Veritas NetBackup™ Administrator's Guide, Volume I

To estimate the disk space that is required for a catalog backup

1. Estimate the maximum number of files that each schedule for each policy backs up during a single backup of all its clients.
2. Determine the frequency and the retention period of the full and the incremental backups for each policy.
3. Use the information from steps 1 and 2 to calculate the maximum number of files that exist at any given time.

   For example:

   Assume that you schedule full backups to occur every seven days. The full backups have a retention period of four weeks. Differential incremental backups are scheduled to run daily and have a retention period of one week.

   The number of file paths you must allow space for is four times the number of files in a full backup. Add to that number one week's worth of incremental backups.

   The following formula expresses the maximum number of files that can exist for each type of backup (daily or weekly, for example):

   Files per Backup × Backups per Retention Period = Max Files

   For example:

   A daily differential incremental schedule backs up 1200 files and the retention period for the backup is seven days. Given this information, the maximum number of files that can exist at one time are the following:

   1200 × 7 days = 8400

   A weekly full backup schedule backs up 3000 files. The retention period is four weeks. The maximum number of files that can exist at one time are the following:

   3000 × 4 weeks = 12,000

   Obtain the total for a server by adding the maximum files for all the schedules together. Add the separate totals to get the maximum number of files that can exist at one time. For example, 20,400.

   For the policies that collect true image restore information, an incremental backup collects catalog information on all files (as if it were a full backup). This changes the calculation in the example: the incremental changes from 1200 × 7 = 8400 to 3000 × 7 = 21,000. After 12,000 is added for the full backups, the total for the two schedules is 33,000 rather than 20,400.

4. Obtain the number of bytes by multiplying the number of files by the average number of bytes per file record.

   If you are unsure of the average number of bytes per file record, use 132. The results from the examples in step 3 yield:

   (8400 × 132) + (12,000 × 132) = 2692800 bytes (or about 2630 kilobytes)

5. Add between 10 megabytes to 15 megabytes to the total sum that was calculated in step 4. The additional megabytes account for the average space that is required for the error logs. Increase the value if you anticipate problems.
6. Allocate space so all the data remains in a single partition.