On Tue, Aug 08, 2006 at 12:26:59PM -0400, Ian Turner wrote: > On Tuesday 08 August 2006 12:11, Jon LaBadie wrote: > > Just to be clear, suppose I have a large tape library with functioning > > barcode reader and associated changer database. > > I neglected to mention that if there is a barcode reader, Amanda will start > by > loading the least recently used reusable tape (if it is in the reader), and > then run the algorithm from there. > > > It is the > > responsibility of the administrator to ensure that the changer gets > > back to a correct slot position before amanda runs. Because otherwise > > it will use whichever of the 21 eligible tapes it hits first. Bleech! > > No, because 20 of the 21 eligible tapes are in I but not in P.
I misunderstood these two line from your original explanation. >> -- The most recently used "tapecycle" number of tapes is in A. >> -- Any remaining tapes are in I. The single least recently used >> of these is also in P. When I first read it, I was in a mindset of # tapes in rotation matches tapecycle and I applied the last sentence to that situation, i.e. the least recently used of tapecycle # of tapes is in P. Which would have to be true, otherwise you'd never cycle through them again. I failed to consider you meant single most recently used of A+I is in P. If at least tapecycle tapes have been used, would it be correct to say P is all unused, labeled tapes plus 1 previously used tape? > So, Amanda will > only use tapes out of order if the proper tape is not available. No bleach > required. ;-) > -- Jon H. LaBadie [EMAIL PROTECTED] JG Computing 4455 Province Line Road (609) 252-0159 Princeton, NJ 08540-4322 (609) 683-7220 (fax)
