RAID Data Recovery

RAID 5 Systems

RAID 5 is a RAID configuration that combines striping with redundancy. The striping portion of RAID 5 is very similar that of RAID 0, but the redundancy portion is quite different from RAID 1.

RAID 5 systems create redundancy by calculating parity blocks and distributing these parity blocks among all disks in the array. A minimum of three disks is required for a RAID 5 system.

The maximum number of disks is limited by the RAID controller. RAID 5 systems are very popular since they have the performance benefits of striping with the added security of redundancy. Even better is that the storage efficiency (the ratio of the RAID system capacity to the total capacity of all individual disks) is much higher than that of RAID 1 (which is 50%).

Before getting into the details of how data is stored in a RAID 5 system, let's take a look at parity and how exactly redundant storage efficiency can exceed 50%. Calculating parity is nothing more than applying the XOR binary operator to the data stored on the disks. XOR stands for "exclusive OR" meaning that the output will equal 1 if and only if the two bits being XOR'd are different. The following is a truth table for the XOR function that illustrates this clearly.

The XOR function has a very unique property that lends itself to efficient data redundancy. If XOR is applied twice in a row, it negates itself. So if we have A and XOR it twice with B, we get A as the result:

A XOR B XOR B = A

The following example will demonstrate how to get 80% redundant storage efficiency out of an array of 5 disks. For 80% efficiency, we must store real data on 4 disks and use only 1 disk for redundancy. Let's store the string "RAID" on our 4 non-redundant disks.

Here's how our string is represented in binary:

Here is our string in our disk array at 80% efficiency with the redundancy portion (parity) taking just 20%. The data on Disk E is the parity and is calculated by applying XOR to all the other data like this:
E = A XOR B XOR C XOR D

Now say that Disk C fails and we're left with the data from Disk A, B, D, and E. We can rebuild the data from Disk C by applying XOR to the data on the remaining disks. Since the data on Disk E = A XOR B XOR C XOR D, applying XOR to all remaining data reduces to

(A XOR B XOR C XOR D) XOR A XOR B XOR D

Now, reformat to

(A XOR A) XOR (B XOR B) XOR (D XOR D) XOR C Since XOR is applied twice in a row to the data from A, B, and D, all we are left with at the end is C. This is of course the data from our failed disk representing the letter "I". The following chart illustrates the calculation:

Read the RAID 5 Example for more information.