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A: Method DDM LS First Cost 140000 480000 M&O 30000 35000 Salvage Value 6000 29000 Life 2 4
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Q: A new antitheft system incorporating MEMS technology is being separately evaluated economically by…
A: Present value, is a calculation that actions the value of a future amount of cash or stream of…
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Q: Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with…
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Q: A remotely located air sampling station can be powered by solar cells or by running an electric line…
A: Future value can be calculated by using the following formula.
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A: *Answer: Given: Process T: First cost =$ 750,000 , Operating cost =$ 60,000 per year , Salvage…
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A: The present worth of each option is the discounted value of future cash flows. It is the present…
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Q: Two methods can be used to produce expansion anchors. Method A costs $90,000 initially and will have…
A: Present worth analysis is a method which compare the cash flow of each alternatives and compare…
Q: Two processes can be used for producing a polymer that reduces friction loss in engines. Process T…
A: Net Present Worth (NPW) serves as a crucial financial metric utilized in capital budgeting,…
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Q: If a project is evaluated using annual worth method, then we know that: Select one: a. Present…
A: If a project is evaluated using annual worth method, then we know that: Select one: a. Present worth…
Q: Two processes can be used for producing a polymer that reduces friction loss in engines. Process T…
A: In the above question, the following equation for present worth calculation Where PVAF= Present…
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TRUE OR FALSE:
- If there is a past estimation error, there is a need to replace the property immediately.
- The physical life is always greater than all the other “life” factors under replacement analysis.
- If breakeven analysis is conducted with PW analysis for different MEAs, it doesn’t guarantee that the fastest
alternative to reach its breakeven point has also the highest equivalent worth.
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Solved in 3 steps
- A portable concrete test instrument used in construction for evaluating and profiling concrete surfaces (MACRS-GDS 5-year property class) is under consideration by a construction firm for $25,500. The instrument will be used for 6 years and be worth $2,500 at that time. The annual cost of use and maintenance will be $13,000. Alternatively, a more automated instrument (same property class) available from the manufacturer costs $26,000, with use and maintenance costs of only $7,500 and salvage value after 6 years of $3,000. The marginal tax rate is 25%, and MARR is an after-tax 12%. Determine which alternative is less costly, based upon comparison of after-tax annual worth. Show the AW values used to make your decision.A new antitheft system incorporating MEMS technology is being separately evaluated economically by three engineers at Dragon Technologies. The first cost of the equipment will be $90,000, and the life is estimated at 6 years with a salvage value of $9000. The engineers made different estimates of the net savings that the equipment might generate. Jacob made an estimate of $9,000 per year. Susan states that this is too low and estimates $18,000, while Tyler estimates $21,000 per year before tax. If the MARR is 8% per year, use PW to determine if these different estimates will change the decision to purchase the equipment. The present worth of the pessimistic estimate is $ The present worth of the most likely estimate is $ The present worth of the optimistic estimate is $ The equipment purchase by the pessimistic estimate is not justified The equipment purchase by the most likely estimate is not justified The equipment purchase by the optimistic estimate is justifiedTwo processes can be used for producing a polymer that reduces friction loss in engines. Process T will have a first cost of $660,000, an operating cost of $90,000 per year, and a salvage value of $80,000 after its 2-year life. Process W will have a first cost of $1,100,000, an operating cost of $25,000 per year, and a $120,000 salvage value after its 4-year life. Process W will also require updating at the end of year 2 at a cost of $90,000. Which process should be selected on the basis of a present worth analysis at a MARR of 12% per year? The present worth of process T is $- and the present worth of process W is $- The process selected on the basis of the present worth analysis is process (Click to select) ✓
- Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with low friction properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 16% per year, which alternative has the better present worth and what is that value (select the closest value)? Method First Cost AOC, per Year Salvage Value Life DDM $170,000 $65,000 $4,000 2 years LS $350,000 $40,000 $29,000 4 years DDM with a PW--$473,000 LS with a PW --$445,900 LS with a PW --$222,055 DDM with a PW--$109,300An IE works in the automation department of a surgical equipment manufacturing company that produces specially ordered equipment for hospitals. To upgrade the quality of the assembly process of the camera used in laparoscopic surgery probes, two approaches are available: make and buy. The make alternative has an initial equipment cost of $175,000, a life of 5 years, a $25,000 salvage value, a processing cost of $3,000 per camera, and an M&O cost of $90,000. The buy alternative requires contracting the assembly operation externally at a cost of $7,250 per camera. If the MARR is 12% per year, how many cameras per year must be assembled to justify the make alternative? 27 cameras must be assembled per year to justify the make alternative.Dexcon Technologies, Inc., is evaluating two alternatives to produce its new plastic filament with tribological (ie.. low friction) properties for creating custom bearings for 3-D printers. The estimates associated with each alternative are shown below. Using a MARR of 10% per year, which alternative has the lower present worth? Method First Cost M&O Cost, per Year Salvage Value Life DDM $-190,000 $-55,000 $4,000 2 years The present worth for the DDM method is $ The present worth for the LS method is $ The LS method is selected. LS $-430,000 $-25,000 $39,000 4 years
- Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $18,000 salvage value after 3 years. The operating cost with this method will be $29,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $130,000, an operating cost of $6000 in year 1, increasing by $6000 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 14% per year, what are the present worth of Method A and Method B? a. PWA = $-133,655 PWB = $-129,647 PWB = $-125,178 PWB = $-129,647 PWB = $-125,178 O b. PWA $-133,655 = O C.PWA $-150,260 O d. PWA = $-150,260 =An industrial machine costing $10,000 will produce net cash savings of $4,000 per year. The machine has a five-year useful life but must be returned to the factory for major repairs after three years of operation. These repairs cost $5,000. The company's MARR is 10% per year. What IRR will be earned on the purchase of this machine? Analyze the sensitivity of IRR to ± $2,000 changes in the repair cost. Perform the sensitivity analysis. Fill-in the table below. (Round to one decimal place.) Change in the repair cost - $2,000 $0 IRR 21.05% 9.61 % Is the project acceptable? Yes NoBased on company records of similar equipment, a consulting aerospace engineer at Aerospatiale estimated AW values for a presently owned, highly accurate steel rivet inserter as shown. A challenger has ESL = 2 years and AWC= $−41,300 per year. The MARR is 12% per year. If Retained ThisNumber of Years The AW Value Is,$ per Year 1 −62,000 2 −51,000 3 −49,000 4 −53,000 5 −70,000 When should the next replacement evaluation take place, and under what assumption? The next replacement evaluation should take place in (Click to select) 4 2 3 5 years, and under the assumption that the (Click to select) defender challenger estimates do not change.
- A remotely located air sampling station can be powered by solar cells or by running an electric line to the site and using conventional power. Solar cells will cost $7000 to install and will have a useful life of 4 years with no salvage value. Annual costs for inspection, cleaning, etc. are expected to be $1600. A new power line will cost $14000 to install, with power costs expected to be $1300 per year. Since the air sampling project will end in 4 years, the salvage value of the line is considered to be zero. At an interest rate of 8.00% per year, which alternative should be selected on the basis of a future worth analysis? (Include a minus sign if necessary.) The future worth of solar cells is $[ and that of electric line is $ (Click to select) :) should be selected on the basis of a future worth analysis.A portable concrete test instrument used in construction for evaluating and profiling concrete surfaces (MACRS-GDS 5-year property class) is under consideration by a construction firm for $24,000. The instrument will be used for 6 years and be worth $1,000 at that time. The annual cost of use and maintenance will be $11,000. Alternatively, a more automated instrument (same property class) available from the manufacturer costs $28,500, with use and maintenance costs of only $7,500 and salvage value after 6 years of $2,500. The marginal tax rate is 25%, and MARR is an after-tax 12%.Determine which alternative is less costly, based upon comparison of after-tax annual worth. select an alternative Show the AW values used to make your decision:Alternative 1: $enter a dollar amount Alternative 2:A textile processing company is evaluating whether it should retain the current bleaching process that uses chlorine dioxide or replace it with a proprietary oxypure process. The relevant information for each process is shown. Use an interest rate of 15% per year to perform the replacement study. Additionally, write the PMT functions that display the information necessary to make the decision. Process Current Oxypure Original cost 6 years ago, $ 450,000 — Investment cost now, $ — 600,000 Current market value, $ 25,000 — Annual operating cost, $/year 190,000 70,000 Remaining life, years 3 10 Salvage value, $ 0 50,000