Temporary space for backup verification

When you verify backups, 15-25 MB of temporary space must be available.

During backup verification, the archecker utility requires about 15 MB of temporary space for a medium-size system (40-50 GB) and 25 MB for a large system. This temporary space is stored on the file system in the directory that the AC_STORAGE parameter specifies, not in the dbspaces. The temporary files contain bitmap information about the backup and copies of partition pages, free pages in a chunk, reserved pages, and optionally, free pages in a blobspace and debugging information. The archecker utility must have permissions to the temporary directory.

If the backup is verified successfully, these files are deleted. If the backup fails verification, these files remain. Copy them to another location so that IBM® Software Support can review them.

If your database server contains only dbspaces, use the following formula to estimate the amount of temporary space in KB for the archecker temporary files:

space = (130 KB * number_of_chunks) + (pagesize * number_of_tables) + 
(.05 KB * number_of_logs)

For IBM Informix®, if your database server contains blobspaces or sbspaces, use the following formula to estimate the amount of temporary space for the archecker temporary files:

space = (130 KB * number_of_chunks) + (pagesize * number_of_tables) + 
(.05 KB * number_of_logs) + (pagesize * (num_of_blobpages/252))

number_of_chunks: The maximum number of chunks that you estimate for the database server.
pagesize: The system page size in KB.
number_of_tables: The maximum number of tables that you estimate for the database server.
number_of_logs: The number of logical logs on the database server.
num_of_blobpages: The number of blobpages in the blobspaces or the number of sbspaces. (If your database server contains sbspaces, substitute num_of_blobpages with the number of sbspaces.)

For example, you would need 12.9 megabytes of temporary disk space on a 50-gigabyte system with a page size of 2 KB. This system does not contain any blobspaces, as the following statement shows:

13,252 KB = (130 KB * 25 chunks) + (2 KB * 5000 tables) + 
             (.05 KB * 50 logs) + (2 KB * 0)

To convert KB to MB, divide the result by 1024:

12.9 MB = 13,252/1024