Hard-raid and soft-raid

RAID (Redundant Arrays of Independent Disks) disk array

Hard-raid

Raid-0: (equivalent mode, stripe): optimal performance

When a file is to be written into raid, the file will be cut based on the size of the block, and then the data will be staggered to each Hard Disk in sequence. Therefore, when your data is to be written into raid, the data is stored on each hard disk in the same way. For example, you have two hard disks in raid-0. When you have 100 MB of data to write, each hard disk is allocated to 50 MB of storage space.

650) This. width = 650; "src =" http://vbird.dic.ksu.edu.tw/linux_basic/0420quota_files/raid0.gif "width =" 293 "Height =" 208 "alt =" raid0.gif "/>

The more hard disks, the higher the raid-0 performance, but because it is a cross-storage, if there is a problem with a hard disk, the data will be lost.

It is best to use two raid-0 groups with the same capacity. Otherwise, a small group will only have a large one after it is used up, resulting in reduced efficiency.

Raid-1: (ing mode, mirror): Full backup

Store the same copy of data on two hard disks. Therefore, the overall raid capacity is almost 50% less.

650) This. width = 650; "src =" http://vbird.dic.ksu.edu.tw/linux_basic/0420quota_files/raid1.gif "width =" 293 "Height =" 208 "alt =" raid1.gif "/>

Therefore, when a large number of RAID-1 writes, the write efficiency may be very poor (because there is only one South Bridge ).

However, if the disk array card is used, it does not occupy the system Io, and the performance will not be greatly affected.

Although the write efficiency of RAID-1 is poor, the Read efficiency is acceptable because the data has two copies on different hard disks. If multiple processes read the same data, raid performs the best read balancing on its own.

Raid-2:

This is an improved version of RAID 0. It uses hamming code to encode the data, partition the data into independent bits, and write the data to the hard disk separately. Because the Error Correction Code (ECC, Error Correction Code) is added to the data, the overall data capacity is larger than the original data, and raid2 requires at least three disk drives to operate.

Raid-3:

Bit-interleaving (Data staggered storage) technology is used. It needs to separate data bits by encoding and then store them in the hard disk separately, and the same bit is checked and then stored separately in the hard disk, however, because the bits in the data are scattered on different hard disks, even if you want to read a small segment of data, all hard disks may need to work, therefore, this type is suitable for reading large amounts of data.

Raid-4:

It differs from RAID 3 in that it exists in hard disks in blocks when partitioning, however, each data access request must retrieve the corresponding same-bit data from the disk with the same-bit check. Due to the frequent use, the hard disk loss may increase. (Block interlace technology, block interleaving)

RAID 5: balanced performance and Data Backup

Raid-5 requires at least three hard disks to form this type of disk array.

The data written to this disk array is similar to raid-0. However, during each cycle of writing, a same-bit check data (parity) is added to each hard disk ), this data records the backup data of other hard disks for rescue when the hard disk is damaged.

650) This. width = 650; "src =" http://vbird.dic.ksu.edu.tw/linux_basic/0420quota_files/raid5.gif "width =" 398 "Height =" 189 "alt =" raid5.gif "/>

The same-bit check code is recorded on Different Hard Disks each time. Therefore, when a hard disk is damaged, the data in the original hard disk can be rebuilt by the check code of other hard disks.

The total raid-5 capacity is equal to the total number of hard disks.

Raid-5 supports damage to only one hard disk by default (three hard disk groups ).

Raid-5 read performance is similar to raid-0, but the storage efficiency is poor because of data check.

Raid-6:

Compared with RAID 5, raid 6 adds the second independent parity information block. Two independent parity systems use different algorithms, and the data reliability is very high. Even if the two disks are invalid at the same time, the data usage will not be affected. However, raid 6 needs to allocate more disk space to the parity check information, which has a greater "Write loss" compared with RAID 5. Therefore, the "Write Performance" is very poor. Poor performance and complex implementation methods make raid 6 rarely applied.

At most two disks in the same array can be damaged. After a new disk is changed, the data is recalculated and written to the new disk. According to the design theory, raid 6 must have more than four disks to take effect.

Raid-7:

Raid 7 is not a public raid standard, but a patented hardware product name of storage computer Corporation. Raid 7 is developed based on RAID 3 and raid 4, however, it has been strengthened to solve some of the original restrictions. In addition, raid 7 can handle a large number of Io requirements simultaneously by using a large number of high-speed caches and dedicated real-time processors for asynchronous array management in the actual operation, therefore, the performance even surpasses many other raid standards. But because of this, the price is very high.

RAID 0 + 1, RAID 1 + 0:

Raid-0 has good performance but data is not secure. Raid-1 has poor data security.

RAID 0 + 1:

(1) first let two hard disks form RAID 0, and such configuration has two groups;

(2) group the two RAID 0 groups into a group of RAID 1.

Raid 1 + 0 represents raid-1 and raid-0.

650) This. width = 650; "src =" http://vbird.dic.ksu.edu.tw/linux_basic/0420quota_files/raid01.gif "width =" 437 "Height =" 197 "alt =" raid01.gif "/>

It has the advantage of RAID 0, so the efficiency is improved. As it has the advantage of RAID 1, data is backed up.

RAID 5 + 0:

The combination of RAID 5 and RAID 0 is RAID 5 first, and then RAID 0, that is, stripe access is formed for each group of RAID 5. Raid 50 is based on RAID 5 and RAID 5 requires at least three hard disks. Therefore, a raid 50 consisting of multiple groups of RAID 5 requires at least six hard disks. Taking the configuration of 6 hard disks with the minimum raid 50 as an example, we first divide the six hard disks into two groups, each of which forms RAID 5. In this way, we can get two groups of RAID 5, then, the two groups of RAID 5 constitute RAID 0.

Raid 50 can still operate when one hard disk is damaged in any group or multiple groups of RAID 5 at the underlying layer. However, if two or more hard disks are damaged in any RAID 5 group, the entire raid 50 is invalid.

Raid 50 combines multiple groups of RAID 5 into a stripe on the upper layer. Its performance is higher than RAID 5 alone, and its capacity utilization is lower than RAID 5. For example, for 9 hard disks, raid 50 is composed of 3 RAID 5 RAID 0. Each RAID 5 wastes a hard disk with a utilization rate of (1-3/9 ), RAID 5 is (1-1/9 ).

RAID 5 + 3:

It is a raid 3 hard disk array consisting of three or more RAID 5 groups.

Raid 6 + 0:

A combination of RAID 6 and RAID 0: Raid 6 and RAID 0. In other words, it is to perform stripe access on raid 6 of more than two groups. Raid 6 requires at least four hard disks, so the minimum requirement for raid 60 is eight hard disks.

Because the underlying layer is composed of RAID 6, raid 60 can allow any group of RAID 6 to damage up to two hard disks, and the system can still operate; however, as long as three hard disks are damaged in any raid 6 group at the underlying layer, the entire raid 60 group will become invalid. Of course, the probability of this situation is quite low.

Compared with raid 6 alone, the upper layer of RAID 60 combines multiple groups of RAID 6 to form stripe access, so the performance is high.

However, the usage threshold is high and the capacity utilization is low.

650) This. width = 650; "src =" http://s3.51cto.com/wyfs02/M01/47/ AE /wKiom1P-DbqhpwDMAAIukUxQdVw098.jpg "Title =" qq20140828005426.jpg "alt =" wKiom1P-DbqhpwDMAAIukUxQdVw098.jpg "/>

Hard-raid and soft-raid