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

Chapter Outline Basic Aspects of Depreciation Straight-Line Depreciation Declining Balancing Modified Accelerated Cost Recovery System (MACRS) Unit of Production Unit of Operating Time Engineering Economics and...

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 Engineering Economics and...

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. Engineering Economics and...

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. Engineering Economics and...

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. Engineering Economics and...

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 Engineering Economics and...

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 Engineering Economics and...

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) Engineering Economics and...

Accelerated Cost Recovery System 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. Engineering Economics and...

Modified Accelerated Cost Recovery System (MACRS) 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 Engineering Economics and...

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) Engineering Economics and...

Straight-Line Depreciation Example: B = $10,500 n = 6 years SV6 = $500 Engineering Economics and...

Straight-Line (SL) Depreciation End of Year t Depreciation (Dept) Book value (BVt ) 10500.00 1 1666.67 8833.33 2 7166.67 3 5500.00 4 3833.33 5 2166.67 6 500.00 Book valuet = Book valuet-1 - Depreciationt, t = 1, 2, …, n and BV0 = B Engineering Economics and...

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). Engineering Economics and...

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. Engineering Economics and...

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. Engineering Economics and...

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 Engineering Economics and...

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 Engineering Economics and...

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

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)(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] 5.76 6 [SL] ½ (11.52)= ½[17.28/1.5] Why the “1/2”? Why the “1/2”? Engineering Economics and...

Example 11-6 Office equipment Purchase price: $150,000 Salvage value: $30,000 (at end of depreciable life) Find yearly depreciations and book values Engineering Economics and...

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. Engineering Economics and...

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 Engineering Economics and...

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 Engineering Economics and...

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. Engineering Economics and...

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. Engineering Economics and...

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 Engineering Economics and...

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 Engineering Economics and...

Usage Depreciation Yr Usage (hrs) Depreciation Schedule Dept = (P – SV) Qt / Q Yr 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 Engineering Economics and...

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. Engineering Economics and...

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 5,000 ($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 Engineering Economics and...

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 Engineering Economics and...

Problem 11-30 - 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 Engineering Economics and...

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 Engineering Economics and...

End of Chapter 11 Engineering Economics and...