When alternatives under consideration have unequal lives, approach is to use an analysis period that is the least common multiple of the alternative lives. For example, in this case, alternative C has a 10-year life and D has a 5-year life, then a 10-year analysis period is used. In such cases, we assume that each alternative can be identically replaced at the end of its service life.
Calculate the Present Worth (PW), taking into account possible different time horizons and compare the values for each alternative.
The PW for the alternative C is:
Year | Costs ($) | P/F @ 10% | PW ($) |
Day 1 | 40,000 | 1.0000 | 40,000 |
Year 1 | 6,000 | 0.9091 | 5,455 |
Year 2 | 6,200 | 0.8264 | 5,124 |
Year 3 | 6,400 | 0.7513 | 4,808 |
Year 4 | 6,600 | 0.6830 | 4,508 |
Year 5 | 6,800 | 0.6209 | 4,222 |
Year 6 | 7,000 | 0.5645 | 3,951 |
Year 7 | 7,200 | 0.5132 | 3,695 |
Year 8 | 7,400 | 0.4665 | 3,452 |
Year 9 | 7,600 | 0.4241 | 3,223 |
Year 10 | 7,800 | 0.3855 | 3,007 |
Salvage (10th year) | -7,000 | 0.3855 | -2,699 |
Total PW | 78,747 |
The PW for the alternative D is:
Assuming that alternative D is identically replaced at end of year 5.
Year | Costs ($) | P/F @ 10% | PW ($) |
Day 1 | 22,000 | 1.0000 | 22,000 |
Year 1 | 3,000 | 0.9091 | 2,727 |
Year 2 | 3,300 | 0.8264 | 2,727 |
Year 3 | 3,500 | 0.7513 | 2,630 |
Year 4 | 3,700 | 0.6830 | 2,527 |
Year 5 | 3,900 | 0.6209 | 2,422 |
Salvage (5th year) | -200 | 0.6209 | -124 |
Year 5 (End)- Replace | 22,000 | 0.6209 | 13,660 |
Year 6 | 3,000 | 0.5645 | 1,693 |
Year 7 | 3,200 | 0.5132 | 1,642 |
Year 8 | 3,400 | 0.4665 | 1,586 |
Year 9 | 3,600 | 0.4241 | 1,527 |
Year 10 | 3,800 | 0.3855 | 1,465 |
Salvage (10th year) | -200 | 0.3855 | -77 |
Total PW | 56,405 |
The present worth of alternative C is $ 78,747 and alternative D is $ 56,405,
Hence, alternative D offers the lower present worth.
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis...
Problem 05.023 Alternative Comparison - Different Lives Compare the alternatives C and D on the basis of a present worth analysis using an interest rate of 12% per year and a study period of 10 years. с $-44,000 $-12,000 $-34,000 $-7,000 Alternative First Cost AOC, per Year Annual Increase in Operating Cost, per Year Salvage Value Life, Years $-1,500 $-1,200 $5,000 10 $1,200 5 The present worth of alternative C is $ -134497.32 and that of alternative D is $...
Compare 10 years the alternatives C and D on the basis of a present worth analysis using an interest rate of 10% per year and a study period of Alternative rst Cost Annual Increase in Operating Cost, per Year Salvage Value $-1,000 $-1.200 0 The present worth of alternative C is S?m) and that of altern Click to select)offers the lower present worth analysis ative Dis $
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Solve it in spreadsheet money, anu Juuay pu mom. Alternative Comparison-Different Lives 5.17 Dexcon Technologies, Inc. is evaluating two alter- natives to produce its new plastic filament with tribological (i.e., low friction) properties for creat- ing custom bearings for 3-D printers. The esti- mates associated with each alternative are shown below. Using a MARR of 20% per year, which al- ternative has the lower present worth? Method First cost, $ M&O cost, $/year Salvage value, $ Life, years DDM -...
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