B-tree node split strategy in SQL Server for monotonically increasing value

Question

Consider a B-tree index on a value that will always increase monotonically, e.g. a column of type IDENTITY. With a conventional B-tree implementation, whenever a node is full, it will be split 50%/50% and we end up with a B-tree in which (almost) all nodes will be only 50% full.

I know that Oracle discovers when a value is ever-increasing and in these cases Oracle performs a 90%/10% split instead. That way, (almost) all nodes will be 90% full and a far better page utilization is obtained for these, quite common, cases.

I have not been able to find documentation for a similar feature in SQL Server. However, I have performed two experiments in which I inserted N random integers, and N consecutive integers in an index, respectively. The former case used far more pages the latter.

Does SQL Server provide a similar functionality? If so: can you point me to some documentation on this feature?

UPDATE: It seems, by the experiments provided below, that leaf nodes are kept un-splitted and internal nodes are split 50%/50%. That makes B-trees on increasing keys more compact than on random keys. However, the 90%/10%-approach by Oracle is even better, and I still look for some official documentation that can verify the behavior seen in experiments.

score 4 · Answer 1 · answered Apr 29 '11 at 21:00

If it is adding a row at the end of the index it will just allocate a new page for the row rather than split the current end page. Experimental evidence for this is below (uses the %%physloc%% function which requires SQL Server 2008). See also the discussion here.

CREATE TABLE T
(
id int identity(1,1) PRIMARY KEY,
filler char(1000)
)
GO

INSERT INTO T
DEFAULT VALUES
GO 7

GO
SELECT sys.fn_PhysLocFormatter(%%physloc%%)
FROM T

GO

INSERT INTO T
DEFAULT VALUES

GO

SELECT sys.fn_PhysLocFormatter(%%physloc%%)
FROM T
GO

DROP TABLE T

Returns (Your results will vary)

(1:173:0) /*File:Page:Slot*/
(1:173:1)
(1:173:2)
(1:173:3)
(1:173:4)
(1:173:5)
(1:173:6)
(1:110:0) /*Final insert is on a new page*/

This does only appear to apply to leaf nodes though. This can be seen by running the below and adjusting the TOP value. For me 622/623 was the cut off point between requiring one and two first level pages (might vary if you have snapshot isolation enabled?). It does split the page in a balanced manner leading to wasted space at this level.

USE tempdb;

CREATE TABLE T2
(
id int identity(1,1) PRIMARY KEY CLUSTERED,
filler char(8000)
)

INSERT INTO T2(filler)
SELECT TOP 622 'A'
FROM master..spt_values v1,  master..spt_values v2

DECLARE @index_info  TABLE
(PageFID  VARCHAR(10), 
  PagePID VARCHAR(10),   
  IAMFID   tinyint, 
  IAMPID  int, 
  ObjectID  int,
  IndexID  tinyint,
  PartitionNumber tinyint,
  PartitionID bigint,
  iam_chain_type  varchar(30),    
  PageType  tinyint, 
  IndexLevel  tinyint,
  NextPageFID  tinyint,
  NextPagePID  int,
  PrevPageFID  tinyint,
  PrevPagePID int, 
  Primary Key (PageFID, PagePID));

INSERT INTO @index_info 
    EXEC ('DBCC IND ( tempdb, T2, -1)'  ); 

DECLARE @DynSQL nvarchar(max) = 'DBCC TRACEON (3604);'
SELECT @DynSQL = @DynSQL + '
DBCC PAGE(tempdb, ' + PageFID + ', ' + PagePID + ', 3); '
FROM @index_info     
WHERE IndexLevel = 1

SET @DynSQL = @DynSQL + '
DBCC TRACEOFF(3604); '

EXEC(@DynSQL)


DROP TABLE T2

B-tree node split strategy in SQL Server for monotonically increasing value

1 Answers1