Re: ASM rebalance estimates

  • From: "Martin Bach" <development@xxxxxxxxxxxxxxxxx>
  • To: "Herring Dave - dherri" <Dave.Herring@xxxxxxxxxx>,"Oracle-L Freelists" <oracle-l@xxxxxxxxxxxxx>
  • Date: Thu, 09 Sep 2010 20:24:02 +0100

Hi Dave,

The duration of the rebalance operation differs between Oracle 10 and 11. 10g 
is slower and moves more data around. You van see this in v$asm_operation. And 
more importantly-you can't simply compare figures. What worked quickly for me 
doesn't mean it does for you.

If you have a short maintenance window you might want to consider a standby 
database and a switchover operation. 
Simply create the standby database on the new storage array and dispose of the 
old primary after the switchover.

The copying of LUNs between arrays is also a good way of doing it, but you 
might not have the software and/or license to do so.

Hope this helps,

Martin

Martin Bach

Oracle Certified Master 10g
http://martincarstenbach.wordpress.com
http://www.linkedin.com/in/martincarstenbach

----- Reply message -----
From: "Herring Dave - dherri" <Dave.Herring@xxxxxxxxxx>
Date: Thu, Sep 9, 2010 16:18
Subject: ASM rebalance estimates
To: "Oracle-L Freelists" <oracle-l@xxxxxxxxxxxxx>

Folks,

Has anyone done any detailed investigations on how to estimate the time needed 
for a disk add and rebalance operation with ASM?  Athough everything can be 
done "hot", our client (like probably most) requires us to perform the adding 
of any disks to ASM during our maintenance window and also asks for an estimate 
on the time involved.  Their expectation has been that the time involved is 
only affected by the # of disks added.  My contention is it's that plus the 
amount of data to be rebalanced, basically existing disk fullness or size of 
the diskgroup.

Assuming everything stays consistent (power of 11 for the rebalance operation, 
load on the diskgroup, and each LUN or "disk" is physically made up of the same 
# and capacity of physical disks) here's what I believe to be true:

* If you had 10 x 100 GB disks that were 80% full (or 800 GB of total data) and 
added 2 x 100 GB disks, the final set should be 12 x 100 GB disks with 800 GB 
of data balanced across them, so each would have 66.67 GB of data.  To me that 
means, at a minimum, you'd have to move 10(80-66.67) = 133  GB of data.

* Take the same scenario but this time the disks are 60% full (or 600 GB of 
total data), the final set should be 12 x 100 GB disks with 600 GB of data 
balanced across them, so each would have 50 GB of data.  To me that means, at a 
minimum, you'd have to move 10(60-50) = 100 GB of data.

Unfortunately I don't have any system to test this on and existing data is way 
off from my assumption on what's being done.  Here's a listing of what I've 
collected:

                                Size of
                        # of    Disks           # of                            
Used GB Est. GB GB              "Total
        Elapsed Disks   Added           Existing        "Total  in              
to be           Moved           Work" -
Server (Min.)   Added (GB)              Disks           Work"           
Diskgroup       Moved           per Min Est. Work
----------------------------------------------------------------------------------------------------------
LnxA    412             9       118             51              1,763,618       
5,747           862.05  2.09            880,879
LnxB    228             3       269             13                534,693       
2,814           527.63  2.31             -5,595
LnxB    166             4       118             51                792,866       
5,662           411.78  2.48            371,201
LnxC    202             4       118             51              1,199,042       
5,820           423.27  2.10            765,611
LnxC    216             7       118             46                931,080       
5,339           705.15  3.26            209,005
LnxD    98              3       118             11                332,504       
1,218           261.00  2.66             65,240
LnxD    98              1       118             10                174,605       
1,152           104.73  1.07             67,364
LnxE    76              3       118             11                321,145       
1,176           252.00  3.32             63,097

"Total Work" refers to the column in V$ASM_OPERATION and the value listed is 
the last one displayed from this view before the operation completed.

Has anyone done any research into this or recorded similar data?

Thx.

Dave Herring  | DBA, Global Technology Services
A c x i o m  C o r p o r a t i o n
630-944-4762 office | 630-430-5988 cell | 630-944-4989 fax
1501 Opus Pl | Downers Grove, IL, 60515 | U.S.A. | www.acxiom.com
Service Desk: 888-243-4566, https://servicedesk.acxiom.com, GSCA@xxxxxxx


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