> Will I get markedly better performance with 5 drives (2^2+1) or 6 drives > 2*(2^1+1) because the parity calculations are more efficient across 2^N > drives?
If only parity calculations stand to benefit, then it wouldn't make a difference because your CPU is more than powerful enough to take care of it either way ;) This message posted from opensolaris.org _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss