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Book Depreciation Lecture No.22 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.

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Presentation on theme: "Book Depreciation Lecture No.22 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005."— Presentation transcript:

1 Book Depreciation Lecture No.22 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005

2 Chapter 8 Accounting for Depreciation and Income Taxes  Asset Depreciation  Book Depreciation  Tax Depreciation  How to Determine “Accounting Profit”  Corporate Taxes

3 Depreciation Definition: Loss of value for a fixed asset Example: You purchased a car worth $15,000 at the beginning of year 2000. Depreciation End of Year Market Value Loss of Value 012345012345 $15,000 10,000 8,000 6,000 5,000 4,000 $5,000 2,000 1,000

4 Why Do We Consider Depreciation? Gross Income -Expenses: (Cost of goods sold) (Depreciation) (operating expenses) Taxable Income - Income taxes Net income (profit) Business Expense: Depreciation is viewed as a part of business expenses that reduce taxable income.

5 Depreciation Concept  Economic Depreciation Purchase Price – Market Value (Economic losses due to both physical deterioration and technological obsolescence)  Accounting Depreciation A systematic allocation of the cost basis over a period of time.

6 Asset Depreciation Depreciation Economic depreciation the gradual decrease in utility in an asset with use and time Accounting depreciation The systematic allocation of an asset’s value in portions over its depreciable life—often used in engineering economic analysis Physical depreciation Functional depreciation Book depreciation Tax depreciation

7 Factors to Consider in Asset Depreciation  Depreciable life (how long?)  Salvage value (disposal value)  Cost basis (depreciation basis)  Method of depreciation (how?)

8 What Can Be Depreciated? Assets used in business or held for production of income Assets having a definite useful life and a life longer than one year Assets that must wear out, become obsolete or lose value A qualifying asset for depreciation must satisfy all of the three conditions above.

9 Cost Basis Cost of a new hole-punching machine (Invoice price) $62,500 + Freight725 + Installation labor2,150 + Site preparation3,500 Cost basis to use in depreciation calculation $68,875

10 Cost Basis with Trade-In Allowance Old hole-punching machine (book value)$4,000 Less: Trade-in allowance5,000 Unrecognized gains$1,000 Cost of a new hole-punching machine$62,500 Less: Unrecognized gains(1,000) Freight725 Installation labor2,150 Site preparation3,500 Cost of machine (cost basis)$67,875

11 Asset Depreciation Range ADR (years) Assets UsedLower LimitMidpoint LifeUpper Limit Office furniture, fixtures, and equipment 81012 Information systems (computers) 567 Airplanes 567 Automobiles, taxis 2.533.5 Buses 7911 Light trucks 345 Heavy trucks (concrete ready-mixer) 567 Railroad cars and locomotives 121518 Tractor units 567 Vessels, barges, tugs, and water transportation system 14.51821.5 Industrial steam and electrical generation and or distribution systems 17.52226.5 Manufacturer of electrical and nonelectrical machinery 81012 Manufacturer of electronic components, products, and systems 567 Manufacturer of motor vehicles 9.51214.5 Telephone distribution plant 283542

12 Types of Depreciation Book Depreciation  In reporting net income to investors/stockholders  In pricing decision Tax Depreciation  In calculating income taxes for the IRS  In engineering economics, we use depreciation in the context of tax depreciation

13 Book Depreciation Methods Purpose: Used to report net income to stockholders/investors Types of Depreciation Methods:  Straight-Line Method  Declining Balance Method  Unit Production Method

14 Straight – Line (SL) Method Principle A fixed asset as providing its service in a uniform fashion over its life Formula Annual Depreciation D n = (I – S) / N, and constant for all n. Book Value B n = I – n (D) where I = cost basis S = Salvage value N = depreciable life

15 Example 8.2 – Straight-Line Method D1 D2 D3 D4 D5 B1 B2 B3 B4 B5 $10,000 $8,000 $6,000 $4,000 $2,000 0 1 2 3 4 5 Total depreciation at end of life nD n B n 11,6008,400 21,6006,800 31,6005,200 41,6003,600 51,6002,000 I = $10,000 N = 5 Years S = $2,000 D = (I - S)/N Annual Depreciation Book Value n

16 Declining Balance Method Principle: A fixed asset as providing its service in a decreasing fashion Formula Annual Depreciation Book Value where 0 <  < 2(1/N) Note: if  is chosen to be the upper bound,  = 2(1/N), we call it a 200% DB or double declining balance method.

17 Example 8.3 – Declining Balance Method D1 D2 D3 D4 D5 B1 B2 B3 B4 B5 $10,000 $8,000 $6,000 $4,000 $2,000 0 1 2 3 4 5 Total depreciation at end of life $778 Annual Depreciation Book Value n012345n012345 D n $4,000 2,400 1,440 864 518 B n $10,000 6,000 3,600 2,160 1,296 778 n

18 Example 8.3 – Declining Balance Method (if B<salvage value) D1 D2 D3 D4 B1 B2 B3 B4 B5 $10,000 $8,000 $6,000 $4,000 $2,000 0 1 2 3 4 5 Total depreciation at end of life $778 Annual Depreciation Book Value n012345n012345 D n $4,000 2,400 1,440 160 0 B n $10,000 6,000 3,600 2,160 2000 n

19 Example 8.4 DB Switching to SL SL Dep. Rate = 1/5 see eq8.2 for any year calculations  (DDB rate) = (200%) (SL rate) = 0.40 Asset:Invoice Price $9,000 Freight 500 Installation 500 Depreciation Base$10,000 Salvage Value 0 Depreciation200% DB Depreciable life5 years

20 nDepreciation Book Value 1234512345 10,000(0.4) = 4,000 6,000(0.4) = 2,400 3,600(0.4) = 1,440 2,160(0.4) = 864 1,296(0.4) = 518 $6,000 3,600 2,160 1,296 778 n Book Depreciation Value 1234512345 4,000 $6,000 6,000/4 = 1,500 < 2,4003,600 3,600/3 = 1,200 < 1,4402,160 2,160/2 = 1,080 > 8641,080 1,080/1 = 1,080 > 518 0 (a) Without switching(b) With switching to SL Note: Without switching, we have not depreciated the entire cost of the asset and thus have not taken full advantage of depreciation’s tax deferring benefits. Case 1: S = 0

21 Case 2: S = $2,000 End of Year DepreciationBook Value 10.4($10,000) = $4,000$10,000 - $4,000 = $6,000 20.4(6,000) = 2,4006,000 – 2,400 = 3,600 30.4(3,600) = 1,4403,600 –1,440 = 2,160 40.4(2,160) = 864 > 1602,60 – 160 = 2,000 502,000 – 0 = 2,000 Note: Tax law does not permit us to depreciate assets below their salvage values.

22 Units-of-Production Method Principle Service units will be consumed in a non time-phased fashion Formula Annual Depreciation D n = Service units consumed for year total service units (I - S)

23 Example 8.5 Given: I = $55,000, S = $5,000, Total service units = 250,000 miles, usage for this year = 30,000 miles Solution:


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