On Mon, 5 Jul 1999 [EMAIL PROTECTED] wrote:
> On Sun, 4 Jul 1999, Casey Bralla wrote:
>
> > Ok, but I'm still confused. How is the load averages calculated? Is
> > it (processor time per task)/ (total processor time)?
> >
> >
> If I understand it right myself, it is the average number of processes
> that are waiting for the processor. A load average of 0 means any time
> a process is ready to run, there is a processor free to run it. A load
> average of 1 means on average, it is going to have to wait for one other
> process to give up an interrupt, or block waiting for sometheng. A load
> average of .1 means 9 times out of 10 ther will be a processor free, and
> it can get right to work, but once in awhile it has to wait.
AFAIK, this is mostly right. I'd like to add that the "perfect" load
average is 1, since if it is smaller, you have some free processor , so
you paid too much, and if it is larger, you have more processing to do
than available.
Frank
>
> OOn some mainframes it was called dispatcher queue depth, and it often
> ran 3-10 on good days.
>
> Lawson
> >< Microsoft free environment
>
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