Partitioning Large MS SQL-Server FACT table

  • Scenario:

    •Data should be retained for 5 years or 60 months.

    •Table will have about 800 million records

    •Have adapted monthly partition.

    •Each File group have 8 files to match the Number of CPU improve parallelism.

    •Each partition will be kept in one File group.

    •The above scenario requires 62 file groups need to be created (60 for partition and two for window sliding)

    Question:

    Do I need to create 62 file groups and required data files using DDLs or any better way of handling this situation? Any ideas or scripts would be great?

  • rajesh999 (12/6/2010)


    Scenario:

    •Data should be retained for 5 years or 60 months.

    •Table will have about 800 million records

    •Have adapted monthly partition.

    •Each File group have 8 files to match the Number of CPU improve parallelism.

    •Each partition will be kept in one File group.

    •The above scenario requires 62 file groups need to be created (60 for partition and two for window sliding)

    Question:

    Do I need to create 62 file groups and required data files using DDLs or any better way of handling this situation? Any ideas or scripts would be great?

    Either you create the DDL manually or you create a script that would create your DDL 🙂 something/somebody has to do it

    _____________________________________
    Pablo (Paul) Berzukov

    Author of Understanding Database Administration available at Amazon and other bookstores.

    Disclaimer: Advice is provided to the best of my knowledge but no implicit or explicit warranties are provided. Since the advisor explicitly encourages testing any and all suggestions on a test non-production environment advisor should not held liable or responsible for any actions taken based on the given advice.
  • If you want to get really snazzy and generate the 'optimum' partitions for the data set, you could try using the SolverFoundation sample (description found here, search the page for GenerateDataPartition).

    Steve.

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