Harry, the owner of an automobile battery distributorship in Atlanta, Georgia, performed an economic analysis 3 years ago when he decided to place surge protectors in line for all his major pieces of testing equipment. The estimates used and the annual worth analysis at MARR = 15% are summarized below. Two different manufacturers’ protectors were compared.

The spreadsheet in Figure 6–9 is the one Harry used to make the decision. Lloyd’s was the clear choice due to its substantially larger AW value. The Lloyd’s protectors were installed.

During a quick review this last year (year 3 of operation), it was obvious that the maintenance costs and repair savings have not followed (and will not follow) the estimates made 3 years ago. In fact, the maintenance contract cost (which includes quarterly inspection) is going from $300 to $1200 per year next year and will then increase 10% per year for the next 10 years. Also, the repair savings for the last 3 years were $35,000, $32,000, and $28,000, as best as Harry can determine. He believes savings will decrease by $2000 per year thereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in 7 years is zero, not $3000.

Case Study Exercises

1. Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years.

2. With these new estimates, what is the recalculated AW for the Lloyd’s protectors? Use the old first cost and maintenance cost estimates for the first 3 years. If these estimates had been made 3 years ago, would Lloyd’s still have been the economic choice?

3. How has the capital recovery amount changed for the Lloyd’s protectors with these new estimates?

Answer rating: 100% (QA)

1 Plot of the newly estimated maintenance costs and repair savings projections 2 Re... View the full answer