StoreVirtual Calculator

[cursur949]

I am currently trying to figure out the math behind calculating storage capacities for the new StoreVirtual line, I know there is a calculator out there for the P4000 series gear but it does not have the 4330 part numbers loaded up yet and I kind of just want to understand it so I it is being done right.

Basically what I am trying to do is calculate how much usable storage I would have if I had two B7E17A or three B7E17A or fore B7E17A

B7E17A = 4330 w/8 450GB drives

Thanks in advanced for your help

[oikjn]

depends on your hardware raid level you want to run and your network raid level you want to run.  Typically its raid5 with two raid groups per shelf and network raid10.

If you can't calculate the raid5 space, its the # of hdds in the group minus one and then multiplied by the HDD capacity.  Network raid works just like regular raid for space, so NR10 gives you 1/2 the available space you have left after the raid5 groups are combined.

I like http://www.ibeast.com/content/tools/RaidCalc/RaidCalc.asp as a raid calculator, but it doesn't do raid6.

Assuming hardware raid5 w/ 4 per group(including a hot spare per group), you would get 838GB usable storage per node assuming NR10 and you have two or more nodes.  If you do one large raid5 group, you would get 1257GB, but you would have a 8 drive group(including one spare) raid5 which might be a bit long for rebuild on a drive failure.  You could also get 1047GB out of a raid6 group with one spare.

Any way you look at it, its always sad when you think about how much usable space you actually have after you take into account hardware and network raid.  That said... such is the cost of the kind of availability you get from network raid...  raw capacity is relitively cheap... availability is not.

[cursur949]

Oikjn and all –

Thanks for responding. My question is less about how raid works and network raid works but more about the actual math works as I am extremely familiar with how raid works. More specific questions below –

How does the network raid math work when adding a third node to a two node cluster? The confusion here come from if you have two drive raid you are technically using raid 1 if you have a three drive raid you are using raid 5 so the same thing should apply for network raid. In this example the raid 1 I would lose the capacity of 50% of the nodes (after hardware raid) and with raid I would lose the capacity of 1 total drives (or one node)

How does HP calculate usable drive capacity after right loss?

The way I think the equation should work out is below though I am not sure if I am right.

Drive size = DS

Write loss = WL (not sure how to calculate this)

Total number of drive in an node = ND

Raid loss = RL (this assumes you know how to calculate hardware raid loss)

Total node capacity of a node = TNC

Total Cluster Nodes = TCN

Total usable capacity = TUC

(DS-WL)*(ND-RL)= (TNC)

Then to calculate the total usable storage of a two node cluster

TNC/TCN=TUC

Then to calculate the total storage of a three node cluster

TNC/???=TUC

[oikjn]

I'm a little lost there.   If you know how hardware raid works, you should know how to calculate your usable space.

I can't remember the formula for conversion from HDD manufacturer's capacity claims to usable space, but that link I gave has it built in.  For 450GB hdd you get 420GB usable per drive.  The usable space for the raid is dependent on the raid level you select, but don't forget about making sure you have hot-spares in your setup (I guess you could make the decision that you would rather get the capacity and are ok that if a drive fails you have lost your hardware redundancy and would rely on your network redundancy for availability until a new drive is shipped in).  The max node capacity you could get would be an 8 drive raid5 group which gives you 2933GB usable space per node with no hot-spare according to the raid calc link I gave.  If you do NR10, that would give you 1466GB usable space on the san per node.

Assuming you use NR10 and all nodes are equal capacity, the total capacity of your san would be the sum of the total usable capacity of each node divided by TWO... this works even for odd numbers of nodes.

[Tedh256]

Network raid 10 is network raid 10 - whether you have 2 nodes or 5 or 7 or 8

Basically SanIQ ensures that there are 2 of each chunk of data, spread across the nodes. (I don't think that it puts chunks of any particular volume on more than 2 nodes - but it always ensures that there are 2 chunks and that each chunk is on a different node)

I wish there were more clearly articulated documentation about exactly how it works - but for sizing questions it is:

right-sized drive size * number of drives - number of drives used for parity / 2

[oikjn]

I believe the manual does spell out exactly the pattern that NR10 gives you.  I know for an even number of nodes, (and assuming you label your nodes 1,2,3,4 and they are ordered in CMC that same way), the data is mirrored between the mirror is every other node and the stripe is the adjacent node, so 1+3 are mirrors and 1+2 are stripes.

For NR10 the data has exactly two copies of the data, if you do NR10+1, the data has three copies, NR10+2 has four.

[Tedh256]

---

@oikjn wrote:

the data is mirrored between the mirror is every other node and the stripe is the adjacent node, so 1+3 are mirrors and 1+2 are stripes.

I'm not sure what you mean by "1+2 are stripes"? Could you try to elaborate? I usually think of a "strip set" as being essentially the opposite of mirroring: where a "volume" is striped across 2 (or more) physical devices with no parity or mirroring, so that if one of the devices fails, the whole "stripe set" or volume fails.

I don't think that SanIQ does that when using nr10 does it? Or what am I missing?

[oikjn]

good call.  Sorry about that.  Mental fart.  every time I said stripe I meant mirror.

NR10 has two copies of the same data

NR10+1 has three copies fo the same data

NR10+2 has four copies of the same data.

[Tedh256]

So instead of " so 1+3 are mirrors and 1+2 are stripes" you meant:

 so 1+3 are mirrors and 2+4 are mirrors. I *think*

;-)

[oikjn]

damm I confused things up with my words :)

1+3 are stripes and 1+2 are mirros.  1+3 are the Raid0 part and 1+2 are the Raid1 part..  If node one goes down, your LUN availability will be dependent on node two staying available. 

[Tedh256]

oh wow I don't get this at all!

Here is how I think about it:

at the nr10 level each chunk of data is kept twice: call it chunk1a and chunk1b.

Chunk1a is on one node, chunk1b is on another node - that is the nr10 (it is not really traditional raid 10 which is a mirror of mirrors - it is simple raid 1 mirroring, but abstracted to the network-across nodes level).

I do not think striping comes into it at all?

Maybe I am completely lost though - it has happened before!

[oikjn]

If you have only two nodes, it is NR10 is really just NR1, but if you add a third or more node to the system you will get the stripes as well.

For example.  If you have:  Node1 = 100GB, Node2=100GB, Node3=100GB, Node4=100GB and you see the following in CMC:

Cluster-Whateverthenameis

     -Node1

     -Node2

     -Node3

     -Node4

Your available space will show as 400GB for the cluster

If you make a LUN that is NR10 and is 50GB is size, that will consume 100GB of cluster storage, and the data will be stored on all the nodes with Node1's data mirrored on Node2 and Node3's data mirrored on Node4, so if node2 and node4 were to both shut off, your LUN would still be available.

If you do a LUN that is 50GB NR0, that would consume only 50GB space on the cluster and the data would be spread across all four nodes, but if ANY node were to go offline, that LUN would become unavailable.

If you do a LUN that is 50GB NR10+1, that would consume 150GB space on the cluster.  The data layout is a little more complicated with that where I think the data would go 134, 241, 312, 423, but I'm not 100% on that one.

[Tedh256]

I still think something is not right - but short of having time to dig into deep tech detail, I don't know that we can make more certainty here - at least not for myself.

I'll add only that: I believe your nr10 scenario would in fact be 4 copies of each chunk of data, so that a 50GB volume would consume 200GB instead of 100GB of space. I do though agree 100% on your description of nr0 :-)

As I said - I have no confidence that I can describe how it actually works, and while I also don't have confidence that what you describe is actually correct, I find your effort laudable. And perhaps you are correct and I'm just not "getting" it.

I really wish there were better/more detailed descriptions. It's been a couple years since I looked at the documentation, I'll try to do that when I have a few moments.

Thank you again for spending all this time - now, I hope the original poster has had HIS question answered satisfactorily!

[Bart_Heungens]

Hi all,

Like they say, a picture says more than 1000 words...

Hopefully this explains network RAID:

http://www.bitcon.be/networkraid.png

It is explained until 3 copies, I assume you can imagine yourself for 4 copies...

Kr,

Bart

[Tedh256]

lol - ok so no lables. Sigh:

what are the "dynamic pools"? Are these volumes?
The picture seems to show 4 nodes - but what are the B1, B2 items, and why are there 6 of them spread unevenly between the "dynamic pools" and the nodes?

Sorry, I still don't get it. Perhaps I'm dense, but ....

[Bart_Heungens]

Hi,

Don't focus on the Dynamic Pool notation. You can see it as a cluster where you will put your storage nodes in...

B1, B2 and so on are data blocks that you will write from your server onto the nodes. It is just to give you the idea how the data is spread across all nodes.

Yes this is an example of a cluster with 4 nodes, if you would have 6 nodes, data blocks are spread across in the same manner.

The most chosen RAID level is network RAID 10 which is the middle picture. So every block of data (B1, B2 and so on) will be written onto 2 nodes. In the way like it is on the picture.

This is not important if all nodes are in the same server room. However if you want to spread the storage nodes across 2 server rooms, be aware that nodes 1 and 3 should be in 1 server room, and nodes 2 and 4 in the other. Otherwise you would not survive a datacenter failure...

Kr,

Bart

[Tedh256]

AHHHH - excellent, yes!

This is exactly as I thought. There are no stripes, and any one NR10 volume has 2 copies of each chunk of data: 1 copy on 2 different nodes. If I have 2 total nodes in the cluster, or 4 or 6 there are still only 2 copies of each chunk of data, stored on 2 particular nodes.

Thank you ...

[Bart_Heungens]

Correct that is how it works.

You're welcome...

[Tedh256]

so looking back over this monster thread, I can see now that the OP question still has not been answered (sorry if I hijacked your thread!).

I think it is abundantly clear now how to calculate useable storeage when using NR10 volumes and there are 2 nodes:

assuming 1 rg per node, no stand-by drives, then you multiply "right-sized" drive * 7 = total useable storage (this is data drives in one node only, because nr10 is mirroring so the other nodes disks are "lost" to redundancy).

But as you add nodes - you don't lose the same amount of disk space for the 3rd node as you did the 2nd, and I assume that for the 4th node you lose even less, etc .... (again, this is all assuming all volumes are nr10).

So - how do we calculate useable disk space for different cluster sizes?

whew - it took FOREVER to get the question even figured out!!!

lol!

[Bart_Heungens]

Hi,

I don't see the issue, for sure not for NR10, since all the blocks are spread across the nodes... See the picture...

So even if there are 7 nodes with NR10 there are 2 data blocks written for every original data block. So you will loose always 50%... For 2, 3, 4, 5, 6 and so on nodes...

Kr,

Bart

[Tedh256]

so for a B7E17A = 4330 w/8 450GB drives we get 2933 per node / 2 = 1466GB useable. If we have 2 nodes then 2933. If three nodes then 2933+1466= 4399. If four nodes then 4399+1466 ...

Hmm that makes sense - I have always "heard" though that for some reason as you add nodes you lose "less" per node. But thinking about it I guess I'm not sure why you would ....

[Bart_Heungens]

A lot of people makes mistakes here... For NR10 you always loose 50%... Because that is network RAID.

What a lot of people forget is that on the underlying hard disks there is (by default - can be changed to RAID 1) RAID 5 on the disks to protect the data against a hard disk failure, which is still always the case with spinning disks...

So on top of the NR10 (50%) you will loose also the size of 1 physical disk for parity data...

I know some customers who choose for performance and not size, and so set also on the physical disks RAID 1, which ment that every block of data was written 4 times across the cluster... 2 times because of hardware RAID and 2 times because of network RAID...

Kr,

Bart