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- A piece of equipment has a first cost of $75000, a maximum useful life of 4 years, and a market (salvage) value described by the relation Sk = 60000 – 10500k, where k is the number of years since it was purchased. The AOC series is estimated using AOC = 30000 + 4500k. The interest rate is 9% per year. When should the company replace this asset?From the given values below, which machine should be selected using (a) Present worth Method and (b) Future worth Method if interest rate is 10% per year? Machine Initial Cost (Php) Annual Operating Cost A В 146,000 15,000 220,000 10,000 75,000 25,000 Annual Revenue 80,000 10,000 Salvage Value Useful Life (years) 6An asset which has a first cost of RM 40,000 is expected to have an annual operating cost of RM 15,000 per year. It will provide the needed service for a maximum of 6 years. If the salvage value changes as shown below , determine the economic life of the asset at 20 % per year.
- With the estimates shown below, Sarah needs to determine the trade-in (replacement) value of machine X that will render its AW equal to that of machine Y at an interest rate of 13% per year. Determine the replacement value. Market Value, $ Annual Cost, $ per Year Salvage Value Life, Years The replacement value is $ Machine X ? -58,000 14,500 3 Machine Y 84,000 -40,000 for year 1,increasing by 2000 per year thereafter. 20,000 5Consider palletizer at a bottling plant that has a first cost of $134,338, has operating and maintenance costs of $17.980 per year, and an estimated net salvage value of $37,509 at the end of 30 years. Assume an interest rate of 8%. What is the future worth of this project? Enter your answer as follow: 123456.78A company is considering two methods for obtaining a certain part on annual basis. Method A will involve purchasing a machine for $60K with a life of 5 years, a $3K salvage value and a fixed annual operating cost of $10K. Additionally, each part produced by the method will cost $10. Method B will involve purchasing the part from a subcontractor for $27 per part. At an interest rate of 10% per year, What is the number of parts (in x thousands) per year required for the two methods to break even ? A. 1480 B.1490 C. 1500 which is it and why
- With the estimates shown below, Sarah needs to determine the trade-in (replacement) value of machine X that will render its AW equal to that of machine Y at an interest rate of 8% per year. Determine the replacement value. Machine X Machine Y 90,000 Market Value, S -40,000 for year 1,increasing by 2000 per year thereafter. 24,000 Annual Cost, $ per Year -59,500 Salvage Value Life, Years 19,500 The replacement value is $Alternative X has a first cost of 22000 an annual operating cost of 4100 , and a salvage value of 6825 after 27 year. Alternative Y has a first cost of 23000 an annual operating cost of 4200 , and a salvage value of 16560 after 27 year. If MARR of 27% per year, approximately what is the PW of each alternative?An equipment which can be purchase for P700,000 is expected to generate a net cash flow of P200,000 annually for five years which is the estimated service life of the equipment. Its salvage value at the end of the service life is estimated to be 5% of its purchased cost. a. What is the rate of return of the initial investment? b. What is the simple pay-back period? c. If the company's minimum attractive rate of return(MARR) is set at 15%, using NPW is this investment acceptable? d. What is the internal rate of return(IRR) of this machine? e. What is the external rate of return(ERR) at the 15% MARR?
- A company wants to purchase a piece of equipment that costs $18,000. In 8 years, that same piece of equipment is expected to have a salvage value of $5,000. The estimated annual maintenece cost is $1,000 in the first year, but is expected to increase by $300 each year thereafter. What is the present worth of the project, using a 10% interest rate.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.A firm is trying to decide between two equipment for a production activity. The following information is available. Equip 1 Equip 2 Initial investment $25,000 $17,000 7 years 7 years $0 Useful life Salvage value Operating cost per hour $0 $4.30 $5.00 At a MARR of 8% per year, determine the breakeven number of operating hours per year. Click the icon to view the interest and annuity table for discrete compounding when i = 8% per year. The breakeven number of operating hours per year is O A. 1,866 O B. 2,195 O C. 1,756