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Compare 10 years the alternatives C and D on the basis of a present worth analysis...
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 $...
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 12% per year and a study period of 10 years. с $-44,000 $-12,000 $-34,000 $-7,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-1,500 $-1,200 $5,000 10 $1,200 5 The present worth of alternative C is $ -134497.32 and that of alternative D is $...
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.
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
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year....
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 First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Inve...
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...
Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of y...
Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
spreadsheet solution? 5.35 Compare the alternatives shown on the basis of their capitalized costs using an...
spreadsheet solution?
5.35 Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per year. Alternative M -150,000 -50.000 Alternative N -800,000 -12.000 First cost, S Annual operating cost, $ per year Salvage value, $ Life, years 8,000 1,000,000
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 $...
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 12% per year and a study period of 10 years. с $-44,000 $-12,000 $-34,000 $-7,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-1,500 $-1,200 $5,000 10 $1,200 5 The present worth of alternative C is $ -134497.32 and that of alternative D is $...
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.
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
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 First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
Compare alternatives A and B with the present worth method if the MARR is 10% per year. Which one would you recommend? Assume repeatability and a study period of 20 years $15,000 $45,000 Capital Investment Operating Costs $4,000 at end of year 1 and increasing by $400 per year thereafter $4,000 every 5 years 20 years $8,000 at end of year 1 and increasing by $800 per year thereafter None Overhaul Costs Life 10 years Salvage Value $8,000 if just...
Compare alternatives A and B with the present worth method if the MARR is 11% per year. Which one would you recommend? Assume repeatability and a study period of 12 years. $25,000 $10,000 at end of year 1 and increasing by $1,000 per year thereafter None Capital Investment Operating Costs $55,000 $5,000 at end of year 1 and increasing by $500 per year thereafter $5,000 every 3 years 12 years $10,000 if just overhauled Overhaul Costs Life 6 years negligible...
spreadsheet solution?
5.35 Compare the alternatives shown on the basis of their capitalized costs using an interest rate of 10% per year. Alternative M -150,000 -50.000 Alternative N -800,000 -12.000 First cost, S Annual operating cost, $ per year Salvage value, $ Life, years 8,000 1,000,000
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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