In: Accounting
CASE STUDY
ANNUAL WORTH ANALYSIS—THEN AND NOW
Background and Information
Harry, 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.
PowrUp |
Lloyd’s |
|
Cost and installation, $ |
−26,000 |
−36,000 |
Annual maintenance cost, $ per year |
−800 |
−300 |
Salvage value, $ |
2,000 |
3,000 |
Equipment repair savings, $ |
25,000 |
35,000 |
Useful life, years |
6 |
10 |
The spreadsheet in below sheet 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.
MARR | 15% | |||||
PoweUp | Lloyd's | |||||
Investment | Annual | Repair | Investment | Annual | Repair | |
Years | and salvage | maintenance | savings | and salvage | maintenance | savings |
0 | -$26,000 | $0 | $0 | -$36,000 | $0 | $0 |
1 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
2 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
3 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
4 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
5 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
6 | $0 | -$800 | $25,000 | $0 | -$300 | $35,000 |
7 | $2,000 | -$800 | $25,000 | $0 | -$300 | $35,000 |
8 | $0 | -$300 | $35,000 | |||
9 | $0 | -$300 | $35,000 | |||
10 | $3,000 | -$300 | $35,000 | |||
AW element | -$6,642 | -$800 | $25,000 | -$7,025 | -$300 | $35,000 |
Total AW | $17,558.31 | $27,674.68 |
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 7% per year for the next 10 years. Also, the repair savings for the last 3 years were $28,482, $31,506, and $32,429, as best as Harry can determine. He believes savings will decrease by $1,577 per year hereafter. 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
Submission details:
1-Each student must submit an excel file with the required solution (graphs and tables). answers of Q1 & Q2 above.
2- Each Student to submit one file of a hand written verification solution using the factors/tables to calculate the AW value of the revised LIoyds scenario. To be uploaded as image or PDF.verification of Q2, answers Q3 &Q4 above.
please answer clearly Q1,Q2,Q3,Q4