Setting up a RAID 5 Array

external hard driveraidraid-5software-raid

I am currently trying to convert my drives from just a bunch of external disks, each one individually shared from my (CentOS) server into a RAID array (ideally, RAID 5).

At current, my drives are:

2.7TB (0TB used)
2.7TB (2.6TB used)
2.7TB (1.1TB used)
0.9TB (0.93TB used)
0.9TB (0.26TB used)
0.9TB (0.8TB used)
460GB (400GB used, this one I am considering binning though)

Now, what would be the best approach in creating a RAID array so that I can create a single redundant disk without formatting all of them and starting from scratch (I will need to preserve the data, so is there some method of slowly creating the array then extending it drives as they get their contents emptied into the new array)? Or is RAID not the way to go?

Also, will I be able to create an array comprised of the 3x 3TB drives and the 3x 1TB drives to create a single 9TB drive with 3TB of redundancy?

Best Answer

Windows does not like to mix drive sizes with raid.

You could put the first three in raid 5 (2.7*2 for total storage ) You could put the second three in raid 5 (0.9*2 for total storage ) and 'binning'="in american English 'throw away' " the 7th.

This leaves less total storage then you are currently using, so you would have to throw stuff away.

That's about the best you could do with Windows.

There are oodles of other options in Linux.

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Can someone explain RAID-0 in plain English

A RAID-0 configuration utilizes 2 separate hard disks and writes "chunks" of data to each one to minimize actuator movement and read/write data faster (each arm does half the work in a sense). The caveat with this is that there is a lack of fault tolerance - if one drive dies, all of your data is gone. And since there is 2 drives handling your data, the chances of losing your data is basically doubled. You may want to look into RAID1+0, which give you speed benefits as well as fault tolerance. You can swap in another disk upon failure and have the other disks rebuild it.

The RAID-0 image from wikipedia domonstrates it well:

alt text

imagine the disks are showing a file called A.txt. Now, imagine each section as a 4096-byte cluster for example. Each disk is holding roughly 16384 bytes of this text file, and if one disk dies, there is no way to recover the other half.

If you are limiting yourself to 2 disks, I'd suggest only doing the RAID-0 configuration on the operating system. Any files important to you should be stored on separate media such as an external drive, or if it was a possibility, 1 more hard drive. If it meant the difference between having a fast machine without fault tolerance, or having a slower machine with my files safe, I'd choose a slower machine. For me at least, I'd rather be safe than sorry.

RAID-5 Array – How Parity Works

It just XORs each corresponding bit from each drive - If you lose any drive, you can re-build the missing data.

For background:

A B (A XOR B)
0 0    0
1 1    0
0 1    1
1 0    1

Assume that D is the XOR of the other columns, then as long as you only lose one drive, you can figure out what you lost.

Some times the stripe bit will be distributed across the drives, but the concept is the same.

So for RAID-5, no matter how many drives, you only need 1 drive for parity equal or bigger than the smallest drive in the array you want to RAID.

RAID-5 for personal use is probably best as computational complexity is much lower than RAID-6.

RAID-6 is more complicated using Galois Fields to compute parity. And that can tax parity computations. However, you can lose more drives, but if you rebuild your array as soon as you get a single failure, you should be fine sticking with RAID-5.

Best Answer

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Can someone explain RAID-0 in plain English

RAID-5 Array – How Parity Works

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