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 _______________________________________________ freebsd-stable@freebsd.org mailing list http://lists.freebsd.org/mailman/listinfo/freebsd-stable To unsubscribe, send any mail to "[EMAIL PROTECTED]"
