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 values 2,000 3,000 25,000 35,000 Equipment repair savings, 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 Year SSSSS PoweUp Lloyd's Investment Annual Repair Investment Annual Repair and salvage maintenance and salvage maintenance Savings -526.000 50 30 536,000 50 10 10 -5800 525.000 $300 535,000 2 50 1800 525,000 $300 535,000 3 50 -1000 125,000 -5300 535 000 50 $25,000 -5300 $35.000 > 50 3800 525.000 -5300 $35.000 30 1000 525.000 SO $100 $35.000 7 $2,000 525.000 -5300 $35.000 10 -5300 $35.000 50 $35.000 -5300 135,000 AWewens -5800 125,000 -17025 -$300 $35.000 $18.131 127,674.65 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 cout (which includes quarterly inspection) is going from $200 to $1200 per year next year and will then increases per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25.137, and $33,416, as best as Harry can determine. He believes savings will decrease by $3,845 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. 10 $1.000 Total AW 4 $300 10 SUUU 52 000 50 -$300 315.000 50 -5300 $25,000 50 -$300 $35.000 50 -5800 525.000 50 -$300 $35,000 30 -5800 525.000 50 -5300 $35.000 50 -5800 $25.000 50 -$300 $35,000 52.000 -5800 $25.000 50 -$300 $35.000 50 $35.000 50 -5300 $35.000 TO $3,000 -$300 $35.000 AWlement -56,066 -$800 525.000 -57,025 -$300 $35.000 Total AW $18,131.35 $27,674.68 Daring 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 11 per year for the next 10 years. Also, the repair savings for the last 3 years were $29.055, S5,137, and $33,416, as best as Harry can determine. He believes savings will decrease by $3,845 per year hereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in years is zero, not $3000. Case Study Exercises . Qi-Plot a graph of the newly estimated maintenance costs and repair saving projections, assuming the protectors last for seven more years. Q2. 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, type your answer in the answer box 23. If these estimates had been made 3 years ago, would Lloyd's still have been the economie choice Q4-How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details 1-Each student must submit an excel file with the required solution (graphs and tables). answers of Qi & Qa 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 Lloyds scenario. To be uploaded as image or PDF verification of Qa, answers Q3 &Q4 above. 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 pleces of testing equipment. The estimates used and the annual worth analysis at MARR - 15% are summarized below. Two different manufacturers' protectors were compared. Powrl'p Lloyd's -26,000 -36,000 -800 Cost and installation, Annul maintenance cost, S per year Salvage value, $ Equipment repair savings, 5 --300 3,000 2,000 25,000 35,000 Useful life, years 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 Repair Years PoweUp Investment Annual and salvage maintenance g -526.000 50 1 50 3800 2 10 21 10 1800 30 1800 $0 -1000 50 -1800 7 12.000 -1800 30 525.000 525,000 125.000 $25,000 125,000 525.000 325,000 Lloyd's Investment Annual Repair and salvege maintenance Savings -516,000 SO 30 10 -1100 $35.000 10 -$300 $35.000 10 $100 $35.000 10 $300 $35.000 19 $300 $35.000 VO $300 125.000 $300 135,000 10 -5300 $35.000 30 -5300 335,000 $3,000 $35,000 -17,025 $100 $35.000 127,6746 9 10 AW element Total AW -56,060 -1800 525.000 516,131.35 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 11% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25,137, and $33.416, as best as Harry can determine. He believes savings will decrease by $3,845 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. . 02- 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, type your answer in the answer box . 03-If these estimates had been made 3 years ago, would Lloyd's still have been the economie choice . 04- How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details: 1-Each student must submit an excel file with the required solution (graphs and tables). answers of Qi& Q2 above. Each Student to submit one file of a hand written verification solution using the factors/tables to calculate the AW value of the revised Lloyds scenario. To be uploaded as image or PDF. verification of Q2, answers Q3 &Q4 above. 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. Powrp Lloyd's -26,000 -36,000 -800 -300 Cost and installation, Annual maintenance cost, $ per year Salvage value, Equipment repair savings, Useful life, years 2,000 3,000 25,000 35,000 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 and salvage maintenance savings o -$26,000 50 50 -536.000 50 50 1 50 -5800 $25,000 -5300 $35.000 2 $0 5800 $25,000 50 -5300 $35.000 3 50 -5800 $25.000 50 -$300 $35.000 50 -5800 525,000 50 -5300 $35,000 $0 -5800 $25,000 -5300 $35.000 6 50 -5800 $25,000 50 -$300 $35.000 7 3800 $25,000 -$300 $0 -5300 $35,000 . SO -5300 $35.000 10 $3.000 -5300 $35,000 AW element -56,068 -5800 $25,000 -$7.025 -5300 $35,000 Total AW $18.131.15 $27.67466 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 11% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25,137, and $33.416, as best as Harry can determine. He believes savings will decrease by $3,845 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 $2.000 Case Study Exercises Qu- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years. . 02- 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, type your answer in the answer box 03- If these estimates had been made 3 years ago, would Lloyd's still have been the economic choice 04- How has the capital recovery amount changed for the Lloyd's 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 Lloyds scenario. To be uploaded as image or PDF. verification of Qa, answers Q3 &Q4 above. 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 values 2,000 3,000 25,000 35,000 Equipment repair savings, 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 Year SSSSS PoweUp Lloyd's Investment Annual Repair Investment Annual Repair and salvage maintenance and salvage maintenance Savings -526.000 50 30 536,000 50 10 10 -5800 525.000 $300 535,000 2 50 1800 525,000 $300 535,000 3 50 -1000 125,000 -5300 535 000 50 $25,000 -5300 $35.000 > 50 3800 525.000 -5300 $35.000 30 1000 525.000 SO $100 $35.000 7 $2,000 525.000 -5300 $35.000 10 -5300 $35.000 50 $35.000 -5300 135,000 AWewens -5800 125,000 -17025 -$300 $35.000 $18.131 127,674.65 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 cout (which includes quarterly inspection) is going from $200 to $1200 per year next year and will then increases per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25.137, and $33,416, as best as Harry can determine. He believes savings will decrease by $3,845 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. 10 $1.000 Total AW 4 $300 10 SUUU 52 000 50 -$300 315.000 50 -5300 $25,000 50 -$300 $35.000 50 -5800 525.000 50 -$300 $35,000 30 -5800 525.000 50 -5300 $35.000 50 -5800 $25.000 50 -$300 $35,000 52.000 -5800 $25.000 50 -$300 $35.000 50 $35.000 50 -5300 $35.000 TO $3,000 -$300 $35.000 AWlement -56,066 -$800 525.000 -57,025 -$300 $35.000 Total AW $18,131.35 $27,674.68 Daring 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 11 per year for the next 10 years. Also, the repair savings for the last 3 years were $29.055, S5,137, and $33,416, as best as Harry can determine. He believes savings will decrease by $3,845 per year hereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in years is zero, not $3000. Case Study Exercises . Qi-Plot a graph of the newly estimated maintenance costs and repair saving projections, assuming the protectors last for seven more years. Q2. 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, type your answer in the answer box 23. If these estimates had been made 3 years ago, would Lloyd's still have been the economie choice Q4-How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details 1-Each student must submit an excel file with the required solution (graphs and tables). answers of Qi & Qa 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 Lloyds scenario. To be uploaded as image or PDF verification of Qa, answers Q3 &Q4 above. 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 pleces of testing equipment. The estimates used and the annual worth analysis at MARR - 15% are summarized below. Two different manufacturers' protectors were compared. Powrl'p Lloyd's -26,000 -36,000 -800 Cost and installation, Annul maintenance cost, S per year Salvage value, $ Equipment repair savings, 5 --300 3,000 2,000 25,000 35,000 Useful life, years 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 Repair Years PoweUp Investment Annual and salvage maintenance g -526.000 50 1 50 3800 2 10 21 10 1800 30 1800 $0 -1000 50 -1800 7 12.000 -1800 30 525.000 525,000 125.000 $25,000 125,000 525.000 325,000 Lloyd's Investment Annual Repair and salvege maintenance Savings -516,000 SO 30 10 -1100 $35.000 10 -$300 $35.000 10 $100 $35.000 10 $300 $35.000 19 $300 $35.000 VO $300 125.000 $300 135,000 10 -5300 $35.000 30 -5300 335,000 $3,000 $35,000 -17,025 $100 $35.000 127,6746 9 10 AW element Total AW -56,060 -1800 525.000 516,131.35 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 11% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25,137, and $33.416, as best as Harry can determine. He believes savings will decrease by $3,845 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. . 02- 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, type your answer in the answer box . 03-If these estimates had been made 3 years ago, would Lloyd's still have been the economie choice . 04- How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details: 1-Each student must submit an excel file with the required solution (graphs and tables). answers of Qi& Q2 above. Each Student to submit one file of a hand written verification solution using the factors/tables to calculate the AW value of the revised Lloyds scenario. To be uploaded as image or PDF. verification of Q2, answers Q3 &Q4 above. 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. Powrp Lloyd's -26,000 -36,000 -800 -300 Cost and installation, Annual maintenance cost, $ per year Salvage value, Equipment repair savings, Useful life, years 2,000 3,000 25,000 35,000 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 and salvage maintenance savings o -$26,000 50 50 -536.000 50 50 1 50 -5800 $25,000 -5300 $35.000 2 $0 5800 $25,000 50 -5300 $35.000 3 50 -5800 $25.000 50 -$300 $35.000 50 -5800 525,000 50 -5300 $35,000 $0 -5800 $25,000 -5300 $35.000 6 50 -5800 $25,000 50 -$300 $35.000 7 3800 $25,000 -$300 $0 -5300 $35,000 . SO -5300 $35.000 10 $3.000 -5300 $35,000 AW element -56,068 -5800 $25,000 -$7.025 -5300 $35,000 Total AW $18.131.15 $27.67466 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 11% per year for the next 10 years. Also, the repair savings for the last 3 years were $29,655, $25,137, and $33.416, as best as Harry can determine. He believes savings will decrease by $3,845 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 $2.000 Case Study Exercises Qu- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years. . 02- 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, type your answer in the answer box 03- If these estimates had been made 3 years ago, would Lloyd's still have been the economic choice 04- How has the capital recovery amount changed for the Lloyd's 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 Lloyds scenario. To be uploaded as image or PDF. verification of Qa, answers Q3 &Q4 above
