| Re: Help with calculating fair allocations [message #7935 is a reply to message #7922] |
Tue, 31 October 2006 06:39   |
Vasily Tarasov
Messages: 1345
Registered: January 2006
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Senior Member |
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Hello,
you wrote:
| Quote: |
So..if we consider:
Example "C"-248MB RAM --kmemsize 16384000:18022400
What would:
"A"-62MB RAM --kmemsize ?
"B"-124MB RAM --Kmemsize ?
and a monster:
Example "D" -496MB RAM --Kmemsize ?
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kmemsize parameter is measured in bytes, as all parameters, which name is terminated by ...size.
So if you want to set kmemsize limit to 248MB, limit/barrier should equal 248 * 1024 * 1024 = 260046848. I suppose you'll be able to calculate appropriate numbers for 62/124/496MB.
Also, I would like to add, that kmemsize is not RAM. To get full information on user beancountes (UBC) visit this page and read detailed information about all parameters: http://wiki.openvz.org/UBC.
Sorry, if I misunderstood you.
HTH,
vass.
[Updated on: Tue, 31 October 2006 06:39] Report message to a moderator |
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