# 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 _______________________________________________ Users mailing list [email protected] https://openvz.org/mailman/listinfo/users Send instant messages to your online friends http://uk.messenger.yahoo.com _______________________________________________ Users mailing list [email protected] https://openvz.org/mailman/listinfo/users
