In: Finance
Because it fumes at room temperatures, hydrochloric acid creates a very corrosive work environment. A machine working in that environment is deteriorating quickly and can be used for only one more year, at which time it will be scrapped with no salvage value. It was purchased 3 years ago for $88,000, and its operating cost for the next year is expected to be $54,000. A more corrosion-resistant challenger will cost $208,000 with an operating cost of $54,000 per year. It is expected to have a $45,000 salvage value after its 10-year ESL. At an interest rate of 11% per year, what minimum replacement value would render the challenger attractive? The minimum replacement value that would render the challenger attractive is $
Since the 2 alternatives here have unequal lives, we will go for annualized cost approach which means that we will calculate per annum outflow of each alternative and choose the alternative with lower per annum outflow.
Alternative 1 is the machine with 1 year life and zero salvage, the operating cost for the next year = $ 54,000
Present value of outflow of $ 54000 next year = 54000/1.11 = $ 59,940
Alternative 2 is the challenger with 10 year life, $ 208000 initial outflow, $ 45000 salvage at the end of year 10
Present value of outflow = initial outflow + present value of annual cost of 54000 - present value of 45000 salvage
Present value of outflow = 2,08,000 + 346,554 - 19008 = $ 535,546
Annualised cost = present value of outflow / sum of discounting factor at 11% for 10 years
Annualised cost = 535546 / 5.8892 = $ 90937
Now since the alternative 1 has lower outflow we will not want to buy the challenger so to render the challenger attractive, its annualised cost must go down by (90937 - 59,940) = $ 30997
So if the cash outflow in replacing becomes 2,08,000-30997 = $ 177003 this price will make the challenger attractive as in that case the annual outflow under both the alternatives will be the same. Minimum replacement value that would render the challenger attractive is $ 177003