On Tuesday 08 August 2006 14:29, Jon LaBadie wrote: > >> -- 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.
Unless you have used fewer than tapecycle number of tapes, in which case the only tapes in P are new tapes. > I failed to consider you meant single most recently used of A+I is in P. Any tape in P is never in A. Consider that P is a subset of I, and A and I are disjoint sets. > 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? Yes, unless you have set some tapes as no-reuse. Cheers, --Ian -- Forums for Amanda discussion: http://forums.zmanda.com/
