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V-series i/o alignment

Hi

Quick question regarding how write i/o is aligned with v-series backend disks. I assume the v-series uses the product-id to calculate the correct i/o offset?

Cheers

Shane

Re: V-series i/o alignment

Also

From  the v-series maintenance guide

For any given LUN, Data ONTAP will assign one path as the active, and the other path as the

passive/failover path. For tier 1(active-active) arrays, the “active” path will be split as evenly as possible

between available paths for any group of LUNs. For tier 2 arrays (active-passive) the array will tell us which

path is preferred for any given LUN.

I assume this refers to how the backend array is configured? a tier 1 array has both controllers active?

Re: V-series i/o alignment

Hi Shane,

When the array LUNs are created, a "host" type is chosen (per the V-Series Implementation Guides in the DataONTAP documentation).  This ensures the correct settings (including offsets if needed) are chosen.

Generally speaking, a Tier 1 array will have symmetric paths.  In ALUA terms, both paths would be Active/Optimized.  These have equal cost, and array LUN access would be evenly split (half the LUNs on one path, the other half on the other path.)


A Tier 2 array may have asymmetric paths.  In ALUA terms, one path is Active/Optimized, and the other is Active/Non-Optomized.  In this case, all the array LUNs in that group would use the A/O path, and the A/N path would be the failover path

In both cases we depend on the array to tell us which paths are active or preferred.

Hope that helps!

Daniel Isaacs
Technical Marketing Engineer
V-Series

NetApp
Daniel.Isaacs@netapp.com
www.netapp.com

Re: V-series i/o alignment

Thanks for that, we were trialling an unsupported config with a HDS HUS150 behind a V6280. Dont ask why

One of the things i notices was we weren't able to get it running Active/active even though the HUS supports that kind of I/O my assumption was it was defaulting to active/passive due to it not quite understanding what array was connected to it. Which makes a certain amount of sense.