Run all three schemes and measure the restore cost of each
Task
Perform full, incremental and differential backups of the same data on the guest, then restore from each and measure what it took. The arithmetic is simple and the point only lands once you have restored from a five-member incremental chain.
Steps
- Create a data set of at least 200 MB on the guest and record its size and file count in
lab/schemes/data.txt. - Take a full backup, then change some files each day for five simulated days, taking an incremental each time. Record each in
lab/schemes/incremental.csvwith headerday,type,size_mb,duration_sec. - Revert, repeat with differentials, and record in
lab/schemes/differential.csvusing the same header. - Restore to a fresh location from each scheme, and record in
lab/schemes/restore.csvwith headerscheme,sets_needed,duration_sec,files_correct. - Write
lab/schemes/notbackup.mdexplaining why a snapshot, a synchronised folder and a redundant array are each not a backup, with the single test that disposes of all three.
Verify
grep -c . lab/schemes/data.txt
awk -F, 'NR>1 && NF>=4 {n++} END {print n" incremental run(s)"}' lab/schemes/incremental.csv
awk -F, 'NR>1 && NF>=4 {n++} END {print n" differential run(s)"}' lab/schemes/differential.csv
awk -F, 'NR>1 && NF>=4 {n++} END {print n" restore(s) measured"}' lab/schemes/restore.csv
awk -F, 'NR>1 {print $2}' lab/schemes/restore.csv | sort -u | wc -l
grep -Eic 'same (disk|storage)|independent copy|separate' lab/schemes/notbackup.md
Five runs of each scheme, restores measured from each, and at least two distinct set counts across the restores. If the incremental and the differential restore needed the same number of sets, one of the two backup chains was not what you think.
Notes
The single test that disposes of snapshot, sync and redundant array: if it fails when the original storage fails, it is not a backup. It is worth being able to state in one sentence, because it is the answer to a family of exam questions.
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