comparison of different raid types

Features	RAID 0		RAID 1		RAID 1E		RAID 5		RAID 5EE
Minimum # Drives	2		2		3		3		4
Data Protection	No Protection		Single-drive failure		Single-drive failure		Single-drive failure		Single-drive failure
Read Performance	High		High		High		High		High
Write Performance	High		Medium		Medium		Low		Low
Read Performance (degraded)	N/A		Medium		High		Low		Low
Write Performance (degraded)	N/A		High		High		Low		Low
Capacity Utilization	100%		50%		50%		67% - 94%		50% - 88%
Typical Applications	High End Workstations, data logging, real-time rendering, very transitory data		Operating System, transaction databases		Operating system, transaction databases		Data warehousing, web serving, archiving		Data warehousing, web serving, archiving
Features		RAID 6		RAID 10		RAID 50		RAID 60
Minimum # Drives		4		4		6		8
Data Protection		Two-drive failure		Up to one disk failure in each sub-array		Up to one disk failure in each sub-array		Up to two disk failure in each sub-array
Read Performance		High		High		High		High
Write Performance		Low		Medium		Medium		Medium
Read Performance (degraded)		Low		High		Medium		Medium
Write Performance (degraded)		Low		High		Medium		Low
Capacity Utilization		50% - 88%		50%		67% - 94%		50% - 88%
Typical Applications		Data archive, backup to disk, high availability solutions, servers with large capacity requirements		Fast databases, application servers		Large databases, file servers, application servers		Data archive, backup to disk, high availability solutions, servers with large capacity requirements

The write penalty of RAID 5

By Rickard Nobel | August 2, 2011

14 Comments

Compared to other RAID levels we have a higher write overhead in RAID 5. In this article we will see in some detail why there is a larger “penalty” for writing to RAID 5 disk systems.

In a RAID 5 set with any number of disks we will calculate a parity information for each stripe. See this article on how the RAID 5 parity works. In short, we use the XOR operation on all binary bits on all disks and save the result on the parity disk. For example if we have an eight disk set the actual data is saved on seven disks and parity on the last disk, see picture above.

A disadvantage with RAID 5 is how to write small IOs against the disk system. Even if the write IO will only affect the data on one disk, we still need to calculate the new parity. Since the parity, as explained in the other article, is created by using XOR on all disks this could now be done in two ways. We could either do a read against all the other disks and then XOR with the new information. This would however cause a very large overhead and it is not reasonable to block all other disks for just one write.

There is however a quite clever way to calculate the new parity with a minimum of disk IO.

Assume we have the following eight disks and a write should be done at the fifth disk, which should be changed to, say, 1111. (For simplicity we will only look at four bits at each disk, but this could be of any size.)

To get the new parity some actions has to be done. First we read the old data on the blocks that should be changed. We can call this “Disk5-Old” and will be the first IO that must be done. The data that should be written, here 1111, can be called Disk5-New.

Disk5-0ld = 0110
Disk5-New = 1111

We will now use XOR on the old and the new data, to calculate the difference between the old and new. We can call this Disk5-Delta.

Disk5-Delta = Disk5-Old XOR Disk5-New = 0110 XOR 1111 = 1001

When we know the “delta” we will have to commit another read. This is against the old parity. We call this Parity-Old, in this example the old parity is 0010. We will now XOR the old parity with the Disk5-Delta. What is quite interesting is that this will create the new parity, but without the need to read the other six disks.

Parity-New = Parity-Old XOR Disk5-Delta = 0010 XOR 1001 = 1011

When we know the new parity we can write both the new data block and the new parity. This causes two write IOs against the disks and makes up the last of the “penalty”.

So in summary this disk actions that must be done:

1. Read the old data
2. Read the old parity
3. Write the new data
4. Write the new parity

This means that each write against a RAID 5 set causes four IOs against the disks where the first two must be completed before the last two could be performed, which introduces some additional latency.