Question:

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I am wondering how the Salvage, EAC Capital & Maintenance values (seen above in the table) were calculated for each of the 8 years?

I know EAC(capital) = (P + I S)(A/P, i, N) + Si and the straight-line depreciation is BV = P - n[(P-S)/N]

A new bottle-capping machine costs 68,000, including 6,000 for installation. The machine is expected to have a useful life of eight years with no salvage value at that time (assume straight-line depreciation). Operating and maintenace costs are expected to be 4,500 for the first year, increasing by 1,500 each year thereafter. Interest is 11%. Construct a spreadsheet that has the following headings: Year, Salvage Value, Maintenance Costs, EAC (Capital Costs), EAC (Operating Costs), and EAC (Total Costs). Compute the EAC (Total Costs) if the bottle capper is kept for n years, n1.,8 What is the economic life of the bottle capper (in years)? EAC EAC Year Salvage Maintenance Capital MaintenanceTotal 0 62,000 154,250 2 46,500 338,750 4 31,000 5 23,250 6 15,500 77,750 4,500 21,230 17,669 16,232 15,336 14,666 14,115 13,639 13,214 4,500 5,211 5,896 6,555 7,189 7,797 8,380 8,938 25,730 22,880 22,128 21,891 21,855 21,912 22,019 22,152 6,000 9,000 10,500 12,000 13,500 15,000 The economic life of the project is chosen as the life with the least equivalent annual cost. Based on the table above, the economic life of the project is 5 years 23,250 is salvage in right year