Raid 5 Configuration Clarification

Posted on 2006-05-15
Last Modified: 2010-04-03
My HP ACU is showing this RAID 5 configuration (which I inherited):

Physical Drives:

#1 - 18.2 gb
#2 - 18.2 gb
#3 - 18.2 gb
#4 - 18.2 gb
#5 - 18.2 gb  (spare)
#6 - 18.2 gb

Logical Drives:

Disks 1-4 (50.2 gb total space)  << partitioned as drive C:
Disk   6    (16.9 gb total space)  << partitioned as drive E:


One array with the above two logical drives.

So my question is this.  Is Disk #6 (partitioned as drive E:) part of the RAID 5 configuration?  I'm sure the answer is yes.  I don't want it to be a simple volume (stand alone drive) without redundancy.
Question by:mcnuttlaw
    LVL 7

    Accepted Solution

    Raid 5 spreads the information accross all drives.  You have 5 drives in the array, the total usable space is 72.8 Gigs.  Looks like it is partitioned ini 2, one of 50.2 and the other 16.9.  I think it is all the same array.  If the 1-4 were in their own Raid 5 array the usable disk would be 54.59.  Plus most raid adapters will not allow a stand alone drive not designated as a spare to be used.  You should be able to verify in the disk manager in administrative tools, or the raid software that HP provides.
    LVL 2

    Author Comment

    My raid software shows one array, 5 physical drives, two logical drives (50.2 and 16.9).

    So thanks for helping me verify that my sixth drive is in fact part of the array.
    LVL 70

    Assisted Solution

    Just to be sure you understand the disparity in the numbers ...

    Your array is, as noted, comprised of 5 18.2gb drives, which yields a total available space of 4 x 18.2 = 72.8gb   But those are "disk manufacter's" gigabytes (= 1,000,000,000 bytes/gigabyte), NOT "computerese" gigabytes (= 1024x1024x1024 = 1,073,741,824 bytes/gigabyte).   Note that 72.8gb in "disk drive-ese" is 67.8gb in "computerese."

    Your RAID software is reporting the sizes of your partitions in "computerese,"  so that's why you only see 50.2gb and 16.9gb ==> a total of 67.1gb.   The difference between this and 67.8 is easily accounted for in rounding errors (look at more digits than just 50.2 and 16.9) and the fact that the drives most likely don't have "exactly" 18.2gb -- and the actual array size will be 4 x the SMALLEST drive's capacity.
    LVL 2

    Author Comment

    Thanks garycase.

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