"Kevin Grittner" <Kevin.Grittner@xxxxxxxxxxxx> writes: > >>> On Tue, Jan 29, 2008 at 9:52 AM, in message > <877ihsvdcb.fsf@xxxxxxxxxxxxxxxxxx>, Gregory Stark <stark@xxxxxxxxxxxxxxxx> > 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? 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. -- Gregory Stark EnterpriseDB http://www.enterprisedb.com ---------------------------(end of broadcast)--------------------------- TIP 5: don't forget to increase your free space map settings
