DEPRECIATION & TAX SOLUTION1. The “Big-Deal” Company has purchased new furniture for their offices at a retail price of $125,000. An additional $20,000 has been charged for insurance, shipping, and handling. The company expects to use the furniture for 10 years (useful life = 10 years) and then sell it at a salvage (market) value of $15,000. Use the SL method of depreciation to answer these questions. (7.6) a. What is the depreciation during the second year? b. What is the BV of the asset at the end of the first year? c. What is the BV of the asset after 10 years? SOLUTION: Basis = $145,000 = 125,000 +20,000 (a) d2 = ($145,000 − $15,000)/10 = $13,000 (b) BV1 = $145,000 − $13,000 = $132,000 (c) BV10 = $145,000 − $13,000(10) = $15,000 2. Cisco Systems is purchasing a new bar code– scanning device for its service center in San Francisco. The table that follows lists the relevant cost items for this purchase. The operating expenses for the new system are $10,000 per year, and the useful life of the system is expected to be five years. The SV for depreciation purposes is equal to 25% of the hardware cost. (7.7) Cost Item Cost Hardware $160,000 Training $15,000 Installation $15,000 a. What is the BV of the device at the end of year three if the SL depreciation method is used? b. Suppose that after depreciating the device for two years with the SL method, the firm decides to switch to the double declining balance depreciation method for the remainder of the device’s life (the remaining three years). What is the device’s BV at the end of four years? SOLUTION: Basic cost = $160,000 + $15,000 + $15,000 = $190,000 SV = 25% * 160,000 = $40,000 (a) d = ($190,000 − $40,000)/5 = $30,000 BV3 = $190,000 – (3)($30,000) = $100,000 (b) BV2 = $190,000 – (2)($30,000) = $130,000 R = 2/3 for the double declining balance method BV4 = $130,000(1 – 2/3)2 = $14,444.44 3. An asset for drilling was purchased and placed in service by a petroleum production company. Its cost basis is $60,000, and it has an estimated MV of $12,000 at the end of an estimated useful life of 14 years. Compute the depreciation amount in the third year and the BV at the end of the fifth year of life by each of these methods: (7.8) a. The SL method. b. The 200% DB method with switchover to SL. Its estimated SV at that time is expected to be $5. What is the depreciation in the amount of depreciation that would be claimed in year five? (7.000 at the time of the trade-in. The basis is the original purchase price of the bus plus the BV of the old bus that was traded in.000. However.000 + $10. buses have a nine-year class recovery period. First.000. Thus.SOLUTION: 4. The old bus has a BV of $10.000.000 − $8. The La Salle Bus Company has decided to purchase a new bus for $85. and the expected salvage value for depreciation purposes is $8. we must calculate the cost basis. The new bus will be kept for 10 years before being sold.12) SOLUTION: Straight-line depreciation in year five would be ($100.500. The company will use the truck for eight years and will depreciate it over this period of time with the SL method. A construction company is considering depreciation to the SL method for a general purpose hauling truck.000.000)/8 = $11. Construct the tabular for Depreciation & Book Value using: Straight line method DB method DB with Switchover to SL Depreciation .000 with a trade-in of their old bus. the basis is $85. or $95. 5.000. The cost basis of the truck is $100. 2222) = $21.2222.111. d1 = $95.SOLUTION: Straight line method DB method: R = 2/9 = 0.000(0. . At the end of year four. A highly specialized piece of equipment has a first cost of $50. The firm’s effective income tax rate is 40%.000. If this equipment is purchased. If the after-tax MARR is 7% per year.DB with Switchover to SL Depreciation 6. the equipment will be sold for a negligible amount. Estimated annual expenses for upkeep are $3. should the equipment be purchased? Note: Depreciation is given in this case as below: SOLUTION: .000 during each of the four years. it will be used to produce income (through rental) of $20.000 per year for only four years. 912.000) will be disposed of for $0 at the end of year five.000 per year for five years. (a) Set up a table and determine the ATCF for this project. are treated as nondepreciable capital investments whose MVs at the end of year five are estimated to equal their first costs .000. The company would need to purchase two or more acres of land for $275. A tax credit of 0.5%.000 (total). and its estimated amount is $10.000.000. The facility would cost $60. and operating expenses are estimated to be $8.715 at MARR = 12% per year.000.000. Acquisitions of land.912. An increment of working capital would be required. this integrated circuit production line should be recommended because it appears to be quite attractive. The facility will be depreciated using in five years.000 will be claimed at the end of year five. . Because the selling price (MV) is zero.800 is created at the end of year five.000 and have no net MV at the end of five years.912. The depreciable property ($60.7. Gross income is expected to increase by $30.000. and a loss on disposal of $6. as well as additional working capital. The Ajax Semiconductor Company is attempting to evaluate the profitability of adding another integrated circuit production line to its present operations. .000 per year for five years. Based on economic considerations.40($6. (b) Is the investment worthwhile when the after-tax MARR is 12% per year? SOLUTION: .764. The after-tax PW equals $936.000.000) = $2. the loss on disposal equals our BV of $6. The after-tax IRR is obtained from entries in column E of Table 7-8 and is found to be 12. . The firm’s effective income tax rate is 40%. A firm must decide between two system designs. which design should be chosen? DESIGN S1 DESIGN S2 SOLUTION: . whose estimated cash flows are shown in the following table. The effective income tax rate is 40%. If the after-tax desired return on investment is 10% per year. 8. S1 and S2. Both designs have a recovery period of five years. so if a system is not required. 10%. don’t recommend either one. 7) = −$1. 10%.411(0. Neither design however makes money. 6) = −$16. directly compare the AWs of the ATCFs by using the repeatability assumption from Chapter 6. however. .2296) = −$3. AWS1(10%) = PWS1(A/P.681(0. Design S1 is preferred since it has the greater (less negative) AW.830 Based on an after-tax annual worth analysis.2054) = −$290 AWS2(10%) = PWS2(A/P. We can. We can’t directly compare the PW of the after-tax cash flows because of the difference in the lives of the alternatives.
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