Question: Full Sail Ale in Hood River currently uses a conveyor (C) to move its
ID: 1107080 • Letter: Q
Question
Question: Full Sail Ale in Hood River currently uses a conveyor (C) to move its products to ongoing trucks. The conveyor originally cost $18,000 six years ago. Because of high maintenance costs and dropping 4 cases of product every hour, Full Sail is evaluating whether to replace the conveyor (C) with a new Kubota (K) or new John Deere (J) forklift. Full Sail only needs one forklift and has the budget to buy either, but not both.
The remaining useful life of the conveyor belt is 12 years, which is thesame for the Kubota (K), and John deere (J) forklift. At the end of the twelve years, all three options have zero salvage value. Even if Full Sail replaces the conveyor today with a forklift, the conveyor has no salvage value.
A: Using this table, Determine what mutually exclusive system Full Sail should select TODAY using the Internal Rate of Return (IRR) incremental analysis with a MARR of 6% per year compounded annually. Show all work, including interest equations, interpolations from tables and IRR to closest 1/2%
This Question is for an Engineering Economy Class, and was given during a practice test.. Please answer in detail, with steps shown. I'm not looking for an EXCEL answer, as during our tests we won't have a computer. I will thumbs down any excel answer as that is NOT what I'm asking for. Thank you for your help.
Conveyor ( C) Kubota (K) John Deere (J) Intial cost (price + freight) $18,000 $52,000 $20,000 Annual Operating Cost $12,000 $8,000 $10,000 Annual Maintenance Cost $8,000 $4,100 $6,200 Remaining useful life 12 years 12 years 12 yearsExplanation / Answer
We need to calculatePresent Worth of Cost in todays term ( of projects using IRR incremental analysis
We have MARR equals to 6%
Remaining useful life is 12 years
Lets say Initial Cost is given for different projects is for today
( If we are allowed to assume the Initial Cost (IC) of Conveyor for 6 years ago we would consider it as 18000*(1.06)^6)
Calculating PW for Cost for each project as follows (Salvage value is 0)
PW for any project= Initial Cost +(Annual Operating Cost + Annual Maintainance Cost)discounted for today + Annual Benefit (Discounted for PV)
PW for Conveyor = -[18000 + (12000+8000) Discounted each year cost for today]
Calculation would be to find PV of cost
20000[(1/1.06)+(1/1.06)^2+....+(1/1.06)^12]=229,398
PW for Cost of Conveyor = -247398
PW for Cost of K = -52000 -PV (12100) = -5200-138786 =-143,986
PW for Cost of J = -20000-PV (16200) = -20000-185812 = -205812
PW for Cost of K< PW for Cost of J < PW for Cost of C
Hence Kubata is preferable with PW for Cost is considerd.
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