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ender can be replaced with a challenger which will cost $70,000 to install, have operating expenses of S8,50 per year, and have a final market value ot 512,0XX) at the end of ts 20-year economic lite. I reciated with the straight line method over a 20-year life with an estimated salvage value of 512,000 at EOY 20. Such equipment (defender or challenger will be needed indefinitely. If the after-tax MAR is 25%, should the defender or the challenger be recommended? n the inlerest and anuity lable for discrele compounding when MARR = 10% per yeur. More Info To Find F To Find P To Find F To Find P To Find A To Find A ender is $ (Round to the nearest dellar. Given P Given F Given A Given A Given F Given P PIF FIA PIA A/F A/P allenger is $ (Rounid lo the nourest dollar.) 1 1.1000 0.9091 1.0000 0.9091 1.0000 1.1000 1.2100 0.0264 2.1000 1.7355 0.4762 0.5762 3 1.3310 0.7513 3.31D0 2.4869 0.3021 0.4021 4 1.4641 0.6830 4.0410 3.1000 0.2155 0.3155 1.6105 O.6209 6.1051 3.7908 0.1638 0.2638 7.7156 1.7716 1.9487 0.5645 0.6132 4.3553 4.8684 0.1296 0.1054 0.2206 replaced immediately 8.4872 0.2054 2.1436 0.1605 11.4359 5.3319 0.0874 0.1874 kept tor at least ene more year 2.3579 0.4241 13.5795 5.7590 0.0735 0.1736 10 2.5037 0.3865 15.9374 6.1446 0.0627 0.1627 2.8531 3.1384 18.5312 21.3843 0.1540 0.1468 11 0.3505 6.4951 0.0540 0.3106 6.0137 0.0468 a 0408 0.0357 12 13 3.4523 0.2897 24.5227 7.1034 0.1408 3.7975 0.2033 27.0750 7.3007 0.1367 0.1315 14 a 0315 0.0276 0.0247 15 4.1772 0.2394 31.7725 7.6061 16 4.5050 0.2176 35.9497 7.8237 0.1270 17 5.0545 0.1978 40.5447 8.0216 0.1247 18 5.5500 0.1799 45.5902 8.2014 0.0219 0.1219 19 6.1159 0.1635 51.1591 8.3649 0.0195 0.1195 20 6.7275 0.1480 57.2750 8.6130 0.0175 0.1176

ender can be replaced with a challenger which will cost $70,000 to install, have operating expenses of S8,50 per year, and have a final market value ot 512,0XX) at the end of ts 20-year economic lite. I reciated with the straight line method over a 20-year life with an estimated salvage value of 512,000 at EOY 20. Such equipment (defender or challenger will be needed indefinitely. If the after-tax MAR is 25%, should the defender or the challenger be recommended? n the inlerest and anuity lable for discrele compounding when MARR = 10% per yeur. More Info To Find F To Find P To Find F To Find P To Find A To Find A ender is $ (Round to the nearest dellar. Given P Given F Given A Given A Given F Given P PIF FIA PIA A/F A/P allenger is $ (Rounid lo the nourest dollar.) 1 1.1000 0.9091 1.0000 0.9091 1.0000 1.1000 1.2100 0.0264 2.1000 1.7355 0.4762 0.5762 3 1.3310 0.7513 3.31D0 2.4869 0.3021 0.4021 4 1.4641 0.6830 4.0410 3.1000 0.2155 0.3155 1.6105 O.6209 6.1051 3.7908 0.1638 0.2638 7.7156 1.7716 1.9487 0.5645 0.6132 4.3553 4.8684 0.1296 0.1054 0.2206 replaced immediately 8.4872 0.2054 2.1436 0.1605 11.4359 5.3319 0.0874 0.1874 kept tor at least ene more year 2.3579 0.4241 13.5795 5.7590 0.0735 0.1736 10 2.5037 0.3865 15.9374 6.1446 0.0627 0.1627 2.8531 3.1384 18.5312 21.3843 0.1540 0.1468 11 0.3505 6.4951 0.0540 0.3106 6.0137 0.0468 a 0408 0.0357 12 13 3.4523 0.2897 24.5227 7.1034 0.1408 3.7975 0.2033 27.0750 7.3007 0.1367 0.1315 14 a 0315 0.0276 0.0247 15 4.1772 0.2394 31.7725 7.6061 16 4.5050 0.2176 35.9497 7.8237 0.1270 17 5.0545 0.1978 40.5447 8.0216 0.1247 18 5.5500 0.1799 45.5902 8.2014 0.0219 0.1219 19 6.1159 0.1635 51.1591 8.3649 0.0195 0.1195 20 6.7275 0.1480 57.2750 8.6130 0.0175 0.1176

ENGR.ECONOMIC ANALYSIS
ENGR.ECONOMIC ANALYSIS
14th Edition
ISBN:9780190931919
Author:NEWNAN
Publisher:NEWNAN
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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A defender was installed 10 years ago at a capital investment cost of S67,000. It presently has a market value of S13,000. If kept, the defender has an economic life of three years, operating expenses of $15,000 per year, and a market value of S9,500 at the end of year (EOY) three. The detender is being depreciated by the straight line method using a 15-year wnte-off pericd with an estimated salvage value tor depreciation purposes of $10,50D. As an alternative, the detender can be replaced with a challenger wnich will cost $70,000 to Install, have operating expenses of SR,500 per year, and have a final market value ot 512,000 at the end of its 20 year economic Ilte. It the replacement is made, the challenger wil be depreciated with the straight line method over a 20-year life with an estimated salvage value of 512,000 at EOY 20. Such equipment (defender or challenger; will be needec indefinitely. If the after-tax MARR is 10% per year and the effective incorne tax rate is 25%, should the defender or the challeıger be recommended? E Click the icon to view the interesl and arınuity lable for discrele cormpounding when MARR = 10% per year. - X More Info To Find F To Find P To Find F To Find P To Find A To Find A The AW value for the defender is $. Round to the nearest dellar. Given P Given F Given A Given A Given F Given P FP PIF FIA PIA A/F A/P The AW value for the challenger is $. (Round to the nearest dollar.) 1.1000 0.9091 1.0000 0.9091 1.0000 1.1000 1 2 1.2100 0.0264 2.1000 1.7355 0.4782 0.5762 The defender shoukd be 3 1.3310 0.7513 3.31DD 2.4869 0.3021 0.4021 4 1.4641 0.6830 4.6410 3.1600 0.2155 0.3155 5 1.6105 0.6209 6.1051 3.7908 0.1638 0.2638 1.7716 0.5645 7.7156 4.3553 0.1298 0.2296 replaced immediately 1.9487 0.5132 9.4872 4.8684 0.1054 0.2054 8 2.1436 0.1006 11.1359 5.3349 0.0871 0.1874 kent for at keast one more year 2.3579 0.4241 13.5795 5.7590 0.0735 0.1736 10 2.5937 0.3855 15.9371 6.1446 0.0627 0.1627 11 2.8531 0.3506 18.5312 6.4951 0.0540 0.1540 12 3.1384 0.3106 21.3843 6.0137 0.0488 0.1460 13 3.4523 0.2897 24.5227 7.1034 a 0408 0,1408 14 3.7976 0.2033 27.9750 7.3607 0.0357 0.1357 15 4.1772 0.2394 31.7725 7.6061 0.0315 0.1315 16 4.5950 0.2176 35.9497 7.8237 0.0278 0.1278 17 5.0545 0.1978 40.5447 8.0216 0.0247 0.1247 18 5.5599 0.1799 45.5002 8.2014 0.0219 0.1219 19 6.1159 0.1635 51.1591 8.3649 0.0195 0.1195 20 6.7275 0.1486 57.2750 8.5130 0.0175 0.1175
Transcribed Image Text:A defender was installed 10 years ago at a capital investment cost of S67,000. It presently has a market value of S13,000. If kept, the defender has an economic life of three years, operating expenses of $15,000 per year, and a market value of S9,500 at the end of year (EOY) three. The detender is being depreciated by the straight line method using a 15-year wnte-off pericd with an estimated salvage value tor depreciation purposes of $10,50D. As an alternative, the detender can be replaced with a challenger wnich will cost $70,000 to Install, have operating expenses of SR,500 per year, and have a final market value ot 512,000 at the end of its 20 year economic Ilte. It the replacement is made, the challenger wil be depreciated with the straight line method over a 20-year life with an estimated salvage value of 512,000 at EOY 20. Such equipment (defender or challenger; will be needec indefinitely. If the after-tax MARR is 10% per year and the effective incorne tax rate is 25%, should the defender or the challeıger be recommended? E Click the icon to view the interesl and arınuity lable for discrele cormpounding when MARR = 10% per year. - X More Info To Find F To Find P To Find F To Find P To Find A To Find A The AW value for the defender is $. Round to the nearest dellar. Given P Given F Given A Given A Given F Given P FP PIF FIA PIA A/F A/P The AW value for the challenger is $. (Round to the nearest dollar.) 1.1000 0.9091 1.0000 0.9091 1.0000 1.1000 1 2 1.2100 0.0264 2.1000 1.7355 0.4782 0.5762 The defender shoukd be 3 1.3310 0.7513 3.31DD 2.4869 0.3021 0.4021 4 1.4641 0.6830 4.6410 3.1600 0.2155 0.3155 5 1.6105 0.6209 6.1051 3.7908 0.1638 0.2638 1.7716 0.5645 7.7156 4.3553 0.1298 0.2296 replaced immediately 1.9487 0.5132 9.4872 4.8684 0.1054 0.2054 8 2.1436 0.1006 11.1359 5.3349 0.0871 0.1874 kent for at keast one more year 2.3579 0.4241 13.5795 5.7590 0.0735 0.1736 10 2.5937 0.3855 15.9371 6.1446 0.0627 0.1627 11 2.8531 0.3506 18.5312 6.4951 0.0540 0.1540 12 3.1384 0.3106 21.3843 6.0137 0.0488 0.1460 13 3.4523 0.2897 24.5227 7.1034 a 0408 0,1408 14 3.7976 0.2033 27.9750 7.3607 0.0357 0.1357 15 4.1772 0.2394 31.7725 7.6061 0.0315 0.1315 16 4.5950 0.2176 35.9497 7.8237 0.0278 0.1278 17 5.0545 0.1978 40.5447 8.0216 0.0247 0.1247 18 5.5599 0.1799 45.5002 8.2014 0.0219 0.1219 19 6.1159 0.1635 51.1591 8.3649 0.0195 0.1195 20 6.7275 0.1486 57.2750 8.5130 0.0175 0.1175
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