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a) Determine the annual depreciation cost by the straight line method. b) Determine the annual depreciation cost by the sinking fund method. Assume interest at 612% compounded annually. Solution: CO= P53,000 + P1,500 = P54,500 CL= P5,000 a) ? =??−???= ?54,500−?5,00010= P4,950 b) ? =??−???[?,𝑖%,?]= ?54,500−?5,000(1+.065)10−1.065= ?49,50013.49= P3,668.18 Example A firm bought an equipment for P56,000. Other expenses including installation amounted to P4,000. The equipment is expected to have a life of 16years with a salvage value of 10% of the original cost. Determine the book
Engineering Economy Page 134 value at the end of 12years by (a) the straight line method and (b) sinking fund method at 12% interest. Solution:CO= P65,000 + P4,000 = P60,000 CL= P60,000(0.10) =P6,000 L = 16 n=12 i-12% a) Straight line method ? =??−?𝐿?= ?60,000−?6,00016= P3,375 D12= (d) (n) = P3,375 (12) = P40,500 C12= CO–D12= P60,000 –P40,500 = P19,500 b) Sinking fund method ? =??−???[?,𝑖%,?]= ?60,000−?6,000(1+0.12)16−10.12= ?54,00042.7533= P1,263.06 D12= ?(1+0.12)12−10.12= P1,263.06(14.1331) = P30,481.60 C12= CO–D12= P60,000 –P30,481.60 = P29,518.40 Problem Solve the above problem using sinking fund method at i =5% Given: CO = ₱ 4,000SV = 0 L = 10 years
Engineering Economy Page 135 Declining Balance Method In this method, sometimes called the constant percentage method or the Matheson Formula, it is assumed that the annual cost of depreciation is a fixed percentage of the salvage value at the beginning of the year. The ratio of the depreciation in any year to the book value at the beginning of that year is constant throughout the life of the property and is designed by k, the rate of depreciation C0C1C2C3Cn-1CnCL-1CL0 1 2 3 n-1 n-1 L-1 L d1d2d3dndLdn= depreciation during the nthyear Year Book Value at beginning of year Depreciation during the year Book Value at the end of the year 1 C0 d1= kC0 C1= C0–d1= C0(1-k) 2 C0(1-k) d2= kC1C2= C1–d2= C0(1-k)2 3 C0(1-k)2 d3= kC2C3= C2–d3= C0(1-k)3 n C0(1-k)n-1 dn= kCn-1Cn= Cn-1–dn= C0(1-k)n L C0(1-k)L-1dL-1= kCLCL= CL-1–dL= C0(1-k)L dn= C0(1-k)n-1k (3 - 7)Cn= C0(1-k)n= ?0[???0]??(3 - 8) CL= C0(1-k)L(3 - 9) k = 1 - √?𝑛?0𝑛= 1 − √?𝐿?0𝐿(3 - 10)
Engineering Economy Page 136 This method does not apply , if the salvage value is zero, because k will be equal to one and d1will be equal to C0. Example A certain type of machine loses 10% of its value each year. The machine costs P2,000 originally. Make out a schedule showing the yearly depreciation, the total depreciation and the book value at the end of each year for 5years. Solution:Year Book value at beginning of year Depreciation during the year 10% Total depreciation at end of year Book value at end of year 1 P2,000.00 P200.00 P200.00 P1,800.00 2 P1,800.00 P180.00 P380.00 P1,620.00 3 P1,620.00 P162.00 P542.00 P1,458.00 4 P1,458.00 P145.80 P687.80 P1,312.20 5 P1,312.20 P131.22 P819.02 P1,180.98 Double Declining Balance(DDB) Method This method is very similar to the declining balance method except that thw rate of depreciation k is replaced by 2/L. ??= ?0(1 −2?)?−12?(3 - 11) ??= ?0(1 −2?)?(3 - 12) ??= ?0(1 −2?)?(3 - 13) When the DDB method is used, the salvage value should not be subtracted from the first cost when calculating the depreciation charge.