! Required information For equipment that has a first cost of $19,000, the estimated operating costs and year-end salvage values are as shown. Year Operating Cost, $ Salvage Value, $ 1 -1,000 7,000 12 -1,200 5,000 3 -1,300 4,500 -2,000 3,000 -3,000 2,000 Determine the economic service life ESL, at i= 10% per year using factors. The economic service life ESL, is years with the AW = $-
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- For equipment that has a first cost of $13,000, the estimated operating costs and year-end salvage values are as shown. Year Operating Cost, $ Salvage Value, $ 1 -1,000 7,000 2 -1,200 5,000 3 -1,300 4,500 4 -2,000 3,000 5 |-3,000 2,000 Determine the economic service life ESL, at /= 10% per year using factors. The economic service life ESL, is [ years with the AW = $-|For equipment that has a first cost of $11,500, the estimated operating costs and year-end salvage values are as shown. Year 1 2345 Operating Cost, $ Salvage Value, $ -1,000 7,000 -1,200 5,000 -1,300 4,500 -2,000 3,000 -3,000 2,000 etermine the economic service life ESL, at /= 10% per year using factors. he economic service life ESL, is years with the AW= $-|5.24 For the cash flows below, use an annual worth comparison to determine which alternative is best at an interest rate of 1% per month. First cost, $ M&O costs, $/month Overhaul every 10 years, $ Salvage value, $ Life, years X -90,000 -400,000 -30,000 -20,000 - Y 7000 3 - 25,000 10 Z -900,000 -13,000 -80,000 200,000 8
- The Briggs and Stratton Commercial Division designs and manufacturers small engines for golf turf maintenance equipment. A robotics-based testing system with support equipment will ensure that their new signature guarantee program entitled "Always Insta-Start" does indeed work for every engine produced. First cost of equipment AOC per Year Salvage Value Estimated Life Pull System $-1,800,000 $-620,000 $90,000 8 years Push System $-2,500,000 $-640,000 $130,000 8 years Determine the salvage value for the push system that will make the company indifferent to the two systems. Also, MARR = 13% per year. The salvage value for the push system is determined to be $ in $1000 units.New microelectronics testing equipment was purchased 2years ago by Mytesmall Industries at a cost of $600,000. Atthat time, it was expected to be used for 5 years and thentraded or sold for its salvage value of $75,000. Expandedbusiness in newly developed international markets is forcingthe decision to trade now for a new unit at a cost of$800,000. The current equipment could be retained, ifnecessary, for another 2 years, at which time it would have a$5000 estimated market value. The current unit is appraisedat $350,000 on the international market, and if it is used foranother 2 years, it will have M&O costs (exclusive ofoperator costs) of $125,000 per year. Determine the valuesof P, n, S, and AOC for this defender if a replacementanalysis were performed today. P = market value =$350,000AOC = $125,000 per yearn = 2 yearsS = $5,000A contractor has a 4-year concrete mixer whose first cost was $6,000, having 3 more years to live before being scrapped and sold at $801. Itcould now be sold for $11,922. It has an annual cost for operation and maintenance of $9,352. Its replacement is being proposed with a newmachine whose first cost will be $8,000 having a life of 9 years and salvage value $1,600. It has an operating cost of $800 per year andmaintenance cost of $320 per year. Ifthe interest is 20% cpd-a, what is the Annual Equivalent Cost of the Old Machine? 14,792
- For th below two machines and based on AW analysis which machine we should select? MARR=10% Machine A Machine B 26612 12417 4135 Life, years 3 Answer the below question: B- the AW for machine B= First cost, $ Annual cost, $/year Salvage value, $ 140454 7170 infiniteTwo lathes are being considered in the manufacture of certain machine parts. Data is given below, all cost in peso: LATHE A LATHE B First Cost 40,000 56,000 Salvage Value 5,000 7,000 Annual Maintenance 2,000 2,800 Operation, Cost/hour 4 3.5 Life, in years 10 12 Time per part (hours) 0.40 0.25 REQUIRED: Determine the number of machine parts/year that could be produced so that 2 lathes will be equally economical if the MARR is 18%. Use AWM If the number of parts is 10,000 units, which lathe will you recommend? Use ROR If the number of parts is 10,000 units, which lathe will you recommend? Use PWM If the number of parts is 10,000 units, which lathe will you recommend? Use EUAC. Calculate the present worth of a machine which costs $105000 initially and has a 10 year life with a $20000 salvage value. The operating cost of the machine is expected to be $6000 in year 1(end_of_year) and $6600 in year 2, and amounts increasing by the 10% through its 10-year life. Use an interest rate of 12% per year. Please don't used excel I again says don't used excel
- How to get to the answer? Please show full working. Answer should be C) Replacement every 3 years, $130, 364/year. Hand written plzAn industrial plant has hired a contractor to develop a solar farm nearby for the industrial plant's electricity consumption. During the 3-year construction period, the contractor will need to run a generator to power the construction. Cost analysis from previous projects indicates: Generator Brand Installed Cost Cost per Hour A $22M $500 B $23M $400 C $25M $250 D $30M $150 At the end of 3 years, the generator will have a salvage value equal to the cost of removing them. The generator will operate 6,000 hours per year. The lowest interest rate at which the contractor is willing to invest money is 7%. (The minimum required interest rate for invested money is called the minimum attractive rate of return, or MARR.) Select the alternative with the least present worth of cost. O Choice "C" with $29,872,900 O Choice "B" with $29,298,320 O Choice C with $28,936,450 O Choice "A" with $29,872,900Question 3 For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A Answer the below questions: A- PW for machine A= 23,979 8,679 4,000 Machine B 30000 6,000 5,000 Machine C 10000 4,000 1,000