1. Contemporary Engineering Economics, 4 th edition, © 2007 Book Depreciation Lecture No. 33 Chapter 9 Contemporary Engineering Economics Copyright © 2006

2. Contemporary Engineering Economics, 4 th edition, © 2007 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

3. Contemporary Engineering Economics, 4 th edition, © 2007 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

4. Contemporary Engineering Economics, 4 th edition, © 2007 Example 9.3 – Straight-Line Method 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 n

5. Contemporary Engineering Economics, 4 th edition, © 2007 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 (DDB) method.

6. Contemporary Engineering Economics, 4 th edition, © 2007 Example 9.4 – Declining Balance Method n012345n012345 D n $4,000 2,400 1,440 864 518 B n $10,000 6,000 3,600 2,160 1,296 778

7. Contemporary Engineering Economics, 4 th edition, © 2007 SL Dep. Rate = 1/5  (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 Example 9.5 DB Switching to SL

8. Contemporary Engineering Economics, 4 th edition, © 2007 Adjustments to the DB Method Switch from DB to SL after n’ No further depreciation allowances are available after n”

9. Contemporary Engineering Economics, 4 th edition, © 2007 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

10. Contemporary Engineering Economics, 4 th edition, © 2007 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. Case 2: S = $2,000

11. Contemporary Engineering Economics, 4 th edition, © 2007 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)

12. Contemporary Engineering Economics, 4 th edition, © 2007 Example 9.7 Units-of-Production Given: I = $55,000, S = $5,000, Total service units = 250,000 miles, usage for this year = 30,000 miles Solution: