1. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 ENGINEERING ECONOMICS Depreciation

2. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Physical assets lose value with passage of time, it is said that they depreciate in value. With the possible exception of land, this phenomenon is the characteristics of all physical assets Depreciation is the loss in value of asset over time

3. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Depreciation The allowance for wear and tear on equipment and machineryThe allowance for wear and tear on equipment and machinery You can depreciate an item only if it meets the following requirements:You can depreciate an item only if it meets the following requirements: –It is used in business or held for the production of income –It must have a useful life that extends substantially beyond the year it was placed in service –It wears out, decays, gets used up, becomes obsolete, or looses value from natural causes

4. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Types Physical depreciationPhysical depreciation –The physical depreciation is the total loss in market value which is caused due to some of the physical damage –In this depreciation there are two kinds, one that is curable and the other that is incurable –The curable is the one, which can be corrected economically –The incurable, involves huge amounts of cost –Deterioration due use of parts, corrosion, rotting, breakage –Wear and tear

5. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Types Functional DepreciationFunctional Depreciation –is the total loss caused by inadequate design –Work with the current machine is not profitable –Change in the need of an asset –Availability of superior asset in the market –Inadequacy or inability of the current asset to meet the demand Functional depreciation is more serious than physical

6. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Types of Assets Tangible – which can be seen, touched and quantified (machines, land, goods)Tangible – which can be seen, touched and quantified (machines, land, goods) Intangible – which can not be seen, touched and quantified (software, trademarks etc)Intangible – which can not be seen, touched and quantified (software, trademarks etc)

7. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Methods Depreciation methods based on timeDepreciation methods based on time –Straight line method –Declining balance method –Declining balance method –Sum-of-the-years'-digits method depreciation is the reduction in the value of an asset due to usage, passage of time, wear and tear, technological outdating or obsolescence, depletion, inadequacy, rot, rust, decay or other such factors

8. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Straight Line Depreciation Method Value of the asset decreases at the constant rateValue of the asset decreases at the constant rate D(t) = {First cost (P) – salvage Value (F)}D(t) = {First cost (P) – salvage Value (F)} Time (n) Declining balance method Asset depreciate faster in early age and slower in the laterAsset depreciate faster in early age and slower in the later Rate of depreciation is determinedRate of depreciation is determined D(t) =  (1-  ) t-1 P, where  is the rate of depreciationD(t) =  (1-  ) t-1 P, where  is the rate of depreciation Sum of the year digits is the combine effect of time

9. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Straight Line Depreciation Method On April 1, 2006, Company A purchased an equipment at the cost of $140,000. This equipment is estimated to have 5 year useful life. At the end of the 5th year, the salvage value (residual value) will be $20,000. Calculate the depreciation expenses for 2006, 2007 and 2008 using straight line depreciation method.On April 1, 2006, Company A purchased an equipment at the cost of $140,000. This equipment is estimated to have 5 year useful life. At the end of the 5th year, the salvage value (residual value) will be $20,000. Calculate the depreciation expenses for 2006, 2007 and 2008 using straight line depreciation method. Depreciation for 2006 = ($140,000 - $20,000) x 1/5 x 9/12 = $18,000 Depreciation for 2007 = ($140,000 - $20,000) x 1/5 x 12/12 = $24,000 Depreciation for 2008 = ($140,000 - $20,000) x 1/5 x 12/12 = $24,000 Depreciation for 2006 = ($140,000 - $20,000) x 1/5 x 9/12 = $18,000 Depreciation for 2007 = ($140,000 - $20,000) x 1/5 x 12/12 = $24,000 Depreciation for 2008 = ($140,000 - $20,000) x 1/5 x 12/12 = $24,000

10. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Declining Balance Depreciation Method Depreciation = Book value x Depreciation rate Book value = Cost - Accumulated depreciationDepreciation = Book value x Depreciation rate Book value = Cost - Accumulated depreciation Depreciation rate for double declining balance method = Straight line depreciation rate x 200%Depreciation rate for double declining balance method = Straight line depreciation rate x 200% Depreciation rate for 150% declining balance method = Straight line depreciation rate x 150%Depreciation rate for 150% declining balance method = Straight line depreciation rate x 150%

11. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Declining Balance Depreciation Method On April 1, 2006, Company A purchased an equipment at the cost of $140,000. This equipment is estimated to have 5 year useful life. At the end of the 5th year, the salvage value (residual value) will be $20,000. Calculate the depreciation expenses for 2006, 2007 and 2008 using double declining balance depreciation method. Useful life = 5 years : Straight line depreciation rate = 1/5 = 20% per year Depreciation rate for double declining balance method = 20% x 200% = 20% x 2 = 40% per year Depreciation for 2006 = $140,000 x 40% x 9/12 = $42,000 Depreciation for 2007 = ($140,000 - $42,000) x 40% x 12/12 = $39,200 Depreciation for 2008 = ($140,000 - $42,000 - $39,200) x 40% x 12/12 = $23,520

12. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Sum-of-the-years'-digits method Depreciation expense = (Cost - Salvage value) x Fraction Fraction for the first year = n / (1+2+3+...+ n) Fraction for the second year = (n-1) / (1+2+3+...+ n) Fraction for the third year = (n-2) / (1+2+3+...+ n)... Fraction for the last year = 1 / (1+2+3+...+ n) n represents the number of years for useful life.Depreciation expense = (Cost - Salvage value) x Fraction Fraction for the first year = n / (1+2+3+...+ n) Fraction for the second year = (n-1) / (1+2+3+...+ n) Fraction for the third year = (n-2) / (1+2+3+...+ n)... Fraction for the last year = 1 / (1+2+3+...+ n) n represents the number of years for useful life.

13. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009 Sum-of-the-years'-digits method Calculation of depreciation expense Sum of the years' digits = 1+2+3+4+5 = 15Calculation of depreciation expense Sum of the years' digits = 1+2+3+4+5 = 15 Depreciation for 2000 = ($100,000 - $10,000) x 5/15 = $30,000 Depreciation for 2001 = ($100,000 - $10,000) x 4/15 = $24,000 Depreciation for 2002 = ($100,000 - $10,000) x 3/15 = $18,000 Depreciation for 2003 = ($100,000 - $10,000) x 2/15 = $12,000 Depreciation for 2004 = ($100,000 - $10,000) x 1/15 = $6,000 Depreciation for 2000 = ($100,000 - $10,000) x 5/15 = $30,000 Depreciation for 2001 = ($100,000 - $10,000) x 4/15 = $24,000 Depreciation for 2002 = ($100,000 - $10,000) x 3/15 = $18,000 Depreciation for 2003 = ($100,000 - $10,000) x 2/15 = $12,000 Depreciation for 2004 = ($100,000 - $10,000) x 1/15 = $6,000

14. Engineering Economics, Lecture # 11, Ejaz Gul, FUIEMS, 2009