protecting data by constantly making backup copies of files moving across the organization’s networks. Due to the internal configuration and implementation difference to meet special needs, RAID systems are categorized by level such as RAID 0, RAID 1, RAID 2, ...., RAID 6, or nested (hybrid) types, such as RAID 01 (i.e., RAID 0 + RAID 1), RAID 10 (RAID 1 + RAID 0), RAID 03 (RAID 0 + RAID 3), RAID 50 (RAID 5 + RAID 0), RAID 7 (RAID 3 + RAID 4), etc. Among all RAID systems, RAID 1 and RAID 5 are currently the most common two RAID levels. Here, in particular, RAID 5 is introduced by an example followed by a few questions for you to answer. A RAID 5 system requires a minimum of three hard disks with no maximum limit. The figure below shows an example of RAID 5 with three disks, numbered

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Chapter1: Introduction
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RAID is a kind of mass storage or a part of enterprise storage system. Every RAID contains several inexpensive hard drives, each is configured to connect to one another to meet specific purposes. RAID delivers high performance by expanding external storageimproving access speedproviding reliable storage, and protecting data by constantly making backup copies of files moving across the organization’s networks.

Due to the internal configuration and implementation difference to meet special needs, RAID systems are categorized by level such as RAID 0RAID 1RAID 2, ...., RAID 6, or nested (hybrid) types, such as RAID 01 (i.e., RAID 0 + RAID 1), RAID 10 (RAID 1 + RAID 0), RAID 03 (RAID 0 + RAID 3), RAID 50 (RAID 5 + RAID 0), RAID 7 (RAID 3 + RAID 4), etc. Among all RAID systems, RAID 1 and RAID 5 are currently the most common two RAID levels.

Here, in particular, RAID 5 is introduced by an example followed by a few questions for you to answer.

RAID 5 system requires a minimum of three hard disks with no maximum limit. The figure below shows an example of RAID 5 with three disks, numbered as 0 ~ 2.

( raid 5 picture below)

 

 

For demonstration, we assume three files, shown in pink, blue, and yellow, are distributed and stored across these three disks. In RAID 5, every file is evenly divided in two disks by using a striping process. Each half of a file is indicated in the figure by a unique number such as 1, 4, 7, etc. For example, 7 and 2 together makes the pink file, 4 and 1 represent the blue file, 5 and 3 are the yellow file. Then, each disk is equally loaded with two halves of two different files. For example, disk 0 stores half of pink file (i.e., 7) and half of blue file (i.e., 4).

When the pink file is stored in Disk 0 and 1, a parity is calculated (for realibility or fault tolerance purpose) by a formula. For instance, if the parity formula is a simple addition, then, 7+2 or 9 is the parity. The parity should be stored in the 3rd disk, i.e., Disk 2. Similarly, the same parity formula applying to blue and yellow files results parity 5 and 8, which are stored in Disk 0 and 1, respectively. It's very important to notice that, for each group of three files, their parities are also evenly distributed and stored in three disks, i.e., every disk maintains only one parity of the file that is stored in the other two disks.

Because files are striped evenly across disks, RAID 5 evenly balances reads and writes with no single disk being a bottleneck. Striping all parities evenly also allow data to be reconstructed in case of a disk failure. Therefore, RAID 5 is considered one of the most secure RAID configurations due to its fast reliable read/write speed plus its ability to tolerate any single drive failure.

Now, suppose a 4th file needs to be stored in this RAID storage with its two halves denoted by 18 and 9. By repeating the above storage process, this file's first half, say, 18, will be stored in Disk 0, second half, 9, in Disk 1, and its parity 27 will store in Disk 2. 

Question A.  Assume Disk 1 of this RAID is crashed with the four files stored and a new disk is ready to replace Disk 1. Describe in detail how you use the data and parities in Disk 0 and Disk 2 to rebuild the entire Disk 1 in the new disk so it contains partial files "2", "5", "9" and the parity 5. This is actually how the RAID controller will do to recover a failed disk in one of its "hot" spare drives. You should find that, without the parity data and striping process, reconstructing a disk is impossible.

Question B.  If more than one disk, say, Disk 1 and 2, are crashed or stolen with no backups, then, is it possible to rebuild them by using only Disk 0 in the system? Yes or no, explain.

Disk 0
7
4
P = 5+3=8
Disk 1
2
P = 4+1=5
5
Disk 2
P = 7+2=9
1
3
Transcribed Image Text:Disk 0 7 4 P = 5+3=8 Disk 1 2 P = 4+1=5 5 Disk 2 P = 7+2=9 1 3
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