Truck A Truck B Initial cost P600,000 P750,000 7 years P85,000 P50,000 P20,000 Economic life 7 years Annual operations cost P120,000 Annual maintenance cost P55,000 Salvage value none
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using an 11% annual interest rate to decide if they should buy Truck A or Truck B. Which truck is more economical to buy, assuming that the two trucks serve similar functions and purposes?
Use (a) Annual Cost Method (b) Present Worth Method, (c)
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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 $12, what is thebreak even point in units? What is the break even pointin dollar sales volume?Solar plant investment or cost establish 40000$, 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: note:round off final answer to 2 decimal ANSWER for ALTERNATIVE A: ANSWER for ALTERNATIVE B:
- 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 2E2 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…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 METHOD
- Please no written by hand and no emage Solve in excel Carp, Inc. wants to evaluate two machines for packaging their products.Machine A:Initial cost is $700,001st year O&M cost is 18,000; this cost increases $900 each year.The annual benefits are $154,000It can be sold at the end of 10 years useful life for $145,000 Machine B:Initial cost is $1,600,001st year O&M cost is 28,000; this cost increases $650 each year.The annual benefits are $300,000It can be sold at the end of 20 years useful life for $210,000The companies uses an interest rate of 15% Use annual cash flow analysis to decide which is the most desirable alternative.A 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=sharingProblem Solving. Solve the following problems completely. 4. Atty. Gacayan invested P280, 000 which will be used in a project that will produce auniform annual revenue of P180,000 for 5 years and then have a salvage value of 16% ofthe investment. Out-of-pocket costs for operation and maintenance will be P80,000 peryear. Taxes and insurance will be 3% of the first cost per year. Atty Gacayan expectscapital to earn not less than 30% before income taxes. Determine if the investment is goodand Calculate the following:a. Calculate using Rate of Return Method.b. Payback period of the investment.
- MARR 20% EOY Cash Flow 0 $(70,000.00) 1 $ 20,000.00 2 $ 19,000.00 3 $ 18,000.00 4 $ 17,000.00 5 $ 16,000.00 6 $ 15,000.00 7 $ 14,000.00 8 $ 13,000.00 9 $ 12,000.00 10 $ 11,000.00 11 $ 10,000.00 12 $ 9,000.00 13 $ 8,000.00 14 $ 7,000.00 15 $ 6,000.00 16 $ 5,000.00 17 $ 4,000.00 18 $ 3,000.00 19 $ 2,000.00 20 $ 1,000.00 Plot a graph of FW versus MARR, where MARR varies from 0 percent to 50 percent by 1 percent increments.FW should be on the y-axis and MARR on the x-axis.MARR 20% EOY Cash Flow 0 $(70,000.00) 1 $ 20,000.00 2 $ 19,000.00 3 $ 18,000.00 4 $ 17,000.00 5 $ 16,000.00 6 $ 15,000.00 7 $ 14,000.00 8 $ 13,000.00 9 $ 12,000.00 10 $ 11,000.00 11 $ 10,000.00 12 $ 9,000.00 13 $ 8,000.00 14 $ 7,000.00 15 $ 6,000.00 16 $ 5,000.00 17 $ 4,000.00 18 $ 3,000.00 19 $ 2,000.00 20 $ 1,000.00 Plot a graph of FW versus MARR, where MARR varies from 0 percent to 50 percent by 1 percent increments.FW should be on the y-axis and MARR on the x-axis. How do you find FW? Please show all steps and formulas used in excel format. Thank you!Ronald McDonald decides to install a fuel storage system for his farm that will save him anestimated 6.5 cents/gallon on his fuel cost. He uses an estimated 20,000 gallons/year on his farm.Initial cost of the system is $10,000 and the annual maintenance the first year is $25 and increasesby $25 each year thereafter. After a period of 10 years the estimated salvage is $3,000. If moneyis worth 12%, is it a wise investment?
