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I just configured 4 3TB drives in a RAID-10 configuration but the total disk space in the RAID is 4TB. I used the following guide to setup the RAID-10, RAID-10 Intall. Shouldn't the size if the RAID-10 array be 6TB? According to this RAID Calculator it should be. Anyone have any idea of what might have gone wrong in the install?

Thanks in advance for your time.

3 Answers 3

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Yes, it should be 6 TB. The 4 TB limit is probably due to using the MSDOS partition table, which is limited to 4 TB. You need to partition the array with GPT to get around that limit.

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  • Actually you will get 5.46TB not the full 6TB. But Psusi is right about using GPT.
    – user121358
    Jan 10, 2013 at 23:14
  • No, 3 x 4 / 2 = 6 TB, not 5.46. Where did you get this number?
    – psusi
    Mar 21, 2014 at 14:06
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Yes, 6TB.

RAID10 is a RAID0 of two RAID1 arrays. You lose 50% of physical disk space to reduncancy but gain speed from RAID0.

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  • At least as far as mdadm is concerned, raid10 is not just a raid0 on top of raid1. It generalizes the combination of striping and mirroring and is cable of building a raid10 out of any number of disks from 2 up. Dumber raid systems do just think of it as raid0 on top of raid1, and so require exactly 4 disks, or sometimes allow a greater number but only if it is an even number.
    – psusi
    Jul 3, 2012 at 13:44
  • And it would actually be RAID1 (mirroring) on top of RAID0 (striping).
    – dobey
    Jan 10, 2013 at 23:32
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I do not see the Raid 10 or 1+0 option in that Raid calculator. Only the 0+1 or Raid 01 is there (apart from the base raid levels). For what I understand the first level of RAID 10 would be Raid 1 which creates the mirror and then Raid 0 on top of that to create the performance effect. To then calculate the space available after everything is done for a RAID 10 we would use the following (Assuming all hard drives are the same size. If not then the smallest is the base space for all the rest):

M = Mirrored Discs N = Number of Discs C = Capacity of Discs

(N x C) / M = Space Available

In your case it would be:

M = 2 N = 4 C = 3TB

(4 x 3) / 4 = 6TB Space Available

So what you reckon about the final size is true. What I might think is happening is that the total REAL space available from each drive is been subtracted from the total final size. So if for example each 3TB hard drive is really a 2.7 or 2.8TB hard drive (The actual usable space you can work with in the hard drive). That is about 1.2TB to 800GB of missing space. So you would be going from the 6TB in theory to the real 4.8GB or 5.2GB. Maybe less, maybe more.

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