>>> On Tue, Jan 29, 2008 at 9:52 AM, in message <[EMAIL PROTECTED]>, Gregory Stark <[EMAIL PROTECTED]> 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
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