A truck was purchased four years ago for $58,000 to move raw materials and finished goods between a production facility and four remote warehouses. This truck (the defender) can be sold at the present time for $40,000 and replaced by a new truck (the challenger) with a purchase price of $73,000. a. Given the MVs and operating and maintenance costs, what is the economic life of the challenger if MARR = 6%? Note: This is a before-tax analysis that does not require any calculations involving the defender. b. Suppose that the defender was set up on a depreciation schedule with a five-year MACRS class life at the time of its purchase (four years ago). The corresponding depreciation rates are 20%, 32%, 19.2%, 11.52%, 11.52% and 5.76%. The tax rate is 25%. The defender can be sold now for $40,000, or a rebuilt engine and transmission can be purchased and installed at a cost of $15,000 (capital investment with three-year depreciable life, straight line, salvage value = 0). If the defender is kept in service, assume that it will have operating and maintenance costs as shown in Part (a) and a MV of $0 at the end of four years. Determine the ATCFs for the defender Click the icon to view the data for both alternatives. Click the icon to view the interest and annuity table for discrete compounding when MARR = 6% per year. More Info - C. a. The economic life of the challenger is year(s). (Round to the nearest whole number.) b. Fill in the following table with the ATCFs for the defender. (Round to the nearest dollar.) ATCH EOY 0 $ N More Info 1 $ 2 $ 3 Challenger EOY Market value 1 2 3 4 5 6 4 EOY 1 2 3 4 Defender Market O&M Costs value $27,000 $9,500 21,000 11,500 11,000 14,000 4,000 16,000 1 2 3 4 $55,000 44,000 33,000 24,500 O&M Costs $5,000 6,200 6,900 8,700 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Discrete Compounding; i = 6% Single Payment Uniform Series Compound Compound Sinking Amount Present Amount Present Fund Factor Worth Factor Factor Worth Factor Factor To Find F To Find P To Find F To Find P To Find a Given P Given F Given A Given A Given F F/P P/F FIA P/A A/F 1.0600 0.9434 1.0000 0.9434 1.0000 1.1236 0.8900 2.0600 1.8334 0.4854 1.1910 0.8396 3.1836 2.6730 0.3141 1.2625 0.7921 4.3746 3.4651 0.2286 1.3382 0.7473 5.6371 4.2124 0.1774 1.4185 0.7050 6.9753 4.9173 0.1434 1.5036 0.6651 8.3938 5.5824 0.1191 1.5938 0.6274 9.8975 6.2098 0.1010 1.6895 0.5919 11.4913 6.8017 0.0870 1.7908 0.5584 13.1808 7.3601 0.0759 1.8983 0.5268 14.9716 7.8869 0.0668 2.0122 0.4970 16.8699 8.3838 0.0593 2.1329 0.4688 18.8821 8.8527 0.0530 2.2609 0.4423 21.0151 9.2950 0.0476 2.3966 0.4173 23.2760 9.7122 0.0430 2.5404 0.3936 25.6725 10.1059 0.0390 2.6928 0.3714 28.2129 10.4773 0.0354 2.8543 0.3503 30.9057 0.0324 3.0256 0.3305 33.7600 11.1581 0.0296 3.2071 0.3118 36.7856 0.0272 3.3996 0.2942 39.9927 11.7641 0.0250 3.6035 0.2775 43.3923 12.0416 0.0230 3.8197 0.2618 46.9958 12.3034 0.0213 4.0489 0.2470 50.8156 12.5504 0.0197 4.2919 0.2330 54.8645 12.7834 0.0182 5.7435 0.1741 79.0582 13.7648 0.0126 Capital Recovery Factor To Find A Given P A/P 1.0600 0.5454 0.3741 0.2886 0.2374 0.2034 0.1791 0.1610 0.1470 0.1359 0.1268 0.1193 0.1130 0.1076 0.1030 0.0990 0.0954 0.0924 0.0896 0.0872 0.0850 0.0830 0.0813 0.0797 Print Done 10.8276 11.4699 21 22 23 24 25 0.0782 30 0.0726 A truck was purchased four years ago for $58,000 to move raw materials and finished goods between a production facility and four remote warehouses. This truck (the defender) can be sold at the present time for $40,000 and replaced by a new truck (the challenger) with a purchase price of $73,000. a. Given the MVs and operating and maintenance costs, what is the economic life of the challenger if MARR = 6%? Note: This is a before-tax analysis that does not require any calculations involving the defender. b. Suppose that the defender was set up on a depreciation schedule with a five-year MACRS class life at the time of its purchase (four years ago). The corresponding depreciation rates are 20%, 32%, 19.2%, 11.52%, 11.52% and 5.76%. The tax rate is 25%. The defender can be sold now for $40,000, or a rebuilt engine and transmission can be purchased and installed at a cost of $15,000 (capital investment with three-year depreciable life, straight line, salvage value = 0). If the defender is kept in service, assume that it will have operating and maintenance costs as shown in Part (a) and a MV of $0 at the end of four years. Determine the ATCFs for the defender Click the icon to view the data for both alternatives. Click the icon to view the interest and annuity table for discrete compounding when MARR = 6% per year. More Info - C. a. The economic life of the challenger is year(s). (Round to the nearest whole number.) b. Fill in the following table with the ATCFs for the defender. (Round to the nearest dollar.) ATCH EOY 0 $ N More Info 1 $ 2 $ 3 Challenger EOY Market value 1 2 3 4 5 6 4 EOY 1 2 3 4 Defender Market O&M Costs value $27,000 $9,500 21,000 11,500 11,000 14,000 4,000 16,000 1 2 3 4 $55,000 44,000 33,000 24,500 O&M Costs $5,000 6,200 6,900 8,700 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Discrete Compounding; i = 6% Single Payment Uniform Series Compound Compound Sinking Amount Present Amount Present Fund Factor Worth Factor Factor Worth Factor Factor To Find F To Find P To Find F To Find P To Find a Given P Given F Given A Given A Given F F/P P/F FIA P/A A/F 1.0600 0.9434 1.0000 0.9434 1.0000 1.1236 0.8900 2.0600 1.8334 0.4854 1.1910 0.8396 3.1836 2.6730 0.3141 1.2625 0.7921 4.3746 3.4651 0.2286 1.3382 0.7473 5.6371 4.2124 0.1774 1.4185 0.7050 6.9753 4.9173 0.1434 1.5036 0.6651 8.3938 5.5824 0.1191 1.5938 0.6274 9.8975 6.2098 0.1010 1.6895 0.5919 11.4913 6.8017 0.0870 1.7908 0.5584 13.1808 7.3601 0.0759 1.8983 0.5268 14.9716 7.8869 0.0668 2.0122 0.4970 16.8699 8.3838 0.0593 2.1329 0.4688 18.8821 8.8527 0.0530 2.2609 0.4423 21.0151 9.2950 0.0476 2.3966 0.4173 23.2760 9.7122 0.0430 2.5404 0.3936 25.6725 10.1059 0.0390 2.6928 0.3714 28.2129 10.4773 0.0354 2.8543 0.3503 30.9057 0.0324 3.0256 0.3305 33.7600 11.1581 0.0296 3.2071 0.3118 36.7856 0.0272 3.3996 0.2942 39.9927 11.7641 0.0250 3.6035 0.2775 43.3923 12.0416 0.0230 3.8197 0.2618 46.9958 12.3034 0.0213 4.0489 0.2470 50.8156 12.5504 0.0197 4.2919 0.2330 54.8645 12.7834 0.0182 5.7435 0.1741 79.0582 13.7648 0.0126 Capital Recovery Factor To Find A Given P A/P 1.0600 0.5454 0.3741 0.2886 0.2374 0.2034 0.1791 0.1610 0.1470 0.1359 0.1268 0.1193 0.1130 0.1076 0.1030 0.0990 0.0954 0.0924 0.0896 0.0872 0.0850 0.0830 0.0813 0.0797 Print Done 10.8276 11.4699 21 22 23 24 25 0.0782 30 0.0726