Hi Milan,

Thank you very much for the detailed explanation!= This is tremendous help to us!

I had already brought this up in our group meetin= g. We will re-arrange out partitions to ensure all have enough space = for default configurations. Thank you very much on that!

May I ask further – (sorry more questions, = I just want to do it right and make the best out from your original design.= )

1. I’m using linux kernel keyring as token for p= assphrase. Do I need to enlarge JSON? (BTW, Why JSON area is st= ored twice, for backup only that area?)

2. &nbs= p; Do we still need to use luksHeaderBackup <= device> and luksHeaderRestore <device> are for entire 16 M header backup?&n= bsp; This means each luks partition needs 32 M for its header!

Now here is our sto= ry : We have storage redundancy on our board, that is, for each component (= for example linux rootfs) we have two partitions to save two copies of the = component. I think with that, we may not need luks header backup. When we detect anything wrong with curr= ent active partition, include luks header, we can switch to use the standby= partition for rootfs for example, and then repair, or simply wipe everythi= ng and redo luks format and copy the data into it.

Should this work?&n= bsp; Can you suggest some ways, or check points, for our background task to= periodically checking to ensure all luks’s are good, in case you hav= e something on top of your head? 8-)

Thank you so much!&= nbsp;

Hualing<= /span>

-----Original Message-----
From: Milan Broz [mailto:gmazyland@gmail.com]
Sent: Sunday, October 20, 2019 6:08 AM
To: Hualing Yu <hualing.yu@jci.com>; dm-crypt@saout.de
Subject: Re: [dm-crypt] 10 M Luks2 header size?

Hi,

this information should be later in FAQ, so I try= to explain it here.

Anyway, stay with defaults, if you can.

On 19/10/2019 21:59, Hualing Yu wrote:=

> May I ask a couple of additional questions a= bout this so that we know how to trade off.

> 1. What the re= encryption can do for us? Could you explain very

> briefly as I’m not sure if we need it?=

In principle it can perform changes that requires= full-device rewrite (change of the volume key).

See man cryptsetup-reencrypt - just for LUKS2 it = is more reliable and mainly online (you can use device while it is in reenc= ryption process).

See slides from Ondra

https://nam02.safelin= ks.protection.outlook.com/?url=3Dhttps%3A%2F%2Fokozina.fedorapeople.org%2Fo= nline-disk-reencryption-with-luks2-compact.pdf&data=3D02%7C01%7Chua= ling.yu%40jci.com%7Ca096abcf38e8483e599808d7554555fc%7Ca1f1e2147ded45b681a1= 9e8ae3459641%7C0%7C1%7C637071628596824108&sdata=3DZn13uT%2B7wsLKex3= r6u3LWAC7xFobCn4PLs10ywQYxeU%3D&reserved=3D0<= /p>

There should be also some online demos=

Reencryption demo: https://nam02.safelin= ks.protection.outlook.com/?url=3Dhttps%3A%2F%2Fasciinema.org%2Fa%2F268573&a= mp;amp;data=3D02%7C01%7Chualing.yu%40jci.com%7Ca096abcf38e8483e599808d75545= 55fc%7Ca1f1e2147ded45b681a19e8ae3459641%7C0%7C1%7C637071628596824108&am= p;sdata=3D6DkH8Bwz699zeGzk25vf8gh4%2FKuImVaMeGEu34qHkCA%3D&reserved= =3D0

Encryption demo: https://nam02.safelin= ks.protection.outlook.com/?url=3Dhttps%3A%2F%2Fasciinema.org%2Fa%2F268574&a= mp;amp;data=3D02%7C01%7Chualing.yu%40jci.com%7Ca096abcf38e8483e599808d75545= 55fc%7Ca1f1e2147ded45b681a19e8ae3459641%7C0%7C1%7C637071628596824108&am= p;sdata=3D8nuvhvj5fBB%2FeH0pu0%2F0qRNd7l47dVMQwzDrNFoeeMA%3D&reserv= ed=3D0

For this we require some reserved area for storin= g temporary encryption data.

> 2. We need onl= y one or at most two keyslots but we do want them

> to be scattered as much as needed just as if= for the default case,

> what we can do? Use –luks2-keysl= ots-size=3D1 M (or whatever size that

> will give two key enough space to scatter)?<= o:p>

There are two areas (see LUKS2 docs) - JSON area = for metadata and binary area.

JSON has small binary header, than JSON data (it = is 16k currently, stored twice).

For the binary area, it depends what you need, ex= act size depends on the stored key size (here the binary keyslot data are s= tored, exactly the same as in LUKS1).

I would expect you are using current default for = disk encryption, AES256-XTS.

Then you need to store 512bit (2x256bit) key in e= ach binary keyslot.

With the LUKS AF filter and 4k alignment it shoul= d be 256KiB of binary data per keyslot.

So for 1M and 512bit key it allows 4 LUKS keyslot= s here.

> 3. What the si= ze of metadata size for default configuration?

> What’s the downside of using 16 K?

The whole LUKS2 default header takes 16MiB.<= /o:p>

For JSON area it is 16k, stored twice (we will in= crease it later, this is for compatibility reasons), for binary area - it i= s "16M - 2x16k" (16M minus JSON areas).

There is only several possible sizes of JSON area= you can use (see LUKS2 docs), binary area is basically arbitrary with maxi= mum 128M, it must be aligned to 4k sectors.

JSON areas allows to store user token metadata, s= o if you do not need it, no need to enlarge it.

Thanks,

Milan