On 01/11/2013 07:31 PM, paul.szabo@sydney.edu.au wrote:
Seems that any i386 PAE machine will go OOM just by running a few
processes. To reproduce:
sh -c 'n=0; while [ $n -lt 19999 ]; do sleep 600 & ((n=n+1)); done'
My machine has 64GB RAM. With previous OOM episodes, it seemed that
running (booting) it with mem=32G might avoid OOM; but an OOM was
obtained just the same, and also with lower memory:
Memory sleeps to OOM free shows total
(mem=64G) 5300 64447796
mem=32G 10200 31155512
mem=16G 13400 14509364
mem=8G 14200 6186296
mem=6G 15200 4105532
mem=4G 16400 2041364
The machine does not run out of highmem, nor does it use any swap.
I think what you're seeing here is that, as the amount of total memory
increases, the amount of lowmem available _decreases_ due to inflation
of mem_map[] (and a few other more minor things). The number of sleeps
you can do is bound by the number of processes, as you noticed from
ulimit. Creating processes that don't use much memory eats a relatively
large amount of low memory.
This is a sad (and counterintuitive) fact: more RAM actually *CREATES*
RAM bottlenecks on 32-bit systems.
On my large machine, 'free' fails to show about 2GB memory, e.g. with
mem=16G it shows:
root@zeno:~# free -l
total used free shared buffers cached
Mem: 14509364 435440 14073924 0 4068 111328
Low: 769044 120232 648812
High: 13740320 315208 13425112
-/+ buffers/cache: 320044 14189320
Swap: 134217724 0 134217724
You probably have a memory hole. mem=16G means "give me all the memory
below the physical address at 16GB". It does *NOT* mean, "give me
enough memory such that 'free' will show ~16G available." If you have a
1.5GB hole below 16GB, and you do mem=16G, you'll end up with ~14.5GB
available.
The e820 map (during early boot in dmesg) or /proc/iomem will let you
locate your memory holes.