Please dont use excel,show me the formula used
Homework Answers
Since the life of the projects are unequal, then LCM method will be used to make lives equal. As per the LCM method, the lives of each project will be considered to be 6 years.
R = 8%
Future value of alternative P = - Future worth of the initial investments - Future worth of the annual operating cost + future worth of the salvage value
Future value of alternative P = (-23000*1.08^6 - 23000*1.08^3) - 4000*(1.08^6 - 1)/.08 + (3000*1.08^3 + 3000)
Future value of alternative P = -$88036.1
Future value of alternative Q = - Future worth of the initial investments - Future worth of the annual operating cost + future worth of the salvage value
Future value of alternative Q = -30000*1.08^6 - 2500*(1.08^6 -1)/.08 + 1000
Future value of alternative Q = -$64946.1
Since the future value is cost effective more of alternative Q than that of alternative P, then alternative Q should be selected.
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Please dont use excel,show me the formula used
7. Compare the alternatives shown below on the...
please dont use excel, show me the formula used 5. Machines that have the following costs...
please dont use excel, show me the formula used
5. Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis. Machine X Machine Y First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years - 250,000 --60.000 70.000 -430.000 -40,000 95.000
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
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...
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 $...
For the below Me alternatives, which machine should be selected based on the future worth analysis....
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
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
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 $
dont use excel 5.14 Tivo processes can be used for producing a polyuner that reduces friction...
dont use excel
5.14 Tivo processes can be used for producing a polyuner that reduces friction loss in en. gines. Process K will have a first cost of $160.000, an operating cost of $7000 per quarter, and a salvage value of $40,000 after its 2-yew life. Process I will have a first cost of $210,000, an operating cost of $5000 per quarier, and a $26,000 salvage value after its 4-year life. Which process should be selected on the basis of...
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 $...
please dont use excel, show me the formula used
5. Machines that have the following costs are under consideration for a robotized welding process. Using an interest rate of 10% per year, determine which alternative should be selected on the basis of a present worth analysis. Machine X Machine Y First cost, $ Annual operating cost, $ per year Salvage value, $ Life, years - 250,000 --60.000 70.000 -430.000 -40,000 95.000
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
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...
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 $...
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
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
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 $
dont use excel
5.14 Tivo processes can be used for producing a polyuner that reduces friction loss in en. gines. Process K will have a first cost of $160.000, an operating cost of $7000 per quarter, and a salvage value of $40,000 after its 2-yew life. Process I will have a first cost of $210,000, an operating cost of $5000 per quarier, and a $26,000 salvage value after its 4-year life. Which process should be selected on the basis of...
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 $...
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