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CASE STUDY THE CHANGING SCENARIO OF AN ANNUAL VALUE ANALYSIS Harry, owner of a battery distributor in Atlanta, Georgia, conducted an economic analysis 3 years ago when he decided to place in-line voltage protectors for all the important parts of his test equipment. The estimates used and the analysis of annual value at a MARR = 15% are summarized here. Two protectors from two different manufacturers are compared. PowerUp($) Lloyd($) Costo de instalacin Costo de mantenimiento anual Valor residual Ahorros de reparacin de equipos Vida til, aos -22,000 -600 2,000 25,000 6 -36,000 -300 3,000 35,000 10 The Excel spreadsheet shown below was used to make the decision. Lloyd's company was the clear decision because of its substantially higher AW value. Lloyd's protectors were installed. MARR = 15% Powrup Annual maint ($600.00) Investment and salvage ($5,584.74) Repair savings $25,000.00 YEAR AW values Lloyd's Annual maint Investment and salvage ($7,025) Repair savings $35,000 AW Powrup $18,815 ($300.00 AW Lloyd's $27,675 0 $ 1 25,000 2 3 $ (22,000.00) $ $ $ $ $ $ 2,000.00 s (600) $ (600) $ (600) $ (600) $ (600) $ (600) $ 4 25,000 25,000 25,000 25,000 25,000 5 (36,000) $ $ $ $ $ $ $ $ $ 3,000 $ (300) $ (300) $ (300) $ (300) $ (300) (300) $ (300) $ (300) $ (300) $ (300) $ 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 6 7 8 9 10 $ During a quick review this past year (year 3 of operation), it was obvious that maintenance costs and repair savings had not been in line with estimates made 3 years ago. In fact, the maintenance contract (which includes quarterly inspection) will increase from $300 to $1200 next year and from there it will 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. Harry believes savings can decrease by $2,000 per year going forward. Finally, these old 3-year-old protectors are worthless in today's market. This makes the residual value in 7 years zero instead of $3,000. Requirements Build the spreadsheet in Excel according to the image shown below. A B D E E F G H J K L - PMT(C3,6, C10,016,0) -- PMT(C3,6,NPV[C3,E11:E:16),0,0) ) MARR 1596 SUM(C8E8) PowrUp Investment Annual Repair and salvage maint savings 155,584.7415600.00 $25,000.00 Lloyd's Annual maint 15 300.00) Investment and salvare 157,025) YEAR AW values Repair savings $35,000 $ AW PowrUp $18,815.26 AW Lloyd's $27,675 S 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 0 1 2 3 $ 122,000.00) $ S S $ 5 S 5 s 2,000.00 S (600) S (600) 5 1600) S (600) S (600) S (600) S 25,000 25,000 25,000 25,000 25,000 25,000 4 5 6 6 7 8 9 10 (36,000) 5 $ $ $ S S $ $ $ $ s 3,000 $ (300) 5 (300) $ (300) $ (300) 5 (300) 5 (300) 5 (300) 5 (300) $ (300) S (5 (300) $ 5 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 35,000 - PMT(C3,6,NPV(C3,011:D:16),0,0) S a. Plot for projections of estimated maintenance costs and repair savings for Lloyd Company, assuming the protectors last 7 more years. For this consider the old estimates (as shown in the Excel file) Ear: This is a graph of money flows where we use arrows and to identify the series we join them with a line from beginning to end of the series. b. With the new estimates of the problem replace the values of the estimates in the Excel file. What is the recalculated AW for Lloyd's protectors? Would Lloyd be the economic choice if we let ourselves be carried away by these new estimates? Why yes or why not? C. How has the amount of capital recovery for Lloyd's protectors changed with these new estimates? You must set the difference in AW (annual value)