On fre, 2007-08-24 at 14:20 -0700, Nicole wrote:
> > l2 = 256
>
> So, this should always be the same size?
Yes, there is not much reason to change L2.
> > L1 = at least cache_dir size * 2 / 256 / 256 / 13KB, or ca cache_dir in
> > GB * 2. (13 KB is the estimated average object size)
>
> ca?
yes? (circa) I rounded it a bit.. it's not an exact math. As long as it
ends up in about those numbers.. L1 * L2 * L2 should be significantly
more than the number of objects you have in the cache, and L2 should not
be too big or too small.
> I guess I am missing something?
> 90000 * 2 / 256 / 256 = 2.746582 / 13000 = .0002112 ??
You are missing an unit.. 90000 in the above should be 90000MB
L1 = 90000MB * 2 / 256 / 256 / 13KB =
900000 * 1024 * 2 / 256 / 256 / 13 = 216
> Could you provide an example or 2?
simplified formula:
L2 = 256
L1 = cache_dir size / 500, rounded upwards on small numbers..
If L2 is changed or you have a singnificantly different object size
distribution then use the equation above. This simplified formula is
only valid for L2 = 256 and average object size of about 13KB.
Regards
Henrik
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