Re: [RFC] Reproducible OOM with just a few sleeps

[Simon Jeons <simon.jeons@gmail.com>]

On 01/14/2013 11:00 PM, Dave Hansen wrote:
> 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

So if he config sparse memory, the issue can be solved I think.

> 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.

Dear Dave, two questions here:

1) e820 map is read from BIOS, correct? So if all kinds of ranges dump 
from /proc/iomem are setup by BIOS?
2) only "System RAM" range dump from /proc/iomem can be treated as real 
memory, all other ranges can be treated as holes, correct?

>
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