Date: Tue, 10 Nov 2009 11:24:12 +0100 From: Ivan Voras <ivoras@freebsd.org> To: freebsd-geom@freebsd.org Subject: Re: gjournal and calculation of the size of journal provider Message-ID: <hdbf04$21k$1@ger.gmane.org> In-Reply-To: <4AF84245.7070108@quip.cz> References: <4AF84245.7070108@quip.cz>
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Miroslav Lachman wrote: > What is the right rule for journal size calculation? > There are two sources stating different things. > > 1] journal size depends on disk write speed > http://lists.freebsd.org/pipermail/freebsd-fs/2006-June/002016.html > > "For example your disk can write > at 60MB/s. Journal switch time is 10 seconds. The journal provider has > to have place to keep two journals (active and inactive). So bascially > you need 60*10*2MB + gjournal headers." > > > 2] journal size depends on RAM size > http://www.freebsd.org/doc/en/articles/gjournal-desktop/article.html#UNDERSTANDING-JOURNALING > > > "Your RAM size should fit in 30% of the journal provider's space. For > example, if your system has 1 GB RAM, create an approximately 3.3 GB > journal provider. (Multiply your RAM size with 3.3 to obtain the size of > the journal)." > > > What's the right size for journal on 143GB 15k rpm SAS disks on machine > with 16GB of RAM? Based on second case, it will be more than 50 GB - one > third of the size of disk. This is insane vasting. It really does depend on the speed of drives but it could be approximated by saying there will not be more data to write than the size of memory (which is probably wrong since you can write from /dev/zero indefinitely). The first advice is sufficient, but you should probably extend the result by 20% to be safer.
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