CN104809035B - The storage system construction method that a kind of quick single-deck of energy is repaired - Google Patents
The storage system construction method that a kind of quick single-deck of energy is repaired Download PDFInfo
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Abstract
The invention discloses a kind of construction method for the storage system that can be rapidly completed single-deck reparation, it is characterized in that building process includes disk group of subarrays into table construction step, disk areas is divided and region in-line coding step, and disk subarray construction step and storage system repair step.Because the present invention utilizes multiple disks to repair single low-quality disk parallel, the storage system is set to be rapidly completed single-deck reparation；Because the present invention adds one group of region interior coding in each region, the storage system is set to hold any three disk mistakes；Stagger the time, repaired merely with interregional coding when occurring single disk, stagger the time, repaired jointly using region interior coding and interregional coding when occurring multiple disks.Compared with traditional storage system construction method, using the inventive method single-deck can be made to repair speed lifting several times, while improving the performance that storage system degeneration is read, improve system reliability.
Description
Technical field
The invention belongs to computer distribution type memory system technologies field, and in particular to can quickly repair point of single low-quality disk
The construction method of cloth storage system.
Background technology
Due to the rapidly development of internet, nowadays mass data can be all produced daily.Distributed memory system needs to utilize
More disk increase capacity, access speed is accelerated using the concurrently access of multiple disks, ensures that data are reliable using data redundancy
Property.What Chinese publishing house of Tsing-Hua University published for 2010《RAID data recovery technology is exposed》The independent magnetic mentioned of page 21 to 29
Disk redundant array -5 (RAID-5) technology is a kind of distributed memory system constructing plan.One RAID-5 system is by multiple magnetic
Disk is constituted, and access speed is accelerated using striping technology, and producing data redundancy using correcting and eleting codes coding method ensures that data are reliable
Property.When single disk failures, RAID-5 systems can read all data of survival disk, to carry out data recovery.In recent years, it is single
The capacity of individual disk increases very fast, and the mismatch of the read or write speed slower development of disk, this disk read-write speed and capacity
The time for causing the reparation of RAID-5 systems single-deck is increasingly longer；Meanwhile, RAID-5 systems can only hold any one disk mistake,
In large-scale distributed storage system, it is impossible to ensure data reliability.
The content of the invention
It is existing to overcome the purpose of the present invention is to propose to a kind of construction method for the storage system that can be rapidly completed single-deck reparation
There is the drawbacks described above of technology, can be rapidly completed single-deck reparation on the premise of it can hold any 3 disks mistake and improve storage system
Performance and system reliability that degeneration is read.
The construction method for the storage system that the quick single-deck of energy of the invention is repaired, it is characterised in that comprise the following steps：
The first step, disk group of subarrays are into table construction step：
From block design parameter list, selection a line obtains five parameters of b, v, r, k, λ respectively, and optimal difference collection is designated as
D={ d0,d1,d2,…,dk-1}；Assignment p is the prime number more than or equal to k；
The matrix of a k rows v row is created, the 0th row are entered as the d that optimal difference is concentrated successively from top to bottom first0、d1、
d2To dk-1；Subsequently since arranging the 1st, the 0th row is entered as (d successively0+ 1)~(v-1), and the 0th row 0 is arranged be entered as again
v；1st row is entered as (d successively1+ 1)~v and 1~(d1- 1), if d1- 1 is less than 1, then is only entered as (d1+ 1)~v；2nd row according to
It is secondary to be entered as (d2+ 1)~v and 1~(d2- 1), if d2- 1 is less than 1, then is only entered as (d2+ 1)~v, by that analogy, per a line all
It is incremented by relation for circulation；IfBe on set V={ 1,2,3 ..., v } one (b, v, r, k,
λ)-BIB DESIGN BIBD, regard k value of the 0th row in matrix as set B0Element, by matrix the 1st row
K value be used as set B1Element, using in matrix the 2nd row k value be used as set B2Element, by that analogy, finally give
(b, v, r, k, λ)-BIBD and its collection of sets
The matrix of k p-1 rows p row is created, number consecutively is No. matrix 0~(k-1)；No. 0 matrix is compiled first
Number, 0~[p × (p-1) -1] is followed successively by from left to right, from top to bottom；Define a right-shift operation：In a matrix, the i-th row
Numbering ring shift right i；The right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix can be obtained；To No. 1 matrix
The right-shift operation is carried out, the numbering of No. 2 matrix can be obtained, by that analogy；If{P0,P1,P2,…,Pp×(p-1)-1It is one
Individual collection of sets, element P thereini(0≤i≤p × (p-1) -1) is ordered set；Since No. 0 matrix, compiled successively
Row where number 0, regard the k value arranged as P successively0Element；Since No. 0 matrix, the place of numbering 1 is obtained successively
Row, regard the value of k row as P successively1Element；Since No. 0 matrix, the row where numbering 2 are obtained successively, by k row
Value successively as P2Element, by that analogy, finally give collection of sets
If collection of setsIn each element be one set；Define oneMiddle member
Element withThe multiplying of middle element：Bi×Pj={ (b0-1)×3+p0,(b1-1)×3+p1,(b2-1)×3+p2,…,(bk-1-
1)×3+pk-1(0≤i≤b-1,0≤j≤p × (p-1) -1), wherein bm(0≤m≤k-1) is followed successively by BiIn element, pn(0
≤ n≤k-1) it is followed successively by PjIn element；Then collection of sets is constructedQi×p×(p-1)+j=Bi×Pj(0≤i≤b-1,0≤j≤p×
(p-1)-1)；Disk group of subarrays is finally built into table, array number is followed successively by 0~[b × p × (p-1) -1], array i correspondences
Disk number be followed successively by collection of setsMiddle QiIn element value；
Second step, disk areas are divided and region in-line 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, build storage system；All disks are divided into v groups, every group of p disk；Each disk is averagely cut into p × r+1 points
Area；For each disk group, p subregion being chosen from top to bottom from each disk, constituting a region, each region includes one
Individual p × p sectionized matrix, each disk group r region of formation and a Free Partition；Finally, whole storage system includes (p
× r+1) × p × v subregion and r × v region；
To each region, region in-line coding is all carried out：If ri,jThe subregion arranged for the i-th row jth in region,<j-i>pTable
Show j-i to p modulus；Will<j-i>pIdentical subregion is encoded, and concrete mode is that the partition data progress of 0~p-2 rows is different
Or computing, produce verification data and be stored in the subregion of p-1 rows；Finally, in all regions of storage system the row of pth -1 subregion
All it is verification subregion；Meanwhile, using last subregion of all disks all as hot standby subregion, do not deposit appoint in the normal mode
What data；Verification subregion and hot standby subregion are marked as using subregion, and remaining subregion is that subregion is not used；
3rd step, disk subarray construction step：
Define a disk subarray and build operation：The disk group of subarrays built from the first step obtains array into table
Number i and the disk number corresponding to it, from the disk corresponding to these disk numbers, obtain untapped subregion, and compile successively
Number be i, then with these subregions build i disk subarrays, and dispose RAID -5 (RAID-5) encode；Profit
Operation, which is built, with disk subarray builds No. disk subarray 0~[b × p × (p-1) -1] successively；
4th step, storage system repair step：
Staggered the time when single disk occurs for storage system, all partition numbers that inquiry low-quality disk is included；For each subregion
Number i, from disk group of subarrays into table, finds out all disk numbers corresponding to array i, and from corresponding to these disk numbers
Survival disk in, search the subregion that partition number is i, then read data from these subregions, carry out xor operation, repair out bad
Data in disk in i subregions；Concomitantly the data of all reparations are write in hot standby subregion temporarily, HotSpare disk is finally write again
In, complete single-deck reparation；Staggered the time when multiple disks occur for storage system, first look for the disk group of one and only one low-quality disk,
Low-quality disk is repaired using region interior coding；The method for recycling single-deck to repair subregion in repairing, repairs remaining low-quality disk.
The storage system construction method of the quick single-deck reparation of the invention described above energy includes disk group of subarrays and built into table
Step, disk areas is divided and region in-line coding step, and disk subarray construction step and storage system repair step.Due to
The present invention repairs single low-quality disk parallel using multiple disks, the storage system is rapidly completed single-deck reparation；Due to this hair
It is bright that one group of region interior coding is added in each region, the storage system is held any three disk mistakes；It is single when occurring
Individual disk is staggered the time, and is repaired merely with interregional coding, is staggered the time when occurring multiple disks, utilizes region interior coding and interregional
Coding is common to be repaired.Compared with traditional storage system construction method, using the inventive method single-deck can be made to repair speed lifting
Several times, while improving the performance that storage system degeneration is read, improve system reliability.
It is of the invention can the storage system construction method repaired of quick single-deck compared with prior art, with advantages below：
1st, the storage system constructed by the present invention has substantial amounts of disk to simultaneously participate in reparation, than existing when single-deck is repaired
The reparation speed of technology improves several times, improves the performance degenerated and read, improves storage system reliability.
2nd, the present invention has just reached the wrong ability of appearance any 3 merely with the coding of XOR, compared with prior art, compiles
Faster, checksum update expense is smaller for code speed.
Brief description of the drawings
Fig. 1 is block design parameter list；
Fig. 2 is for building collection of setsMatrix；
Fig. 3 is for building collection of setsMatrix；
Fig. 4 is disk group of subarrays into table.
Fig. 5 is the storage system general structure schematic diagram built by the inventive method.
Fig. 6 is region interior coding schematic diagram.
Fig. 7 is that single-deck repairs schematic diagram；
Fig. 8 is that polydisc repairs schematic diagram.
Embodiment
The storage system construction method repaired below in conjunction with the accompanying drawings by specific embodiment to the quick single-deck of energy of the invention is made
It is further to describe in detail.
Embodiment 1：
The storage system construction method that the quick single-deck of the present embodiment energy is repaired, specifically includes following steps：
The first step, disk group of subarrays are into table construction step：
Fig. 1 is block design parameter list, wherein the parameter of the 1st row is b=7, v=7, r=3, k=3, λ=1 is optimal poor
Different collection D1={ 0,1,3 }；The parameter of 2nd row is b=13, v=13, r=4, k=4, λ=1, optimal difference collection D2=0,1,3,
9}；The parameter of 3rd row is b=21, v=21, r=5, k=5, λ=1, optimal difference collection D3={ 0,1,4,14,16 }；4th row
Parameter be b=31, v=31, r=6, k=6, λ=1, optimal difference collection D4={ 0,1,3,8,12,18 }；The parameter of 5th row
For b=57, v=57, r=8, k=8, λ=1, optimal difference collection D5={ 0,1,3,13,32,36,43,52 }；The parameter of 6th row
For b=73, v=73, r=9, k=9, λ=1, optimal difference collection D6={ 0,1,3,7,15,31,36,54,63 }；The ginseng of 7th row
Number is b=91, v=91, r=10, k=10, λ=1, optimal difference collection D7={ 0,1,3,9,27,49,56,61,77,81 }.From
Selection the first row parameter, i.e. b=7 in Fig. 1, v=7, r=3, k=3, λ=1, optimal difference collection D={ 0,1,3 }, assignment p is 3.
Fig. 2 is for building collection of setsMatrix.
The matrix as shown in Figure 2 of 3 rows 7 row is created, the 0th row are entered as 0,1,3 successively from top to bottom first.Afterwards
Continue since arranging the 1st, the 0th row is followed successively by 1~6, and the 0th row the 0th row are entered as into 7 again；1st row is entered as 2 successively~
7；2nd row is entered as 4~7 and 1~2 successively.IfIt is one on set V={ 1,2,3 ..., 7 }
(7,7,3,3,1)-BIB DESIGN (BIBD), 3 values that matrix the 0th is arranged are used as set B0Element, by square
3 values of the row of battle array the 1st are used as set B1Element, by that analogy.Finally in collection of setsIn, B0={ 1,3,7 }, B1=1,2,
4 }, B2={ 2,3,5 }, B3={ 3,4,6 }, B4={ 4,5,7 }, B5={ 1,5,6 }, B6={ 2,6,7 }.
Fig. 3 is for building collection of setsMatrix.
The matrix as shown in Figure 3 of three 2 rows 3 row is created, No. 0 matrix, No. 1 matrix and No. 2 matrixes are followed successively by.It is right first
No. 0 matrix is numbered, and from left to right, is followed successively by 0~5 from top to bottom.Define a right-shift operation：In a matrix, i-th
Capable numbering ring shift right i.The right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix is can obtain；To No. 1 square
Battle array carries out the right-shift operation, can obtain the numbering of No. 2 matrix.Finally, in No. 0 matrix the numbering of the 0th row be followed successively by 0,1,
The numbering of 2, the 1st row is followed successively by 3,4,5；The numbering of the 0th row is followed successively by the numbering of the 0,1,2, the 1st row and is followed successively by No. 1 matrix
5、3、4；The numbering of the 0th row is followed successively by the numbering of the 0,1,2, the 1st row and is followed successively by 4,5,3 in No. 2 matrix.IfIt is a collection of sets, element P thereini(0≤i≤5) are ordered set.Opened from No. 0 matrix
Begin, the row where numbering 0 are obtained successively, the value of 3 row are regard as P successively0Element, then P0={ 0,0,0 }.With same side
Method obtains collection of setsIn remaining ordered set：P1={ 1,1,1 }, P2={ 2,2,2 }, P3={ 0,1,2 }, P4={ 1,2,0 }, P5
={ 2,0,1 }.
If collection of setsIn each element be one set；Define oneMiddle element with
The multiplying of middle element：Bi×Pj={ (b0-1)×3+p0,(b1-1)×3+p1,(b2-1)×3+p2}(0≤i≤6,0≤j≤
5), b0、b1、b2It is B successivelyiIn element, p0、p1、p2It is P successivelyjIn element.Collection of sets is constructed using the multiplying
Qi×6+j=Bi×Pj(0≤i≤6,0≤j≤5).Disk group of subarrays is finally built into table, as shown in figure 4, array number is followed successively by
The corresponding disk number of 0~41, array i is followed successively by collection of setsMiddle QiElement value, i.e., array number 0 correspondence disk number 0,6,18, battle array
The correspondence disk number 1,7,19 of row number 1, the correspondence disk number 2,8,20 of array number 2, the correspondence disk number 0,7,20 of array number 3, array number 4
Correspondence disk number 1,8,18, the correspondence disk number 2,6,19 of array number 5, the correspondence disk number 0,3,9 of array number 6, the corresponding magnetic of array number 7
Reel number 1,4,10, the correspondence disk number 2,5,11 of array number 8, the correspondence disk number 0,4,11 of array number 9, the corresponding disk number of array number 10
1st, 5,9, the correspondence disk number 2,3,10 of array number 11, the correspondence disk number 3,6,12 of array number 12, the correspondence of array number 13 disk number 4,
7th, 13, array number 14 correspondence disk number 5,8,14, array number 15 correspondence disk number 3,7,14, array number 16 correspondence disk number 4,8,
12, array number 17 correspondence disk number 5,6,13, array number 18 correspondence disk number 6,9,15, array number 19 correspondence disk number 7,10,
16, the correspondence disk number 8,11,17 of array number 20, the correspondence disk number 6,10,17 of array number 21, the correspondence of array number 22 disk number 7,
11st, 15, the correspondence disk number 8,9,16 of array number 23, the correspondence disk number 9,12,18 of array number 24, the corresponding disk number of array number 25
10th, 13,19, the correspondence disk number 11,14,20 of array number 26, the correspondence disk number 9,13,20 of array number 27, the corresponding magnetic of array number 28
Reel number 10,14,18, the correspondence disk number 11,12,19 of array number 29, the correspondence disk number 0,12,15 of array number 30,31 pairs of array number
Answer disk number 1,13,16, the correspondence disk number 2,14,17 of array number 32, the correspondence disk number 0,13,17 of array number 33, array number 34
Correspondence disk number 1,14,15, the correspondence disk number 2,12,16 of array number 35, the correspondence disk number 3,15,18 of array number 36, array number
37 correspondence disk numbers 4,16,19, the correspondence disk number 5,17,20 of array number 38, the correspondence disk number 3,16,20 of array number 39, array
Numbers 40 correspondence disk numbers 4,17,18, the correspondence disk number 5,15,19 of array number 41.
Second step, disk areas are divided and region in-line coding step：
According to the parameter value selected in the first step, prepare the disk of 21 same sizes.Fig. 5 is to be built by the inventive method
Storage system general structure schematic diagram.As shown in Figure 5, all grids of each row represent a disk altogether, by 21 magnetic
Disk number consecutively is D0~D20, and all disks are divided into 7 groups, every group of 3 disks, and all disks are averagely cut into 10
Subregion, each lattice represents a subregion.For each disk group, 3 are selected successively from top to bottom from each disk
Subregion, constitutes a region, therefore each region is the sectionized matrix of one 3 × 3, each disk group formation 3 regions and one
Individual Free Partition.Finally, whole storage system includes 210 subregions and 21 regions.
Fig. 6 is region interior coding schematic diagram.
As shown in Figure 6, for above-mentioned each region, region in-line coding is all carried out, if ri,jFor the i-th row jth in region
The subregion of row,<j-i>pRepresent j-i to p modulus.Will<j-i>pIdentical subregion (i.e. subregion in Fig. 6 on a diagonal) enters
Row coding, concrete mode is that the partition data of the 0th row and the 1st row is carried out into XOR, produces verification data and is stored in the 2nd
In capable subregion.Finally, as shown in figure 5, the subregion of the 2nd row (is represented in lattice using oblique line in all regions of storage system
Subregion) all be verification subregion.Meanwhile, by last subregion (subregion represented in lattice using round dot) of all disks
All as hot standby subregion, any data are not deposited in the normal mode.Verification subregion and hot standby subregion, which are marked as using, to be divided
Area, is subregion is not used in remaining subregion (being the subregion of blank in lattice), for building disk subarray.
3rd step, disk subarray construction step：
Fig. 4 is disk group of subarrays into table.From Fig. 4 disk group of subarrays into table, obtain successively array i and it
Corresponding disk number, then from the corresponding disk of these disk numbers, obtains untapped subregion successively, and it is i to number.Compile
Number result it is as shown in Figure 5：Disk D0~D20 the 0th subregion number consecutively be：0、1、2、6、7、8、0、1、2、6、7、8、
12、13、14、18、19、20、0、1、2；Disk D0~D20 the 1st subregion number consecutively be：3、4、5、1、9、10、5、3、4、
10、11、9、16、17、15、22、23、21、4、5、3；Disk D0~D20 the 3rd subregion number consecutively be：6、7、8、12、13、
14、12、13、14、18、19、20、24、25、26、30、31、32、24、25、26；Disk D0~D20 the 4th subregion is compiled successively
Number it is：9、10、11、15、16、17、17、15、16、23、21、22、29、27、28、34、35、33、28、29、27；Disk D0~
D20 the 6th subregion number consecutively be：30、31、32、36、37、38、18、19、20、24、25、26、30、31、32、36、37、
38、36、37、38；Disk D0~D20 the 7th subregion number consecutively be：33、34、35、39、40、41、21、22、23、27、
28、29、35、33、34、41、39、40、40、41、39.Take the subregion of identical numbering to build disk subarray, and dispose independent superfluous
Residual magnetism disk array -5 (RAID-5) is encoded, and final whole storage system contains 42 disk subarrays that numbering is 0~41.
4th step, storage system repair step：
Staggered the time when single disk occurs for storage system, it is necessary to carry out single-deck reparation.Fig. 7 is that single-deck is repaired in schematic diagram, figure
XOR is represented with the symbol of the additional circle of cross, is represented to read corresponding partition data with dotted arrow, with disk
" fork " pictograph number represents disk failures.No. 8 disk D8 is damaged in Fig. 7, inquires about all subregions that D8 is included first, i.e., 2,4,
14th, 16,20, No. 23 subregions.For No. 2 subregions, from the disk group of subarrays in Fig. 4 into inquiring about the corresponding magnetic of array number 2 in table
Reel number, i.e., 2,8,20.From survival disk D2 and survival disk D20, No. 2 all partition datas are read, XOR is carried out, and will
The result of computing writes hot standby subregion temporarily, and HotSpare disk is finally write again, you can repair disk D8 No. 2 subregions.With same
Method, reads the subregion represented in lattice using horizontal line, can repair successively in D8 4,14,16,20, No. 23 subregions, it is complete
Into single-deck reparation.
Staggered the time when multiple disks occur for storage system, it is necessary to carry out polydisc reparation.Fig. 8 is that polydisc is repaired in schematic diagram, figure
XOR equally is represented with the symbol of the additional circle of cross, is represented to read corresponding partition data with dotted arrow, uses disk
On " fork " pictograph number represent disk failures.Disk D7, disk D8 and disk D12 in Fig. 8 are damaged.Due to where D12
Disk group only D12 is damaged, and repairs D12 with the region interior coding where D12 first：Disk D13 the 1st partition data can be read
With disk D14 the 2nd partition data, XOR is carried out, D12 the 0th subregion is repaired out, can be successively with same method
Repair out complete D12.Then the method repaired with above-mentioned single-deck, repairs out D7 and D8 data, completes polydisc reparation.
In the single-deck shown in Fig. 7 is repaired, the data of 1 subregion are respectively read from corresponding survival disk, available for simultaneously
Disk D8 6 data are repaired, compared with RAID-5 systems, the former reparation speed is 6 times of the latter.Built in storage system
In the process first step, if selection other parameters, can also further improve the reparation speed of storage system.Meanwhile, the present embodiment
In storage system can hold any 3 disk mistakes, compared with RAID-5 systems, improve system reliability.
Claims (1)
1. the construction method for the storage system that a kind of quick single-deck of energy is repaired, it is characterised in that comprise the following steps：
The first step, disk group of subarrays are into table construction step：
From block design parameter list, selection a line obtains five parameters of b, v, r, k, λ respectively, and optimal difference collection is designated as into D=
{d0,d1,d2,…,dk-1}；Assignment p is the prime number more than or equal to k；
The matrix of a k rows v row is created, the 0th row are entered as the d that optimal difference is concentrated successively from top to bottom first0、d1、d2Arrive
dk-1；Subsequently since arranging the 1st, the 0th row is entered as (d successively0+ 1)~(v-1), and the 0th row 0 is arranged be entered as v again；The
1 row is entered as (d successively1+ 1)~v and 1~(d1- 1), if d1- 1 is less than 1, then is only entered as (d1+ 1)~v；2nd row is assigned successively
It is worth for (d2+ 1)~v and 1~(d2- 1), if d2- 1 is less than 1, then is only entered as (d2+ 1)~v, is to follow per a line by that analogy
Ring is incremented by relation；IfIt is one (b, v, r, k, λ)-balance on set V={ 1,2,3 ..., v }
Incomplete Block Designs BIBD, regard k value of the 0th row in matrix as set B0Element, by matrix the 1st row k be worth
It is used as set B1Element, using in matrix the 2nd row k value be used as set B2Element, by that analogy, finally give (b, v, r,
K, λ)-BIBD and its collection of sets
The matrix of k p-1 rows p row is created, number consecutively is No. matrix 0~(k-1)；No. 0 matrix is numbered first, from
It is left-to-right, be followed successively by 0~[p × (p-1) -1] from top to bottom；Define a right-shift operation：In a matrix, the volume of the i-th row
Number ring shift right i；The right-shift operation is carried out to No. 0 matrix, the numbering of No. 1 matrix can be obtained；No. 1 matrix is carried out
The right-shift operation, can obtain the numbering of No. 2 matrix, by that analogy；If It is a collection of sets,
Element P thereini(0≤i≤p × (p-1) -1) is ordered set；Since No. 0 matrix, the place of numbering 0 is obtained successively
Row, using k row value be used as P successively0Element；Since No. 0 matrix, the row where numbering 1 are obtained successively, by k
The value of row is successively as P1Element；Since No. 0 matrix, the row where numbering 2 are obtained successively, by the value of k row successively
It is used as P2Element, by that analogy, finally give collection of sets
If collection of setsIn each element be one set；Define oneMiddle element withThe multiplying of middle element：Bi×Pj={ (b0-1)×3+p0,(b1-1)×3+p1,(b2-1)×3+p2,…,(bk-1-1)×
3+pk-1(0≤i≤b-1,0≤j≤p × (p-1) -1), wherein bm(0≤m≤k-1) is followed successively by BiIn element, pn(0≤n≤
K-1) it is followed successively by PjIn element；Then collection of sets is constructedQi×p×(p-1)+j=Bi×Pj(0≤i≤b-1,0≤j≤p×(p-
1)-1)；Disk group of subarrays is finally built into table, array number is followed successively by 0~[b × p × (p-1) -1], the corresponding magnetic of array i
Reel number is followed successively by collection of setsMiddle QiIn element value；
Second step, disk areas are divided and region in-line 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, structure
Build storage system；All disks are divided into v groups, every group of p disk；Each disk is averagely cut into p × r+1 subregion；It is right
In each disk group, choose p subregion from top to bottom from each disk, constitute a region, each region comprising a p ×
P sectionized matrix, each disk group r region of formation and a Free Partition；Finally, whole storage system includes (p × r+1)
× p × v subregion and r × v region；
To each region, region in-line coding is all carried out：If ri,jThe subregion arranged for the i-th row jth in region,<j-i>pRepresent j-i
To p modulus；Will<j-i>pIdentical subregion is encoded, and concrete mode is that the partition data of 0~p-2 rows is carried out into XOR fortune
Calculate, produce verification data and be stored in the subregion of p-1 rows；Finally, the subregion of the row of pth -1 is all in all regions of storage system
Verify subregion；Meanwhile, using last subregion of all disks all as hot standby subregion, do not deposit any number in the normal mode
According to；Verification subregion and hot standby subregion are marked as using subregion, and remaining subregion is that subregion is not used；
3rd step, disk subarray construction step：
Define a disk subarray and build operation：The disk group of subarrays built from the first step into table, obtain array i with
And the disk number corresponding to it, from the disk corresponding to these disk numbers, untapped subregion is obtained successively, and it is i to number,
Then i disk subarrays are built with these subregions, and disposes RAID -5 (RAID-5) coding；Utilize disk
Subarray builds operation and builds No. disk subarray 0~[b × p × (p-1) -1] successively；
4th step, storage system repair step：
Staggered the time when single disk occurs for storage system, all partition numbers that inquiry low-quality disk is included；For each partition number i,
From disk group of subarrays into table, finding out all disk numbers corresponding to array i, and from depositing corresponding to these disk numbers
In removable pan, the subregion that partition number is i is searched, data are then read from these subregions, xor operation is carried out, repairs out in low-quality disk
Data in i subregions；Concomitantly the data of all reparations are write in hot standby subregion temporarily, finally write again in HotSpare disk,
Complete single-deck reparation；Staggered the time when multiple disks occur for storage system, first look for the disk group of one and only one low-quality disk, utilize
Region interior coding repairs low-quality disk；The method for recycling single-deck to repair subregion in repairing, repairs remaining low-quality disk.
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