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On Fri, Mar 01, 2002 at 04:29:42PM -0700, Rick Kelly and a few other
people earlier wrote:

> Running 'w' and 'top' simultaneously will show that top thinks the load
> is about half as high as 'w'.

=2E.. and other examples of misunderstanding of the load average computatio=
n.

Try to understand how it works:

- periodically, a timer interupts regular computation.
- it finds out how many processes are on [one of the] CPU[s] or on the=20
  RUN queue

- this is averaged into gliding averages for 1m, 5m, and 15m effective
  averaging time.

Now, on some systems (e.g. NetBSD/i386, as far as I know) this clock interu=
pt
is the same that is used for normal context switching. That is, if you have
a process that wakes up, does very little, and goes to sleep again, and=20
otherwise not much other activity, you might see it _every time_ your=20
statistical interupt service routine is running, and thus get a load average
of 1.00, although in reality the process is only using 0.01 or even less
of the CPU.

As a net result: don't trust the exact value of the load average, especially
below 1.

A better estimate is to run a long-running low-priority cpu-bound process
and look at its accounting information (e.g., time onetrillionloops); and
subtract the total wallclock time from that.

Regards,
	-is

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