well, there's one cpu, nbuf=128, so lofreebuffers will be MIN(nbuf/25+20, 128); nbuf=128, so it'll be 25. hifreebuffers will be .. (3*25)/2, so 37.
-adrian _______________________________________________ svn-src-head@freebsd.org mailing list https://lists.freebsd.org/mailman/listinfo/svn-src-head To unsubscribe, send any mail to "svn-src-head-unsubscr...@freebsd.org"