RAID 0
Also known as stripping mode. Requires at least 2 hard drives. The system is to combine the capacity of multiple hard drives. So logically only "seen" a hard drive with large capacity (number of total disk capacity).
At first, RAID 0, is used to form a partition that is larger than some hard drive in a cost efficient.
For example:We need a partition with a size of 500GB. The price of a 100GB hard drive size is Rp.500.000, - while the price of a 500GB hard drive size is 5.000.000, -. Well, we can set up a partition of a 500GB size of 5 units sized 100GB hard drive using RAID 0. Of course, this scenario is less expensive because it takes less cost: 5 x Rp.500.000, - = Rp.2.500.000, -. Cheaper than having to buy a 500GB hard drive size. That is why at first called redundant arrays of inexpensive disks.
Another example:At this time the size of the largest hard drive available on the market is 500GB, while we need a partition to the size of 2TB. Well, we can buy 4 units with a capacity of 500GB hard drive and configure it with RAID 0, so that we can have a partition without having to wait 2TB hard drive with a capacity for it is available in the market.
Data written to the disk-drive is divided into fragments. Where the fragments are dispersed throughout the disk. Thus, if one hard drive has physical damage, then the data can not be read at all.
But there are advantages in the presence of these fragments: the speed. Data can be accessed faster with RAID 0, because when the computer reads a fragment on one disk, the computer can also read the other fragments in the other hard drive.
RAID 1
Commonly referred to as mirroring mode. Requires at least 2 hard drives. The system is copying the contents of a hard drive to another hard drive with a purpose: if one hard drive is physically damaged, then the data can still be accessed from any other hard drive.
Example:A server has two disk units each with a capacity of 80GB and configured RAID 1. After several years, one of the hard drive has physical damage. However, other data on the hard drive can still be read, so the data can still be saved as long as not all physically damaged hard drive simultaneously.
RAID 2
RAID 2, also using the stripping system. However, three hard drive is added again to Hamming parity, so the data becomes more reliable. Therefore, the number of disks required is at least 5 (n +3, n> 1). These three last hard drive used to store the Hamming code from the calculation of each of bits that exist in any other hard drive.
Example:We have 5 hard drive (let's call it drive A, B, C, D, and E) of the same size, each of 40GB.If we configure the four hard drives with RAID 2, then the capacity is obtained: 2 x 40GB = 80GB (of hard drive A and B). While the hard drive C, D, and E are not used for data storage, but only to store information Hamming parity from the other two hard drives: A and B. When there is physical damage to one of the main hard drive (A or B), then the data can still be read taking into account Hamming parity code on the hard drive C, D, and E.
RAID 3
RAID 3, also using the stripping system. Also use an additional hard drive for reliability, but only added a hard drive for parity again .Therefore, the number of disks required is at least 3 (n +1, n> 1). Last hard drive used to store the parity of the calculation results of each of bits that exist in any other hard drive.
Case in point:We have 4 hard drive (let's call it drive A, B, C, and D) with the same size, each of 40GB.If we configure the four hard drives with RAID 3, the capacity obtained is: 3 x 40GB = 120GB. While the disk D is not used for data storage, but only to store parity information from the three other hard drives: A, B, and C. When there is physical damage to one of the main hard drive (A, B, or C), then the data can still be read taking into account that there is parity in the disk D. However, if the damaged disk D, then the data can still be read from the three other hard drives.
RAID 4
Similar to RAID 3 systems, but using the parity of each disk block, instead of bits.Minimum hard disk requirement is the same, 3 (n +1, n> 1).
RAID 5
RAID 5 is basically similar to RAID 4, but with distributed parity. Namely, do not use special hard drive for storing parity, but parity spread to the entire hard drive.Minimum hard disk requirement is the same, 3 (n +1, n> 1).
This is done to speed up access and avoid the bottleneck that occurs because disk access is not focused on the collection of hard drives containing data only.
RAID 6
In general, an increase of RAID 5, ie, with the addition of parity into 2 (p + q). So the minimum number of hard drives is 4 (n +2, n> 1). With the addition of these secondary parity, then the damage to two hard drives at the same time can still be tolerated. For example, if a damaged disk, hard disk during the exchange process occurs more damage on one another hard drive, then it can still be tolerated and does not cause data loss on RAID 6 disk collection system.
Conclusions and Suggestions: Many benefits are obtained with RAID configurations, namely speed, data reliability, and fault tolerance. But it's not complete without discussing when discussing RAID hot-swappable hard drives, as well as some advanced configurations such as RAID 0 +1 or RAID 1 +0.
Also known as stripping mode. Requires at least 2 hard drives. The system is to combine the capacity of multiple hard drives. So logically only "seen" a hard drive with large capacity (number of total disk capacity).
At first, RAID 0, is used to form a partition that is larger than some hard drive in a cost efficient.
For example:We need a partition with a size of 500GB. The price of a 100GB hard drive size is Rp.500.000, - while the price of a 500GB hard drive size is 5.000.000, -. Well, we can set up a partition of a 500GB size of 5 units sized 100GB hard drive using RAID 0. Of course, this scenario is less expensive because it takes less cost: 5 x Rp.500.000, - = Rp.2.500.000, -. Cheaper than having to buy a 500GB hard drive size. That is why at first called redundant arrays of inexpensive disks.
Another example:At this time the size of the largest hard drive available on the market is 500GB, while we need a partition to the size of 2TB. Well, we can buy 4 units with a capacity of 500GB hard drive and configure it with RAID 0, so that we can have a partition without having to wait 2TB hard drive with a capacity for it is available in the market.
Data written to the disk-drive is divided into fragments. Where the fragments are dispersed throughout the disk. Thus, if one hard drive has physical damage, then the data can not be read at all.
But there are advantages in the presence of these fragments: the speed. Data can be accessed faster with RAID 0, because when the computer reads a fragment on one disk, the computer can also read the other fragments in the other hard drive.
RAID 1
Commonly referred to as mirroring mode. Requires at least 2 hard drives. The system is copying the contents of a hard drive to another hard drive with a purpose: if one hard drive is physically damaged, then the data can still be accessed from any other hard drive.
Example:A server has two disk units each with a capacity of 80GB and configured RAID 1. After several years, one of the hard drive has physical damage. However, other data on the hard drive can still be read, so the data can still be saved as long as not all physically damaged hard drive simultaneously.
RAID 2
RAID 2, also using the stripping system. However, three hard drive is added again to Hamming parity, so the data becomes more reliable. Therefore, the number of disks required is at least 5 (n +3, n> 1). These three last hard drive used to store the Hamming code from the calculation of each of bits that exist in any other hard drive.
Example:We have 5 hard drive (let's call it drive A, B, C, D, and E) of the same size, each of 40GB.If we configure the four hard drives with RAID 2, then the capacity is obtained: 2 x 40GB = 80GB (of hard drive A and B). While the hard drive C, D, and E are not used for data storage, but only to store information Hamming parity from the other two hard drives: A and B. When there is physical damage to one of the main hard drive (A or B), then the data can still be read taking into account Hamming parity code on the hard drive C, D, and E.
RAID 3
RAID 3, also using the stripping system. Also use an additional hard drive for reliability, but only added a hard drive for parity again .Therefore, the number of disks required is at least 3 (n +1, n> 1). Last hard drive used to store the parity of the calculation results of each of bits that exist in any other hard drive.
Case in point:We have 4 hard drive (let's call it drive A, B, C, and D) with the same size, each of 40GB.If we configure the four hard drives with RAID 3, the capacity obtained is: 3 x 40GB = 120GB. While the disk D is not used for data storage, but only to store parity information from the three other hard drives: A, B, and C. When there is physical damage to one of the main hard drive (A, B, or C), then the data can still be read taking into account that there is parity in the disk D. However, if the damaged disk D, then the data can still be read from the three other hard drives.
RAID 4
Similar to RAID 3 systems, but using the parity of each disk block, instead of bits.Minimum hard disk requirement is the same, 3 (n +1, n> 1).
RAID 5
RAID 5 is basically similar to RAID 4, but with distributed parity. Namely, do not use special hard drive for storing parity, but parity spread to the entire hard drive.Minimum hard disk requirement is the same, 3 (n +1, n> 1).
This is done to speed up access and avoid the bottleneck that occurs because disk access is not focused on the collection of hard drives containing data only.
RAID 6
In general, an increase of RAID 5, ie, with the addition of parity into 2 (p + q). So the minimum number of hard drives is 4 (n +2, n> 1). With the addition of these secondary parity, then the damage to two hard drives at the same time can still be tolerated. For example, if a damaged disk, hard disk during the exchange process occurs more damage on one another hard drive, then it can still be tolerated and does not cause data loss on RAID 6 disk collection system.
Conclusions and Suggestions: Many benefits are obtained with RAID configurations, namely speed, data reliability, and fault tolerance. But it's not complete without discussing when discussing RAID hot-swappable hard drives, as well as some advanced configurations such as RAID 0 +1 or RAID 1 +0.
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