2) IBM tool storage modeller gives advise to use 85 % raid physical capacity doesn't matter whether use DRP pool or not
Original Message:
Sent: Mon June 23, 2025 03:04 AM
From: Istvan Buda
Subject: DRAID size
Hi Tomas,
1:
The slice size is not the calculated value, but the strip_size: which is 256 KiB fixed.
The complete explanation can be seen here:
https://www.ibm.com/docs/en/flashsystem-7x00/8.7.x_cd?topic=ac-distributed-raid-array-properties-1
2:
The 85% recommendation is for DRP pool only. For legacy pool there is no such value:
"A general guideline is to ensure that the provisioned capacity with the data reduction pool does not exceed 85% of the total usable capacity of the data reduction pool. "
https://www.ibm.com/docs/en/flashsystem-7x00/8.7.x_cd?topic=c-pools
Regards,
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Istvan Buda
budai88@gmail.com
Original Message:
Sent: Fri June 20, 2025 04:30 AM
From: Tomas Kovacik
Subject: DRAID size
can somebody explain this :
I have STV 7300, DRAID6 with following parameters :
raid_level raid6
strip_size 256
drive_count 24
rebuild_areas_total 1
stripe_width 12
physical_capacity 664.21TB
Drive FCM 104,8TB physical size :
IBM_FlashSystem:sks73017a:tomas>lsdrive 0 | grep physical_capacity
physical_capacity 34.93TB
counting array size :
As I understand stripe_width means that 12 slices are backet by P+Q+S slice so I have 12 data slices on each drive
- slice size = drive_size/12+P+Q+S
slice size = 34.93/15 = 2.32866667 TB
- I have 12*24 data slices so usable data capacity should by 12*24*2.32866667=670,656 TB
- Why array show 664,21 TB instead of 670.565 TB ?
- IBM recommended capacity is 85 % from raid usable is this really necessary ?
Thanks
Tomas
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Tomas Kovacik
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