CHAPTER 11. DEPRECIATION & Depletion DEPRECIATION. Property is Depreciable if it must:
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1 DEPRECIATION & Depletion CHAPTER 11 By: Magdy Akladios, PhD, PE, CSP, CPE, CSHM DEPRECIATION Decrease in value of physical properties with passage of time and use Accounting concept establishing annual deduction against beforetax income to reflect effect of time and use on asset s value in firm s financial statements to match yearly fraction of value used by asset in production of income over asset s economic life Property is Depreciable if it must: be used in business or held to produce income have a determinable useful life which is longer than one year wear out, decay, get used up, become obsolete, or lose value from natural causes not be inventory, stock in trade, or investment property (WIP, etc.).
2 DEPRECIABLE PROPERTY TANGIBLE - can be seen/touched Personal property - includes assets such as machinery, vehicles, equipment, furniture, etc... Real property - anything erected on, growing on, or attached to land (Note that: since land does not have a determinable life itself, it is not depreciable) INTANGIBLE - can NOT be seen/touched personal property, such as copyright, patent, or franchise When does depreciation start? Depreciation STARTS when property is placed in service for use in business or for production of income (ie, year 1) Property is considered in service when ready and available for specific use, even if not actually used yet When does depreciation stop? Depreciation STOPS when cost of placing it in service is removed or it is retired from service
3 DEPRECIATION METHODS 1. <1981: a) Straight-Line (SL) b) Declining Balance (DB) c) Sum-of-the-years-digits (SYD) 2. > 1981, but < 1987 a) Accelerated Cost Recovery System (ACRS) Implemented by Economic Recovery Tax Act (ERTA) of >1986: a) Modified Accelerated Cost Recovery System (MACRS) Brought about by (TRA 86) Tax Reform Act of 1986 Depreciation Definitions Adjusted cost basis allowable adjustment (increase or decrease) to original cost basis, used to calculate depreciation and depletion deductions (eg: cost of improvements to an asset) Basis, or cost basis also called unadjusted cost -- initial cost of acquiring an asset, plus sales tax, transportation, and normal costs of making asset serviceable Depreciation Definitions Book Value (BV) Worth of depreciable property as shown on accounting records Original cost basis of property, including adjustments, less allowable depletion or depreciation deductions Represents amount of capital remaining invested in property and must be recovered in future through accounting Book Value k = Adjusted Cost Basis - k j=1 Depreciation Deduction j
4 Depreciation Definitions Market Value (MV) Amount paid by willing buyer to willing seller for property where no advantage and no compulsion to transact Approximates present value of what will be received through ownership of property, including time-value of money (or profit) Depreciation Definitions Recovery Period Number of years over which basis of property is recovered through accounting processes. Normally the useful life for classical methods Property class for General Depreciation System (GDS) under MACRS Class Life for Alternative Depreciation System (ADS) Recovery Rate Percentage for each year of MACRS recovery period used to calculate an annual depreciation deduction. Depreciation Definitions Salvage Value (SV) Estimated value of property at the end of useful life. expected selling price of property when asset can no longer be used productively net salvage value used when expenses incurred in disposing of property; cash outflows must be deducted from cash inflows for final net salvage value with classical methods of depreciation, estimated salvage value is established and used with MACRS, the salvage value of depreciable property is defined to be zero
5 Depreciation Definitions Useful Life Expected (estimated) period of time property will be used in trade or business or to produce income; sometimes referred to as: Depreciable life < 1981 Terms used in the Classical/Historical depreciation method equations: N = depreciable life of the asset in years B = cost basis, including allowable adjustments d k = annual depreciation deduction in year k (1< k <N) d k* = cumulative depreciation through year k BV k = book value at the end of year k BV N = book value at the end of the depreciable (useful) life SV N = salvage value at the end of year N R = the ratio of depreciation in any one year to the BV at the beginning of the year
6 1.a) STRAIGHT-LINE (SL) METHOD Simplest depreciation method Assumes constant amount is depreciated each year over depreciable/useful life This method requires an estimate of the final SV, and the final BV at EOY N. STRAIGHT-LINE (SL) METHOD d k = (B - SV N )/N d k* = kd k for 1 < k < N BV Value k = B ($) - d k* Cost Basis (B) Salvage Value (SV) Time (N) Example The plant you are working for is purchasing an electric saw at $4,000. The machine has 10-year depreciable life The estimated salvage value at EOY 10 is $0. Determine the annual depreciation amounts using the SL method.
7 Solution d k = (B - SV N )/N d k = ($4,000-0)/10 = $400/year d k* = kd k for 1 <k <N d k* for year 1 = $400 X 1 = $400 d k* for year 2 = $400 X 2 = $800 d k* for year 3 = $400 X 3 = $1,200,... etc. BV k = B - d k* BV k for year 1 = $4,000 - $400 = $3,600 BV k for year 2 = $4,000 - $800 = $3,200 BV k for year 3 = $4,000 - $1,200 = $2,800 Solution Year K d k BV k 1 $400 3,600 2 $400 3,200 3 $400 2,800 4 $400 2,400 5 $400 2,000 6 $400 1,600 7 $400 1,200 8 $ $ $400 0 Total $4,000 1.b) DECLINING BALANCE (DB) METHOD Sometimes called constant percentage method, or Matheson formula Assumes annual cost of depreciation is fixed percentage of BV at beginning of year Annual Depreciation = Depreciation Rate * Book Value at Beginning of Year Does not account for Salvage Values* R is constant: R = 2/N when 200% declining balance used R = 1.5/N when 150% declining balance used
8 DECLINING BALANCE (DB) METHOD d 1 = B(R) d k = B(1 - R) k-1 X (R) d k* = B[1 - (1 - R) k ] BV k = B(1 - R) k BV N = B(1 - R) N Example Rework the previous example using: A. R = 2/N B. R = 1.5/N Solution (A) R = 2/N R = 2/10 = 0.2, or 20% d 1 = B(R) d 1 = $4,000 (0.2) = $800 d k = B(1 - R) k-1 X (R) d 2 = $4,000 (1 0.2) 2-1 X (0.2) = $640 d 3 = $4,000 (1 0.2) 3-1 X (0.2) = $512 d 4 = $4,000 (1 0.2) 4-1 X (0.2) = $409.60, etc.
9 Solution (A) d k* = B[1 - (1 - R) k ] d 1* = $4,000[1 - (1 0.2) 1 ] = $800 d 2* = $4,000[1 - (1 0.2) 2 ] = $1,440 d 3* = $4,000[1 - (1 0.2) 3 ] = $1,952, etc. BV k = B(1 - R) k BV 1 = $4,000(1 0.2) 1 = $3,200 BV 2 = $4,000(1 0.2) 2 = $2,560 BV 3 = $4,000(1 0.2) 3 = $2,048, etc. BV N = B(1 - R) N BV 10 = $4,000(1 0.2) 10 = $ Solution (A) Year K d k BV k 1 $800 3,200 2 $640 2,560 3 $512 2,048 4 $ , $ , $ , $ $ $ $ Total $3, Note that: Declining balance method never reaches BV = 0
10 Solution (B) R = 1.5/N R = 1.5/10 = 0.15, or 15% d 1 = B(R) d 1 = $4,000 (0.15) = $600 d k = B(1 - R) k-1 X (R) d 2 = $4,000 (1 0.15) 2-1 X (0.15) = $510 d 3 = $4,000 (1 0.15) 3-1 X (0.15) = $ d 4 = $4,000 (1 0.15) 4-1 X (0.15) = $368.48, etc. Solution (B) d k* = B[1 - (1 - R) k ] d 1* = $4,000[1 - (1 0.15) 1 ] = $600 d 2* = $4,000[1 - (1 0.15) 2 ] = $1,110 d 3* = $4,000[1 - (1 0.15) 3 ] = $1,543.50, etc. BV k = B(1 - R) k BV 1 = $4,000(1 0.15) 1 = $3,400 BV 2 = $4,000(1 0.15) 2 = $2,890 BV 3 = $4,000(1 0.15) 3 = $2,456.50, etc. BV N = B(1 - R) N BV 10 = $4,000(1 0.15) 10 = $ Solution (B) Year Total d k $ $ $ $ $ $ $ $ $ $ $3, BV k $3, $2, $2, $2, $1, $1, $1, $1, $ $787.50
11 1.c) SUM-OF-THE-YEARS-DIGITS (SYD) METHOD 1. Number for each permissible year of life are listed in reverse order 2. Sum the digits of this reverse order 3. The depreciation factor (df) for any year is the corresponding number from the reverse order listing divided by the sum of those digits, or the following: df = [2(N - k + 1)]/[N (N + 1)] 4. Depreciation for any year is the product of this year s SYD depreciation factor and the difference between B and the final estimated SV d k = (B - SV N ) df SUM-OF-YEARS-DIGITS METHOD (SYD) Book value at the EOY k: BV k = B {[2(B - SV N )/N]K + [(B - SV N )/ N(N + 1)] K(k + 1)} The cumulative depreciation through the k th year d k* = B - BV k Example Rework the previous problem using the SYD method.
12 Solution BV k = B {[2(B - SV N )/N]K + [(B - SV N )/ N(N + 1)] K(k + 1)} BV 1 = $4,000 {[2($4,000-0)/10]1 + [($4,000-0)/10(10 + 1)] 1(1 + 1)} = $3, BV 2 = $4,000 {[2($4,000-0)/10]2 + [($4,000-0)/10(10 + 1)] 2(2 + 1)} = $2,618.18, etc. d k* = B BV k d 1* = $4,000 BV 1 = $4,000 - $3, = $ d 2* = $4,000 BV 2 = $4,000 - $2, = $1,381.82, etc. Solution Year Years in Rev. SYD-factor d k BV k /55 $ $3, /55 $ $2, /55 $ $2, /55 $ $1, /55 $ $1, /55 $ $ /55 $ $ /55 $ /55 $ $ /55 $72.73 $0 Total 55 $4,000 Declining Balance w/switch over to SL Because declining balance method never reaches BV = 0, it s permissible to switch from this to SL method so asset s SV N will be $0, or other desired value
13 Solution Year K BOY BV d k 200% SL 1 $4,000 $800 $ , , , , , , _ Total $3,570.5 $4,000 Therefore ΣDep. Amounts Are Year K d k 1 $ Total $4,000 1.d) UNITS-OF-PRODUCTION METHOD Not based on the idea that decrease in value of property is a function of time Decrease in value is mostly a function of use Method results in cost basis (minus final SV) being allocated equally over the estimated number of units produced during useful life of asset. Depreciation/unit of production = (B - SV N )/(Estimated lifetime production in units)
14 Example B = $50,000 S = $10,000 at 30,000 hours Find: 1. Depreciation value/hour 2. BV after 10,000 hours of operation Solution 1. Depreciation/unit of production = (B - SV N )/(Estimated lifetime production in units) Depreciation/unit of production = ($50,000 - $10,000)/(30,000) = $1.33/hr 2. BV = Basis Depreciation deduction: After 10,000 hrs, BV = $50, /hr (10,000 hrs) = $36,700 < 1981, but < 1986
15 2.a) ACCELERATED COST RECOVERY SYSTEM (ACRS) Recognizes an asset as belonging to one of four (tangible) property classes IRS prescribes the specific series of depreciation per property class Rates are based on 150% Declining Balance depreciation, switching to S-L > a) MODIFIED ACCELERATED COST RECOVERY SYSTEM (MACRS) The principal method for computing depreciation deductions for property in engineering projects. Applies to most tangible depreciable property placed in service after December 31, 1986 SV N is defined to be $0 useful life estimates are not used directly in calculating depreciation amounts Consists of 2 systems for computing depreciation deductions: 1. The General Depreciation System (GDS) 2. The Alternative Depreciation System (ADS)
16 INFORMATION NEEDED TO CALCULATE MACRS DEPRECIATION 1. The cost Basis 2. The date the property was placed in service 3. The property class and recovery period 4. The MACRS depreciation used (GDS or ADS) 5. The time convention that applies (1/2 year) GENERAL DEPRECIATION SYSTEM (GDS) 1. Tangible depreciable property assigned to one of six personal property classes (3, 5, 7, 10, 15 and 20- year) 2. Corresponds to GDS recovery period; personal depreciable property not corresponding to these periods is considered 7-yr property class. 3. Real property assigned to 2 classes: 1. Nonresidential real property; and 2. Residential rental property. 4. GDS recovery period: 39-years for nonresidential real property (31.5 years if in service before May 13, 1993) 27.5-years for residential rental property. ALTERNATIVE DEPRECIATION SYSTEM (ADS) 1. Provides longer recovery period and uses only S-L method of depreciation 2. Assets depreciated under ADS include property placed in any tax-exempt use and property used predominantly outside the U.S. 3. ADS recovery period for tangible personal property is normally the same as the class life of the property, with some exceptions (i.e., asset class & 00.22) 4. Any tangible personal property that does not fit into one of the asset classes is depreciated using a 12- year ADS recovery period 5. ADS recovery period for nonresidential real property is 40-yrs
17 CALCULATING DEPRECIATION DEDUCTIONS UNDER MACRS Depreciation Personal Approach Method Property Class GDS 3-, 5-, 7-200% DB method 10-year with switch to SL when deduction greater GDS 15- & % DB method year with switch to SL when deduction greater GDS residential SL over fixed GDS recovery & real rental periods ADS personal & SL method over fixed ADS real recovery periods 1/2-YEAR TIME CONVENTIONS FOR MACRS DEPRECIATION CALCULATIONS All assets placed in service during the year are treated as if use began in the middle of the year -- 1/2- year depreciation is allowed If asset is disposed of before the full recovery period is used, only 1/2 of the normal depreciation deduction can be taken for that year GDS MACRS DEPRECIATION GDS OR ADS? ADS Find property class; Same as recovery period for personal Find recovery period Obtain recovery rates Compute depreciation amount; Asset s cost basis SL = Recovery period Compute depreciation deduction in year k (d k ) by multiplying cost basis by recovery period. Compute depreciation deduction in year k (d k )
18 Note that: Basis for Depreciation (B) = Actual Cash Cost + Book Value (BV) of tradein Example: You are trading in a scanner (BV = $35,000) for a newer one (cost = $105,000). Fair MV of your older scanner = $45,000 Dealer will take old scanner + $60,000 What is the Basis for Depreciation? Solution: B = Cost in cash + BV B = $60,000 + $35,000 B = $95,000
19 Example: In May 1999, your company traded in a computer (BV = $25,000) for another (MV = $400,000). The vendor accepted the older computer + $325,000 cash. 1. What is the GSD property class of the new computer system? 2. How much depreciation can be deducted each year based on this class life? Solution (1): Asset Class: Class Life = 6-years Therefore, GDS property class/recovery period = 5 years. Solution (2): Cost Basis = $325,000 (cash) + $25,000 (BV of older computer) = $350,000. Use values in table 9-3 for 5-year class, substitute in: d k = r k X B 1999: 0.2 X $350,000 = $70, : 0.32 X $350,000 = $112, : X $350,000 = $67, : X $350,000 = $40, : X $350,000 = $40, : X $350,000 = $20,160_ Total: $350,000
20 Example: Your company purchased a computer for $100,000. Equipment was disposed of at EOY 5 Find: 1. The depreciation charge in the 4 th year 2. The BV at EOY 4 3. The cumulative depreciation deduction at EOY 3 4. The BV at EOY 5 Solution: Asset Class: Class Life = 6-years Therefore, GDS property class/recovery period = 5 years. Use values in table 9-3 for 5-year class, d 4 = X $100,000 = $11,520 Solution: BV at EOY 4 is the cost Basis total depreciation through EOY 4: BV 4 = $100,000 - $100,000 ( ) = $17,280 Accumulated depreciation at EOY 3: d 3 = d 1 + d 2 + d 3 d 3 = $100,000 ( ) d 3 = $71,200 Depreciation deduction at EOY 5 if equipment was disposed of at that time is ½ the allotted depreciation amount: d 5 = ½ X X $100,000 = $5,760
21 Example: Your company purchased a new Motor Vehicle manufacturing system for $3,000,000. They decided to use the ADS systems under MACRS. Calculate the depreciation amounts for this project Solution: From Table 9-2, Asset Class = ADS Class Life = 12 years ADS Recovery Period = 12 years. Start the depreciation w/half year conventions: Depreciation amounts: d 1 = $3,000,000 X 0.5/12 = $125,000 d 2 through d 12 = $250,000 d 13 = $125,000 Note that: Firms naturally prefers depreciation methods that result in larger PW (of depreciation amounts). This will reduce the firm s income tax paid to the government.
22 More notes: For the same project/equipment, PW at the company s MARR will be in the following ascending order: 1. SL 2. DB 3. SYD 4. MACRS Therefore, most profitable companies prefer to use MACRS method of depreciation. Comparison of Depreciation methods: BV ($) PW (SL) < PW (DB) < PW (SYD) < PW (MACRS) SL SYD MACRS DB Year k Depletion only used for natural resources
23 DEPLETION Used to indicate the Decrease in Natural Resources that are being consumed in producing products or services, ex: Mining properties Oil and gas wells Timberlands, etc... Amounts charged as depletion cannot be used to replace sold resources PAYMENTS TO RESOURCE OWNERS 1. Earned profit + 2. Portion of owner s capital returned, marked as depletion TO COMPUTE DEPLETION ALLOWANCE 1. Cost method (similar to Units-of Production method): Applies to all types of property and is more widely used Depletion unit is determined by dividing adjusted cost basis by the number of units to be mined or harvested Depletion allowance for tax year is the product of the number of units sold times the depletion unit
24 Cost Method Example Your Zinc company purchased an ore land for $2,000,000 Recoverable reserves of the mine is estimated to be 500,000 tons. 1. If 75,000 tons of ore were mined during year 1, 50,000 tons were sold, what is the depletion amount for year 1? 2. Suppose reserves were reevaluated and found to be 400,000 tons, and if 50,000 tons were sold in year 2, what is the depletion amount for year 2? Solution (1) Depletion unit = $2,000,000/500,000 tons = $4.00/ton. Therefore, depletion amount based on units sold in year 1 = 50,000 X $4 = $200,000 Solution (2) The adjusted cost basis at BOY 2 = $2,000,000 - $200,000 = $1,800,000 Depletion unit = $1,800,000/400,000 tons = $4.5/ton. Therefore, depletion allowance for year 2 = 50,000 tons X $4.5/ton = $225,000
25 TO COMPUTE DEPLETION ALLOWANCE 2. The Percentage Method Based on percentage of year s Gross Income (provided that amount charged does not exceed 50% of Net Income before deduction of depletion allowance) Can be used for most types of metal mines, geothermal deposits, and coal mines Cannot be used for timber and, in most cases, is not applicable to oil and gas When percentage method applies, depletion allowance must be calculated by both, cost and percentage methods -- the larger of the two is used. To determine which method to use in a given year: Percentage depletion method Compute % depletion (appropriate % X Gross Income) Compute 50% of taxable income w/o depletion allowance Select the smaller as allowed % depletion Compute cost depletion (based on depletion unit) Cost depletion method Select the larger as depletion allowance Percentage Method Example A gold mine is producing 30,000 oz of gold. Mine was purchased for $2,400,000 The gold can be sold for $450/oz It costs $265/oz to mine If 3,500 oz were produced this year Find the depletion amounts for 1. Unit depletion 2. Percentage depletion
26 Solution (1. Cost Method) Depletion unit = $2,400,000 / 30,000 oz = $80/oz Cost depletion = 3,500oz X $80/oz = $280,000 Solution (2. Percentage method) Gross income = $450/oz X 3,500 oz = $1,575,000 Percentage depletion allowance = 15% X $1,575,000 = $236,350 Solution (2)... Cont. Net income = Gross income Total cost Total costs = $265/oz X 3,500 oz = $927,500 Net income = $1,575,000 - $927,500 = $647,500 The percentage depletion allowance cannot exceed 50% of net income of $647,500 (ie, $323,750) Therefore, $236,350 is an acceptable depletion allowance However, since unit depletion is higher, select it. Therefore, depletion allowance = $280,000
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