Given 2 alternatives: dneleviupe erf First Cost 4,000 1,000 2,000 6,000 500 Annual Cost Annual Benefit 2,200 10 Rod tedt pni Life owT dinome Salvage If i 10%, find the better alternative computing NPW of both alternatives Assume alternative A is replacer at the end of its useful life. 3,000 1,000

Consider three mutually exclusive alternatives, each with a 15-year useful life. If the MARR is 12%, which alternative should be selected? Solve the problem by benefit–cost ratio analysis.

A cost analysis is to be made to determine what, if anything, should be done in a situation offering three "do-something" and one "do-nothing" alternatives. Estimates of the costs and benefits are as follows: End-of- Uniform Useful- Life Benefit Salvage (years) Value Useful First Alternatives Cost Annual Life -$500 $135 -600 100 250 5 -700 100 180 10 4 Use Present Worth Analysis with the Least Common Multiple Method to select the best alternative. Use MARR=6%. (Note: You MUST describe your answers with conversion factor notations and you DON'T NEED TO calculate the values of PW for supporting a decision. Simply describe your selection criterion with PW expressions.) 1.

The following information is for five mutually exclusive alternatives that have 20-year useful lives. The decision maker may choose any one of the options or reject them all. Prepare a choice table.

Q.2- if the rate of return is 10 % and all equipment shown in the following table have service life for 20 years with no salvage value, which equipment should be selected using the incremental benefit cost analysis. Equipment First cost,S Y 1600000 1500000 1900000 1200000 Annnal M&O, S/year 50000 60000 35000 so000 Benefits, S'year Dis benefits, S/year 269000 260000 282000 250000 13000 15000 22000 20000 Life, years 20 20 20 20

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0 ave servie if for 20 years with o savage value, Which equipment should be the rate of retum is 10 % and all equipment shown in the following table. slected using the incrementa bencfitcostanalysis

The following three alternatives have been developed and costed to improve the efficiency of aircraft engines. Compare the alternatives using capitalized cost and a MARR of 10% per year compounded annually. Alternative Alternative Alternative G First Cost, $ -1,000,000 -3,000,000 -5,000,000 Net income $ per year 325 000 400,000 600,000 Salvage value, S 50,000 70,000 Life, years 4 8.

solve, dont use excel. dont write the answer in a paper becuse of the handwriting. thanks