Re: [PATCH RFC] pram: persistent over-kexec memory file system

[Vladimir Davydov <vdavydov@parallels.com>]

On 07/28/2013 03:02 PM, Marco Stornelli wrote:
> Il 28/07/2013 12:05, Vladimir Davydov ha scritto:
>> On 07/27/2013 09:37 PM, Marco Stornelli wrote:
>>> Il 27/07/2013 19:35, Vladimir Davydov ha scritto:
>>>> On 07/27/2013 07:41 PM, Marco Stornelli wrote:
>>>>> Il 26/07/2013 14:29, Vladimir Davydov ha scritto:
>>>>>> Hi,
>>>>>>
>>>>>> We want to propose a way to upgrade a kernel on a machine without
>>>>>> restarting all the user-space services. This is to be done with CRIU
>>>>>> project, but we need help from the kernel to preserve some data in
>>>>>> memory while doing kexec.
>>>>>>
>>>>>> The key point of our implementation is leaving process memory 
>>>>>> in-place
>>>>>> during reboot. This should eliminate most io operations the services
>>>>>> would produce during initialization. To achieve this, we have
>>>>>> implemented a pseudo file system that preserves its content during
>>>>>> kexec. We propose saving CRIU dump files to this file system,
>>>>>> kexec'ing
>>>>>> and then restoring the processes in the newly booted kernel.
>>>>>>
>>>>>
>>>>> http://pramfs.sourceforge.net/
>>>>
>>>> AFAIU it's a bit different thing: PRAMFS as well as pstore, which has
>>>> already been merged, requires hardware support for over-reboot
>>>> persistency, so called non-volatile RAM, i.e. RAM which is not 
>>>> directly
>>>> accessible and so is not used by the kernel. On the contrary, what 
>>>> we'd
>>>> like to have is preserving usual RAM on kexec. It is possible, because
>>>> RAM is not reset during kexec. This would allow leaving applications
>>>> working set as well as filesystem caches in place, speeding the reboot
>>>> process as a whole and reducing the downtime significantly.
>>>>
>>>> Thanks.
>>>
>>> Actually not. You can use normal system RAM reserved at boot with mem
>>> parameter without any kernel change. Until an hard reset happens, that
>>> area will be "persistent".
>>
>> Thank you, we'll look at PRAMFS closer, but right now, after trying it I
>> have a couple of concerns I'd appreciate if you could clarify:
>>
>> 1) As you advised, I tried to reserve a range of memory (passing
>> memmap=4G$4G at boot) and mounted PRAMFS using the following options:
>>
>> # mount -t pramfs -o physaddr=0x100000000,init=4G,bs=4096 none 
>> /mnt/pramfs
>>
>> And it turned out that PRAMFS is very slow as compared to ramfs:
>>
>> # dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy
>> bs=4096 count=$[100*1024]
>> 102400+0 records in
>> 102400+0 records out
>> 419430400 bytes (419 MB) copied, 9.23498 s, 45.4 MB/s
>> # dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy
>> bs=4096 count=$[100*1024] conv=notrunc
>> 102400+0 records in
>> 102400+0 records out
>> 419430400 bytes (419 MB) copied, 3.04692 s, 138 MB/s
>>
>> We need it to be as fast as usual RAM, because otherwise the benefit of
>> it over hdd disappears. So before diving into the code, I'd like to ask
>> you if it's intrinsic to PRAMFS, or can it be fixed? Or, perhaps, I used
>> wrong mount/boot/config options (btw, I enabled only CONFIG_PRAMFS)?
>>
>
> In x86 you should have the write protection enabled. Turn it off or 
> mount it with noprotect option.

I tried. This helps, but the write rate is still too low:

with write protect:

# mount -t pramfs -o physaddr=0x100000000,init=4G,bs=4096 none /mnt/pramfs
# dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy 
bs=4096 count=$[100*1024]
102400+0 records in
102400+0 records out
419430400 bytes (419 MB) copied, 17.6007 s, 23.8 MB/s
# dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy 
bs=4096 count=$[100*1024] conv=notrunc
102400+0 records in
102400+0 records out
419430400 bytes (419 MB) copied, 4.32923 s, 96.9 MB/s

w/o write protect:

# mount -t pramfs -o physaddr=0x100000000,init=4G,bs=4096,noprotect none 
/mnt/pramfs
# dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy 
bs=4096 count=$[100*1024]
102400+0 records in
102400+0 records out
419430400 bytes (419 MB) copied, 9.07748 s, 46.2 MB/s
# dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/pramfs/dummy 
bs=4096 count=$[100*1024] conv=notrunc
102400+0 records in
102400+0 records out
419430400 bytes (419 MB) copied, 3.04596 s, 138 MB/s

Also tried turning off CONFIG_PRAMFS_WRITE_PROTECT, the result is the 
same: the rate does not exceed 150 MB/s, which is too slow comparing to 
ramfs:

# mount -t ramfs none /mnt/ramfs
# dd if=/dev/zero of=/mnt/pramfs if=/dev/zero of=/mnt/ramfs/dummy 
bs=4096 count=$[100*1024]
102400+0 records in
102400+0 records out
419430400 bytes (419 MB) copied, 0.200809 s, 2.1 GB/s

>
>> 2) To enable saving application dump files in memory using PRAMFS, one
>> should reserve half of RAM for it. That's too expensive. While with
>> ramfs, once SPLICE_F_MOVE flag is implemented, one could move anonymous
>> memory pages to ramfs page cache and after kexec move it back so that
>> almost no extra memory space costs would be required. Of course,
>> SPLICE_F_MOVE is to be yet implemented, but with PRAMFS significant
>> memory costs are inevitable... or am I wrong?
>>
>> Thanks.
>
> From this point of view you are right. Pramfs (or other solution like 
> that) are out of page cache, so you can't do any memory transfer. It's 
> like to have a disk but it's actually a separate piece of RAM. We 
> could talk about it again when this kind of implementation will be done

Yeah, that's the main difference. PRAMFS lives in a dedicated region, 
while page cache is spread all over the whole RAM, and what is worse, 
some pages can be used at early boot.

I believe dealing with page cache could be wired into PRAMFS, but the 
question is what would be clearer: implement something lightweight and 
standalone, or blow existing code, which was not initially planned to 
handle things like that? I will think about that.

Thanks.

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