Thanks I now know more than I needed, but I love technical reading. :thumbsup:
caudio51 BoM Nov '05; Mar '06 Rating - 96% 32 1 0 Joined Mar 24, 2005 Messages 19,935 Location Jersey Jan 23, 2007 #21 Thanks I now know more than I needed, but I love technical reading. :thumbsup:
Rank_Tyro Rating - 0% 0 0 0 Joined Dec 26, 2005 Messages 889 Location Sacramento Jan 23, 2007 #22 Glad I could help.
Electric Sheep Dsicle - BoM Dec 06 Rating - 100% 58 0 0 Joined Aug 1, 2006 Messages 5,147 Location Dallas, TX Jan 23, 2007 #23 Rank_Tyro said: The usable capacity of a RAID 5 array is (N-1) * \min(S_1, S_2, \dots, S_N) = (N-1) * S_{min}, where N is the total number of drives in the array, Si is the capacity of the ith drive, and Smin is the capacity of the smallest drive in the array. Click to expand... Hmm, let me try to figure this out: RAID 5 array is (N-1) * \min(S_1, S_2, \dots, S_N) = (N-1) * S_{min} = ELEVENTY Is that right? :grin:
Rank_Tyro said: The usable capacity of a RAID 5 array is (N-1) * \min(S_1, S_2, \dots, S_N) = (N-1) * S_{min}, where N is the total number of drives in the array, Si is the capacity of the ith drive, and Smin is the capacity of the smallest drive in the array. Click to expand... Hmm, let me try to figure this out: RAID 5 array is (N-1) * \min(S_1, S_2, \dots, S_N) = (N-1) * S_{min} = ELEVENTY Is that right? :grin:
Rank_Tyro Rating - 0% 0 0 0 Joined Dec 26, 2005 Messages 889 Location Sacramento Jan 23, 2007 #24 "This one goes to ELEVEN".