>>> On Tue, Jan 29, 2008 at 10:45 AM, in message
<873asginrz.fsf@stark.xeocode.com>, Gregory Stark <gsstark@mit.edu> wrote:
> Well consider when you've reached n-1 drives; the expected number of requests
> before you hit the 1 idle drive remaining out of n would be n requests. When
> you're at n-2 the expected number of requests before you hit either of the
> two
> idle drives would be n/2. And so on. The last term of n/n would be the first
> i/o when all the drives are idle and you obviously only need one i/o to hit
> an
> idle drive.
You're right. Perhaps the reason more requests continue to improve
performance is that a smart controller will move across the tracks
and satisfy the pending requests in the most efficient order?
-Kevin
