Charting server load

[Robert Brockway]

On Thu, 1 Feb 2007, Ian Petersen wrote:

> [snip]
>
> OK, that makes a certain amount of sense, but I still feel unconvinced.
>
> The load average is exactly that: an average.  A load of 50 says that,
> over the last minute, every time the run queue was sampled (how often
> does that happen anyway?) there were 50 processes "ready to go".

It doesn't really need to "sample" afaik as the kernels knows how many 
processes are flagged to run each timeslice since it is moving them in and 
out of the run queue.  It only needs to count them and it probably has to 
do that anyway.

> Now, if all of our 50 hypothetical processes need 2.01% of the CPU to
> get all their work done, shouldn't each one show up in roughly 2.01%
> of the samples?  If that's true, then each sample would see, on
> average, one process in the ready state and the load would be one
> (well, 1.05 since we're talking 2.01% not 2%).

This logic sounds spot on to me.  At any given time approximately 49 of 
the 50 processes are not asking for the cpu because they are i/o bound at 
the time.  Thus the load would indeed be approximately 1.05 as you say.

Getting back to William's original question, a load of 50 would mean that 
the system is sustaining a cpu bound situation which translates to an 
overloaded system in my books since processes are getting far less time on 
the cpu than they need to run optimally.

A cpu bound system which was not also heavily i/o loaded may respond quite 
well to interactive use (ie, the shell may seem to be responding fine to 
many commands) but this is because in general interactive use demands 
little of a system.  If you are typing, a modern computer is quite capable 
of servicing many processes between your keystrokes.

Cheers,

Rob

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