Question
can some one help me solving this case study Harry, owner of an automobile battery distributorship in Atlanta, Georgia, performed an economic analysis 3 years
can some one help me solving this case study 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
Lloyds
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. Lloyds was the clear choice due to its substantially larger AW value. The Lloyds 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 -$6,068 -$800 $25,000 -$7,025 -$300 $35,000 Total AW $18,131.35 $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 13% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,753, $20,878, and $37,513, as best as Harry can determine. He believes savings will decrease by $3,266 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
- Q1- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years.
- Q2- With these new estimates, what is the recalculated AW for the Lloyds protectors? Use the old first cost and maintenance cost estimates for the first 3 years. type your answer in the answer box
- Q3- If these estimates had been made 3 years ago, would Lloyds still have been the economic choice
- Q4- How has the capital recovery amount changed for the Lloyds protectors with these new estimates?
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.
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 | Lloyds | |
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. Lloyds was the clear choice due to its substantially larger AW value. The Lloyds 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 | -$6,068 | -$800 | $25,000 | -$7,025 | -$300 | $35,000 |
| Total AW | $18,131.35 | $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 13% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,753, $20,878, and $37,513, as best as Harry can determine. He believes savings will decrease by $3,266 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
- Q1- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years.
- Q2- With these new estimates, what is the recalculated AW for the Lloyds protectors? Use the old first cost and maintenance cost estimates for the first 3 years. type your answer in the answer box
- Q3- If these estimates had been made 3 years ago, would Lloyds still have been the economic choice
- Q4- How has the capital recovery amount changed for the Lloyds protectors with these new estimates?
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.
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Accounting Information Systems
Authors: Marshall B. Romney, Paul J. Steinbart
13th edition
978-0133428537
