Homework Answers
NPW of alternative C = -44000 - 12000 * (P/A, 12%,10) - 1200 * (P/G, 12%,10) + 5000 * (P/F, 12%,10)
= -44000 - 12000 * 5.650223 - 1200 * 20.254089 + 5000 * 0.321973
= -134497.72
NPW of alternative D = -34000 - 7000 * (P/A, 12%,5) - 1500 * (P/G, 12%,5) + 1200 * (P/F, 12%,5) + (-34000 - 7000 * (P/A, 12%,5) - 1500 * (P/G, 12%,5) + 1200 * (P/F, 12%,5))*(P/F, 12%,5)
= -34000 - 7000 * 3.604776 - 1500 * 6.397016 + 1200 * 0.567427 + (-34000 - 7000 * 3.604776 - 1500 * 6.397016 + 1200 * 0.567427)* 0.567427
= -106817.08
Alternative D whould be selected as it has lowr present cost
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Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis...
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis...
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. $ 40,000 $-6,000 D $-22.000 $-3,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-200 $-300 $7,000 10 $200 5 The present worth of alternative C is $ and that of alternative D is $...
Compare 10 years the alternatives C and D on the basis of a present worth analysis...
Compare 10 years the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
I cant get alternative y Problem 14.030: Compare alternatives by calculating their capitalized cost Compare the alternatives below on the basis of their capitalized costs with adjustments made for...
I cant get alternative y
Problem 14.030: Compare alternatives by calculating their capitalized cost Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use 3.2% per year. 10% per year and Alternative First cost,$ -19,000,000 13,500,000 -10,000 82,000 10 AOC, $per year25,000 Salvage value, $ Life, years 105,000 The capitalized cost for alternative X is s -19199107 The capitalized cost for alternative Y is $ 135762 。 Select alternative「-Y- O
Problem 14.030: Compare...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /=...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
Y 0.83 points Compare the alternatives below on the basis of their capitalized costs with adjustments...
Y 0.83 points Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use i=12% per year and f= 3.8% per year. Alternative х First cost, $ -16,000,000 -14,500,000 AOC, $ per year -25,000 -10.000 Salvage value, $ 105,000 82,000 Life, years 10 eBook Hint Print The capitalized cost for alternative X is $ References The capitalized Click to select) IS Select alternativex
show the formula Compare the following alternatives on the basis of Present worth Analysis at an...
show the formula
Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
can you solve it by using Excel Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is...
can you solve it by using Excel
Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20 % per year, which al- ternative has the lower present worth? LS DDM Method -370,000 -164,000 First cost, $ -21,000 -55,000 M&O cost, $/year 30,000...
Solve it in spreadsheet money, anu Juuay pu mom. Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc....
Solve it in spreadsheet
money, anu Juuay pu mom. Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20% per year, which al- ternative has the lower present worth? Method First cost, $ M&O cost, $/year Salvage value, $ Life, years DDM -...
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC)...
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC) at i=10% per year, which is the best alternative in this scenario? • Alternative 1, AW = $87,500 and n = (forever) • Alternative 2, PW = -$895,000 and n = (forever) • Alternative 3, First cost (FC) of $900,000, annual operating savings of 3,000 per year, salvage = $200,000, and n = (forever) Alternative 2 Alternative 3 None of them Alternative 1 QUESTION...
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of 10 years. $ 40,000 $-6,000 D $-22.000 $-3,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-200 $-300 $7,000 10 $200 5 The present worth of alternative C is $ and that of alternative D is $...
Compare 10 years the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
I cant get alternative y
Problem 14.030: Compare alternatives by calculating their capitalized cost Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use 3.2% per year. 10% per year and Alternative First cost,$ -19,000,000 13,500,000 -10,000 82,000 10 AOC, $per year25,000 Salvage value, $ Life, years 105,000 The capitalized cost for alternative X is s -19199107 The capitalized cost for alternative Y is $ 135762 。 Select alternative「-Y- O
Problem 14.030: Compare...
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.
Y 0.83 points Compare the alternatives below on the basis of their capitalized costs with adjustments made for inflation. Use i=12% per year and f= 3.8% per year. Alternative х First cost, $ -16,000,000 -14,500,000 AOC, $ per year -25,000 -10.000 Salvage value, $ 105,000 82,000 Life, years 10 eBook Hint Print The capitalized cost for alternative X is $ References The capitalized Click to select) IS Select alternativex
show the formula
Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
can you solve it by using Excel
Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20 % per year, which al- ternative has the lower present worth? LS DDM Method -370,000 -164,000 First cost, $ -21,000 -55,000 M&O cost, $/year 30,000...
Solve it in spreadsheet
money, anu Juuay pu mom. Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20% per year, which al- ternative has the lower present worth? Method First cost, $ M&O cost, $/year Salvage value, $ Life, years DDM -...
Assume a mutually exclusive scenario. Compare three alternatives on the basis of their capitalized cost (CC) at i=10% per year, which is the best alternative in this scenario? • Alternative 1, AW = $87,500 and n = (forever) • Alternative 2, PW = -$895,000 and n = (forever) • Alternative 3, First cost (FC) of $900,000, annual operating savings of 3,000 per year, salvage = $200,000, and n = (forever) Alternative 2 Alternative 3 None of them Alternative 1 QUESTION...
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