HPE Storage Users Group

With 96 large NL drives you might need to increase the DEFAULT step and growth size. the Default might result in a logical disk that may not INITIALLY span all 96 spindles leaving you with less iops. Filtering = BAD.

For 96 drives in R6 6+2 I estimate the growth to be 72GB and the step size 64k.

When you start out writing data on an empty CPG, there is an initial logical disk. over time and additional writes, more logical disk are created and cover more/different spindles, but under early/near empty configs you first logical drive may not be a full strip of all disks.

One way to check is to look at PD activity. if only some disks are getting activity and others don't have any you have an undersized initial logical disk.

Maurice

Maurice, i'm curious, why do you think this is beneficial?
1. A larger step-size?
That would be totally depended on the IO-size written to disk. Without knowing the IO-size being read and written, it is impossible to say what is the correct value. Default is therefore best (the 64k you mentioned is default for 6+2 by the way)
2. Increasing the growth size beyond the default?
An increased growth-size leads to an increased row-depth for an LD.
With an increased row-depth the growth extends are touching up to all drives behind that node.
Touching all drives also means that each "set of rows" is impacted during a drive failure. Further more it diminishes the flexibility of the system to "re-level" over (new) drives by growing into new empty drives first.
I understand this is complex and a real debatable one. Personally i see little real gain (except really specific circumstances). Yes it gives a more even spread over the drives (initially), but from experience it also gives a bigger impact when there are drive issues.

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