Wednesday, September 5, 2012

"enq: CR - block range reuse ckpt" and the recycle bin

I decided to write this blog entry because we ran into an issue over the weekend.  The process that normally runs fine, was running slow.. "like molasses" was the comment from one of the folks.  After digging into an AWR report I found one of the top waits for the sql was

enq: CR - block range reuse ckpt

OK.. what is that, and why ?   I didn't find a whole lot except that it indiciated contention on blocks from multiple processes trying to update blocks.. hmmm..

I looked further in the AWR and saw the top Segment for Logical reads was "RECYCLEBIN$"

and one of the top queries was.

"delete from RecycleBin$ where bo=:1"

Well since the process was finished I did my own test.. I created a tablespace and created an object in it.. dropped the object, added the to table, then dropped the table.. over and over again, until the number of objects in  the tablespace remained constant.  I then created the table again, and started to let it grow (so it would have to free up the recycle bin to get space)... And what I saw in the top 5 wait events was again.....

enq: CR - block range reuse ckpt

I wanted to document this for others that may hit this.. If a search for this wait event brought you to my blog, Please check your recycle bin and make sure that it isn't cleaning itself out to make room causing this wait event ...

Tuesday, September 4, 2012

Recognize the magic optimizer numbers

Well I figured I document some of the magic numbers that the optimizer uses to help remember them, and help others. The back ground of this is simple.

I was looking through a query that was running for a long, long time, and the cardinality looked wrong.  I know the developers were using a table operation (looping over a LOB that was treated like  table). 

The Cardinality estimate for the step was 8168, and I thought hmmmm I've seen that before when dynamic sampling didn't happen.  Well after some digging I came across this page. Cardinality

The page contained this handy chart below...  These are important numbers to remember because when you see a cardinality matching this chart it is probably because the optimizer couldn't estimate the correct cardinality, and it couldn't dynamically sample.  Below is a snippet from the query I was investigating. Notice the cardinality on the first line.

(cr=0 pr=0 pw=0 time=0 us cost=29 size=16336 card=8168) 0 0 0 HASH JOIN RIGHT OUTER (cr=0 pr=0 pw=0 time=0 us cost=8757 size=233200 card=100) 0 0 0 VIEW (cr=0 pr=0 pw=0 time=0 us cost=8614 size=14 card=1) 0 0 0 HASH UNIQUE (cr=0 pr=0 pw=0 time=0 us cost=8614 size=2069 card=1) 0 0 0 FILTER (cr=0 pr=0 pw=0 time=0 us) 0 0 0 NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us) 0 0 0 NESTED LOOPS (cr=0 pr=0 pw=0 time=0 us cost=8613 size=2069 card=1) 0 0 0 HASH JOIN (cr=0 pr=0 pw=0 time=0 us cost=8612 size=2044 card=1)

Default cardinality for database objects

The following table demonstrates the estimated cardinalities (using a 8K blocksize) of various objects which have had no statistics generated for them :

Object TypeEstimated Cardinality
Heap Table82
Global Temporary Table8168
Index-Organized Table1
System Generated Materialized View
(such as the output of the TABLE operator)

Saturday, September 1, 2012

Exadata sizing updated for 3tb drives 1/2 rack SATA

OK, Now I new Exadata coming in that has 3tb drives, and the first question asked is .. How much disk to I have to configure on it ?  Well I'm going to expand on a previous entry I did on sizing .

1/2 Rack. Sata drives. normal redundancy

This means we have
  • 7 storage cells
  • Each storage cell contains 12 disks
  • each disk is 3tb (which is about 2.794 tb usable)  *** This is calculated using base 1024 
  • The first 2 disks in each storage cell has 29.103g already partitioned for the OS (which is mirrored).
  • The rest of the disks in the group are used for DBFS
Given this, I am going to calculate out the total disk available then subtract out the 29.103g (for OS and DBFS).

First 12 disks * 7 cells x 2.794 = 234.696 tb of total raw storage/
Subtract out 29g* 2 disks * 7 cells = 406g    ----- OS
Subtract out 29g * 10 disks * 7 cells = 2.03tb  -----   DBFS
Available raw is 234.696 - 2.436 = 232.26  

Now I said we were running Normal Redundancy.. This means that we loose 1/2

DBFS = 1.015tb
OS        29g
Remaining for Data and Reco = 116.13

But of course we need to account for cell being off line.  This takes out 1/7 of the storage.

DBFS   === .870 tb (29g * 10 * 6)/2
Everything else  ===  ( 2.765 * 12 disks * 6 cells)/2   == 99.54

So now we have 99.54 raw storage available for Data and Reco.

This is now easy to figure out now.. You have really 100tb raw storage (with normal redundancy) to split up between Data and Reco.

Now a full rack is easy to do.

2.765 * 12 disks * 13 cells) / 2 =  215.67tb