A special milling machine is being installed at a first cost of $10,000. Mainten
ID: 1222642 • Letter: A
Question
A special milling machine is being installed at a first cost of $10,000. Maintenance cost is estimated to be $5,500 for the first year and will increase by 6% each year. If interest is 12% and the salvage value is $2,000 at any time, for what service life will AE(i) cost be a minimum? Answer: 7 years The maintenance cost of a certain machine is zero the first year and increases by $200 per year for each year thereafter. The machine cost $2,000 and has no salvage value at any time. Its annual operating cost is $1,000 per year. If the interest rate is 10% what life will result in minimum AE(i) cost? Solve by trial and error, showing yearly costs in tabular form.Explanation / Answer
20.
Initial cost = $10000
Maintenance cost = $5500 in first year then increasing every year by 6% (g)
Salvage value = $2000
R = 12%
Time of useful life = n
Present value of all cost = initial cost + present value of maintenance cost - Present value of salvage value
Present value of all cost = 10000 + (5500/(.12 - .06))*(1- (1.06/1.12)^n) - 2000/1.12^n
AE Cost = Present value of all cost / ((1-1/1.12^n)/.12)
AE Cost = (10000 + (5500/(.12 - .06))*(1- (1.06/1.12)^n) - 2000/1.12^n) / ((1-1/1.12^n)/.12)
From the above formula, following AE cost has been calculated for different years of useful life.
AE cost for 1 year useful life = $14700
AE cost for 2 year useful life = $10629.25
AE cost for 3 year useful life = $9381.78
AE cost for 4 year useful life = $8839.61
AE cost for 5 year useful life = $8578.94
AE cost for 6 year useful life = $8458.34
AE cost for 7 year useful life = $8417.06
AE cost for 8 year useful life =$8424.61
AE cost for 9 year useful life =$8463.97
Here, lowest AE cost is with useful life of 7 years. Afterwards, it again starts increasing.
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