>>> On Tue, Jan 29, 2008 at 9:52 AM, in message
<877ihsvdcb.fsf@oxford.xeocode.com>, Gregory Stark <stark@enterprisedb.com>
wrote:
> I got this from a back-of-the-envelope calculation which now that I'm trying
> to reproduce it seems to be wrong. Previously I thought it was n(n+1)/2 or
> about n^2/2. So at 16 I would have expected about 128 pending i/o requests
> before all the drives could be expected to be busy.
That seems right to me, based on the probabilities of any new
request hitting an already-busy drive.
> Now that I'm working it out more carefully I'm getting that the expected
> number of pending i/o requests before all drives are busy is
> n + n/2 + n/3 + ... + n/n
What's the basis for that?
-Kevin