Girish Venkatachalam <girishvenkatachalam@...> writes: > On Mon, Feb 21, 2011 at 10:35 AM, Kumaraswamy <kumarrkslinux@...> wrote: > This is actually 4 disks(not 3). What am I missing? > > > hdd is dedicated parity in the above
This is what you are missing; actually, with 3 data disks and a dedicated parity disk, this is raid 4 not raid 5. > > In this case If i have problem in two hard disk > > So even if any one disk crashes no data is lost. Which does not answer the OP question; if one data disk crashes, no data will be lost, but you will have to use the parity disk to reconstruct it. If the parity disk crashes, no data will be lost, but you will have to put a replacement disk and repopulate the parity information. If two disks crash, you will lose data (in general). Also, in this case, total available capacity is not 2/3 but 3/3 = sum of the space on the 3 data disks (as the 4th disk is used for parity). > RAID concepts are not worth the trouble for me at least. It is not that much difficult (wikipedia is your friend); however, with multi TB disks available rather cheaply, for most applications, mirroring (which is raid 1 actually) with two disks gives sufficient disk capacity and reliability. If you want large storage capacities, 2-mirroring gets inexpensive (since available space = 50% of total disk space) and raid 5 becomes more attractive with available space = n-1/n of the total space in the n disks in the array. If you want no data loss with multiple disk failures, you can look at n disk mirroring or raid 6. _______________________________________________ ILUGC Mailing List: http://www.ae.iitm.ac.in/mailman/listinfo/ilugc
