Re: How to estimate the upper bound of the peak memory consumption of cryptsetup itself?

[Coiby Xu <coxu@redhat.com>]

On Sat, Jun 18, 2022 at 05:12:56PM +0200, Milan Broz wrote:
>On 16/06/2022 06:43, Coiby Xu wrote:
>>Hi,
>>
>>Recently, I notice cryptsetup itself consumes significant amount of
>>memory (~256M) when estimating the memory requirement for dumping vmcore
>>to a LUKS-encrypted disk,
>>
>>$ time -v cryptsetup luksOpen encrypted.img volume --key-file mykey.keyfile | grep "Maximum resident set size"
>>          Maximum resident set size (kbytes): 1309828
>>$ cryptsetup luksDump encrypted.img
>>...
>>Keyslots:
>>    0: luks2
>>          PBKDF:      argon2id
>>          Memory:     1048576
>>          ...
>>
>>
>>So is there a way to estimate the upper bound of the peak memory
>>consumption of cryptsetup itself without running cryptsetup?
>
>As you already found, the major memory consumption is by memory-hard KDF.
>But this memory is used only while calculating keyslot encryption key,
>it is released immediately after the Argon call is finished.
>I do not think we have better estimation here.

Thanks for the reply! Sorry I meant the way to estimate the overhead of
crypsetup itself i.e. ~256M in the above example. Previously I only take
the memory consumption by memory-hard KDF into consideration and
neglected the memory consumption of cryptsetup itself. This obviously
leads to an underestimation of the memory requirement of cryptsetup. I
need to overestimate the memory requirement a bit to make sure OOM won't
happen that's why I am asking if there is a way to estimate the
upper bound of memory requirement of cryptsetup itself. 

>
>(Another story is locking all memory, including big areas used by libc,
>but that should not be problem here, I hope.)

Do you mean locking all memory first in order to know the memory
requirement?

>
>Milan
>

-- 
Best regards,
Coiby