Messages in this thread | | | Subject | Re: [PATCH 4/7] ppc64 - Specify amount of kernel memory at boot time | From | Dave Hansen <> | Date | Thu, 23 Feb 2006 10:15:55 -0800 |
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On Thu, 2006-02-23 at 18:01 +0000, Mel Gorman wrote: > On Thu, 23 Feb 2006, Dave Hansen wrote: > > OK, back to the hapless system admin using kernelcore. They have a > > 4-node system with 2GB of RAM in each node for 8GB total. They use > > kernelcore=1GB. They end up with 4x1GB ZONE_DMA and 4x1GB > > ZONE_EASYRCLM. Perfect. You can safely remove 4GB of RAM. > > > > Now, imagine that the machine has been heavily used for a while, there > > is only 1 node's memory available, but CPUs are available in the same > > places as before. So, you start up your partition again have 8GB of > > memory in one node. Same kernelcore=1GB option. You get 1x7GB ZONE_DMA > > and 1x1GB ZONE_EASYRCLM. I'd argue this is going to be a bit of a > > surprise to the poor admin. > > > > That sort of surprise is totally unacceptable but the behaviour of > kernelcore needs to be consistent on both the x86 and the ppc (any any > other ar. How about; > > 1. kernelcore=X determines the total amount of memory for !ZONE_EASYRCLM > (be it ZONE_DMA, ZONE_NORMAL or ZONE_HIGHMEM)
Sounds reasonable. But, if you're going to do that, should we just make it the opposite and explicitly be easy_reclaim_mem=? Do we want the limit to be set as "I need this much kernel memory", or "I want this much removable memory". I dunno.
> 2. For every node that can have ZONE_EASYRCLM, split the kernelcore across > the nodes as a percentage of the node size > > Example: 4 nodes, 1 GiB each, kernelcore=512MB > node 0 ZONE_DMA = 128MB > node 1 ZONE_DMA = 128MB > node 2 ZONE_DMA = 128MB > node 3 ZONE_DMA = 128MB > > 2 nodes, 3GiB and 1GIB, kernelcore=512MB > node 0 ZONE_DMA = 384 > node 1 ZONE_DMA = 128 > > It gets a bit more complex on NUMA for x86 because ZONE_NORMAL is > involved but the idea would essentially be the same.
Yes, chopping it up seems like the right thing (or as close as we can get) to me.
-- Dave
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