QUESTION 3 For the below ME alternatives, which machine should be selected based on the AW analys Machine A Machine B Machine C First cost, $ 15,828 30000 10000 Annual cost, $/year Salvage value, s Life, years 8,753 6,000 4,000 4,000 5,000 1,000 Answer the below questions: A- AW for machine A= QUESTION 4 For the below ME alternatives, which machine should be selected based on the AW analysis. Machine A Machine B Machine C First cost, $ Annual cost, $/year 15000 21,344 10000 8,314 6,000 4,000 Salvage value, $ Life, years 4,000 5,000 1,000 Answer the below questions: B- AW for machine B=
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- Chambers Company has just gathered estimates forconducting a break-even analysis for a new product.Variable costs are $7 a unit. The additional plant willcost $48,000. The new product will be charged $18,000a year for its share of general overhead. Advertisingexpenditures will be $80,000, and $55,000 will be spenton distribution. If the product sells for $12, what is thebreak even point in units? What is the break even pointin dollar sales volume?BASED ON ESTIMATES THE DATA FOR TWO TYPES OF BRIDGES WITH DIFFERENT LIVES ARE AS FOLLOWS. IFTHE MINIMUM RATE OF RETURN IS 9%, DETERMINE W/C PROJECT IS MORE DESIRABLE. TIMBER BRIDGE STEEL BRIDGEFIRST COST P 50,000.00 P 140,000.00SALVAGE VALUE 2,000.00 10,000.00LIFE IN YEARS 12 36ANNUAL MAINTENANCE 6,000.00 2,500.00EVALUATE USING:A.) THE ANNUAL COST METHODB.) PRESENT WORTH COST METHODC.) RATE OF RETURN METHODGiven the two machines’ data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value 0 2,000.00 Annual operation 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B: note:round off final answer to 2 decimal ANSWER for ALTERNATIVE A: ANSWER for ALTERNATIVE B:
- Given EVM data, estimate the cost at completion of a project using the method "EAC based on combined cost and schedule performance." BCWS = $82.2M BCWP = $70.5M ACWP = $95.5M BAC = $105.5M A) $130M B) $150M C) $160M D) $170MYou need to determine whether a project is profitable or not in a long run. Based on the data given, which of theseprojects will be profitable according to engineering economy methods?a. θ = 3 yrs., Net Value: 0b. Accumulated (without interest) net values of the revenues, expenses and investments after 5 years is +300.c. Accumulated (without interest) net values of the revenues, expenses and investments after 4 years is +100.d. θ = 6 yrs., Net Value: +400A fabrication company engaged in production with a capacity of 150, 000 pieces per year. But, it is just operating at 70% of its full capacity. The company has an annual income of P 250, 000.00, annual fixed cost are P 50, 000.00 and variable costs are P 1.00 per unit. How many productions of parts must be produced for break-even point? Given: Required: Solution: refer to this textbook: https://drive.google.com/file/d/1h4ra80IE8IRtYyja16iK6TtjCrTDi73j/view?usp=sharing
- 1. The Present Worth Method A project your firm is considering for implementation has these estimated costs and revenues: an investment cost of $50,000; maintenance costs that start at $5,000 at the end of year (EOY) 1 and increase by $1,000 for each of the next 4 years, and then remain constant for the following 5 years; savings of $20,000 per year (EOY 1–10); and finally a resale value of $35,000 at the EOY 10. If the project has a 10-year life and the firm’s MARR is 10% per year, what is the present worth of the project? Is it a sound investment opportunity?Solar plant investment or cost establish 40000$, Maintenance Cost Expected customer Demand (KWH) Year 0 0 0 350 1500 1 300 1200 2 200 1000 3 200 800 4 150 700 5 What is the cost Consumption Kilowatt Per Hour? please tell me how can I solve it , as you know we have Present worth, annual worth and future worth! please i need correct answer because i have post it before and they give me wrong answer , do not search it because these number is randomly , i am looking for the way of answer in case of the question has change that i can solve it by your clarification . rate 10%Given the two machines’ data Machine A Machine B First Cost P8,000.00 P14,000.00 Salvage value 0 2,000.00 Annual operation 3,000.00 2,400.00 Annual maintenance 1,200.00 1,000.00 Taxes and insurance 3% 3% Life, years 10 15 Money is worth at least 16% Using equivalent uniform annual cost method, determine the value of alternative A and alternative B: ANSWER for ALTERNATIVE A: Blank 1 ANSWER for ALTERNATIVE B: Blank 2
- Alternative 3 is incorrect EUAC=( Equivalent Annaul COst of Initial Investment)+ (Expected Moderate annaual flood damage cost )+ (expected severe annaual flood cost )Consider these two alternatives.Alternative A Alternative BCapital investment OMR 6000 7500Annual revenues OMR 1800 2250Annual expenses OMR 500 750Estimated market valueOMR1200 1600Useful life 10 10MARR 12% 1. Recommend which alternative should be selected.2. How much capital investment of the expensive alternative have to vary so that theinitial decision would be reversed.E2 A steel bridge on Louisiana state highway near the Gulf of Mexico is costing $450.000 yearlyin maintenance large chipping, priming, and painting. It originaly cost $1.600.000 when it wasbuilt 15 years ago. The Louisiana bridge engineers estimate that its remaining life is 10 years,then it will need to be replaced because of increased traffic. Its salvage value at any point intime is zero, because the cost of demolition will most like equal its value as scrap steel.A concrete bridge is considered to be the best challenger. It will cost $3.000.000 to build and$100.000 annually in maintenance costs. Its estimated life is 50 years. Its resale value may becounted as zero at any time during its life.No taxes of any kind will be considered for this government project. All costs are in constantdollars of year 0. Inflation may be ignored. Assume that annual benefits for either structure areexactly the same. A discount rate of 10 percent is to be used in analysis.(a) What is the economic life…