Category Archives: 101 Performance Tuning Tips and Tricks

How to get an Index’s Root Page, Intermediate Pages and Leaf Pages Information? Tip 5: Sql Server 101 Performance Tuning Tips and Tricks

In this article, I will explain you on How to get an Index’s Root Page, Intermediate Pages and Leaf Pages Information. Prior to that let us get some insights into the Index structure.

In Sql Server an index is made up of a set of pages (index nodes) that are organized in a B+ tree structure. This structure is hierarchical in nature. The top node is called the ROOT node (Root Page). The bottom level of nodes in the index are called the leaf nodes (leaf Pages). Any index levels between the root and the leaf nodes are collectively known as intermediate levels. In a clustered index, the leaf nodes contain the data pages of the underlying table. The root and intermediate level nodes contain index pages holding index rows. Each index row contains a key value and a pointer to either an intermediate level page in the B-tree, or a data row in the leaf level of the index. The pages in each level of the index are linked in a doubly-linked list. In case of Clustered Index, data in a table is stored in a sorted order by the Clustered Index key Column. As a result, there can be only one clustered index on a table or view. In addition, data in a table is sorted only if a clustered index has been defined on a table.

Non-Clustered Index structure is similar to that of a B+ Tree Structure of Clustered Index, but the leaf nodes of a Non-Clustered index contain only the values from the indexed columns and row locators that point to the actual data rows, rather than contain the data rows themselves. Because of this in case of Non-Clustered Index, Sql Server takes additional steps to locate the actual data. The row locators structure depends on whether underlying table is a clustered table (Table with Clustered Index) or a heap table (Table without Clustered Index). In case the underlying table is a Clustered table the row locator points to the Clustered Index key value corresponding to the Non-Clustered Key and it is used to look-up into the Clustered Index to navigate to the actual data row. If underlying table is a heap table, the row locator points to the actual data row address. In case of Non-Clustered Index, the data rows of the underlying table are not sorted and stored in order based on their Non-Clustered keys.

B+ Tree Structure of a Clustered Index

Structure of Clustered Index

Let us explain how to get an Index’s Root Page, Intermediate Pages and Leaf Pages Information with an example

To explain how to get an Index’s Root Page, Intermediate Pages and Leaf Pages Information, let us create a Customer table as shown in the below image with 35K records. Execute the following script to Create the Customer Table with Clustered Index on the CustomerId column with sample 35K records.

--Create Demo Database
CREATE DATABASE SqlHints101PerfTips4
GO
USE SqlHints101PerfTips4
GO
--Create Demo Table Customers
CREATE TABLE dbo.Customers (
	CustomerId INT IDENTITY(1,1) NOT NULL,
	FirstName VARCHAR(50), LastName  VARCHAR(50),
	PhoneNumber VARCHAR(10), EmailAddress VARCHAR(50),
	CreationDate DATETIME
)
GO
--Populate 35K dummy customer records
INSERT INTO dbo.Customers (FirstName, LastName, PhoneNumber, EmailAddress, CreationDate)
SELECT TOP 35000 REPLACE(NEWID(),'-',''), REPLACE(NEWID(),'-',''), 
    CAST( CAST(ROUND(RAND(CHECKSUM(NEWID()))*1000000000+4000000000,0) AS BIGINT) AS VARCHAR(10)),
    REPLACE(NEWID(),'-','') + '@gmail.com',     
    DATEADD(HOUR,CAST(RAND(CHECKSUM(NEWID())) * 19999 as INT) + 1 ,'2006-01-01')
FROM sys.all_columns c1
        CROSS JOIN sys.all_columns c2
GO
--Create a PK and a Clustered Index on CustomerId column
ALTER TABLE dbo.Customers 
ADD CONSTRAINT PK_Customers_CustomerId 
PRIMARY KEY CLUSTERED (CustomerId)

As explained in the previous article we can use Dynamic Management Function sys.dm_db_database_page_allocations to get the list of all the pages associated with an Index. Note sys.dm_db_database_page_allocations is an un-documented (i.e. feature that may change or removed without any notice or may produce un-expected result) DMF. So avoid using it in the Production environment.

Get all the Pages Associated with the Clustered Index PK_Customers_CustomerId

To get all the Pages associated with the Index PK_Customers_CustomerId of the Customers table, we need to pass the @DatabaseId, @TableId, @IndexId and @Mode Parameter values of the DMF sys.dm_db_database_page_allocations. We can use sys.indexes catalog view as shown below to get the Index Id for a Index.

SELECT OBJECT_NAME(object_id) table_name, object_id, 
     name index_name, index_id, type, type_desc
FROM sys.indexes
WHERE OBJECT_NAME(object_id) = 'Customers'
	AND name = 'PK_Customers_CustomerId'

RESULT:
Use sys.indexes to get the indexid for the index

As explained in the previous article:

  • Clustered Index will always will have the Index Id as 1.
  • Index Id for the Non-Clustered Index will be >=2.
  • Index Id for the Heap Table is 0.

We can execute the below query to get all the pages associated with the Clustered Index PK_Customers_CustomerId

SELECT DB_NAME(PA.database_id) [DataBase], 
    OBJECT_NAME(PA.object_id) [Table], SI.Name [Index], 
    is_allocated, allocated_page_file_id [file_id], 
    allocated_page_page_id [page_id], page_type_desc, 
    page_level, previous_page_page_id [previous_page_id], 
    next_page_page_id [next_page_id]
FROM sys.dm_db_database_page_allocations 
    (DB_ID('SqlHints101PerfTips5'), 
         OBJECT_ID('Customers'), 1, NULL, 'DETAILED') PA
         LEFT OUTER JOIN sys.indexes SI 
        ON SI.object_id = PA.object_id 
                   AND SI.index_id = PA.index_id
WHERE is_allocated = 1  
	and page_type in (1,2) -- INDEX_PAGE and DATA_PAGE Only
ORDER BY page_level DESC, is_allocated DESC,
         previous_page_page_id

RESULT:
Clustered Index All Pages

As shown in the above image the Root Page File Id is 1 and Page Id is 968. Root Page is the one with maximum Page_level (here in this example it is 2) having previous_page_id and next_page_id as NULL. Intermediate Pages File Id is 1 and Page Id’s are: 2016 and 2032. Intermediate pages are the one whose Page_Level is less than the root page level and greater than 0 (i.e. Leaf Page Level (i.e. 0) < Intermediate Page level (here it is 1) < Root Page level (here it is 2)). And leaf level pages are the the ones whose page level is 0.

We can use the below function to get the pages of the different levels of a B+ Tree Index:

We can use the below function to get the B+ Tree Clustered/Non-Clustered Index Root Page, Intermediate Level Pages and Leaf Level Pages. Note I am using sys.dm_db_database_page_allocations un-documented (i.e. feature that may change or removed without any notice or may produce un-expected result) DMF. So avoid using it in the Production environment.

CREATE FUNCTION dbo.GetPagesOfBPlusTreeLevel(
	@DBName VARCHAR(100), @TableName VARCHAR(100) = NULL, @IndexName VARCHAR(100) = NULL, 
	@PartionId INT = NULL, @MODE VARCHAR(20), @BPlusTreeLevel VARCHAR(20) 
)
RETURNS
@IndexPageInformation TABLE (
	[DataBase] VARCHAR(100), [Table] VARCHAR(100), [Index] VARCHAR(100), 
	[partition_id] INT, [file_id] INT, [page_id] INT, page_type_desc VARCHAR(100), 
	page_level INT, [previous_page_id] INT, [next_page_id] INT)
AS
BEGIN

	DECLARE @MinPageLevelId INT = 0 , @MaxPageLevelId INT = 0, @IndexId INT = NULL
		
	SELECT	@IndexId = index_id
	FROM sys.indexes
	WHERE OBJECT_NAME(object_id) = @TableName AND name = @IndexName

	IF @IndexId IS NULL
		RETURN
	
	IF @BPlusTreeLevel IN ('Root', 'Intermediate') 	
	BEGIN 	
		SELECT  @MaxPageLevelId = (CASE WHEN  @BPlusTreeLevel ='Intermediate' THEN MAX(page_level) - 1 ELSE MAX(page_level) END), 
				@MinPageLevelId = (CASE WHEN  @BPlusTreeLevel ='Intermediate' THEN 1 ELSE MAX(page_level) END)
		FROM sys.dm_db_database_page_allocations 
			(DB_ID(@DBName), OBJECT_ID(@TableName), @IndexId, @PartionId, 'DETAILED') PA
 					LEFT OUTER JOIN sys.indexes SI 
				ON SI.object_id = PA.object_id AND SI.index_id = PA.index_id
		WHERE is_allocated = 1 AND page_type in (1,2)  -- INDEX_PAGE and DATA_PAGE Only

		IF @MaxPageLevelId IS NULL OR @MaxPageLevelId = 0 
			RETURN
	END
 
	INSERT INTO @IndexPageInformation
	SELECT DB_NAME(PA.database_id) [DataBase], OBJECT_NAME(PA.object_id) [Table], SI.Name [Index], 
		[partition_id], allocated_page_file_id [file_id],  allocated_page_page_id [page_id], page_type_desc, 
		page_level, previous_page_page_id [previous_page_id], next_page_page_id [next_page_id]
	FROM sys.dm_db_database_page_allocations 
			(DB_ID(@DBName), OBJECT_ID(@TableName), @IndexId, @PartionId, 'DETAILED') PA
			 LEFT OUTER JOIN sys.indexes SI 
			ON SI.object_id = PA.object_id AND SI.index_id = PA.index_id
	WHERE is_allocated = 1 AND page_type in (1,2) -- INDEX_PAGE and DATA_PAGE Only
			AND page_level between @MinPageLevelId AND @MaxPageLevelId
	ORDER BY page_level DESC, previous_page_page_id

	RETURN
END

How to get an Index’s Root Page Information?

We can use the above function like below to get the root page of the Clustered Index PK_Customers_CustomerId on the Customers Table in the SqlHints101PerfTips5 Database.

SELECT *
FROM dbo.GetPagesOfBPlusTreeLevel 
	('SqlHints101PerfTips5', 'Customers', 
	 'PK_Customers_CustomerId', NULL, 'DETAILED', 'Root')

RESULT:
How to Get Root Page of B Tree Index

There will be always on Root Page. Root Page is the one with maximum Page_level having previous_page_id and next_page_id as NULL. If the number of records in the table are very less and just one page is enough to store it. Then Sql Server doesn’t create the B+ Tree structure. So, in such scenario, the above function will not return the Root Page information.

How to get an Index’s Intermediate Pages Information?

We can use the above function like below to get the Index’s Intermediate pages of the Clustered Index PK_Customers_CustomerId on the Customers Table in the SqlHints101PerfTips5 Database.

SELECT *
FROM dbo.GetPagesOfBPlusTreeLevel 
	('SqlHints101PerfTips5', 'Customers', 
	 'PK_Customers_CustomerId', NULL, 'DETAILED', 'Intermediate')

RESULT:
How to Get Intermediate Pages of a B Tree Index

Intermediate pages are the one whose Page_Level is less than the root page level and greater than Leaf Page Level (i.e. 0) (i.e. Leaf Page Level < Intermediate Page level < Root Page level). In the following situations an Index will not have intermediate pages: 1) If the number of records in the table are very less and just one page is enough to store it. Then Sql Server doesn't create the B+ Tree structure. 2) If the number records stored in the table are very less and it is spanning only couple of pages in such scenario Sql Server can have B+ Tree Index with just Root Page and the Leaf Level Pages. In such scenarios this function will not return any information. If we have a very large Table, then we can multiple Intermediate Levels.

How to get an Index’s Leaf Page Information?

We can use the above function like below to get the Index’s Leaf pages of the Clustered Index PK_Customers_CustomerId on the Customers Table in the SqlHints101PerfTips5 Database.

SELECT *
FROM dbo.GetPagesOfBPlusTreeLevel 
	('SqlHints101PerfTips5', 'Customers', 
	 'PK_Customers_CustomerId', NULL, 'DETAILED', 'Leaf')

RESULT:
How to Get Leaf Pages of a B Tree Index

In case of Clustered Index the Page_Type_Desc of Leaf Pages is DATA_PAGE where as in case of Non-Clustered Index it will be INDEX_PAGE.

In the Next Article I will explain how we can use the DBCC PAGE command to look into what actually Sql Server stores in the Page.

How to find the list of all Pages that belongs to a Table and Index? Tip 4: Sql Server 101 Performance Tuning Tips and Tricks

I was writing an article on when and why we have to use included columns and wanted to explain it by showing how key column and included column values are stored in the Index Pages. That time, I got realized that first I need to explain how we can look into the Sql internal storage. Because of this I have decided to explain: How to find all the pages allocated to a Table and Index? How to identify Root, Intermediate and Leaf Pages of the Index? and How to inspect the Page data?. As a result of this realization I have written this and the next two articles. These articles will be referred in most of the articles in this series of 101 Performance Tuning Tips and Tricks.

From Sql Server 2012 onwards we have an un-documented (i.e. feature that may change or removed without any notice or may produce un-expected result) dynamic management function sys.dm_db_database_page_allocations to get the list of pages associated with one or more Tables and Indexes. This function returns one row for each page associated with the Table/Index.

This dynamic Management function takes following 5 parameters/arguments

  • @DatabaseId: Database ID of the Database whose tables or indexes assocaited pages information is required. This is a mandatory required parameter. We can use the DB_ID() function to get the Database ID of the Database by Database Name
  • @TableId: Table Id of the table whose assocaited pages information is required. We can use the OBJECT_ID() functio to get the table id by table name. It is an optional parameter, if it is Passed as NULL then it returns the pages associated with all tables in the Database and when it is NULL then next two parameters (i.e. @IndexId and @PartionId) values are ignored
  • @IndexId: Index Id of the Index whose associated pages information we are looking for. We can use the sys.indexes catalog view to get the Index Id. It is an optional parameter, if it is passed as NULL then it returns the pages associated with all the indexes.
  • @PartitionId: Id of the partition whose associated pages information we are looking for. It is an optional parameter, if it is passed as NULL then it returns the pages associated with all the partitions.
  • @Mode: This is mandatory Parameter, it’s value can be either ‘LIMITED’ or ‘DETAILED’. ‘LIMITED’ returns less information. However, ‘DETAILED’ returns detailed/more information. Obivously the ‘DETAILED’ mode uses more resources.

Let us understand the usage of this Dynamic Management Function sys.dm_db_database_page_allocations with examples

To understand how we can use this DMF to identify the pages belonging to the Table and Index, let us create a Customer table as shown in the below image with 35K records. Execute the following script to Create the Customer Table with Clustered Index on the CustomerId column and Composite Non-Clustered Index on the FirstName and LastName Column with sample 35K records.

--Create Demo Database
CREATE DATABASE SqlHints101PerfTips4
GO
USE SqlHints101PerfTips4
GO
--Create Demo Table Customers
CREATE TABLE dbo.Customers (
	CustomerId INT IDENTITY(1,1) NOT NULL,
	FirstName VARCHAR(50), LastName  VARCHAR(50),
	PhoneNumber VARCHAR(10), EmailAddress VARCHAR(50),
	CreationDate DATETIME
)
GO
--Populate 35K dummy customer records
INSERT INTO dbo.Customers (FirstName, LastName, PhoneNumber, EmailAddress, CreationDate)
SELECT TOP 35000 REPLACE(NEWID(),'-',''), REPLACE(NEWID(),'-',''), 
    CAST( CAST(ROUND(RAND(CHECKSUM(NEWID()))*1000000000+4000000000,0) AS BIGINT) AS VARCHAR(10)),
    REPLACE(NEWID(),'-','') + '@gmail.com',     
    DATEADD(HOUR,CAST(RAND(CHECKSUM(NEWID())) * 19999 as INT) + 1 ,'2006-01-01')
FROM sys.all_columns c1
        CROSS JOIN sys.all_columns c2
GO
--Create a PK and a Clustered Index on CustomerId column
ALTER TABLE dbo.Customers 
ADD CONSTRAINT PK_Customers_CustomerId 
PRIMARY KEY CLUSTERED (CustomerId)
GO
--Create a Non-Clustered Composite Index on the FirstName and LastName column
CREATE NONCLUSTERED INDEX IX_Customers_FirstName_LastName
    ON dbo.Customers (FirstName, LastName)

Example 1: Get all the Pages Associated with the Table Customers

To get all the Pages associated with the table Customers we need to pass only the @DatabaseId, @TableId and @Mode Parameter values of the DMF sys.dm_db_database_page_allocations. We can write a query like below to get this information:

SELECT DB_NAME(PA.database_id) [DataBase], 
	OBJECT_NAME(PA.object_id) [Table], SI.Name [Index], 
	is_allocated, allocated_page_file_id [file_id], 
	allocated_page_page_id [page_id], page_type_desc, 
	page_level, previous_page_page_id [previous_page_id], 
	next_page_page_id [next_page_id]
FROM sys.dm_db_database_page_allocations 
	(DB_ID('SqlHints101PerfTips4'), 
         OBJECT_ID('Customers'),NULL, NULL, 'DETAILED') PA
	     LEFT OUTER JOIN sys.indexes SI 
		ON SI.object_id = PA.object_id 
                   AND SI.index_id = PA.index_id
ORDER BY page_level DESC, is_allocated DESC,
         previous_page_page_id

RESULT:
How to get all the Pages Associated with a Table in Sql Server

From the above result we can see that, this DMV has returned all the pages associated with all the indexes of the Table when we have not specified (i.e. NULL is passed here) the @IndexId parameter value. In the result we can see the pages associated with the non-clustered index IX_Customers_FirstName_LastName and as well as the clustered index PK_Customers_CustomerId.

The records with page_level as 0 are the leaf pages and INDEX_PAGE with previous_page_id and next_page_id as NULL are the ROOT pages of the Index. And the root pages page_level value will be the maximum value for that index.

Clustered Index leaf pages store the actual table data, so the page type of the leaf pages of the Clustered Index is a DATA_PAGE. The Page Type of the non-leaf page (i.e. Root Page and Intermediate Page) of the Clustered Index is an INDEX_PAGE. Where as for the Non-Clustered index all the pages including leaf pages are of Type INDEX_PAGE, as the leaf page of the Non-Clustered index again points to the Clustered Index where actual Table Data is stored.

allocated_page_file_id (i.e. file_id) and allocated_page_page_id (i.e. page_id) are the frequently used columns from the result set of this DMV.

Example 2: Get all the Pages Associated with an Index

To get all the Pages associated with a specific Index of the table, we need to pass the @DatabaseId, @TableId, @IndexId and @Mode Parameter values of the DMF sys.dm_db_database_page_allocations.

Assume that we want to find all the pages associated with the Non-Clustered Index IX_Customers_FirstName_LastName. Then first question comes in the mind is how to get the Index Id for the Non-Clustered Index IX_Customers_FirstName_LastName. Answer to this question is very simple we can use the sys.indexes catalog view to get the Index Id. We can write a query like below to get Customers tables all the Indexes and their Index Id

SELECT OBJECT_NAME(object_id) table_name, object_id, 
     name index_name, index_id, type, type_desc
FROM sys.indexes
WHERE OBJECT_NAME(object_id) = 'Customers'

RESULT:
How to get Index Id in Sql Server

From the above result we can see that Index Id for the Non_Clustered
Index IX_Customers_FirstName_LastName is 2.

Note:

  • Clustered Index will always will have the Index Id as 1.
  • Index Id for the Non-Clustered Index will be >=2.
  • Index Id for the Heap Table is 0.

We can write a query like below to get all the pages associated with the Non-Clustered Index IX_Customers_FirstName_LastName

SELECT DB_NAME(PA.database_id) [DataBase], 
	OBJECT_NAME(PA.object_id) [Table], SI.Name [Index], 
	is_allocated, allocated_page_file_id [file_id], 
	allocated_page_page_id [page_id], page_type_desc, 
	page_level, previous_page_page_id [previous_page_id], 
	next_page_page_id [next_page_id]
FROM sys.dm_db_database_page_allocations 
	(DB_ID('SqlHints101PerfTips4'), 
         OBJECT_ID('Customers'), 2, NULL, 'DETAILED') PA
	     LEFT OUTER JOIN sys.indexes SI 
		ON SI.object_id = PA.object_id 
                   AND SI.index_id = PA.index_id
ORDER BY page_level DESC, is_allocated DESC,
         previous_page_page_id

RESULT:
How to get all the Pages Associated with an Index

Following are the list of columns returned by this DMF

database_id
object_id
index_id
partition_id
rowset_id
allocation_unit_id
allocation_unit_type
allocation_unit_type_desc
data_clone_id
clone_state
clone_state_desc
extent_file_id
extent_page_id
allocated_page_iam_file_id
allocated_page_iam_page_id
allocated_page_file_id
allocated_page_page_id
is_allocated
is_iam_page
is_mixed_page_allocation
page_free_space_percent
page_type
page_type_desc
page_level
next_page_file_id
next_page_page_id
previous_page_file_id
previous_page_page_id
is_page_compressed
has_ghost_records

Does the order of Columns in a Composite Index matters? Tip 3: Sql Server 101 Performance Tuning Tips and Tricks

The order of the columns in a composite index does matter on how a query against a table will use it or not. A query will use a composite index only if the where clause of the query has at least the leading/left most columns of the index in it.

Let us understand how the order of a column in a composite index matters with an example. Let us create a Customer table as shown in the below image with sample one million records by executing the following script.

Performance Issue Function on Index Column Customers Demo Table

CREATE DATABASE SqlHints101PerfTips3
GO
USE SqlHints101PerfTips3
GO
--Create Demo Table Customers
CREATE TABLE dbo.Customers (
	CustomerId INT IDENTITY(1,1) NOT NULL,
        FirstName VARCHAR(50), 
	LastName  VARCHAR(50),
	PhoneNumber VARCHAR(10),
	EmailAddress VARCHAR(50),
	CreationDate DATETIME
)
GO
--Populate 1 million dummy customer records
INSERT INTO dbo.Customers (FirstName, LastName, PhoneNumber, EmailAddress, CreationDate)
SELECT TOP 1000000 REPLACE(NEWID(),'-',''), REPLACE(NEWID(),'-',''), 
	CAST( CAST(ROUND(RAND(CHECKSUM(NEWID()))*1000000000+4000000000,0) AS BIGINT) AS VARCHAR(10)),
	REPLACE(NEWID(),'-','') + '@gmail.com',		
	DATEADD(HOUR,CAST(RAND(CHECKSUM(NEWID())) * 19999 as INT) + 1 ,'2006-01-01')
FROM sys.all_columns c1
		CROSS JOIN sys.all_columns c2
GO
--Update one customer record with some know values
UPDATE dbo.Customers 
SET FirstName = 'Basavaraj', LastName = 'Biradar', 
    PhoneNumber = '4545454545', EmailAddress = 'basav@gmail.com'
WHERE CustomerId = 100000
GO
--Create a PK and a Clustered Index on CustomerId column
ALTER TABLE dbo.Customers 
ADD CONSTRAINT PK_Customers_CustomerId 
PRIMARY KEY CLUSTERED (CustomerId)
GO

Create a Composite Non-Clustered Index on FirstName and LastName columns

Create a Composite Non-Clustered Index on FirstName and LastName column of the Customers Table by executing following statement:

--Create a non-clustered index on FirstName column
CREATE NONCLUSTERED INDEX IX_Customers_FirstName_LastName
	ON dbo.Customers (FirstName, LastName)
GO

First let us now enable the execution plan in the Sql Server Management Studio by pressing the key stroke CTRL + M

Let us now verify how the Order of the columns in the above composite index impacts it’s usage for the queries against this table

Scanerio 1: Query having FirstName and LastName column in the WHERE clause in the same order

SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE FirstName = 'Basavaraj' AND LastName = 'Biradar' 

Below is the execution plan for the above query execution

Query With FirstName and LastName

From the above execution plan we can see that the composite index IX_Customers_FirstName_LastName is used and it is performing the INDEX SEEK of it.

Scanerio 2: Query having LastName and FirstName column in the WHERE clause in the same order

SELECT CustomerId, FirstName, LastName
FROM dbo.Customers
WHERE LastName = 'Biradar' AND FirstName = 'Basavaraj'

Below is the execution plan for the above query execution

Query With LastName and FirstName

From the above execution plan we can see that the composite index IX_Customers_FirstName_LastName is used and it is performing the INDEX SEEK of it.

Scanerio 3: Query having only FirstName column in the WHERE clause

SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE FirstName = 'Basavaraj'

Below is the execution plan for the above query execution

Query With FirstName Only

From the above execution plan we can see that the composite index IX_Customers_FirstName_LastName is used and it is performing the INDEX SEEK of it.

Scanerio 4: Query having only LastName column in the WHERE clause

SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE LastName = 'Biradar' 

Below is the execution plan for the above query execution

Query With LastName Only

From the above execution plan we can see that the composite index IX_Customers_FirstName_LastName is used but it is performing the INDEX SCAN of it. Also we can see from the execution plan that, Sql Server has given missing index hint and it is suggesting to add Index on the LastName column.

Verify the Cost of the Query

Execute the following four statement in one batch and verify the cost of the query

SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE FirstName = 'Basavaraj' AND LastName = 'Biradar' 
GO
SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE LastName = 'Biradar' AND FirstName = 'Basavaraj'
GO
SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE FirstName = 'Basavaraj'
GO
SELECT CustomerId, FirstName, LastName
FROM dbo.Customers WITH(NOLOCK)
WHERE LastName = 'Biradar' 
GO

Execution Plan Cost

From the above execution plan we can see that the queries which have the leading column FirstName of the Index in the WHERE clause have 0% cost compared to the last query which has 100% cost which has only LastName non-leading column of the index in the WHERE clause.

Conclusion:

In Sql Server the order of the columns in a composite index does matter on how a query against a table will use it or not. A query will use a composite index only if the where clause of the query has at least the leading/left most columns of the index in it. In the example listed in this article we can see that all the queries having the leading Index column FirstName in the WHERE clause are using the Index. And the query having only the non-leading column LastName of the Index in the WHERE clause is not doing the Index Seek