P Q Using the Benefit-Cost Ratio Method, Find what is the best alternative from the projects listed below if u know that (-10%) for both project. Details Initial investment (CU) Annual revenue (CU/year) Annual expense (CU/year) Project life (year) Investment due to replacement of some machines every 3 years. Salvage value (CU) Purchasing a new machine in the seventh year. Alternative 1 300 000 50 000 15 500 5 13 500 Non Non Alternative 2 200 000 75 000 29 000 10 Non 22 000 13 000
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- Compare the following two alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of return (AROR). Alt. Construction cost $ Benefits $/yr Salvage $ Service Life (yrs) A 410,000 55,000 20,000 11 B 250,000 35,000 10,000 11a. Cobre Company is considering the purchase of new equipment that will speed up the process for extracting copper. The equipment will cost $3,800,000 and have a life of 5 years with no expected salvage value. The expected cash flows associated with the project are as follows: Year Cash Revenues Cash Expenses $6,000,000 $4,800,000 6,000,000 4,800,000 3 6,000,000 4,800,000 4 6,000,000 4,800,000 6,000,000 4,800,000 b. Emily Hansen is considering investing in one of the following two projects. Either project will require an investment o $75,000. The expected cash revenues minus cash expenses for the two projects follow. Assume each project is depreciable. Year Project A Project B 1. $2,500 $22,500 2 30,000 30,000 45,000 45,000 4 75,000 22,500 75,000 22,500 c. Suppose that a project has an ARR of 30% (based on initial investment) and that the average net income of the project is $220,000. d. Suppose that a project has an ARR of 50% and that the investment is $250,000. 3.Based on estimates the data for 2 types of bridges with different lives are as follows. If the minimum attractive rate of return is 9%, determine which project is more desirable using Annual Cost Method & Rate of Return Method. Timber Bridge Steel Bridge First Cost Salvage Value Life in yrs Annual maintenance 500k 20k 12 60k 1.4M 100k 36 25k
- Check whether the project is accepted or rejected and recommend the best project, MARR=6% and n=8 years, using Payback period, ROI, NPV, Aw, IRR methods. ID Project I Project II increase Annual Initial Annual cost revenues initial Annual Salvage value annual salvage value running in cost revenues maintenance revnues cost 1910095 280000 60000 2000 28000 400000 72000 2000 40000 12000Two alternative machines will produce the same product, but one is capable of higher-quality work, which can be expected to return greater revenue. The following are relevant data. Determine which is the better alternative, assuming repeatability and using SL depreciation, an income-tax rate of 27%, and an after-tax MARR of 11%. Capital investment Life Calculate the AW value for the Machine A. Machine A $23,000 12 years $4,000 Terminal BV (and MV) Annual receipts Annual expenses Click the icon to view the interest and annuity table for discrete compounding when the MARR is 11%er year. AWA (11%) = $ (Round to the nearest dollar.) Machine B $33,000 6 years $1,500 $197,000 $176,000 $152,000 $130,000Two altemnative machines will produce the same product, but one is capable of higher-quality work, which can be expected to return greater revenue. The following are relevant data. Determine which is the better alternative, assuming repeatability and using SL depreciation, an income-tax rate of 25%, and an after-tax MARR of 10%. Capital investment Life Machine A $20,000 12 years $3,500 Calculate the AW value for the Machine A. AWA (10%) = $(Round to the nearest dollar.) Terminal BV (and MV) Annual receipts Annual expenses Click the icon to view the interest and annuity table for discrete compounding when the MARR is 10% per year. Machine B $34,000 9 years $2,000 $144,000 $142,000 $190,000 $163,000
- You have been asked to evaluate two alternatives, X and Y, that may increase plant capacity for manufacturing high-pressure hydraulic hoses. The parameters associated with each alternative have been estimated. Which one should be selected on the basis of a present worth comparison at an interest rate of 13% per year? Why is yours the correct choice? Alternative First Cost X $-45,000 Maintenance cost, per $-9000 Year Salvage Value $1,000 Life 5 years Y $-55,000 $-4000 $6,500 5 years The present worth of alternative X is $ 13888 and that of alternative Y is $ 44459.4 Alternative X is selected by the company.6. Compare the following three alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of retum method (AROR). Construction cost S Benefits S/yr Salvage S Alt. Service Life (yrs) 510,000 145,000 -10,000 775.000 155,000 15,000 1,075,000 165,000 20,000 6.Use the following alternatives to develop an incremental analysis choice table and answer the following questions. Alternative A Initial Cost - $15,000 Annual Revenue (Year 1-15) - $2500 Salvage Benefit (Year 15) - $2000 Alternative B Initial Cost - $25,000 Annual Revenue (Year 1-15) - $3500 Salvage Benefit (Year 15) - $5000 1. Determine the incremental rate of return IRRHigh-Low. Provide your answer as a percentage and round to the nearest hundredth. 2. Determine the internal rate of return for the higher cost alternative IRRHigh. Provide your answer as a percentage and round to the nearest hundredth. 3. Determine the internal rate of return for the lower cost alternative IRRLow. Provide your answer as a percentage and round to the nearest hundredth. Please answer all 3.
- Using the below informtion answer: 5.1 Payback Period of Project Tan (expressed in years, months and days). 5.2 Net Present Value of Project Tan.5.3 Accounting Rate of Return on average investment of Project Tan (expressed to two decimal places). INFORMATIONThe management of Mastiff Enterprises has a choice between two projects viz. Project Cos and Project Tan, each ofwhich requires an initial investment of R2 500 000. The following information is presented to you: PROJECT COS PROJECT TANNet Profit Net ProfitYear R1 130 000 80 0002 130 000 180 0003 130 000 120 0004 130 000 220 0005 130 000 50 000A scrap value of R100 000 is expected for Project Tan only. The required rate of return is 15%. Depreciation is calculatedusing the straight-line method.Given the financial data for four mutually exclusive alternatives in the table below, determine the best alternative using the incremental rate of return (AROR) analysis. MARR = 10%. A B C D $15,000 $21,200 $36,000 45,000 1,600 700 400 1,000 First cost O &M Cost/ year Benefit/year 8,000 9,000 13,000 Salvage value 3,000 4,600 6,000 Life in years 4 15,000 10,000Phoenix Sky Harbor Airport is considering the following two ME revenue-based projects. Using rate of return analysis and assuming a MARR of 10%, Which alternative should be selected? Cash Flow (Project A) Cash Flow (Project B) First Cost, $ -120,000 -300,000 Annual Cost, S/year -60,000 -40,000 Annual revenue, S/year 80,000 90,000 Salvage Value, S 5,000 10,000 Life, years 10 10