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| File system creation guidelines | ||
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Follow these guidelines when creating VxFS file systems:
Specify the maximum log size when creating file systems for databases.
Except for specifying the maximum log size and support for large files as required, use the VxFS defaults when creating file systems for databases.
Never disable the intent logging feature of the file system.
For redo logs, create a single file system using a simple (and mirrored, if necessary) volume. Put the other tablespaces and database files on separate file systems created on striped, striped and mirrored, or RAID-5 volumes.
When using the command line, use the mount points to name the underlying volumes. For example, if a file system named /db01 is to be created on a mirrored volume, name the volume db01 and the mirrors db01-01 and db01-02 to relate to the configuration objects. If you are using the vxassist command or the GUI, this is transparent.
If Quick I/O is supported, select vxfs as the file system type to take advantage of the Quick I/O feature, online administration, fast recovery of the VxFS file system, and superior reliability features.
Choose a file system block size that matches or is a multiple of the block size of your Oracle database (db_block_size).
It is possible to have a file system block size that is smaller than the database block size because the database block-size limit can be bigger than the file system block size. It is fine if the file system block size is smaller than the database block size because VxFS will not perform multiple I/O operations for each database I/O operation. VxFS is capable of performing I/Os with multiple blocks. For example, if your database block size is 32K and your file system block size is 8k, VxFS can put four 8K blocks together to perform one 32K database I/O operation.
When creating the file system, set the number of file system blocks in the intent log so that the log size is 16MB. For example, if the file system block size is 8K (that is, 8192), it will take 2000 blocks to make a 16MB log (2000 x 8192 = ~16MB). If the file system block size is 4K (that is, 4096), then twice as many blocks as in the 8K case would need to be allocated (4000 in this example).