Partitioning Large Table for 1120 millions Rows with SQL Server 2008R2

Create partitioned for existing table with more than 1120 millions (1120962253) rows today. It was amazing to see the query performance within few seconds with my client database.

Make sure to create a files groups ([PARTITION_FG1], [PARTITION_FG2],[PARTITION_FG3],[PARTITION_FG4] for partitioned table 
Also you can create Separate file group for Non-Clustered Index  on different drive NCINDEX_FG5

USE [TestDB]
GO

--Step 1. Creating a Partition Function
CREATE PARTITION FUNCTION PFSvrId_Left (numeric (10,0))
AS RANGE LEFT FOR VALUES (399, 499, 699, 799);

--The result for this RANGE LEFT assignment is:
--{min … 399}, {400 … 499}, {500 … 699}, {799 … max}

--Step 2. Creating a Partition Scheme
CREATE PARTITION SCHEME SvrIdScheme
AS PARTITION  PFSvrId_Left
TO ([PARTITION_FG1], [PARTITION_FG2],[PARTITION_FG3],[PARTITION_FG4],[PRIMARY])

--Step 3. CREATE CLUSTERED INDEX 1

-- Now create a Partitioned using clustered index based Scheme
CREATE CLUSTERED INDEX [idx_LoadID] ON [dbo].[MyTable]
(
      [M_ID] ASC
) ON SvrIdScheme(s_id)

-- Step 4.  CREATE NONCLUSTERED INDEX 2 on Separate file group  
ALTER TABLE [dbo].[MyTable] ADD  CONSTRAINT [PK_MyTable] PRIMARY KEY NONCLUSTERED
(  [t_id] ASC,  [s_id] ASC
)ON [NCINDEX_FG5] 
GO

--– Check for new partitions
SELECT partition_id, object_id, partition_number, rows
FROM sys.partitions
WHERE object_id = OBJECT_ID('MyTable')
 GO