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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. Lloyd's PowrUp -26,000 Cost and installation, -36,000 Annual maintenance cost, S per year -800 --300 2,000 3,000 Salvage value, s Equipment repair savings, Useful life. years 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 Years PoweUp Investment Annual and salvage maintenance $26.000 50 50 1800 -500 Repair Savings SO $25.000 125.000 Lloyd's Investment Annual Repair and salvage maintenance savings -$36,000 30 50 50 -5300 $35.000 30 $100 335.000 Click Click Save All Areas Save AA Por Maher (24) Praca (21) pdf DPDF Maller (20) pdf PDF Male (20) pul Volunteering for pdf MerBook AH ESC So Question completion Status: 1 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 88888888 PoweUp Lloyd's Investment Annual Repair Investment Annual Repair Years and salvage maintenance savings and salvage maintenance savings O -$25,000 50 $0 -$36,000 50 $0 1 $800 $25,000 50 -$300 $35,000 2 5800 $25,000 50 -$300 $35.000 3 $800 $25,000 $0 $300 $35.000 $ -$800 $25,000 50 -$300 $35,000 5 $0 $800 $25,000 50 -$300 $35.000 -5800 $25,000 $0 -5300 $35,000 7 $2,000 $800 $25,000 $0 -$300 $35,000 $0 $300 $35,000 9 $0 -$300 $35,000 $3,000 -$300 $35,000 AW element -56,068 -5800 $25,000 -$7,025 -5300 $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 5% per year for the next 10 years. Also, the repair savings for the last 3 years were $31,186, $25.508, and $34.259, as best as Harry can determine. He believes savings will decrease by $1,809 per year hereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in 7 years is rero, 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 Click Save and Subovt to save and submit. Click Save All Answers to sow all answers, Save All Answers 10 DPDF Maller (21).pdf DPDF Mailer (21).pdf PDF Mailer (20).pdf DPDF Mailer (20).pdf Volunteering for....pdf 1 9 10 AW element Total AW -56,068 -5800 - $0 -$300 $35,000 $3,000 -$300 $35,000 $25,000 $7,025 -$300 $35,000 $18,131,35 527,674.68 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 5% per year for the next 10 years. Also, the repair savings for the last 3 years were $31,186, $25.508, and $34,259, as best as Harry can determine. He believes savings will decrease by $1,809 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 $3000. Case Study Exercises Q1- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years Q2- With these new estimates, what is the recalculated AW for the Lloyd's protectors? Use the old first cost an maintenance cost estimates for the first 3 years. type your answer in the answer box . 03- If these estimates had been made 3 years ago, would Lloyd's still have been the economic choice 04- How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details: 1-Each student must submit an excel file with the required solution (graphs and tables), answers of Q1 & Q2 above. 2- Each Student to submit one file of a hand written verification solution using the factors/tables to calculate the AW value of the revised Lloyds scenario. To be uploaded as image or PDF. verification of Q2, answers Q3 &Q4 above. Click Save and Submit to see and submit. Click Save All Answers to see alles Save All Answers DPDF Maler (21).pdf POF Moller (21).pdf POF Madler (20).pdf POFMater (20) pdf Volunteering for pdf Case Study 2 Annual Worth Analysis From 2008 to 1200$ with 8% increase repair savings for last 24Rs 24,5928, 20,6528, 30.082$ will decreue by : 3,2248 Q1) Amust pugment F[c#ins Jeans more of inha Annual Payment P[J. Niekerk untee Annual Werth (AW) - P (AIP.in)-A+S(AiFi,n) F Alcantated twn of amount Ant Capinder SO Vine Year 200 5 Armuat Maintenance (x) Annuacel Saving.CX) o - 200 1834591 300 90652 30682 -1900 126858 - 1296 23634 -1299.68 $ 20410 1511 6544 17186 -1632.58 8 13962 -1763-18 1 10738 -1964 22 7 $ 9 10 $ 7514 Yearloos Annual ols Annual A 5 tv .pdf (page 2 of 4) S - 1622:58 - 1762-18 -1957 22 13962 1 10738 7514 10 Year P&s A. 4-X /Pw-Alan -26600 o 2 O 2 2141 CI+-53729818-3 O 0 3 4 0 ito.is Annual Annual Maintenance (A) Souding CY) O 0 O 200 $ 24591 39291 - 300 20652 20352 15 789.03 -300 130082 29782 1952.is - 1200 $26858 25658 14670.09 -1296 $22634 22338 H1os. 92 -1399.68 120410 196 10:32 82 18.68 -1511 (594 17186 15679.35 5892.57 -1632.56 13962 12 329.92 4030 SI -1763.18 $ 10738 8979-22 25$1.20 -1904.23 1714 SLO9.77 1308.65 PW tot 112621.06 5 0 6 0 7 6 6 O 8 9 D 10 0 totul AW Q2) Find the Annual maintenance 8* Year = -1511 65 x = -16 77.9715 Find the AW of P and SL 4.-2400 nois be W 77 1365.65 PW tot = 12627.06 Q2) Find the Annual maintenance 8 Year-1565x = -1677 9715 Find the AW of P and SL A :-?500 nalo AW All --2600 (ISCOS = -7173.07 Find the PW of A toid of PW = 1/2627.06 -) 115/ ha16 PWA PWT ((1+i) CHIS (10.19 = 1/2627:06 erretje) = 22441.174 Find Net AW: CAW of PR ) + (Pword) (-9172.07) + C22441-174) Net = [15268 104 AU P/F 9 A MacBook Air 3823 16460 P/F 3459 74571 0696 $20080.11 20.52 X 0.7561 f 15615 10082 % 0-6575 2625705717 115357 27634 x 0.4972 $1751 2040 X 0.4323 17176 *6.3759 (13962 x 0.3269 - $4564 10728 x 0.2843 7514 X 0.2472 total $117340.37 tot 36000 $82 340.37 hot-360001925). 16 207.07 Q3) The AW of Lioyals is more than that of power up so this would had been still selected Q4) The old estimation from Capellad recrecy of Liny do -70 25 Now Capital recovering estimations 5-7500 CAIP 570) Bourse of mil salvage value --96099(014) 7172 has increase - 365? 9 30) sty MacBook Air CCIO OOG 4 00 F5 F& DI 52 FU FO EVO % ies, and improvements, choose check for Updates Check for . B C D E G H and a M Years o manne saving 50 waving Total M & Repair BO and a 0.0001 so 50 $40.000 1 SO 12.000 3341 2 125.000 50 to 10 om $30.000 > SO 10002 $20,000 4 30 (1800 30 50 525.000 125,000 325.000 000 525.000 1.00 1.200 $10,000 BO $23.634 $20.410 30 50 $34.291 $20,352 $29,782 $25,658 $22,338 $19,010 $15,674 $12.329 $8,975 $5,510 $22.445 7 1 5800 . $17.68 10 7 11. $10,0001 . 30 $13.900 $10,730 57,514 Annual mainte- -Repair Total M 10 We IST.173 323.500 12.000 517,55831 AW (54,694.43) (5935.37) $15,271.68 $117,339.45 $23,380.13 Shoot + Screen She 2021-03...5.3. A 9 tv P 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. Lloyd's PowrUp -26,000 Cost and installation, -36,000 Annual maintenance cost, S per year -800 --300 2,000 3,000 Salvage value, s Equipment repair savings, Useful life. years 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 Years PoweUp Investment Annual and salvage maintenance $26.000 50 50 1800 -500 Repair Savings SO $25.000 125.000 Lloyd's Investment Annual Repair and salvage maintenance savings -$36,000 30 50 50 -5300 $35.000 30 $100 335.000 Click Click Save All Areas Save AA Por Maher (24) Praca (21) pdf DPDF Maller (20) pdf PDF Male (20) pul Volunteering for pdf MerBook AH ESC So Question completion Status: 1 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 88888888 PoweUp Lloyd's Investment Annual Repair Investment Annual Repair Years and salvage maintenance savings and salvage maintenance savings O -$25,000 50 $0 -$36,000 50 $0 1 $800 $25,000 50 -$300 $35,000 2 5800 $25,000 50 -$300 $35.000 3 $800 $25,000 $0 $300 $35.000 $ -$800 $25,000 50 -$300 $35,000 5 $0 $800 $25,000 50 -$300 $35.000 -5800 $25,000 $0 -5300 $35,000 7 $2,000 $800 $25,000 $0 -$300 $35,000 $0 $300 $35,000 9 $0 -$300 $35,000 $3,000 -$300 $35,000 AW element -56,068 -5800 $25,000 -$7,025 -5300 $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 5% per year for the next 10 years. Also, the repair savings for the last 3 years were $31,186, $25.508, and $34.259, as best as Harry can determine. He believes savings will decrease by $1,809 per year hereafter. Finally, these 3-year-old protectors are worth nothing on the market now, so the salvage in 7 years is rero, 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 Click Save and Subovt to save and submit. Click Save All Answers to sow all answers, Save All Answers 10 DPDF Maller (21).pdf DPDF Mailer (21).pdf PDF Mailer (20).pdf DPDF Mailer (20).pdf Volunteering for....pdf 1 9 10 AW element Total AW -56,068 -5800 - $0 -$300 $35,000 $3,000 -$300 $35,000 $25,000 $7,025 -$300 $35,000 $18,131,35 527,674.68 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 5% per year for the next 10 years. Also, the repair savings for the last 3 years were $31,186, $25.508, and $34,259, as best as Harry can determine. He believes savings will decrease by $1,809 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 $3000. Case Study Exercises Q1- Plot a graph of the newly estimated maintenance costs and repair savings projections, assuming the protectors last for seven more years Q2- With these new estimates, what is the recalculated AW for the Lloyd's protectors? Use the old first cost an maintenance cost estimates for the first 3 years. type your answer in the answer box . 03- If these estimates had been made 3 years ago, would Lloyd's still have been the economic choice 04- How has the capital recovery amount changed for the Lloyd's protectors with these new estimates? Submission details: 1-Each student must submit an excel file with the required solution (graphs and tables), answers of Q1 & Q2 above. 2- Each Student to submit one file of a hand written verification solution using the factors/tables to calculate the AW value of the revised Lloyds scenario. To be uploaded as image or PDF. verification of Q2, answers Q3 &Q4 above. Click Save and Submit to see and submit. Click Save All Answers to see alles Save All Answers DPDF Maler (21).pdf POF Moller (21).pdf POF Madler (20).pdf POFMater (20) pdf Volunteering for pdf Case Study 2 Annual Worth Analysis From 2008 to 1200$ with 8% increase repair savings for last 24Rs 24,5928, 20,6528, 30.082$ will decreue by : 3,2248 Q1) Amust pugment F[c#ins Jeans more of inha Annual Payment P[J. Niekerk untee Annual Werth (AW) - P (AIP.in)-A+S(AiFi,n) F Alcantated twn of amount Ant Capinder SO Vine Year 200 5 Armuat Maintenance (x) Annuacel Saving.CX) o - 200 1834591 300 90652 30682 -1900 126858 - 1296 23634 -1299.68 $ 20410 1511 6544 17186 -1632.58 8 13962 -1763-18 1 10738 -1964 22 7 $ 9 10 $ 7514 Yearloos Annual ols Annual A 5 tv .pdf (page 2 of 4) S - 1622:58 - 1762-18 -1957 22 13962 1 10738 7514 10 Year P&s A. 4-X /Pw-Alan -26600 o 2 O 2 2141 CI+-53729818-3 O 0 3 4 0 ito.is Annual Annual Maintenance (A) Souding CY) O 0 O 200 $ 24591 39291 - 300 20652 20352 15 789.03 -300 130082 29782 1952.is - 1200 $26858 25658 14670.09 -1296 $22634 22338 H1os. 92 -1399.68 120410 196 10:32 82 18.68 -1511 (594 17186 15679.35 5892.57 -1632.56 13962 12 329.92 4030 SI -1763.18 $ 10738 8979-22 25$1.20 -1904.23 1714 SLO9.77 1308.65 PW tot 112621.06 5 0 6 0 7 6 6 O 8 9 D 10 0 totul AW Q2) Find the Annual maintenance 8* Year = -1511 65 x = -16 77.9715 Find the AW of P and SL 4.-2400 nois be W 77 1365.65 PW tot = 12627.06 Q2) Find the Annual maintenance 8 Year-1565x = -1677 9715 Find the AW of P and SL A :-?500 nalo AW All --2600 (ISCOS = -7173.07 Find the PW of A toid of PW = 1/2627.06 -) 115/ ha16 PWA PWT ((1+i) CHIS (10.19 = 1/2627:06 erretje) = 22441.174 Find Net AW: CAW of PR ) + (Pword) (-9172.07) + C22441-174) Net = [15268 104 AU P/F 9 A MacBook Air 3823 16460 P/F 3459 74571 0696 $20080.11 20.52 X 0.7561 f 15615 10082 % 0-6575 2625705717 115357 27634 x 0.4972 $1751 2040 X 0.4323 17176 *6.3759 (13962 x 0.3269 - $4564 10728 x 0.2843 7514 X 0.2472 total $117340.37 tot 36000 $82 340.37 hot-360001925). 16 207.07 Q3) The AW of Lioyals is more than that of power up so this would had been still selected Q4) The old estimation from Capellad recrecy of Liny do -70 25 Now Capital recovering estimations 5-7500 CAIP 570) Bourse of mil salvage value --96099(014) 7172 has increase - 365? 9 30) sty MacBook Air CCIO OOG 4 00 F5 F& DI 52 FU FO EVO % ies, and improvements, choose check for Updates Check for . B C D E G H and a M Years o manne saving 50 waving Total M & Repair BO and a 0.0001 so 50 $40.000 1 SO 12.000 3341 2 125.000 50 to 10 om $30.000 > SO 10002 $20,000 4 30 (1800 30 50 525.000 125,000 325.000 000 525.000 1.00 1.200 $10,000 BO $23.634 $20.410 30 50 $34.291 $20,352 $29,782 $25,658 $22,338 $19,010 $15,674 $12.329 $8,975 $5,510 $22.445 7 1 5800 . $17.68 10 7 11. $10,0001 . 30 $13.900 $10,730 57,514 Annual mainte- -Repair Total M 10 We IST.173 323.500 12.000 517,55831 AW (54,694.43) (5935.37) $15,271.68 $117,339.45 $23,380.13 Shoot + Screen She 2021-03...5.3. A 9 tv P

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