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Table Of Contents  CertiGuide to Network+
 9  Chapter 0110: Network Operating Systems (NOS)
      9  IX  RAID (Server Disk Arrays)

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RAID 1
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RAID Volumes
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RAID 5

RAID 5 is the final option commonly used today. RAID 5 requires a minimum of three drives.

This RAID version begins its operations looking like RAID 0, in that is splits the data across two drives.

The third drive (or last drive, if you have more than three in your RAID volume set) however gets a different type of data set, called parity, which contains enough information to replace the data on the either of the first two drives.

What separates RAID 5 from other RAID levels using parity is the fact that the parity information ‘floats’ across all the drives in the RAID set.

Not only does the parity information become distributed, the actual data also plays ‘round-robin’ across all drives. The end result is if you could see the data written across the drives it would look like:

Data-Data-Parity Data-Parity-Data Parity-Data-Data
Data-Data-Parity Data-Parity-Data Parity-Data-Data…

Figure 57: RAID 5

 


 

This RAID level doesn’t use as much overhead (extra) disk space for redundancy as RAID 1, and it uses more available drive space than RAID 0 for the parity information that ensures redundancy.

The math formula for calculating RAID 5 is N=(N-1).

For example, if N were equal to a 9Gig drive, the total available storage space would be 18Gigs. When you see a math formula with the ( ) in the equation, do the math inside the ( ) first. Therefore 18=(27-9) is the formula yield.

RAID 5

If multiple drives have different amounts of available storage, calculate the above formula using the smallest amount of space available on any of the drives.

RAID 5 needs a minimum of 3 drives.



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RAID 1
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RAID Volumes
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