EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1 Engineering Economics and Chapter Outline Basic Aspects of Depreciation StraightLine Depreciation Declining Balancing ID: 535573
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Chapter 11 DepreciationEGN 3615ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS
1
Engineering Economics and...Slide2
Chapter OutlineBasic Aspects of DepreciationStraight-Line DepreciationDeclining BalancingModified Accelerated Cost Recovery System (MACRS) Unit of ProductionUnit of Operating Time
2
Engineering Economics and...Slide3
Basic Aspects of DepreciationDepreciation 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 taxation3
Engineering Economics and...Slide4
Basic Aspects of DepreciationDepreciable property (by 3 requirements):Used for the production of income Determinable useful life > 1 yearSomething 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.4
Engineering Economics and...Slide5
Amortization of Intangible AssetsIntangible assets: nonphysical and long lived; useful life is greater than one year.Copyrights- legal rights to written or other creative worksTrademarks- legal rights to names and logos. Patents- legal rights to inventions, designs, and processes.5Engineering Economics and...Slide6
Amortization of Intangible AssetsGoodwill—economic value of the reputation and profitability of a business.Franchise—contractual …Leasehold improvements—made by the tenantRental property can not be depreciated by tenant.Only property owner can claim depreciation of a property.6Engineering Economics and...Slide7
Examples of DepreciationExample: 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 toppingsExpensed, life<1 year; lose value immediatelyCost to pay wages for janitorExpensed, life<1 year; lose value immediately
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Engineering Economics and...Slide8
Example of DepreciationCost of a new baking ovenDepreciatedCost of new delivery vanDepreciatedCost of furnishings in dining roomDepreciated
Utility costs for soda refrigeratorExpensed , life<1 year; lose value immediately
8
Engineering Economics and...Slide9
Depreciation MethodsPrior to 1981Three basic methods to choose from. Much flexibility.Straight-line – Uniform write-off (simplest)Sum-of-years Digits (SOYD)Declining Balance & Double Declining Balance (DDB)
9
Engineering Economics and...Slide10
Depreciation MethodsOwner’s choice of method, recovery period and salvage valuesAccelerated Cost Recovery System 1981 -- 1986 (ACRS)Development of recovery property classes; zero salvage value.
10
Engineering Economics and...Slide11
Depreciation MethodsModified Accelerated Cost Recovery System (MACRS)1987 -- PresentUses modified property classes, and the general depreciation system (GDS)May elect the alternative depreciation system (
ADS) when appropriate
11
Engineering Economics and...Slide12
Straight-Line DepreciationStraight – Line Uniform write-off (still used today in other countries, but not for US taxes)Depreciation per year
Book value (unrecovered investment, EOY t)
12
Engineering Economics and...Slide13
Straight-Line DepreciationExample:B = $10,500 n = 6 years SV6 = $500
13
Engineering Economics and...Slide14
Straight-Line (SL) DepreciationEnd of Year t Depreciation (Dept)Book value (BVt )
0
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
value
t
=
Book value
t-1
-
Depreciation
t
, t = 1, 2, …, n and BV
0
= B
14
Engineering Economics and...Slide15
Declining Balance DepreciationDeclining Balance Accelerated write – offDepreciate a fixed %-age (f) of remaining book value each yearDt = f*BV
t-1
=>
D
t
= f*B*(1 – f)
t-1
=>
BV
t
=
B*
(1 – f)
t
15
Engineering Economics and...
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).Slide16
DDB SetupInitial cost: $100Salvage value: $10Recovery period: 5 yearsThe 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.16Engineering Economics and...Slide17
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
.
17
Engineering Economics and...
Initial cost: $100
Salvage value: $10
Recovery period: 5 years
SL depreciation %age: 1/5 (20%)Slide18
MACRS DepreciationModified Accelerated Cost Recovery System (MACRS)General Depreciation System (GDS)Alternative Depreciation System (ADS) –rarely usedDetermine 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
18
Engineering Economics and...Slide19
MACRS DepreciationDetermine the property class and recovery periodUse property class given in problemMatch 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
19
Engineering Economics and...Slide20
MACRS SetupInitial cost: $100Salvage value: $10Recovery period: 5 years (we have been told this)20Engineering Economics and...Slide21
MACRS CALC.
BV (*1000)
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
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]
5.76
0
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Engineering Economics and...
Initial cost: $100
Salvage value: $10
Recovery period: 5 years
Why the “1/2”?
Why the “1/2”?Slide22
Example 11-6Office equipmentPurchase price: $150,000Salvage value: $30,000 (at end of depreciable life)
Find yearly depreciations and book values
22
Engineering Economics and...Slide23
Example 11-6Solution1. The assets qualify as depreciable property2. 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
d
t
= B*
r
t
, t = 1, 2, … , 8, (11-5)
where
r
t
= MACRS depreciation rate in year t,
given in Tables 11-3 and 11-4.
23
Engineering Economics and...Slide24
Example 11-6 yr t rt dt ∑dt
BVt
0 $150,000
1 .1429 $21435 $21435 $128,565
2 .2449 36735 58170 91,830
3 .1749 26235 84405 65,595
4 .1249 18375 103140 46,860
5 .0893 13395 116535 33,465
6 .0892 13380 129915 29,085
7 .0893 13395 143310 6,690
8 .0446 6690 150000
0
sum
1.0000 $150000
24
Engineering Economics and...Slide25
Use of EXCELStraight line depreciationsln(B, S, n) - returns the constant annual depreciationDouble 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
25
Engineering Economics and...Slide26
Use of EXCELMACRS depreciationvdb(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.
26
Engineering Economics and...Slide27
Other Methods of DepreciationThese methods are used for depreciating equipment used in exploring natural resources, such as mines, wells, etc…1. Units of Production
Dep
t
= (B - SV)
U
t
/ U
w
here:
U
t
= # units produced in year
t
U
= total units produced during useful life.
27
Engineering Economics and...Slide28
Other Methods of Depreciation2. Units of Operating Time (usage depreciation)
Dep
t
= (B - SV) Q
t
/ Q
where: Q
t
= # hours (days) used in year
t
Q = total # hours (days) used during useful life
28
Engineering Economics and...Slide29
Other Methods of DepreciationExample: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
29
Engineering Economics and...Slide30
Usage DepreciationYr Usage (hrs) Depreciation Schedule
5,000 ($50,000 – 0) x (5,000 / 12,000) =
$20,833
5,000 (50,000 – 0) x (5,000 / 12,000
) =
$20,833
2,000 (50,000 – 0) x (2,000 / 12,000
) =
$8,334
Dep
t
=
(P – SV) Q
t
/ Q
30
Engineering Economics and...Slide31
Other Methods of Depreciation3. Units of Depletion Dept = (B - SV)
Ut
/ U
w
here:
U
t
= quantity produced in year
t
U = total quantity which
is expected to be
produced during useful life.
31
Engineering Economics and...Slide32
Depletion Dep ExampleYr Usage (Mbbls) Depreciation Schedule
5,000 ($960 M – 0) x (5,000 / 12,000) =
$400 M
5,000
($960 M – 0)
x (5,000 / 12,000) =
$400 M
2,000
($960 M – 0)
x (2,000 / 12,000) =
$160 M
Dep
t
=
(P – SV) Q
t
/ Q
32
Engineering Economics and...
Init.
Capacity: 12
Mbbls
Init.
Value: $960 MSlide33
Problem 11-30B = $200,000 S = $20,000 n = 10 yearsAt r = 5%, which depreciation is preferred?(a) Straight-line depreciation(b) Sum-of-years’-digits depreciation
(c) MACRS depreciation
(d) Double declining balance depreciation
Dep
t
=
(P – SV) Q
t
/ Q
33
Engineering Economics and...Slide34
Problem 11-30 - SolutionAssume equipment is in the 5-yr MACRS property class34Engineering Economics and...yr
SL
SOYD
MACRS
DDB
DDB w SL
1
18000
32727
$40,000
$40,000
$40,000
2
18000
29455
$64,000
$32,000
$32,000
3
18000
26182
$38,400
$25,600
$25,600
4
18000
22909
$23,040
$20,480
$20,480
5
18000
19636
$23,040
$16,384
$16,384
6
18000
16364
$11,520
$13,107
$13,107
7
18000
13091
0
$10,486
$10,486
8
18000
9818
0
$8,389
$8,389
9
18000
6545
0
$6,711
$6,777
10
18000
3273
0
$5,369
$6,777
sum
180000
180000
200000
$178,525
$180,000 Slide35
Problem 11-30 - SolutionPresent worth (PW) at interest rate of 5%35Engineering Economics and...year
SL
SOYD
MACRS
DDB
DDB w SL
1
$17,143
$31,169
$38,095
$38,095
$38,095
2
$16,327
$26,716
$58,050
$29,025
$29,025
3
$15,549
$22,617
$33,171
$22,114
$22,114
4
$14,809
$18,847
$18,955
$16,849
$16,849
5
$14,103
$15,386
$18,052
$12,837
$12,837
6
$13,432
$12,211
$8,596
$9,781
$9,781
7
$12,792
$9,303
$0
$7,452
$7,452
8
$12,183
$6,645
$0
$5,678
$5,678
9
$11,603
$4,219
$0
$4,326
$4,369
10
$11,050
$2,009
$0
$3,296
$4,160
PW
$138,991
$149,123
$174,920
$149,453
$150,361Slide36
End of Chapter 11 Engineering Economics and...36