dump(8): how many bytes written to tape?

Ulrich Spoerlein uspoerlein at gmail.com
Wed Nov 15 02:57:45 PST 2006


Hi,

I'm trying to figure out how much bytes were written to a tape by
dump(8). I'm using a blocksize of 64kB to maximize throughput to the
tape drive. Initially, I thought I could just add up the number of
"tape blocks" written by dump and multiply by 64kB. But it looks like
dump is still reporting those values as 1kB blocks.

Here's some sample output:

  DUMP: Date of this level 1 dump: Wed Nov 15 09:46:37 2006
  DUMP: Date of last level 0 dump: the epoch
  DUMP: Cache 256 MB, blocksize = 65536
  DUMP: DUMP: 30676 tape blocks on 1 volume
  DUMP: finished in 1 seconds, throughput 30676 KBytes/sec

  DUMP: Date of this level 1 dump: Wed Nov 15 10:25:38 2006
  DUMP: Date of last level 0 dump: the epoch
  DUMP: DUMP: 4650864 tape blocks on 1 volume
  DUMP: finished in 132 seconds, throughput 35233 KBytes/sec

  DUMP: Date of this level 1 dump: Wed Nov 15 10:50:36 2006
  DUMP: Date of last level 0 dump: the epoch
  DUMP: DUMP: 328548 tape blocks on 1 volume
  DUMP: finished in 14 seconds, throughput 23467 KBytes/sec

  DUMP: Date of this level 1 dump: Wed Nov 15 11:00:14 2006
  DUMP: Date of last level 0 dump: the epoch
  DUMP: DUMP: 36925423 tape blocks on 1 volume
  DUMP: finished in 973 seconds, throughput 37950 KBytes/sec

If I add the time*throughput, I get 41GB. If I add the number of tape
blocks and assume a block size of 1kB, I get 41GB, too.

So, how exactly is the '-b64' parameter to dump(8) affecting the block
size on tape?

Uli


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