{"id":973,"date":"2010-09-03T14:06:41","date_gmt":"2010-09-03T14:06:41","guid":{"rendered":"http:\/\/blog.open-e.com\/?p=973"},"modified":"2025-07-03T07:49:56","modified_gmt":"2025-07-03T07:49:56","slug":"how-does-raid-5-work","status":"publish","type":"post","link":"https:\/\/www.open-e.com\/blog\/how-does-raid-5-work\/","title":{"rendered":"How does RAID 5 work? The Shortest and Easiest explanation ever!"},"content":{"rendered":"

\t\t\t\tWe all have limited time to study long and complicated information about RAID theories, but you may be interested as to how RAID 5 works. We made it simple for you by providing the shortest and easiest explanation ever. <\/strong><\/p>\n

RAID 5 – how it works?
\n<\/strong><\/h3>\n

First we need to remind you XOR definition:<\/p>\n

XOR function result is equal 1 if both arguments are different.<\/p>\n

XOR (0, 1) = 1
\nXOR (1, 0) = 1<\/p>\n

XOR function output is equal 0 if both arguments are same.<\/p>\n

XOR (0, 0) = 0
\nXOR (1, 1) = 0<\/p>\n

Now let us assume we have 3 drives with the following bits:<\/p>\n

| 101\u00a0| 010 | 011 |<\/p>\n

And we calculate XOR of those data and place it on 4th<\/sup> drive<\/p>\n

XOR (101, 010, 011) = 100\u00a0\u00a0\u00a0\u00a0 (XOR (101,010) = 111 and then XOR (111, 011) = 100<\/p>\n

So the data on the four drives looks like this below:<\/p>\n

| 101 | 010 | 011 | 100<\/strong> |<\/p>\n

Now let\u2019s see how the XOR MAGIC works. Let\u2019s assume the second drive has failed. When we calculate XOR all the remaining data will be present from the missing drive.<\/p>\n

| 101 | 010<\/span><\/em> <\/strong><\/span> | 011 | 100 |<\/p>\n

XOR (101, 011, 100) = 010<\/p>\n

You can check the missing other\u00a0drives and XOR of the remaining data will always give you exactly the data of your\u00a0missing drive.<\/p>\n

| 101 | 010 | 011<\/span><\/em><\/strong><\/span> | 100 |<\/p>\n

XOR (101, 010, 100) = 011<\/p>\n

What works for 3 bits and 4 drives only, works for any number of bits and any number of drives. Real RAID 5 has the most common stripe size of 64k (65536 * 8 = 524288 bits )<\/p>\n

So the real XOR engine only needs to deal with 524288 bits and not 3 bits as in our exercise. This is why the RAID 5 needs a very efficient XOR engine in order to calculate it fast.<\/p>\n

So when adding one drive for parity you will be able to rebuild the missing data in case of any drive failure.<\/p>\n

In our example we have explained RAID 4 where parity is on a dedicated drive. RAID 5 will distribute parities evenly between all drives. Distributed parity provides a slight increase in performance but the XOR magic is the same.<\/p>\n

To learn more about RAID 5 please go to Wikipedia article<\/a>.<\/p>\n

RAID – The Series. Check out:<\/h3>\n