CN104809035A - Storage system establishing method capable of repairing single disk quickly - Google Patents
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Abstract
The invention discloses a storage system establishing method capable of repairing a single disk quickly. The storage system establishing method is characterized by comprising a disk sub-array composition table establishing step, a disk area dividing and area internal encoding step, a disk sub-array establishing step and a storage system repairing step. According to the storage system establishing method, a single bad disk is repaired by utilizing a plurality of disks in parallel, so that single disk repairing can be completed quickly by the storage system; a group of area internal codes are added to each area, so that the storage system can tolerate any three disk faults; when a single disk fault is caused, the single disk is repaired by only utilizing inter-area encoding; when a plurality of disk faults are caused, the disk is repaired by utilizing the intra-area encoding and inter-area encoding. Compared with a traditional storage system establishing method, the single disk repairing speed can be improved by several times by adopting the method disclosed by the invention; meanwhile, the read degradation performance of the storage system is improved, and the system reliability is improved.
Description
Technical field
The invention belongs to computer distribution type memory system technologies field, be specifically related to the construction method repairing the distributed memory system of single low-quality disk fast.
Background technology
Due to the development rapidly of internet, nowadays all mass data can be produced every day.Distributed memory system needs to utilize more disk to increase capacity, utilizes the Concurrency Access of many disks to accelerate access speed, utilizes data redundancy to ensure data reliability." exposition of RAID data recovery technique " the 21 to 29 page of Redundant Array of Independent Disks (RAID) mentioned-5 (RAID-5) technology that publishing house of Tsing-Hua University of China publishes for 2010 is a kind of distributed memory system constructing plan.A RAID-5 system is made up of multiple disk, utilizes striping technology to accelerate access speed, utilizes correcting and eleting codes coding method to produce data redundancy and ensures data reliability.When single disk failures, RAID-5 system can read all data of survival dish, carries out date restoring.In recent years, the capacity of single disk increases very fast, and the read or write speed slower development of disk, and it is more and more longer that not the mating of this disk read-write speed and capacity causes the time that RAID-5 system single-deck repairs; Meanwhile, it is wrong that RAID-5 system can only hold any one disk, in large-scale distributed storage system, cannot ensure data reliability.
Summary of the invention
The object of the invention is to propose a kind of construction method that can complete the storage system that single-deck is repaired fast, to overcome the above-mentioned defect of prior art, single-deck reparation can be completed fast and improve storage system under the prerequisite holding any 3 disk mistakes and to degenerate the performance and system reliability read.
The present invention can the construction method of storage system repaired of single-deck fast, it is characterized in that comprising the following steps:
The first step, disk subarray composition table construction step:
From block design parameter list, select a line to obtain b, v, r, k, λ five parameters respectively, best difference collection is designated as D={d
0, d
1, d
2..., d
k-1; Assignment p is the prime number being more than or equal to k;
Create the matrix of the capable v row of k, first arranging from top to bottom assignment successively by the 0th is the d that best difference is concentrated
0, d
1, d
2to d
k-1; Follow-up all from the 1st row, the 0th row successively assignment is (d
0+ 1) ~ (v-1), and be v by the new assignment of the 0th row 0 column weight; 1st row successively assignment is (d
1+ 1) ~ v and 1 ~ (d
1-1); 2nd row successively assignment is (d
2+ 1) ~ v and 1 ~ (d
2-1), by that analogy; If
set V={1,2,3 ..., (b, v, r, k, λ)-BIB DESIGN (BIBD) on v}, using k value of the 0th row in matrix as set B
0element, using in matrix the 1st row k be worth as set B
1element, using in matrix the 2nd row k be worth as set B
2element, by that analogy, finally obtain (b, v, r, k, λ)-BIBD and collection of sets thereof
Create the matrix of k the capable p row of p-1, number consecutively is No. matrix 0 ~ (k-1); First No. 0 matrix is numbered, is followed successively by 0 ~ [p × (p-1)-1] from left to right, from top to bottom; Define a right-shift operation: in a matrix, the numbering ring shift right i position of the i-th row; This right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix can be obtained; This right-shift operation is carried out to No. 1 matrix, the numbering of No. 2 matrix can be obtained, by that analogy; If
a collection of sets, element P wherein
i(0≤i≤p × (p-1)-1) is all ordered set; From No. 0 matrix, obtain the row at numbering 0 place successively, using k the value arranged successively as P
0element; From No. 0 matrix, obtain the row at numbering 1 place successively, using k the value arranged successively as P
1element; From No. 0 matrix, obtain the row at numbering 2 place successively, using k the value arranged successively as P
2element, by that analogy, finally obtain collection of sets
If collection of sets
in each element be a set; Define one
middle element with
the multiplying of middle element: B
i× P
j={ (b
0-1) × 3+p
0, (b
1-1) × 3+p
1, (b
2-1) × 3+p
2..., (b
k-1-1) × 3+p
k-1(0≤i≤b-1,0≤j≤p × (p-1)-1), wherein b
m(0≤m≤k-1) is followed successively by B
iin element, p
n(0≤n≤k-1) is followed successively by P
jin element; Then collection of sets is constructed
finally build disk subarray composition table, array number is followed successively by 0 ~ [b × p × (p-1)-1], and the disk number that array i is corresponding is followed successively by collection of sets
middle Q
iin element value;
Second step, disk areas divide and intra-zone coding step:
According to the parameter value selected in the first step, prepare the disk of v × p same size, number consecutively is 0 ~ v × p-1, builds storage system; All disks are divided into v group, often organize p disk; Each disk is on average cut into p × r+1 subregion; For each disk group, choose p subregion from each disk from top to bottom, form a region, each region comprises the sectionized matrix of a p × p, and each disk group forms r region and a Free Partition; Finally, whole storage system comprises (p × r+1) × p × v subregion and r × v region;
To each region, all carry out intra-zone coding: establish r
i,jfor the subregion of the i-th row jth row in region, <j-i>
prepresent that j-i asks mould to p; By <j-i>
pidentical subregion is encoded, and concrete mode is that partition data capable for 0 ~ p-2 is carried out XOR, produces checking data and leaves in the capable subregion of p-1; Finally, the subregion that in all regions of storage system, p-1 is capable is all verification subregion; Meanwhile, using last subregion of all disks all as hot standby subregion, any data are not deposited in the normal mode; Verification subregion and hot standby subregion are marked as and use subregion, and all the other subregions are not for use subregion;
3rd step, disk subarray construction step:
Define a disk subarray and build operation: from the disk subarray composition table that the first step builds, obtain array i and the disk number corresponding to it, from the disk corresponding to these disk numbers, obtain untapped subregion successively, and be numbered i, then build i disk subarray with these subregions, and dispose Redundant Array of Independent Disks (RAID)-5 (RAID-5) coding; Utilize disk subarray to build operation and build No. disk subarray 0 ~ [b × p × (p-1)-1] successively;
4th step, storage system repair step:
Stagger the time when single disk occurs storage system, all partition number that inquiry low-quality disk comprises; For each partition number i, from disk subarray composition table, find out all disk numbers corresponding to array i, and from the survival dish corresponding to these disk numbers, search the subregion that partition number is i, then sense data from these subregions, carries out xor operation, repairs out the data in i subregion in low-quality disk; Concomitantly the data of all reparations are write in hot standby subregion temporarily, finally write again in HotSpare disk, complete single-deck reparation; Stagger the time when multiple disk occurs storage system, first search and only have the disk group of a low-quality disk, to utilize in region coding by low-quality disk reparation; Repair the method for subregion in the reparation of recycling single-deck, repair residue low-quality disk.
The storage system construction method of the quick single-deck reparation of the invention described above energy includes disk subarray composition table construction step, and disk areas divides and intra-zone coding step, and disk subarray construction step and storage system repair step.Because the present invention utilizes, multiple disk is parallel repairs single low-quality disk, enables this storage system complete single-deck reparation fast; Encode because the present invention adds in one group of region in each region, enable this storage system hold any three disk mistakes; When the single disk of generation is staggered the time, only utilize interregional coding to repair, when the multiple disk of generation is staggered the time, utilize coding and interregional coding in region jointly to repair.Compared with traditional storage system construction method, adopt the inventive method that single-deck speed of repairing can be made to promote several times, improve storage system simultaneously and to degenerate the performance read, improve system reliability.
The present invention can fast single-deck repair storage system construction method compared with prior art, have the following advantages:
1, the storage system constructed by the present invention, when single-deck is repaired, has a large amount of disks to participate in repairing simultaneously, improves several times, improve the performance of degenerating and reading, improve storage system reliability than the reparation speed of prior art.
2, the present invention only utilizes the coding of XOR just to reach the ability holding any 3 mistakes, and compared with prior art, coding rate is faster, and checksum update expense is less.
Accompanying drawing explanation
Fig. 1 is block design parameter list;
Fig. 2 is for building collection of sets
matrix;
Fig. 3 is for building collection of sets
matrix;
Fig. 4 is disk subarray composition table.
Fig. 5 is the storage system general structure schematic diagram built by the inventive method.
Fig. 6 is schematic diagram of encoding in region.
Fig. 7 is that single-deck repairs schematic diagram;
Fig. 8 is that schematic diagram repaired by polydisc.
Embodiment
The present invention can the storage system construction method of single-deck reparation be fast described in further detail by specific embodiment below in conjunction with accompanying drawing.
Embodiment 1:
The storage system construction method that the quick single-deck of the present embodiment energy is repaired, specifically comprises the following steps:
The first step, disk subarray composition table construction step:
Fig. 1 is block design parameter list, and wherein the parameter of the 1st row is b=7, v=7, r=3, k=3, λ=1, best difference collection D
1={ 0,1,3}; The parameter of the 2nd row is b=13, v=13, r=4, k=4, λ=1, best difference collection D
2={ 0,1,3,9}; The parameter of the 3rd row is b=21, v=21, r=5, k=5, λ=1, best difference collection D
3=0, Isosorbide-5-Nitrae, 14,16}; The parameter of the 4th row is b=31, v=31, r=6, k=6, λ=1, best difference collection D
4={ 0,1,3,8,12,18}; The parameter of the 5th row is b=57, v=57, r=8, k=8, λ=1, best difference collection D
5={ 0,1,3,13,32,36,43,52}; The parameter of the 6th row is b=73, v=73, r=9, k=9, λ=1, best difference collection D
6={ 0,1,3,7,15,31,36,54,63}; The parameter of the 7th row is b=91, v=91, r=10, k=10, λ=1, best difference collection D
7={ 0,1,3,9,27,49,56,61,77,81}.From Fig. 1, select the first row parameter, i.e. b=7, v=7, r=3, k=3, λ=1, best difference collection D={0,1,3}, assignment p are 3.
Fig. 2 is for building collection of sets
matrix.
Create the matrix as shown in Figure 2 that 3 row 7 arrange, first arranging from top to bottom assignment successively by the 0th is 0,1,3.Follow-up all from the 1st row, the 0th row is followed successively by 1 ~ 6, and is 7 by the new assignment of the 0th row the 0th column weight; 1st row successively assignment is 2 ~ 7; 2nd row successively assignment is 4 ~ 7 and 1 ~ 2.If
set V={1,2,3 ..., (7,7,3,3,1)-BIB DESIGN (BIBD) on 7}, 3 values arranged by matrix the 0th are as set B
0element, using matrix the 1st arrange 3 value as set B
1element, by that analogy.Final in collection of sets
in, B
0={ 1,3,7}, B
1={ 1,2,4}, B
2={ 2,3,5}, B
3={ 3,4,6}, B
4={ 4,5,7}, B
5={ 1,5,6}, B
6={ 2,6,7}.
Fig. 3 is for building collection of sets
matrix.
Create the matrix as shown in Figure 3 that three 2 row 3 arrange, be followed successively by No. 0 matrix, No. 1 matrix and No. 2 matrixes.First No. 0 matrix is numbered, from left to right, is from top to bottom followed successively by 0 ~ 5.Define a right-shift operation: in a matrix, the numbering ring shift right i position of the i-th row.This right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix can be obtained; This right-shift operation is carried out to No. 1 matrix, the numbering of No. 2 matrix can be obtained.Finally, in No. 0 matrix, the numbering of the 0th row is followed successively by 0,1,2, and the numbering of the 1st row is followed successively by 3,4,5; In No. 1 matrix, the numbering of the 0th row is followed successively by 0,1,2, and the numbering of the 1st row is followed successively by 5,3,4; In No. 2 matrix, the numbering of the 0th row is followed successively by 0,1,2, and the numbering of the 1st row is followed successively by 4,5,3.If
a collection of sets, element P wherein
i(0≤i≤5) are all ordered set.From No. 0 matrix, obtain the row at numbering 0 place successively, using 3 values arranged successively as P
0element, then P
0={ 0,0,0}.Use the same method and obtain collection of sets
in remaining ordered set: P
1={ 1,1,1}, P
2={ 2,2,2}, P
3={ 0,1,2}, P
4={ 1,2,0}, P
5={ 2,0,1}.
If collection of sets
in each element be a set; Define one
middle element with
the multiplying of middle element: B
i× P
j={ (b
0-1) × 3+p
0, (b
1-1) × 3+p
1, (b
2-1) × 3+p
2(0≤i≤6,0≤j≤5), b
0, b
1, b
2b successively
iin element, p
0, p
1, p
2p successively
jin element.This multiplying is utilized to construct collection of sets
finally build disk subarray composition table, as shown in Figure 4, array number is followed successively by 0 ~ 41, and the disk number that array i is corresponding is followed successively by collection of sets
middle Q
ielement value, i.e. array number 0 corresponding disk number 0,6,18, array number 1 corresponding disk number 1, 7, 19, array number 2 corresponding disk numbers 2, 8, 20, array number 3 corresponding disk numbers 0, 7, 20, array number 4 corresponding disk numbers 1, 8, 18, array number 5 corresponding disk numbers 2, 6, 19, array number 6 corresponding disk numbers 0, 3, 9, array number 7 corresponding disk numbers 1, 4, 10, array number 8 corresponding disk numbers 2, 5, 11, array number 9 corresponding disk numbers 0, 4, 11, array number 10 corresponding disk numbers 1, 5, 9, array number 11 corresponding disk numbers 2, 3, 10, array number 12 corresponding disk numbers 3, 6, 12, array number 13 corresponding disk numbers 4, 7, 13, array number 14 corresponding disk numbers 5, 8, 14, array number 15 corresponding disk numbers 3, 7, 14, array number 16 corresponding disk numbers 4, 8, 12, array number 17 corresponding disk numbers 5, 6, 13, array number 18 corresponding disk numbers 6, 9, 15, array number 19 corresponding disk numbers 7, 10, 16, array number 20 corresponding disk numbers 8, 11, 17, array number 21 corresponding disk numbers 6, 10, 17, array number 22 corresponding disk numbers 7, 11, 15, array number 23 corresponding disk numbers 8, 9, 16, array number 24 corresponding disk numbers 9, 12, 18, array number 25 corresponding disk numbers 10, 13, 19, array number 26 corresponding disk numbers 11, 14, 20, array number 27 corresponding disk numbers 9, 13, 20, array number 28 corresponding disk numbers 10, 14, 18, array number 29 corresponding disk numbers 11, 12, 19, array number 30 corresponding disk numbers 0, 12, 15, array number 31 corresponding disk numbers 1, 13, 16, array number 32 corresponding disk numbers 2, 14, 17, array number 33 corresponding disk numbers 0, 13, 17, array number 34 corresponding disk numbers 1, 14, 15, array number 35 corresponding disk numbers 2, 12, 16, array number 36 corresponding disk numbers 3, 15, 18, array number 37 corresponding disk numbers 4, 16, 19, array number 38 corresponding disk numbers 5, 17, 20, array number 39 corresponding disk numbers 3, 16, 20, array number 40 corresponding disk numbers 4, 17, 18, array number 41 corresponding disk numbers 5, 15, 19.
Second step, disk areas divide and intra-zone coding step:
According to the parameter value selected in the first step, prepare the disk of 21 same sizes.Fig. 5 is the storage system general structure schematic diagram built by the inventive method.As shown in Figure 5, all grids of each row represent a disk altogether, be D0 ~ D20, all disks are divided into 7 groups by 21 disk number consecutivelies, and often organize 3 disks, and all disks are on average cut into 10 subregions, each lattice represents a subregion.For each disk group, from each disk, select 3 subregions successively from top to bottom, form a region, therefore each region is the sectionized matrix of 3 × 3, and each disk group forms 3 regions and a Free Partition.Finally, whole storage system comprises 210 subregions and 21 regions.
Fig. 6 is schematic diagram of encoding in region.
As shown in Figure 6, for above-mentioned each region, all intra-zone coding is carried out, if r
i,jfor the subregion of the i-th row jth row in region, <j-i>
prepresent that j-i asks mould to p.By <j-i>
pidentical subregion (subregion namely in Fig. 6 on a diagonal line) is encoded, and concrete mode is that the partition data of the 0th row and the 1st row is carried out XOR, produces checking data and leaves in the subregion of the 2nd row.Finally, as shown in Figure 5, in all regions of storage system, the subregion (adopting the subregion that oblique line represents in lattice) of the 2nd row is all verification subregion.Meanwhile, last subregion (adopting the subregion that round dot represents in lattice) of all disks, all as hot standby subregion, is not deposited any data in the normal mode.Verification subregion and hot standby subregion are marked as and use subregion, all the other subregions (being blank subregion in lattice) for not use subregion, for building disk subarray.
3rd step, disk subarray construction step:
Fig. 4 is disk subarray composition table.From the disk subarray of Fig. 4 composition table, obtain array i and the disk number corresponding to it successively, then from disk corresponding to these disk numbers, obtain untapped subregion successively, and be numbered i.Numbering result as shown in Figure 5: the 0th the subregion number consecutively of disk D0 ~ D20 is: 0,1,2,6,7,8,0,1,2,6,7,8,12,13,14,18,19,20,0,1,2; 1st the subregion number consecutively of disk D0 ~ D20 is: 3,4,5,1,9,10,5,3,4,10,11,9,16,17,15,22,23,21,4,5,3; 3rd the subregion number consecutively of disk D0 ~ D20 is: 6,7,8,12,13,14,12,13,14,18,19,20,24,25,26,30,31,32,24,25,26; 4th the subregion number consecutively of disk D0 ~ D20 is: 9,10,11,15,16,17,17,15,16,23,21,22,29,27,28,34,35,33,28,29,27; 6th the subregion number consecutively of disk D0 ~ D20 is: 30,31,32,36,37,38,18,19,20,24,25,26,30,31,32,36,37,38,36,37,38; 7th the subregion number consecutively of disk D0 ~ D20 is: 33,34,35,39,40,41,21,22,23,27,28,29,35,33,34,41,39,40,40,41,39.The subregion getting identical numbering builds disk subarray, and disposes raid-array-5 (RAID-5) coding, and final whole storage system contains 42 the disk subarrays being numbered 0 ~ 41.
4th step, storage system repair step:
Stagger the time when single disk occurs storage system, need to carry out single-deck reparation.Fig. 7 is that single-deck repairs schematic diagram, represents XOR in figure with the symbol of the additional circle of cross, represents and reads corresponding partition data, represent disk failures with " fork " pictograph number on disk with dotted arrow.In Fig. 7, No. 8 disk D8 damages, and first inquires about all subregions that D8 comprises, i.e. 2,4,14,16,20, No. 23 subregions.For No. 2 subregions, from the disk subarray composition table Fig. 4, inquire about the disk number of array number 2 correspondences, namely 2,8,20.From survival dish D2 and survival dish D20, read No. 2 all partition datas, carry out XOR, and the result of computing is write hot standby subregion temporarily, finally write HotSpare disk again, No. 2 subregions of disk D8 can be repaired.Use the same method, read in lattice the subregion adopting horizontal line to represent, 4,14,16,20, No. 23 subregions in D8 can be repaired successively, complete single-deck reparation.
Stagger the time when multiple disk occurs storage system, need to carry out polydisc reparation.Fig. 8 is that schematic diagram repaired by polydisc, represents XOR equally in figure with the symbol of the additional circle of cross, represents and reads corresponding partition data, represent disk failures with " fork " pictograph number on disk with dotted arrow.Disk D7 in Fig. 8, disk D8 and disk D12 damage.Due to the disk group only D12 damage at D12 place, first D12 is repaired with coding in the region at D12 place: the 1st partition data of disk D13 and the 2nd partition data of disk D14 can be read, carry out XOR, repair out the 0th subregion of D12, use the same method and can repair out complete D12 successively.Then by the method that above-mentioned single-deck is repaired, repair out the data of D7 and D8, complete polydisc reparation.
In the single-deck shown in Fig. 7 is repaired, each data reading 1 subregion from corresponding survival dish, can be used for 6 data of simultaneously repairing disk D8, compared with RAID-5 system, the former reparation speed is 6 times of the latter.In the storage system building process first step, if select other parameter, the reparation speed of storage system can also be improved further.Meanwhile, the storage system in the present embodiment can hold any 3 disk mistakes, compared with RAID-5 system, improves system reliability.
Claims (1)
1. a construction method for the storage system of the quick single-deck reparation of energy, is characterized in that comprising the following steps:
The first step, disk subarray composition table construction step:
From block design parameter list, select a line to obtain b, v, r, k, λ five parameters respectively, best difference collection is designated as D={d
0, d
1, d
2..., d
k-1; Assignment p is the prime number being more than or equal to k;
Create the matrix of the capable v row of k, first arranging from top to bottom assignment successively by the 0th is the d that best difference is concentrated
0, d
1, d
2to d
k-1; Follow-up all from the 1st row, the 0th row successively assignment is (d
0+ 1) ~ (v-1), and be v by the new assignment of the 0th row 0 column weight; 1st row successively assignment is (d
1+ 1) ~ v and 1 ~ (d
1-1); 2nd row successively assignment is (d
2+ 1) ~ v and 1 ~ (d
2-1), by that analogy; If
set V={1,2,3 ..., (b, v, r, k, λ)-BIB DESIGN (BIBD) on v}, using k value of the 0th row in matrix as set B
0element, using in matrix the 1st row k be worth as set B
1element, using in matrix the 2nd row k be worth as set B
2element, by that analogy, finally obtain (b, v, r, k, λ)-BIBD and collection of sets thereof
Create the matrix of k the capable p row of p-1, number consecutively is No. matrix 0 ~ (k-1); First No. 0 matrix is numbered, is followed successively by 0 ~ [p × (p-1)-1] from left to right, from top to bottom; Define a right-shift operation: in a matrix, the numbering ring shift right i position of the i-th row; This right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix can be obtained; This right-shift operation is carried out to No. 1 matrix, the numbering of No. 2 matrix can be obtained, by that analogy; If
a collection of sets, element P wherein
i(0≤i≤p × (p-1)-1) is all ordered set; From No. 0 matrix, obtain the row at numbering 0 place successively, using k the value arranged successively as P
0element; From No. 0 matrix, obtain the row at numbering 1 place successively, using k the value arranged successively as P
1element; From No. 0 matrix, obtain the row at numbering 2 place successively, using k the value arranged successively as P
2element, by that analogy, finally obtain collection of sets
If collection of sets
in each element be a set; Define one
middle element with
the multiplying of middle element: B
i× P
j={ (b
0-1) × 3+p
0, (b
1-1) × 3+p
1, (b
2-1) × 3+p
2..., (b
k-1-1) × 3+p
k-1(0≤i≤b-1,0≤j≤p × (p-1)-1), wherein b
m(0≤m≤k-1) is followed successively by B
iin element, p
n(0≤n≤k-1) is followed successively by P
jin element; Then collection of sets is constructed
0≤j≤p × (p-1)-1); Finally build disk subarray composition table, array number is followed successively by 0 ~ [b × p × (p-1)-1], and the disk number that array i is corresponding is followed successively by collection of sets
middle Q
iin element value;
Second step, disk areas divide and intra-zone coding step:
According to the parameter value selected in the first step, prepare the disk of v × p same size, number consecutively is 0 ~ v × p-1, builds storage system; All disks are divided into v group, often organize p disk; Each disk is on average cut into p × r+1 subregion; For each disk group, choose p subregion from each disk from top to bottom, form a region, each region comprises the sectionized matrix of a p × p, and each disk group forms r region and a Free Partition; Finally, whole storage system comprises (p × r+1) × p × v subregion and r × v region;
To each region, all carry out intra-zone coding: establish r
i,jfor the subregion of the i-th row jth row in region, <j-i>
prepresent that j-i asks mould to p; By <j-i>
pidentical subregion is encoded, and concrete mode is that partition data capable for 0 ~ p-2 is carried out XOR, produces checking data and leaves in the capable subregion of p-1; Finally, the subregion that in all regions of storage system, p-1 is capable is all verification subregion; Meanwhile, using last subregion of all disks all as hot standby subregion, any data are not deposited in the normal mode; Verification subregion and hot standby subregion are marked as and use subregion, and all the other subregions are not for use subregion;
3rd step, disk subarray construction step:
Define a disk subarray and build operation: from the disk subarray composition table that the first step builds, obtain array i and the disk number corresponding to it, from the disk corresponding to these disk numbers, obtain untapped subregion successively, and be numbered i, then build i disk subarray with these subregions, and dispose Redundant Array of Independent Disks (RAID)-5 (RAID-5) coding; Utilize disk subarray to build operation and build No. disk subarray 0 ~ [b × p × (p-1)-1] successively;
4th step, storage system repair step:
Stagger the time when single disk occurs storage system, all partition number that inquiry low-quality disk comprises; For each partition number i, from disk subarray composition table, find out all disk numbers corresponding to array i, and from the survival dish corresponding to these disk numbers, search the subregion that partition number is i, then sense data from these subregions, carries out xor operation, repairs out the data in i subregion in low-quality disk; Concomitantly the data of all reparations are write in hot standby subregion temporarily, finally write again in HotSpare disk, complete single-deck reparation; Stagger the time when multiple disk occurs storage system, first search and only have the disk group of a low-quality disk, to utilize in region coding by low-quality disk reparation; Repair the method for subregion in the reparation of recycling single-deck, repair residue low-quality disk.
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