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DEPRECIATION - UNIT V - MG6863

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- 1. Depreciation • Any equipment which is purchased today will not work for ever. – Due to Wear & Tear, Becoming out of Date • Successful company has to take the decision, when to replace the old equipment • Replacement Involves Money. – This should be internally generated from the earnings of the equipment. (depreciation Fund). • Depreciation means decrease in value of any physical asset with the passage of time. MG6863 S.BALAMURUGAN, AP/MECH AAACET 2
- 2. • Fixed sum is charged as the depreciation amount throughout the life time. • Accumulated sum at the end of life is equal to the purchase value of the asset. MG6863 S.BALAMURUGAN, AP/MECH AAACET 3
- 3. Straight Line Method of Depreciation • Himalaya Drug Company has just purchased a capsulizing machine for Rs. 10, 00,000. The plant engineer estimates that the machine has a useful life of 5 years and a salvage value of Rs. 10,000 at the end of its useful life. Compute the depreciation schedule for the machine by each of the following depreciation methods : i) Straight line method of depreciation. • Solution: • P = Rs. 10,00,000 n=5 years F = Rs. 10,000 • Depreciation Dt = 𝑷−𝑭 𝒏 = = 10,00,000 − 10,0000 𝟓 = Rs. 1,98,000 End of Year Depreciation (Dt ) Book Value (Bt = Bt-1 - Dt ) 0 10,00,000 1 1,98,000 8,02,000 2 1,98,000 6,04,000 3 1,98,000 4,06,000 4 1,98,000 2,08,000 5 1,98,000 10,000 MG6863 S.BALAMURUGAN, AP/MECH AAACET 4
- 4. • More realistic approach, the depreciation charges decreases with the life of the asset which matches the earning potential of the asset. • Constant percentage of the book value of the previous period of the asset will be charged as the depreciation amount for the current period. • Depreciation, Dt = K × Bt-1 Book Value (Bt) = Bt-1 - Dt • Fixed Percentage K = limited to 2/n. • Generally Assume K = 0.2 Declining Balance method of depreciation MG6863 S.BALAMURUGAN, AP/MECH AAACET 5
- 5. Declining Balance method of depreciation • A company has purchased an equipment whose first cost is Rs.1,00,000 with an estimated life of eight years. The estimated salvage value of the equipment at the end of its lifetime is Rs.20,000. Determine the depreciation charge and book value at the end of various years using decline balance method of depreciation by assuming K=0.2 • Solution: P = Rs.1,00,000 F = Rs.20,000 n = 8 Years K=0.2(Assume) Depreciation, Dt = K × Bt-1 Book Value (Bt) = Bt-1 - Dt End of Year (n) Depreciation (Dt ) (Rs.) Book Value (Bt) (Rs.) 0 1,00,000 1 1,00,000 × 0.2 = 20,000 10,00,000 - 2,00,000= 80,000 2 80,000 × 0.2 = 16,000 80,000 – 16,000 = 64,000 3 64,000 × 0.2 = 12,800 64,000 – 12,800 = 51,200 4 51,200 × 0.2 = 10,240 51,200-10,240 = 40,960 5 40,960 × 0.2 = 8,192 40,960 – 8,192 = 32,768 6 32,768 × 0.2 = 6,553.60 32,768 – 6,553.60 = 26,214.40 7 26,214.40 × 0.2 = 5,242.88 26,214.40 – 5,242.88 = 20,971.52 8 20,971.52 × 0.2 = 4,194.30 20,971.52 – 4,194.30 = 16,777.22 MG6863 S.BALAMURUGAN, AP/MECH AAACET 6
- 6. Sum of Years Digit Method of Depreciation • In this method, it is assumed that the book value of the asset decreases at a decreasing rate. • Step 1 – Find Sum of Years – Example – 5 Years life – Sum of Years = 1+2+3+4+5 = 15 = 𝒏(𝒏+𝟏) 𝟐 • Rate of depreciation for the years 1-5 = 5/15, 4/15, 3/15, 2/15, 1/15 • Step 2 – Depreciation, Dt = Rate × (P-F) – Book Value (Bt) = Bt-1 - Dt MG6863 S.BALAMURUGAN, AP/MECH AAACET 7
- 7. Sum of Years Digit Method of Depreciation • The cost of the machine is Rs.1,00,000 and its scrap value is Rs.40,000. Estimate life 5 years. Using sum of years digit method, determine depreciation charges for each year. • Solution – 5 Years life – Sum of Years = 1+2+3+4+5 = 15 = 𝒏(𝒏+𝟏) 𝟐 – Rate of depreciation for the years 1-5 = 5/15, 4/15, 3/15, 2/15, 1/15 • Depreciation, Dt = Rate × (P-F) Book Value (Bt) = Bt-1 - Dt End of Year Depreciation (Dt ) (Rs.) Book Value (Bt = Bt-1 - Dt ) (Rs.) 0 - 1,00,000 1 Rate (5/15) = (5/15) × (1,00,000 – 40,000) = 20,000 =1,00,000 –20,000=80,000 2 Rate (4/15) = (4/15) × (1,00,000 – 40,000) = 16,000 = 80,000 – 16,000 = 64,000 3 Rate (3/15) = (3/15) × (1,00,000 – 40,000) = 12,000 = 64,000 – 12,000 = 52,000 4 Rate (2/15) = (2/15) × (1,00,000 – 40,000) = 8,000 = 52,000 – 8,000 = 44,000 5 Rate (1/15) = (1/15) × (1,00,000 – 40,000) = 4,000 = 44,000 – 4,000 = 40,000 MG6863 S.BALAMURUGAN, AP/MECH AAACET 8
- 8. Annuity Method or Sinking Fund Method of Depreciation • In this method of depreciation, the book value decreases at increasing rates with respect to the life of the asset. • Step 1 – Find Equal depreciation amount for Loss in value of asset (P – F) – A = (P – F) × [A/F, i ,n] • Step 2 – Depreciation Dt = (P – F) × (A/F, i ,n) × (F/P, i, t – 1) – Book Value (Bt) = P – (P – F) × (A/F, i ,n) × (F/A, i, t ) MG6863 S.BALAMURUGAN, AP/MECH AAACET 9
- 9. Annuity Method or Sinking Fund Method of Depreciation • Find the depreciation annuity by Annuity method after three years, when the initial cost of the machine is Rs.8,00,000 and the salvage value at the end of three years is Rs.4,00,000. Rate of interest 10%. • Solution: • A = (P – F) × (A/F, i ,n) = (8,00,000 – 4,00,000) × (A/F, 10% ,3) =(4,00,000 × 0.3021) A = Rs.1,20,840 • Method 1 • Depreciation Dt = (P – F) × (A/F, i ,n) × (F/P, i, t – 1) At Year 1, D1 = 1,20,840 × (F/P, i, 1 – 1) = Rs. 1,20,840 At Year 2, D2 = 1,20,840 × (F/P, i, 2 – 1) = 1,20,840 × (F/P, i, 1) = 1,20,840 × (1.100) = Rs. 1,32,924 At Year 2, D2 = 1,20,840 × (F/P, i, 3 – 1) = 1,20,840 × (F/P, i, 2) = 1,20,840 × (1.210) = Rs. 1,46,216.4MG6863 S.BALAMURUGAN, AP/MECH AAACET 10
- 10. Annuity Method or Sinking Fund Method of Depreciation • Find the depreciation annuity by Annuity method after three years, when the initial cost of the machine is Rs.8,00,000 and the salvage value at the end of three years is Rs.4,00,000. Rate of interest 10%. • Solution: • A = (P – F) × (A/F, i ,n) = (8,00,000 – 4,00,000) × (A/F, 10% ,3) =(4,00,000 × 0.3021) A = Rs.1,20,840 • Method 2 End of Year Fixed Depreciation (Rs.) Net Depreciation (Dt) (Rs.) Book Value (Bt = Bt-1 - Dt ) (Rs.) 0 8,00,000 1 1,20,840 1,20,840 =8,00,000 – 1,20,840 = 679160 2 1,20,840 1,20,840 + 1,20,840(0.10) = 132924 =679160 – 1,32,924 = 546236 3 1,20,840 1,20840 + (1,20,840 + 1,32,924) (0.10) = 1,46,216.4 =5,46,236 – 1,46,216.4 = 4,00,019.6 MG6863 S.BALAMURUGAN, AP/MECH AAACET 11
- 11. Service Output Method of Depreciation • This method of depreciation based on Service provided by the asset. Not based on Time Period. • Depreciation / unit of service = 𝑷−𝑭 𝑿 • Depreciation for x units of service in a period = 𝑷−𝑭 𝑿 (x) • Example – Printer Toner – Initial Cost (P) – Salvage Value (F)– Capacity (𝑿)1000 Pages • At present 500 pages(x) completed, MG6863 S.BALAMURUGAN, AP/MECH AAACET 12
- 12. • The first coat of a road laying machine is Rs.80,00,000. its salvage value after five years is Rs.50,000. the length of road that can be laid by the machine during its life time is 75,000 km. in its third year of operation, the length of road laid is 2,000 km. Find the depreciation of the equipment for that year. • Solution: • P = 80,00,000, F = Rs.50,000, X = 75,000km, x = 2,000 km • Depreciation for x units of service in a period = 𝑷−𝑭 𝑿 (x) • Depreciation for 3 years (or) for 2,000 km = 𝟖𝟎,𝟎𝟎,𝟎𝟎𝟎−𝟓𝟎,𝟎𝟎𝟎 𝟕𝟓,𝟎𝟎𝟎 (2000) • Depreciation for 3 years (or) for 2,000 km = Rs.2,12,000 Service Output Method of Depreciation MG6863 S.BALAMURUGAN, AP/MECH AAACET 13
- 13. • For private organizations – the selection of alternative is based on Maximum profit. The main goal is to provide goods/service as per specifications with Maximum Profit. • For Public Organizations / Government Organizations / Govt. Projects • The same condition is not applicable for Evaluation of Public Alternatives. • The main objective of any public alternative is to provide goods/service to the public at the minimum cost. • In this case, the benefits of the public alternative are at least equal to its costs. (BC ratio = 1 or >1) If yes, then the public alternative can be taken for implementation. • BC Ratio = 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑩𝒆𝒏𝒆𝒇𝒊𝒕𝒔 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑪𝒐𝒔𝒕𝒔 = 𝑩 𝑷 𝑷+𝑪𝑷 = 𝑩 𝑭 𝑷 𝑭 +𝑪𝑷 = 𝑩 𝑨 𝑷 𝑨 +𝑪 Evaluation of Public Alternatives MG6863 S.BALAMURUGAN, AP/MECH AAACET 14
- 14. • Two mutually exclusive projects are being considered for investment. Project A1 requires an initial outlay of Rs. 30, 00,000 with net receipts estimated as Rs. 9, 00,000 per year for the next 5 years. The initial outlay for the project A2 is Rs. 60, 00,000, and net receipts have been estimated at Rs. 15, 00,000 per year for the next seven years. There is no salvage value associated with either of the projects. Using the benefit cost ratio, which project would you select? Assume an interest rate of 10%. • Solution: Evaluation of Public Alternatives Alternative A1 Alternative A2 Initial Cost (P) = Rs.30,00,000 Net Benefit / year (B) = Rs.9,00,000 Life (n) = 5 years Annual Equivalent of Initial Cost = P (A/P, i, n) = 30,00,000 (A/P, 10%, 5) = 30,00,000 (0.2638) = Rs.7,91,400 Benefit – Cost ratio = 𝑨𝒏𝒏𝒖𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑩𝒆𝒏𝒆𝒇𝒊𝒕𝒔 𝑨𝒏𝒏𝒖𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑪𝒐𝒔𝒕𝒔 = 𝟗,𝟎𝟎,𝟎𝟎𝟎 𝟕,𝟗𝟏,𝟒𝟎𝟎 = 1.137 Initial Cost (P) = Rs.60,00,000 Net Benefit / year (B) = Rs.15,00,000 Life (n) = 7 years Annual Equivalent of Initial Cost = P (A/P, i, n) = 60,00,000 (A/P, 10%, 7) = 60,00,000 (0.2054) = Rs.12,32,400 Benefit – Cost ratio = 𝑨𝒏𝒏𝒖𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑩𝒆𝒏𝒆𝒇𝒊𝒕𝒔 𝑨𝒏𝒏𝒖𝒂𝒍 𝑬𝒒𝒖𝒊𝒗𝒂𝒍𝒆𝒏𝒕 𝑪𝒐𝒔𝒕𝒔 = 𝟏𝟓,𝟎𝟎,𝟎𝟎𝟎 𝟏𝟐,𝟑𝟐,𝟒𝟎𝟎 = 1.217 BC Ratio, Alternative A2 > Alternative A1, So Alternative A2 is selected. 15MG6863 S.BALAMURUGAN, AP/MECH AAACET
- 15. Inflation • The rise in prices of goods & services in a given period of a country is called Inflation. MG6863 S.BALAMURUGAN, AP/MECH AAACET 16
- 16. deflation is a decrease in the general price level of goods and services. It occurs when the inflation rate falls below 0% (a negative inflation rate). Inflation reduces the value of currency over time, but deflation increases it. Deflation MG6863 S.BALAMURUGAN, AP/MECH AAACET 17
- 17. MG6863 S.BALAMURUGAN, AP/MECH AAACET 18
- 18. MG6863 S.BALAMURUGAN, AP/MECH AAACET 19

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