In comparing alternatives with different lives by the present worth (PW) method,

Question

In comparing alternatives with different lives by the present worth (PW) method, it is necessary to: A. Compare them over a period equal to the life of the longer-lived alternative B. Find them over one life cycle for each alternative C. Compare them over a period equal to the life of the shorter-lived alternative D. Compare them over a time period of equal service. Use the data from the following table, and an interest of 10% per year. In comparing the machines on a present worth basis, PW for machine B is closes to: In comparing the machines on a present worth basis, PW for machine A is approximately: Which machine should be selected? The five alternatives are being evaluated by the rate of return method, using the following information: If alternative W and X are mutually exclusive and MARR - 13% per year, which one of the two alternatives is preferable? If all five alternative are independent, MARR = 13% per year and the budget is limited to $140,000, which alternatives should be selected? Alternative V is contingent on Z: and W, X, Y are independent, which alternative (s) should be selected if MARR = 15% per year.? If the alternatives are mutually exclusive and MARR = 15 % per year, the alternative (s) to select is (are);

Explanation / Answer

1) Whenever we are comparing two alternatives on the basis of present worth method it is really necessary to compare them on the basis of the same year criteria.

So, option d

2)

case 2

2) option 'b'

3) option 'c'

4) option 'b'

Investment year 1 year2 year3 investment year 4 year 5 year6 Total Money cash flow 20,000 20,000 20,000 20000 20000 20000 salvage year 10,000 10000 total 20,000 20,000 30,000 20000 20000 30000 present factor (1/(1+r)^n) 0.909 0.826 0.751 0.683 0.683 0.621 0.565 total -80,000 18180 16520 22530 -54640 13660 12420 16950 -34380

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