On Fri, Jul 9, 2010 at 5:18 PM, Brandon High <bh...@freaks.com> wrote:
> I think that DDT entries are a little bigger than what you're using. The > size seems to range between 150 and 250 bytes depending on how it's > calculated, call it 200b each. Your 128G dataset would require closer to > 200M (+/- 25%) for the DDT if your data was completely unique. 1TB of unique > data would require 600M - 1000M for the DDT. > Using 376b per entry, it's 376M for 128G of unique data, or just under 3GB for 1TB of unique data. A 1TB zvol with 8k blocks would require almost 24GB of memory to hold the DDT. Ouch. -B -- Brandon High : bh...@freaks.com
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