Engineering Econ question

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mobikwa
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Engineering Econ question

Postby mobikwa » Wed Jul 29, 2009 4:24 pm UTC

Hey folks, I am helping a friend with his homework since it has stumped him. I have taken the course he is in already but this question is driving me insane!

A hospital bought a machine at a cost of $40000. Maintenance cost is expected to remain constant throughout the life of this machine at $2000 per year. The salvage value is estimated to be "0" at the end of its useful life of 10 years. Determine the economic life of this machine, MARR = 10%
A. 5 years
B. 1 year
C. 10 years
D. 7 years

I started this problem as doing a comparison between a defender and a new option, but there is no new option so that was futile. Then I realized that I may be over thinking this.
I did a straight line depreciation to find the salvage value at the end of each year, I figure it goes down 4000 per year (40000-0)/10=4000

Would the solution be the year that the sum of the marginal costs exceeds the salvage value at that year? Can anyone point me in the right direction?

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Yakk
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Re: Engineering Econ question

Postby Yakk » Thu Jul 30, 2009 3:57 am UTC

http://en.wikipedia.org/wiki/Minimum_ac ... _of_return

In order to do a MARR based decision, don't you need a measure of the value of what the machine is producing? Is this truly all you are told?

Making the linear depreciation decision is assuming facts not in evidence. This may be par for the course in your course, but in that case you are asking "what are the standards of argument in my particular course".
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Seraph
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Re: Engineering Econ question

Postby Seraph » Thu Jul 30, 2009 10:14 am UTC

Unless you are using a very different definition of "economic life" then I've seen I don't see how the machines economic life would differ any from it's stated 10 years of "useful life". A simple version the definition I've used is something along the lines of "The length of time it is cost effective to maintain a piece of equipment rather then replace it, while using it for it's primary purpose". In your example it's always cheaper to maintain it (fixed maintnence costs) right up to the point where it breaks (on the magical 10 year date).

EDIT: Actually Yakk is right. I was assuming linear depreciation, if you don't there may be a point where it is a better deal to sell the equipment for it's residual value then to maintain it.


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