Engineering economy lorie m. Cabanayan francisco d. Cuaresma
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COMPILED LECTURE IN ENGINEERING ECONOMY
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- Sample problem 2.
4. Sinking Fund Method
This method depreciates an asset as if the firm were to make a series of equal deposits whose value at the end of the asset‟s useful life just equaled the cost of replacing the asset. It is assumed that a sinking fund is established in which funds will accumulate for replacement purposes and will bear interest. The total depreciation which has occurred up to any given time is assumed to equal the amount in the sinking fund at that time. If A‟ = sinking fund deposit C = purchase price of replacement asset – net salvage value of current asset N = useful life of current asset i = annual interest rate then A‟ = C (A/F, i%, N) The amount in the sinking fund at the end of year k (k = 1,2….N) is the accumulated depreciation through k, thus: d* k = A‟ (F/A, i%, k) = C (A/F, i%, N) (F/A, i%,k) The depreciation in year k, which includes interest earned at that time, is given as: d k = d* k – d* k-1 = C (A/F, i%, N) [(F/A, i%,k) - (F/A, i%,k-1)] The book value, BV, as usual is defined as BV k = B – d* k 38 Sample problem 2. A contractor imported a bulldozer for his job, paying P250,000 to the manufacturer. Freight and insurance charges amounted to P18,000; customs, broker‟s fee, P8,500; taxes, permits and other expenses, P25,000. If the contractor estimates the life of the bulldozer to be 10 years with a salvage value of P20,000 ,tabulate the annual depreciation amounts and the book value of the bulldozer at the end of each year using sinking fund formula at 8%. B = 250,000 + 18,000 + 8,500 + 25,000 = 301,500 SV = 20,000 A‟ = C (A/F, i%, N) = 301,500 – 20,000 (A/F, 8%, 10) = 281,500 (0.0690) = 19,423.50 d* 2 = A‟ (F/A, 8%, 2) = 19,423.50 (2.0800) = 40,400.88 d* 6 = A‟ (F/A, 8%, 6) = 19,423.50 (7.3359) = 142,488.85 BV 6 = B – d* 6 = 301,500 - 142,488.85 = 159,011.15 d 6 = C (A/F, i%, 10) [(F/A, i%,6) - (F/A, i%,6-1)] = 19, 423.50 (7.3359 – 5.8666) = 19, 423.50 (1.4693) = 28,538.95 or simply: d 6 = d* 6 – d* 5 = 142,488.85 - 113,949.90 = 28,538.95 The depreciation and book value amounts for each year are shown below. EOY, k d k d* k BV k 0 - - 301,500.00 1 19,423.50 19,423.50 282,076.50 2 20,977.38 40,400.88 261,099.12 3 22,655.57 63,056.45 238,443.55 4 24,467.78 87,524.23 213,975.77 5 26,425.67 113,949.90 187,550.09 6 28,583.95 142,488.85 159,011.15 7 30,823.15 173,312.00 128,188.00 8 33,288.00 206,600.00 94,900.00 9 35,952.89 242,552.89 58,947.11 10 38,827.59 281,380.48 20,119.52 Download 436.52 Kb. Do'stlaringiz bilan baham: |
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