1 2 Engineering Economics (2+0) Depreciation of Assets Prof. Dr. Attaullah Shah Lecture # 6 Department of Civil Engineering City University of Science and IT Peshawar Definitions The assets such as equipment and machines purchased for projects have certain limited service life and it cannot remain in use for ever due to wear and tear etc. The assets such as equipment and machines purchased for projects have certain limited service life and it cannot remain in use for ever due to wear and tear etc. Such equipment is procured through huge investment called Capital Cost/Capital Expenditure ( Capex). Such investment is recovered as expenses from the revenue of the firm, during the life of the asset. Hence the book value of the assets is reduced by the amount recovered. Such equipment is procured through huge investment called Capital Cost/Capital Expenditure ( Capex). Such investment is recovered as expenses from the revenue of the firm, during the life of the asset. Hence the book value of the assets is reduced by the amount recovered. Depreciation is the process of recovering the price/investment of an asset/equipment for its replacement in due course of time. Depreciation is the process of recovering the price/investment of an asset/equipment for its replacement in due course of time. Depreciation means decrease in value of physical asset with the passage of time. Depreciation means decrease in value of physical asset with the passage of time. Methods of Depreciation Straight line method of depreciation Straight line method of depreciation Declining balance method of depreciation Declining balance method of depreciation Sum of the yearsdigits method of depreciation Sum of the yearsdigits method of depreciation 4. Sinking-fund method of depreciation 4. Sinking-fund method of depreciation 5. Service output method of depreciation 5. Service output method of depreciation Straight Line Method Here Fixed sum is charged every year over the life of the equipment to recover the investment. Here Fixed sum is charged every year over the life of the equipment to recover the investment. The accumulated sum of the depreciation at the end of the assumed period/life will become equal to the purchase value of the equipment The accumulated sum of the depreciation at the end of the assumed period/life will become equal to the purchase value of the equipment It is assumed that there is no inflation It is assumed that there is no inflation The salvage value of the asset is the selling price, when the asset is retired/disposed off. The salvage value of the asset is the selling price, when the asset is retired/disposed off. P = first cost of the asset, P = first cost of the asset, F = salvage value of the asset, F = salvage value of the asset, n = life of the asset, n = life of the asset, The formula for depreciation of the asset is: D = (P F)/n The formula for depreciation of the asset is: D = (P F)/n B t = book value of the asset at the end of the period t, B t = book value of the asset at the end of the period t, D t = depreciation amount for the period t. D t = depreciation amount for the period t. D = (P F)/n, B t = B t1 D t = P t [(P F)/n] D = (P F)/n, B t = B t1 D t = P t [(P F)/n] t Example 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 the straight line 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 the straight line method of depreciation. Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years Dt = (P F)/n = (1,00,000 20,000)/8 = Rs. 10,000 Dt = (P F)/n = (1,00,000 20,000)/8 = Rs. 10,000 The depreciation is the same for all the periods. The depreciation is the same for all the periods. Declining Balance Method Depreciation amount changes and does not remain constant remain constant for all years. Depreciation amount changes and does not remain constant remain constant for all years. For Example, the depreciation of computer and IT equipment is initially very high For Example, the depreciation of computer and IT equipment is initially very high In this method a constant percentage of the book value of previous period is charged instead of constant value. In this method a constant percentage of the book value of previous period is charged instead of constant value. More realistic for short life products and assets More realistic for short life products and assets P = first cost of the asset, F = salvage value of the asset, P = first cost of the asset, F = salvage value of the asset, n = life of the asset, Bt = book value of the asset at the end of the period t, K = a fixed percentage, and Dt = depreciation amount at the end of the period t. n = life of the asset, Bt = book value of the asset at the end of the period t, K = a fixed percentage, and Dt = depreciation amount at the end of the period t. D = K B t1 D = K B t1 Bt = Bt-1 Dt = Bt-1 K x (Bt-1) = Bt-1 (1-K) Bt = Bt-1 Dt = Bt-1 K x (Bt-1) = Bt-1 (1-K) The formulae for depreciation and book value in terms of P are as follows: Dt = K(1 K) t1 x P or Bt = (1 K) t P The formulae for depreciation and book value in terms of P are as follows: Dt = K(1 K) t1 x P or Bt = (1 K) t P Example Consider previous Example and demonstrate the calculations of the declining balance method of depreciation by assuming 0.2 for K. Consider previous Example and demonstrate the calculations of the declining balance method of depreciation by assuming 0.2 for K. Solution Solution P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, K = 0.2, P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, K = 0.2, Dt = K x B t-1 and Bt = B t1 Dt Dt = K x B t-1 and Bt = B t1 Dt If we are interested in computing Dt and B for a specific period t, the respective formulae can be used. If we are interested in computing Dt and B for a specific period t, the respective formulae can be used. Consider previous Example and calculate the depreciation and the book value for period 5 using the declining balance method of depreciation by assuming 0.2 for K. Consider previous Example and calculate the depreciation and the book value for period 5 using the declining balance method of depreciation by assuming 0.2 for K. Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, K = 0.2 Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, K = 0.2 D t = K(1 K) t 1 P D t = K(1 K) t 1 P D5= 0.2(1 0.2) 4 * 1,00,000 = Rs. 8,192 D5= 0.2(1 0.2) 4 * 1,00,000 = Rs. 8,192 Bt = (1 K) n P or B5= (1 0.2) 5 x 1,00,000 = Rs. 32,768 Bt = (1 K) n P or B5= (1 0.2) 5 x 1,00,000 = Rs. 32,768 Sum-of-the-Years-Digits Method of Depreciation Here the book value of the asset is assumed to decrease at the decreasing rate. The depreciation at the initial year is highest and then it is gradually reduced till the end of the assets life. Here the book value of the asset is assumed to decrease at the decreasing rate. The depreciation at the initial year is highest and then it is gradually reduced till the end of the assets life. If an asset has life of 8 years, then the sum of its years of life is = = 36 = n(n + 1)/2 If an asset has life of 8 years, then the sum of its years of life is = = 36 = n(n + 1)/2 Consider Example 1 and demonstrate the calculations of the sum-of-the-years- digits method of depreciation. Consider Example 1 and demonstrate the calculations of the sum-of-the-years- digits method of depreciation. Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years Sum = n(n + 1)/2 = 8 ( 9)/2 = 36 Sum = n(n + 1)/2 = 8 ( 9)/2 = 36 The rates for years 18, are respectively 8/36, 7/36, 6/36, 5/36, 4/36, 3/36, 2/36 and 1/36. The rates for years 18, are respectively 8/36, 7/36, 6/36, 5/36, 4/36, 3/36, 2/36 and 1/36. The calculations of Dt, and B for different values of t are summarized using the following formulae: The calculations of Dt, and B for different values of t are summarized using the following formulae: Dt =Rate (P F), Bt1 Dt Dt =Rate (P F), Bt1 Dt Consider Example 1 and find the depreciation and book value for the 5th year using the sum-of-the-years-digits method of depreciation. Consider Example 1 and find the depreciation and book value for the 5th year using the sum-of-the-years-digits method of depreciation. P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years 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. Let In this method of depreciation, the book value decreases at increasing rates with respect to the life of the asset. Let P = first cost of the asset, F = salvage value of the asset, P = first cost of the asset, F = salvage value of the asset, n = life of the asset, i = rate of return compounded annually, n = life of the asset, i = rate of return compounded annually, A = the annual equivalent amount, B = the book value of the asset at the end of the period t, and Dt= the depreciation amount at the end of the period t. A = the annual equivalent amount, B = the book value of the asset at the end of the period t, and Dt= the depreciation amount at the end of the period t. The loss in value of the asset (P F) is made available an the form of The loss in value of the asset (P F) is made available an the form of cumulative depreciation amount at the end of the life of the asset by setting up an equal depreciation amount (A) at the end of each period during the lifetime of the asset. A = (P F) [A/F, i, n] cumulative depreciation amount at the end of the life of the asset by setting up an equal depreciation amount (A) at the end of each period during the lifetime of the asset. A = (P F) [A/F, i, n] The fixed sum depreciated at the end of every time period earns an interest at the rate of i% compounded annually, and hence the actual depreciation amount will be in the increasing manner with respect to the time period. A generalized formula for D is Dt = (P F) (A/F, i, n) (F/P, i, t 1) The fixed sum depreciated at the end of every time period earns an interest at the rate of i% compounded annually, and hence the actual depreciation amount will be in the increasing manner with respect to the time period. A generalized formula for D is Dt = (P F) (A/F, i, n) (F/P, i, t 1) The formula to calculate the book value at the end of period t is The formula to calculate the book value at the end of period t is B= P (P F) (A/F, i, n) (F/A, i, t) B= P (P F) (A/F, i, n) (F/A, i, t) t EXAMPLE Consider Example and give the calculations regarding the sinking fund method of depreciation with an interest rate of 12%, compounded annually. EXAMPLE Consider Example and give the calculations regarding the sinking fund method of depreciation with an interest rate of 12%, compounded annually. Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, i = 12% Solution: P = Rs. 1,00,000, F = Rs. 20,000, n = 8 years, i = 12% A = (P F) [A/F, 12%, 8] = (1,00,000 20,000) = Rs. 6,504 A = (P F) [A/F, 12%, 8] = (1,00,000 20,000) = Rs. 6,504 Depreciation at the end of the year 3 (D)= = 6,504 + (6, ,284.48).12 = Rs. 8, Depreciation at the end of the year 3 (D)= = 6,504 + (6, ,284.48).12 = Rs. 8, Depreciation at the end of year 4 (D)4= 6,504 + (6, , ,158.62) 0.12 = Rs. 9, Depreciation at the end of year 4 (D)4= 6,504 + (6, , ,158.62) 0.12 = Rs. 9,137.65