which of the alternatives is the lowest cost based on present worth (PW)? Alternative A has...
which of the alternatives is the lowest cost based on present worth (PW)? Alternative A has initial cost of $5, daily maintence of 0.25% and salvage value of .75 at the end of two weeks. The interest for Alt A is .25%. Alt B has initial cost of 5.50, weekly maintence of 1.50 an salvage value of 1.00 at the end of three weeks. the interest rate of AltB is 1 3/4% per week.
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which of the alternatives is the lowest cost based on present
worth (PW)? Alternative A has...
Perform a present worth (PW)-based evaluation of the two alternatives below using a spreadsheet. The after-tax...
Perform a present worth (PW)-based evaluation of the two alternatives below using a spreadsheet. The after-tax minimum acceptable rate of return (MARR) is 8% per year, Modified Accelerated Cost Recovery System (MACRS) depreciation applies, and Te = 40%. The (GI-OE) estimate is made for the first 3 years; it is zero in year 4 when each asset is sold. Alternative X Y First Cost, $ -8,000 -13,000 Salvage Value, Year 4, $ 0 2,000 GI-OE, $ per Year 3,500 5,000...
Given the following two alternatives, the present worth (PW) of E2 is closest to: Hint: Use...
Given the following two alternatives, the present worth (PW) of E2 is closest to: Hint: Use positive sign for cash outflow and negative sign for cash inflow. Alt. E1 Alt. E2 Capital Investment, $ 6,000 12,000 Annual Expenses, $ 150 175 Useful life, years Salvage value, $ none 4,000 Less than $10,000 Between $10,000 - $10,500 Between $10,500 - $11,000 O Greater than $11,000
In comparing alternatives, I and J by the present worth method, the equation that yields the...
In comparing alternatives, I and J by the present worth method, the equation that yields the present worth of alternative Jis Alternativel Alternative Initial cost, -150,000 - 250.000 Annual Income, $ per year 20,000 40,000 Annual expenses, $ per year --9,000 -14,000 Salvage value, $ 25,000 35,000 Life, years The interest rate is 15% per year. (PWJ--250,000 + 40,000(PIA ,15%.6) + 35,000(P/F.15%,6 (PWJ --250,000 + 26,000(PIA ,15%,6) +35,000(P/F.15%,6 (PWJ - -250,000 - 26,000(PLA 15%,6) +35,000(P/F ,15%,6 (PW J--250,000 - 26,000(PIA...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an initial cost of $6,000. The salvage value after 7 years is expected to be $200. The operating costs including operator wages, routine maintenance, overhauls, etc., is expected to be $2,000 per year. It is expected that this machine will encourage the purchase of an additional 50 drinks per week costing $2.00 apiece to produce and for which $6.00 can be charged. Alternatively, a completely...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an initial cost of $6,000. The salvage value after 7 years is expected to be $200. The operating costs including operator wages, routine maintenance, overhauls, etc., is expected to be $2,000 per year. It is expected that this machine will encourage the purchase of an additional 50 drinks per week costing $2.00 apiece to produce and for which $6.00 can be charged. Alternatively, a completely...
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 $
Compare the two following alternatives in terms of present worth using MARRz 6% for a study...
Compare the two following alternatives in terms of present worth using MARRz 6% for a study period of 2 years Assume that the salvage value does not change depending on year sold. Alternative 1: First cost: 30,000 Yearly cost 6,000 Salvage value: 9,000 Lifetime: 4 Alternative 2 First cost: 75,000 Yearly cost 5,000 Salvage value: 13,000 Lifetime: 6 Alt 1: $-80,826, Alt 2: S-105,266 Alt 1: S-74,228, Alt 2: S-112,525 Alt 1 S-32,990, Alt 2: S-72,597 Alt 1: S-102,270, Alt...
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...
Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are...
Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 3% per year, compounded annually. Alternative A 340 60 Alternative B 870 182 Initial Cost Annual Benefit Salvage Value Useful Life (yrs) 78 106 Alternative A, because it costs $65.43 less than Alternative B, in terms of present worth Alternative B, because it costs $65.43 more than Alternative A, in...
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.
Given the following two alternatives, the present worth (PW) of E2 is closest to: Hint: Use positive sign for cash outflow and negative sign for cash inflow. Alt. E1 Alt. E2 Capital Investment, $ 6,000 12,000 Annual Expenses, $ 150 175 Useful life, years Salvage value, $ none 4,000 Less than $10,000 Between $10,000 - $10,500 Between $10,500 - $11,000 O Greater than $11,000
In comparing alternatives, I and J by the present worth method, the equation that yields the present worth of alternative Jis Alternativel Alternative Initial cost, -150,000 - 250.000 Annual Income, $ per year 20,000 40,000 Annual expenses, $ per year --9,000 -14,000 Salvage value, $ 25,000 35,000 Life, years The interest rate is 15% per year. (PWJ--250,000 + 40,000(PIA ,15%.6) + 35,000(P/F.15%,6 (PWJ --250,000 + 26,000(PIA ,15%,6) +35,000(P/F.15%,6 (PWJ - -250,000 - 26,000(PLA 15%,6) +35,000(P/F ,15%,6 (PW J--250,000 - 26,000(PIA...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an initial cost of $6,000. The salvage value after 7 years is expected to be $200. The operating costs including operator wages, routine maintenance, overhauls, etc., is expected to be $2,000 per year. It is expected that this machine will encourage the purchase of an additional 50 drinks per week costing $2.00 apiece to produce and for which $6.00 can be charged. Alternatively, a completely...
SITUATION: Two alternatives for a margarita mixer are under consideration. One system, the Mixer-Plus has an initial cost of $6,000. The salvage value after 7 years is expected to be $200. The operating costs including operator wages, routine maintenance, overhauls, etc., is expected to be $2,000 per year. It is expected that this machine will encourage the purchase of an additional 50 drinks per week costing $2.00 apiece to produce and for which $6.00 can be charged. Alternatively, a completely...
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 $
Compare the two following alternatives in terms of present worth using MARRz 6% for a study period of 2 years Assume that the salvage value does not change depending on year sold. Alternative 1: First cost: 30,000 Yearly cost 6,000 Salvage value: 9,000 Lifetime: 4 Alternative 2 First cost: 75,000 Yearly cost 5,000 Salvage value: 13,000 Lifetime: 6 Alt 1: $-80,826, Alt 2: S-105,266 Alt 1: S-74,228, Alt 2: S-112,525 Alt 1 S-32,990, Alt 2: S-72,597 Alt 1: S-102,270, Alt...
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...
Use Present Worth Analysis to determine whether Alternative A or B should be chosen. Items are identically replaced at the end of their useful lives. Assume an interest rate of 3% per year, compounded annually. Alternative A 340 60 Alternative B 870 182 Initial Cost Annual Benefit Salvage Value Useful Life (yrs) 78 106 Alternative A, because it costs $65.43 less than Alternative B, in terms of present worth Alternative B, because it costs $65.43 more than Alternative A, in...
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.
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