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I have shown my work below for the problem and would appreciate if someone can let me know if I'm on the right track or point me in the right direction if not.

An 8 processor file server handles a million operations each day. If each processor remains idle 5% of the time what is the utilization of the entire file server?

I believe the utilization of the entire file server is 95% because it says all processors at idle 5% of the time. Is this correct?

If the file server is fully utilized how much more work could it do?

We know the file server does 10^6 operations each day and we know each processor is utilized 95% of the time. Thus 10^6 / .95 = 1,052,632 operations / day would be the total number of operation if it were fully utilized.

The last question has me a little stumped: If 10% of the operations are used to move files between processors, how efficient is the file server?

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What does efficient even mean? If I add some while(true){} to my code I can get high CPU utilization, but I wouldn't say that is efficient. – adrianN Mar 22 '13 at 11:17

2 Answers 2

The problem with adding another 5% load to the server is that its queues will overflow and performance decreases again.

If you're interested in the topic I suggest you to read the The art of computer systems performance analysis

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This does not seem to address the question at all. – Raphael Oct 31 '12 at 9:55

Not necessarily. If each processor remains idle 5% of the time such that there is some other processor readily available to service the requests then your server as a whole is still running. Hence your server is being utilized 100% of the time. On the other hand if all of the are idle at the same time your server cannot be utilized for 5% of the time. So the utilization in this case is 95%. The reality could be somewhere in between. Next if the server could do 10^6 operations per day and has a utilization factor of 95% it does 0.95*10^6 operations per day not 10^6/0.95. Out of these 0.95*10^6 operations 10% are file transfers and are chores (stuff not beneficial for user). Hence, the remaining 90% i.e. 0.9*0.95*10^6 are useful operations.

  • Efficiency = Useful operations performed/Capacity of the server = 0.9*0.95*10^6/10^6 = 0.855
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"Not necessarily." Not necessarily what? It's not at all clear which part of the question you're addressing, here. Also, I disagree that a system is at 100% utilization if every processor is idle 5% of the time but they're all idle at different times. By that logic, a 100-processor system in which each processor is idle 99% of the time is at 100% utilization as long as 99 of the processors are idle at any one time. That's nonsense: the machine could do 100 times as much work! – David Richerby Sep 29 at 20:41

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