Misaligned Pools and Lost Space...

[Bidule0hm]

It's not better, it's different (plus a 10 drives RAID-Z2 has the same problem as a 8 drives one...). Did you read the first posts in this thread?

 

[RichTJ99]

Why the 2^n + p rule is broken

If we take the example of a 10 drives RAID-Z2, who follows the 2^n + p rule, we have a 5 % overhead (32 + 8 = 40 sectors --> 40 / 3 = 13.333 so we need a 2 sectors padding) so the rule is obviously broken. Now we can't do anything about that, this rule is just a (over) simplification so it's not exact.

Sorry - I am still a little confused but I will take your advice - is it still bad to have an 8 disk RaidZ2 Vdev? Do I end up with less than 2 disks failing as security because it goes over a single sector? Is it the same with a 10 drive setup.

 

[Bidule0hm]

It's not bad. It's just that you lose some space (a few %), that's all. But you can recover it by using recordsize = 1M at the dataset creation instead of the default 128k value.

Anyway , ZFS cares about blocks, not sectors.

Please read the first posts entirely, take the time to understand them, and if you're lost then read some ZFS documentation too to understand how ZFS works regarding the blocks, parity, etc... :)

 

[RichTJ99]

So in terms of data integrety and safetey, a raidz2 with 8 drives will be fine. If I use the default settings I will lose a small percentage of space in the 8 drive format vs the 10 drive format?

 

[Bidule0hm]

Yes, almost 100 % exact ;)

The 10 drives RAID-Z2 also has some overhead (remember, the 2^n + p rule is broken...). AFAIK is you want a RAID-Z2 with no misalignment overhead it needs to be 4, 6 or 18 drives (but the 18 drives one isn't recommended...) OR you need to chose a recordsize value of 1M at the dataset creation and you can use any number of drives you want without losing space ;)

 

[jgreco]

"Without losing space" seems to be an overly optimistic description. It's just that the complexity of determining the space lost, combined with the variability of data stored, makes this a pointless exercise.

 

[Bidule0hm]

Yeah I know; I added more details in the first posts about that but here I just über simplified the thing to make it über clear :)

 

[joeschmuck]

You gotta love all the math brought into this topic.

 

[jgreco]

You gotta love all the math brought into this topic.

Figuring out space lost in ZFS due to RAIDZ allowing for compression efficiency:

 

[Bidule0hm]

If you take some reasonable assumptions on the data you put on the pool it's not that complex once you've figured it out :)

Of course if you want 10 decimal places accuracy, etc. it's not simple anymore...

 

[snicke]

Figuring out space lost in ZFS due to RAIDZ allowing for compression efficiency:

Sorry to ruin the party, but you've clearly made a calculation error in the upper right...

Sent from my Galaxy S6 Edge

 

[jgreco]

Feel free to correct it.

 

[nightshade00013]

I prefer the simplified version of the above equation.

 

[snicke]

Feel free to correct it.

I would have loved to but I'm afraid I'm too busy. I'm about to start painting my submarine and then I have to vacuum clean my duck: https://youtu.be/-n8fn7k9NiE

Sent from my Galaxy S6 Edge

 

[Ericloewe]

Figuring out space lost in ZFS due to RAIDZ allowing for compression efficiency:

I believe I am qualified to say this image is pure gibberish.

 

[jgreco]

I believe I am qualified to say this image is pure gibberish.

As are the calculations required to determine the space lost in ZFS...

 

[Ericloewe]

As are the calculations required to determine the space lost in ZFS...

I believe that was implicit. ;)

 

[jgreco]

In mathematics, it's better to be explicit.

Sorry, spent too much time running computer systems for college math departments. ;-)

 

[RichTJ99]

Boy I am lost... If I just wipe the current 6x 6TB volume/dataset, and recreate it with 8x 6TB's will I be running into problem in the future if i just create a default volume?

 

[Bidule0hm]

Short answer: no.