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A RAID primer for IT consultants

However the drives are RAID configured is how the computer runs.(Computer>adapter card>drives)In the BIOS of the adapter card is where you set up the RAID.In Dell it's Ctrl-A at boot up.There's even a format or erase utility there.I know that there's adapter card flash updates but this one works is fine for me.You'll need the adapter card driver file from Adaptec to install an OS.You should be able to copy it to a pen drive.The numbers are in the card's BIOS.

Isn't the efficiency of a three-disk Raid 5 set .67? After all (at least in the retail world), 1-1/3 = 2/3 = .67.

And, even allowing for the math error, how does the efficiency relate to the disk cost as you state it? If you want a three-disk array, you are going to have to buy three disks. Are you referring to monetary cost or to the cost in lost disk space? There's no explanation of that.

RAID 10 is a stripeset made up of individual mirrorsets.

You take a mirrored pair (efficiency: .50) plus another mirrored pair (efficiency: .50) and stripe them (overall efficiency: .50)

Add mirrored pairs to the stripset as needed, maintaining .50 efficiency.

The growth is flat:

4 - disk RAID 10 (disk size = n)
each mirrorset is 2 drives, yielding n effective storage per mirrorset
2 mirrors are striped, yielding 2n effective storage, 2n effective cost
efficiency: .50
cost: .50

6 - disk RAID 10 (disk size n)
each mirrorset is 2 drives, yielding n effective storage per mirrorset
3 mirrors are striped, yielding 3n effective storage, 3n effective cost
efficiency: .50
cost .50

100 - disk RAID 10 (disk size n)
each mirrorset is 2 drives, yielding n effective storage per mirrorset
50 mirrors are striped, yielding 50n effective storage, 50n effective cost
efficiency: .50
cost .50

Or have I done the math wrong?

We have several small individual Solaris servers all with software raid 1 sets
Performance is good on these as with the overall system loads we can not tell a difference with or without hardware raid on these servers. Same with later versions of Linux, all depends on the role of the servers.
Actually, until later and more recent versions of some of the top three RAID hardware cards available, at one point were were getting better performance under specific configurations and loads with the Linux software raid than using two of the common hardware vendor's RAID cards a few years ago.
Today more and more systems are using SAN, all Windows uses SAN so they are all Hardware and running on Blade centers fiber connected to the SAN's so never have tried any software RAID on windows, I do believe would present performance issues on the more limited OS
Today's serer RAID options (top three) offer more higher performance devices now, 6GB SATA/SAS

We had an IBM server with a RAID-10 array that stored or main database. It functioned well for months and then suddenly the logical volume disappeared overnight. This was caused by a controller malfunction. Despite claims that the RAID config is kept in both the software and on the controller (battery), IBM was not able to restore the array without data loss, because it was RAID-10 (RAID-5 would have worked I was told). Luckily a data recovery specialist was able to rebuild our volume without any data loss, but this took time and money! Were back on HP servers with good old RAID-5.

RAID-6 (like RAID-5, but with two parity drives) is now commonplace. Disks are also cheap enough that we do a lot of these, now.

I believe Raid 10 costs would increase geometrically not exponentially. If you had nine luns this would require 18 disk devices. If you required a tenth lun this would increase the disk devices to 20.

If you had 10 luns and needed to double the amount of disk space required then you would increase your luns to 20. This means that the 10 luns of 20 disk devices would increase to 20 luns of 40 disk devices. Which simply doubles the number of physical disk devices required.

The statement: (1-1/3 = .33) is obviously incorrect. That's .66. And .66 is not the overhead, but the available storage.

1-1/3 = 2/3 = 0.667. There must be a typo up in the text, or I forgot my fractions

see the clone at:
http://vpsmart.org/a-raid-primer-for-it-consultants/#more-2585

someone ripped off your stuff from this article at:
http://www.techrepublic.com/blog/project-management/a-raid-primer-for-it-consultants/4185?tag=nl.e101

What is TL-DNR???

A primer should spell out the words behind things like this (at least when 1st used).

I looked on the web and found "too long, did not read" and I also found a clone/rip-off of the complete article.

I still don't know what TL-DNR means. Any help on this???

RAID 5 section has errors and is not clear;
please consider this rewrite:

RAID 5 ??? potentially the most commonly deployed RAID version when seeking a balance between performance and redundancy ??? stripes data across three or more disks. Performance increases because read/write operations can be performed simultaneously across multiple disks. Because parity information is spread across multiple disks, too, a single disk can fail and the array should continue operating properly (albeit at reduced performance during the failure).

Disk cost proves reasonable, too. Consultants can calculate RAID 5 efficiency using the equation 1 - (1/n) in which n equals the number of disks in the RAID array and n must be 2 or greater.

For example, if there are three 500 GB disks in a RAID 5 array, the disk yield (i.e., available disk capacity as a fraction of the total raw disk capacity) will equal (1 - (1/3) = .667), i.e., one 500 GB disk is lost for protection from single drive failure.

Note, given my preference, I deploy an extra drive within RAID 5 arrays for configuration as a hot online spare. Using a hot spare increases disk cost but helps minimize downtime, as a server configured in this manner does not need to be powered down when it???s necessary to replace a failed disk.

As an added bonus, some RAID controllers will automatically start rebuilding a failed drive on the hot spare drive without any operator intervention required (the specific example M. Carr has seen is a Dell EDGE 2800e with Dell's house brand RAID controller).

The general formula for "hot spare" with "parity" disk yield (i.e., available disk capacity as a fraction of the total raw disk capacity) is:
1 - (2/n) in which n equals the number of disks in the RAID array, n must be 3 or greater, and the 2 accounts for the "one hot spare" and "one parity" disk.

A couple of "hot spare" with "parity" drive examples:

o For the four raw drive case: the disk yield will equal (1 - (2/4) = .500), i.e., two (of the four) 500 GB (or whatever) disks are lost for protection from failure while also allowing quicker recovery when one drive dies.

o For the six raw drive case: the disk yield will equal (1 - (2/6) = .667), i.e., two (of the six) 500 GB (or whatever) disks are lost for protection from failure while also allowing quicker recovery when one drive dies.

RAID 5, with its balance of redundancy vs. performance, is a good fit well for many server environments. Occasionally additional redundancy (or tolerance for multiple simultaneous disk failures) is required. In such cases, RAID 10 may prove best.

How could 1-1/3 = .33?

You calculate the disk cost of a RAID 5 array by solving the following equation: 1 - 1 / n where n = the number of disks. In my example, I said there were three 500GB disks. Thus, the equation becomes 1 - 1 / one 500GB disk, or 500GB. That's one disk, which equals .33 % of the total hard disk space. Here's an online array calculator you may find makes the calculations easier: http://www.z-a-recovery.com/art-raid-estimator.htm.

EDIT: Removing note regarding rewrite; I wish to ensure disk cost information is clear. Note, another online calculator to assist RAID 5 calculations can be found here: http://www.raid-calculator.com/

is the amount of overhead required to maintain the fault tolerance information. Got it.

Your results may be correct, but the math is still wrong. Inferring from results provided by the calculator you linked to:

For a Raid 5 array with 3 500GB disks, 500 GB is is used to store fault tolerance information. Working backwards, 500GB is one-third of the total drive space of 1500GB, therefore the disk cost must be 1/3.

For an array of 4 500GB disks, 500GB again stores fault tolerance information; 500GB is 1/4 of the 2000GB total available space, therefore the disk cost must be 1/4.

Assuming the pattern holds true for arrays of all sizes, the disk cost equation must be 1/n, where n is the number of disks.

The equation you provide (1 - 1/n) yields the ratio of usable drive space to total drive space, or the efficiency of the array with respect to total drive space. A 3-disk array makes 2/3 of the total drive space available, so it's 67% efficient; a 4-disk array makes 3/4 of the total drive space available, so it's 75% efficient.

There's still some confusion here. The disk cost is 1 - 1 / n. If you calculate 1 - 1 /n, that equation provides the resulting total usable space. That resulting total usable space, of course, must be compared to the original total disk space allocation.

Thus, if you have three 500GB disks, and you perform the calculation, you receive .66, or 1TB. Readers are correct in posting that number in comments here. But you must subtract that resulting number from the total of all the disks (100%) to determine the disk cost. Sorry that wasn't clear but I thought it was understood (because it's a cost or liability). I'll get an edit added to the piece.

All companies which i tried used software RAID. I think it`s ok, because i dont have any problem with it.