RAID

does raid technology combine multiple disks to one volume? for example, combine 3 20gb disks for a total of 60gb on one volume.

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2lazybutsmart

Obviously all 3 disks will be working as one single unit; so the space will be the sum. But then when the disk is writing data, it'll either use mirroring or stripping (if that's the right word

). Now u'll loose all your data if one hard disk failes while using the latter method. Most RAID disks use the former method.

ooops u asked about the volume....

Now the funny part in this whole RAID stuff (and I actually laugh about to this day), is the fact that the ID in RAID stands for Inexpensive disks. Now all I see is people using really expensive Compaq or Segate hardrives and calling it RAID. shouldn't they call it RAED (=joke

)

I've used RAID only once in my life; so that's all I've got to say.

pandimus

if you mirror, you will have one volume of 30gb. (if one drive fails, you can replace it and not lose any data.

in stripeing, you will have almost 60 gb, i think it reserves a little bit for the table or something. At least mine does.

Webmaster

The answer to your questions depends entirely on the type of RAID being used.

If you use a stripe set with parity (RAID 5) the total volume size is the sum of all disks minus one disk. Ie. a raid 5 set with 5 30GB disks results in 120 GB disk space.

A RAID 1 (mirror) with 2 30 GB disks results in a max volume of 30GB

In a regular stripe set (without parity, RAID 0) the size is the sum of all disks.

2lazybutsmart wrote:

Now the funny part in this whole RAID stuff (and I actually laugh about to this day), is the fact that the ID in RAID stands for Inexpensive disks. Now all I see is people using really expensive Compaq or Segate hardrives and calling it RAID. shouldn't they call it RAED (=joke )

The actual meaning of the abbriviation RAID is often disputed, some refer to it as Redundant Array of Independent Disks, which now is a more proper term.

Sartan

Webmaster wrote:

If you use a stripe set with parity (RAID 5) the total volume size is the sum of all disks minus one disk. Ie. a raid 5 set with 5 30GB disks results in 120 GB disk space.
proper term.

I thought parity information was 1/N, where N was the amount of disks?
IE: You have 4 drives. You are using 25% of the available space for storage.

You have 7 drives. You are using 14% of the available space....
One drive for Windows..

Drive 0 is 12.5GB
Drive 1 is 3.5GB
Drive 2 is 4.6GB
Drive 3 is 37.2GB
Drive 4 is 10.6GB
Drive 5 is 1.4GB
Drive 6 is 15.0GB

12.5+3.5+4.6+37.2+10.6+1.4+15.0=84.8 GB

14% of 84.8GB is 11.9GB. (84.8*(1/7))
This leaves you with 72.9GB.

Or so as I see it. Perhaps I'm thinking dynamic disks?

Webmaster

I thought parity information was 1/N, where N was the amount of disks?

That calculates the storage required for parity info per disk... multiply that with the amount of disks and you'll get the amount of storage required for parity information for the entire RAID set, which equals the size of one disk again

If you use a stripe set with parity (RAID 5) the total volume size is the sum of all disks minus one disk.

As this is for the Network+ exam, we are not talking about MS implementation of stripe volumes with parity, rather RAID 5 in it's basic form such as in most hardware implementations. Hence, when I say the total volume size is the sum of all disks minus one disk, I'm referring to a RAID 5 array with equal size disks.

When using striped volumes with parity (which is RAID 5 on Dynamic Disks) either you can use several disk with different amount of available (volume-)space, as in your example, the total volume size is (N-1)*s where N is the number of disks and s is the smallest volume. So instead of 72.9GB, your example would leave 6 x 1.4GB =8.4 GB.

www.acnc.com/04_01_05.html

www.winnetmag.com/Article/ArticleID/14675/14675.html

pmann

Ed Tittel says that you lose a whole disk to parity data. (This does seem a bit excessive if you only have a few disks, but I believe the reason for this is that the parity information needs to be written asynchronously).

For the Net+ exam a couple of other points to remember;

RAID 0 is disk striping without parity. So maximal disk space, fast, but no fault tolerance.

RAID 1, mirroring plus duplexing (i.e. an array of disk drives would be in mirrored pairs). This setup will have 50% redundancy, so only 50% of the available disk space, but it is fault tolerant.
- For mirroring, you can tolerate the loss of a disk.
- For duplexing, you can tolerate the loss of a drive controller and/or its associated disk. So duplexing is more fault tolerant in that two things can fail.

RAID 5 is disk striping with parity, you can have from 2-32 disks but one disk will be consumed by parity data. So if you had ten drives of 20G, you would end up with 180G of space.

Note that for W2K Server, the system and boot partition cannot exist on a RAID volume, so you would need to set a separate disk aside.

Webmaster

pmann wrote:

RAID 5 is disk striping with parity, you can have from 2-32 disks but one disk will be consumed by parity data.

Note that it is not 'one disk', but the 'size of one disk'. The parity information is distributed over all disk in the array/set. This is actually the main difference with RAID 4, which reserves 1 disk as a parity disk.

Sartan

Ah, okay.

It's always confusing when you read more than one thing on the same topic for different exams. Gotchya, though. Thanks webmaster