On x86 and x86-64 systems pages are measured in 4k - OVZ uses this these pages
for resource allocation.
I believe you can also use M and G suffixes on recent OVZ package releases to
set privvmpages/oomguarpages/vmguarpages. For example, "vzctl set $veid
--vmguarpages 512M:600M" will translate to 131072:153600 (barrier:limit) pages
automatically so you don't need to do to the math.
Regards,
Andrew Cranson
Layershift Limited
www.layershift.com
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Markus Hardiyanto wrote:
# 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 <[EMAIL PROTECTED]>
To: [email protected]
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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