849 5
$14,103
$15,386
$18,052
$12,837 6
$13,432
$12,211
$8,596
$9,781 7
$12,792
$9,303 $0
$7,452 8
$12,183
$6,645
$5,678 9
$11,603
$4,219
$4,326
$4,369 10
$11,050
$2,009
$3,296
$4,160 PW
$138,991
$149,123
$174,920
$149,453
$150,361
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36. End of Chapter 11
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