Two methods can be used to produce expansion anchors. Method A costs $ 70, 000 initially and will have a $15,000 salvage value after 3 years. The operating cost with this method will be $36,000 in year 1, increasing by $3200 each year. Method B will have a first cost of $ 116,000, an operating cost of $10000 in year 1, increasing by $10000 each year, and a $46, 000 salvage value after its 3 year life. At an interest rate of 8% per year, which method should be used on the basis of a present worth analysis?
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- Two techniques can be used to produce expansion anchors. Technique A costs $90,000 initially and will have a $12,000 salvage value after 3 years. The operating cost with this method will be $33,000 in year 1, increasing by $2600 each year. Technique B will have a first cost of $113,000, an operating cost of $7000 in year 1, increasing by $7000 each year,and a $43,000 salvage value after its 3-year life. At an interest rate of 13% per year, which technique should be used on the basis of a present worth analysis? Notice that there are no revenues. Please work out and do not use excel, however if you use excel please show how to input everything needed down to the formula, thank you!Two methods can be used to produce expansion anchors. Method A costs $70,000 initially and will have a $11,000 salvage value after 3 years. The operating cost with this method will be $22,000 in year 1, increasing by $2600 each year. Method B will have a first cost of $123,000, an operating cost of $8000 in year 1, increasing by $6500 each year, and a $33,000 salvage value after its 3-year life. At an interest rate of 15% per year, which method should be used on the basis of a present worth analysis? The present worth for method A is $ The present worth for method B is $[ Method (Click to select) is used to produce expansion anchors.Two methods can be used to produce expansion anchors. Method A costs $80,000 initially and will have a $15,000 salvage value after 3 years. The operating cost with this method will be $30,000 in year 1, increasing by $4000 each year. Method B will have a first cost of $120,000, an operating cost of $8000 in year 1, increasing by $6500 each year, and a $40,000 salvage value after its 3-year life. At an interest rate of 12% per year, which method should be used on the basis of a present worth analysis?
- One of two methods must be used to produce expansion anchors. Method A costs $80,000 initially and will have a $15,000 salvage value after 3 years. The operating cost with this method will be $30,000 per year. Method B will have a first cost of $120,000, an operating cost of $8000 per year, and a $40,000 salvage value after its 3-year life. At an interest rate of 12% per year, which method should be used on the basis of a present worth analysis? Also, write the two spreadsheet functions to perform the PW analysis.One of two methods must be used to produce expansion anchors. Method A costs $75,000 initially and will have a $19,000 salvage value after 3 years. The operating cost with this method will be $23,000 per year. Method B will have a first cost of $110,000, an operating cost of $19,000 per year, and a $31,000 salvage value after its 3-year life. The interest rate for both the methods is 11%. Which method should be used on the basis of a present worth analysis? The present worth of method A is (about -117,000) and that of method B is (is not about -172,000)Trailer Treasures Inc. is considering two projects and must do one of them. Project A requires an investment of $35,000. Estimated annual receipts for 5 years are $14,000; estimated annual costs are $4,500. Alternatively, Project B requires an investment of $70,000 has annual receipts for 5 years of $20,000, and has annual costs of $4,500. Assume both proejcts have $10,000 salvage value and that MARR IS 15%/year. 1. What is the annual worth of Project A? 2. What is the annual worth of Project B? 3. Which project should be recommended? Why? Please show work through excel
- A "standard" model of a dozer costs $20,000 and has an annual operating expense of $450. The dozer will be replaced in 6 years when the salvage value is expected to be $2,000. A "super" model can be purchased for $25,000, but will have a salvage value of $7,000 when retired in 6 years. Its operating expenses are also $450 a year. The purchaser's other investment opportunities are 5%. Compare these alternatives by using the annual equivalent method.A project has an initial investment of $45,000. This project needs an annual spending of $6,000 to generate an annual revenue of $18,000 for six years. Moreover, the project is expected to return $12,000 as a salvage value at the EOY 6. Calculate the AW using MARR of 10%Machine “A” has a starting cost of $ 50,000 with an estimated period of 12 years and a salvage value of $ 14,000. Annual O&M costs are $ 6,000 the first year, $ 6,300 the second year, and so on. Another Alternative is machine “B” which has an initial cost of $ 30,000 and a recovery value of $ 3,000 at the end of the 12-year service period, the annual operating and maintenance expenses are $ 8,000 the first year, from $ 8,500 the second year and so on. Using an interest rate of 13% per year. Do a Present Value analysis, and determine the value of machine A Possible results $88,760.84 $89,543.09 $88,543.09
- A new production system for a factory is to be purchased and installed for $135331 . This system will save approximately 300,000kWh of electric power each year for a 6-year period. Assume the cost of electricity is $0.10 per kWh , and factory MARR is 15% per year, and the salvage value of the system will be $8662 at year 6 . Using the PW method to analyzes if this investment is economically justified A- calculate the PW of the above investment and insert the result below. _______________________An injection molding system has a first cost of $200,000 and an annual operating cost of $85,000 in years 1 and 2, increasing by $4,500 per year thereafter. The salvage value of the system is 25% of the first cost regardless of when the system is retired within its maximum useful life of 5 years. Using a MARR of 9% per year, determine the ESL and the respective AW value of the system. The ESL is year(s) and AW value of the system is $Two different mutually exclusive alternatives can be used for producing a certain machine part. Alternative A will have a first cost of $60,000 with a $15,000 salvage value after its 3 year life. The operating cost with this method will be $35,000 per year. Alternative B will cost $45,000, but it will last only 2 years. Its salvage value is $10,000 with an operating cost of $25,000 per year. Using an MARR value of 12% per year, which method should be selected?