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                  jon AT jgcomp DOT com
 JG Computing
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