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Chapter 11 Depreciation EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1Engineering Economics and...

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Presentation on theme: "Chapter 11 Depreciation EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1Engineering Economics and..."— Presentation transcript:

1 Chapter 11 Depreciation EGN 3615 ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS 1Engineering Economics and...

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 2Engineering Economics and...

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 3Engineering Economics and...

4 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. 4Engineering Economics and...

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. 5Engineering Economics and...

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. 6Engineering Economics and...

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 Expensed, life<1 year; lose value immediately 7Engineering Economics and...

8 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 8Engineering Economics and...

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) 9Engineering Economics and...

10 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. 10Engineering Economics and...

11 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 11Engineering Economics and...

12 Straight-Line Depreciation Straight – Line Uniform write-off (still used today in other countries, but not for US taxes) Depreciation per year Book value (unrecovered investment, EOY t) 12Engineering Economics and...

13 Straight-Line Depreciation Example: B = $10,500n = 6 years SV 6 = $500 13Engineering Economics and...

14 Straight-Line (SL) Depreciation End of Year tDepreciation (Dep t )Book value (BV t ) Book value t = Book value t-1 - Depreciation t, t = 1, 2, …, n and BV 0 = B 14Engineering Economics and...

15 Declining Balance Depreciation Declining Balance Accelerated write – off Depreciate a fixed %-age (f) of remaining book value each year D t = f*BV t-1 => D t = f*B*(1 – f) t-1 => BV t = B*(1 – f) t 15Engineering 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).

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. 16Engineering Economics and...

17 DDB Example BV (*1000) DDB (2/5)(100) (2/5)(60) (2/5)(36) (2/5)(21.6) (2/5)(12.96) Note that the final value does not match the salvage value. In fact, it never goes to zero. 17Engineering Economics and... Initial cost: $100 Salvage value:$10 Recovery period: 5 years SL depreciation %age: 1/5 (20%)

18 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 18Engineering Economics and...

19 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 19Engineering Economics and...

20 MACRS Setup Initial cost: $100 Salvage value:$10 Recovery period: 5 years (we have been told this) 20Engineering Economics and...

21 MACRS CALC. BV (*1000) MACRS ½(2/5)(100-0) (2/5)( ) (2/5)( ) (2/5)( ) = [SL] 28.8/ [SL] = [17.28/1.5] [SL] ½ (11.52)= ½[17.28/1.5] Engineering Economics and... Initial cost: $100 Salvage value:$10 Recovery period: 5 years Why the “1/2”?

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 22Engineering Economics and...

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 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 Engineering Economics and...

24 Example 11-6 yr t r t d t ∑d t BV t 0$150, $21435 $21435$128, , , , , , , sum $ Engineering Economics and...

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 25Engineering Economics and...

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 & Engineering Economics and...

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 Dep t = (B - SV) U t / U where: U t = # units produced in year t U = total units produced during useful life. 27Engineering Economics and...

28 Other Methods of Depreciation 2. 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 28Engineering Economics and...

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. Dep t = (B – SV) Q t / Q 29Engineering Economics and...

30 Usage Depreciation Yr Usage (hrs) Depreciation Schedule 1 5,000 ($50,000 – 0) x (5,000 / 12,000) = $20, ,000 (50,000 – 0) x (5,000 / 12,000) = $20, ,000 (50,000 – 0) x (2,000 / 12,000) = $8,334 Dep t = (P – SV) Q t / Q 30Engineering Economics and...

31 Other Methods of Depreciation 3. Units of Depletion Dep t = (B - SV) U t / U where: U t = quantity produced in year t U = total quantity which is expected to be produced during useful life. 31Engineering Economics and...

32 Depletion Dep Example Yr Usage (Mbbls) Depreciation Schedule 1 5,000 ($960 M – 0) x (5,000 / 12,000) = $400 M 2 5,000 ($960 M – 0) x (5,000 / 12,000) = $400 M 3 2,000 ($960 M – 0) x (2,000 / 12,000) = $160 M Dep t = (P – SV) Q t / Q 32Engineering Economics and... Init. Capacity: 12 Mbbls Init. Value: $960 M

33 Problem B = $200,000S = $20,000n = 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 Dep t = (P – SV) Q t / Q 33Engineering Economics and...

34 Problem Solution Assume equipment is in the 5-yr MACRS property class 34Engineering Economics and... yrSLSOYDMACRSDDBDDB w SL $40, $64,000$32, $38,400$25, $23,040$20, $23,040$16, $11,520$13, $10, $8, $6,711$6, $5,369$6,777 sum $178,525$180,000

35 Problem Solution Present worth (PW) at interest rate of 5% 35Engineering Economics and... yearSLSOYDMACRSDDBDDB 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$0$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$150,361

36 End of Chapter 11 Engineering Economics and...36


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