| Re: setting memory allocation [message #12567 is a reply to message #12531] |
Thu, 03 May 2007 03:05   |
Markus Hardiyanto
Messages: 27
Registered: April 2007
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Junior Member |
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# RAM is 4k pages, so 131072*4k = 512M
vzctl set 1 --save --vmguarpages 131072
vzctl set 1 --save --oomguarpages 131072
vzctl set 1 --save --privvmpages 131072:196608
i thought all the numbers in beancounters was in bytes.. so why you divided it with 4k?
Best Regards,
Markus
----- Original Message ----
From: Gregor Mosheh <gregor@hostgis.com>
To: users@openvz.org
Sent: Wednesday, May 2, 2007 9:46:52 PM
Subject: Re: [Users] setting memory allocation
I'd like to take a shot at answering this, to "quiz myself" on how well I
understand this stuff. So if my answers are incorrect or incomplete,
please speak up!
> I have a computer with P4 2,4Ghz processor and 1GB of RAM. I'm planning to
> split it into 3 VEs with this memory allocation:
> VE1: 512MB
> VE2: 256MB
> VE3: 256MB
If by "memory allocation" you mean "the amount of RAM they're guaranteed
to have available for use by apps" then try this:
# RAM is 4k pages, so 131072*4k = 512M
vzctl set 1 --save --vmguarpages 131072
vzctl set 1 --save --oomguarpages 131072
vzctl set 1 --save --privvmpages 131072:196608
But, if you really have only 1 GB of RAM, it may not be wise to allocate
all of it. If all of the VEs really use all their RAM, the system will
start swapping to make up more RAM (e.g. for the HN's own use) and nobody
enjoys a system that's swapping.
-Gregor
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