JP2003223285A - Data storage device, data management method, and program - Google Patents

Abstract

<P>PROBLEM TO BE SOLVED: To solve the problem that it is impossible to maximize the reliability of the entire system at a mirroring constitution unless taking due account that which disks are to be coupled to write original data and duplicate data. <P>SOLUTION: A data storage means allocates data dispersedly with the uniform number of blocks by a data block unit to a plurality of storage devices by dividing the data into a plurality of data blocks of every uniform data quantity and stores the data block having identical contents redundantly at every prescribed number of the storage devices and a readout means reads out the data stored in a plurality of storage devices dispersedly by single or plural data block unit for every prescribed number of the storage devices are provided. <P>COPYRIGHT: (C)2003,JPO

Description

Detailed Description of the Invention

[0001]

BACKGROUND OF THE INVENTION 1. Field of the Invention The present invention creates a group of a plurality of secondary storage devices according to the redundancy multiplicity of data,
The present invention relates to a data storage device, a data management method and a program for writing data in each group.

[0002]

2. Description of the Related Art In recent years, the capacity of a secondary storage device (hereinafter referred to as a disk) has been increased, and a method of transferring data from the disk at high speed and a high reliability of the disk are required. There is a disk array device as one of the methods to meet these demands. Several data management methods have been proposed for this disk array device.

One of data management methods for disk array devices
One is mirroring. This mirroring is RAI
D1 (RAID: Redundant Arrays)
Also called ofInexpensive Disks). The outline is to improve the reliability of the disks by writing the same data to the two disks.

FIG. 27 is a diagram showing an example of data storage using mirroring. In the figure, # D1 to # D4 represent physical disks, and B1 to B24 represent data stored in the physical disks # D1 to # D4. Here, physical disk # D1 and physical disk #D
2. Physical disk # D3 and physical disk # D4 form a pair. A pair of these physical disks constitutes one logical disk. The same data is written on the paired physical disks. That is, the data B1 to B12 are stored in the physical disks # D1 and # D2, and the data B13 is stored in the physical disks # D3 and # D4.
To B24 are stored.

With this configuration, physical disk # D1
When a failure occurs in the physical disk # D2, the physical disk # D2 is accessed, and when a failure occurs in the physical disk # D2, the physical disk # D1 is accessed. As a result, data can be accessed as long as the paired physical disks do not fail at the same time.

The above-mentioned mirroring has a load distribution function for improving the overall data transfer performance of the disks by selecting the disks to transfer the data so that the loads on the physical disks are equalized during the data transfer. There is. Specifically, when a data transfer request is issued to a logical disk with a mirroring configuration,
During the waiting time when the requested data is transferred from the physical disk, the next data transfer request is issued to the other physical disk pair. As a result, the data transfer is performed in parallel at the same time, which improves the data transfer efficiency.

As described above, the mirroring is a method of storing the same data in two disks, and realizes high reliability and high speed transfer of the disks.

Further, a method other than forming the pair of disks and storing the original data and the duplicated data as in ordinary mirroring has been proposed. For example, JP-A-6-18710
Japanese Patent No. 1 discloses a conventional data storage device (disk array) that stores a copy of original data in a plurality of disks.

FIG. 28 shows the above-mentioned Japanese Patent Laid-Open No. 6-18710.
FIG. 3 is a diagram showing an example of data storage by the conventional disk array disclosed in Japanese Patent Publication No. 1-A, FIG. 1A shows an example in which a copy of each disk is arranged for all other disks, and FIG.
Shows an example in which a copy of the original data is divided and stored in two or more copy disks.

Next, an outline will be described. FIG. 28
In the data management methods (a) and (b), when the data stored in the disks is transferred, the data is transferred in parallel from each disk, so that high-speed transfer can be realized. Also,
Even when data transfer requests are concentrated on a specific disk, it is possible to distribute the load to other disks. Furthermore, when a disk fails and data is backed up, data can be transferred from a plurality of disks in parallel and written to a spare disk. Therefore, it has the advantage of faster failure recovery and faster backup than the normal mirroring system.

[0011]

Since the conventional data storage device is configured as described above, in the mirroring configuration, it is necessary to fully consider which disk is paired with the original data and the duplicated data for writing. Otherwise, there was a problem that the reliability of the entire system could not be maximized.

The above problem will be described in detail. For example, in the mirroring configuration of FIG. 27, physical disk # D1
And the physical disk # D2 are placed in the same machine. At this time, if a failure occurs at the machine level, the physical disks # D1 and # D2 cannot be simultaneously accessed. The same applies to the data management method shown in FIG. 28B. That is, such a mirroring configuration cannot withstand failures other than disk failures.

In normal mirroring, all data can be accessed as long as the disk storing the original data and the disk storing the duplicated data do not fail at the same time. In other words, in a mirroring system that is normally composed of N disks, it operates even if a maximum of N / 2 disks fails. However, in the example shown in FIG. 28A, there is a problem that inaccessible data is always generated when a failure occurs in any two disks.

Further, in the configuration of FIG. 28 (b),
Depending on the number of disks, the data arrangement is not always uniform, which causes a problem that the load on each disk during data transfer is not uniform.

As described above, in the conventional mirroring configuration in which the load is distributed at the time of data transfer, the data transfer disks are selected so that the data transfer amounts from the respective disks are equalized unless the disks fail. However, if a disk fails, the amount of data transferred to the physical disk paired with the failed disk increases. Therefore, there is a problem that the data transfer time of the disk becomes a bottleneck and the throughput of the entire system is reduced.

The present invention has been made to solve the above problems, and by preparing a plurality of disks and equalizing the amount of data stored in each disk and the amount of data transferred from each disk, A data storage device that can minimize the decrease in overall system throughput,
The purpose is to obtain a data management method and program.

Further, according to the present invention, the disks held in the system are grouped by grouping the disks which are less likely to cause a failure at the same time on the basis of the distance information between the disks, and writing the same data in each group to perform mirroring. An object of the present invention is to obtain a data storage device, a data management method, and a program that can improve the reliability of the.

A further object of the present invention is to obtain a data storage device, a data management method and a program for constructing a multiple redundant configuration of disks.

[0019]

A data storage device according to the present invention divides data into a plurality of data blocks of the same data amount, and disposes the same number of blocks in a plurality of storage devices in units of the data blocks. In addition, the data storage means stores the data blocks of the same content in duplicate for each of the predetermined number of storage devices, and the data stored in the plurality of storage devices in a distributed manner is one or more for each of the predetermined number of storage devices. A reading means for reading in data block units is provided.

The data storage device according to the present invention uses the distance information, which is an index of the interaction between the storage devices of the plurality of storage devices, to form a redundant group in which the storage devices having a small interaction are put together. Data in which data is distributed and stored in a plurality of storage devices, the data is divided into a plurality of data blocks of the same data amount, and data blocks of the same content are redundantly stored in each storage device in the redundancy group. The storage device includes a storage unit and a read unit that reads out data stored in a distributed manner in a plurality of storage devices in units of one or a plurality of data blocks for each redundancy group.

The data storage device according to the present invention uses the distance information, which is an index of the interaction between the storage devices of the plurality of storage devices, to form a redundant group in which the storage devices having a small interaction are put together. When data is distributed and stored in a plurality of storage devices, the data is divided into a plurality of data blocks of the same data amount, and the data blocks of the same content are redundantly stored in each storage device in the redundancy group. Is.

In the data storage device according to the present invention, the data storage means stores data blocks in a plurality of storage devices in a round robin manner.

In the data storage device according to the present invention, the data storage means uses hash distribution to determine the storage device in which the data block should be stored.

In the data storage device according to the present invention, the data storage means arranges a plurality of data blocks read at once from each storage device adjacently in each storage area.

The data storage device according to the present invention stores the data block group to be read from the storage device in a distributed manner when the data reading is impossible in any of the storage devices. The data is evenly read from a plurality of other storage devices.

A data storage device according to the present invention obtains distance information based on a hierarchical structure formed by a plurality of storage devices.

The data storage device according to the present invention obtains the distance information based on the position information in which each storage device is arranged.

The data storage device according to the present invention obtains distance information based on the failure rate determined according to the connection path between the storage devices.

The data storage device according to the present invention sets the multiplicity, which is the degree to which data blocks having the same content are overlapped, for each redundancy group.

The data storage device according to the present invention sets the reliability level for each redundancy group based on the distance information.

In the data management method according to the present invention, the data is divided into a plurality of data blocks of the same data amount, and the same number of blocks are distributed and arranged in the plurality of storage devices in units of the data blocks. A data storage step of redundantly storing data blocks of the same content for each storage device, and reading of data stored in a distributed manner in a plurality of storage devices in units of one or a plurality of data blocks for each predetermined number of storage devices And steps.

In the data management method according to the present invention, in the data storing step, the storage devices having a small interaction are put together by using the distance information which is an index of the interaction between the storage devices of the plurality of storage devices. When a redundant group is formed and data is distributed and stored in a plurality of storage devices, the data is divided into a plurality of data blocks of the same data amount, and data blocks having the same content are stored in each storage device in the redundancy group. It is stored in duplicate.

A program according to the present invention divides data into a plurality of data blocks of the same data amount, disperses and arranges the same number of blocks in a plurality of storage devices in units of the data blocks, and a predetermined number of storage devices. A computer as a data storage unit that stores data blocks of the same content in duplicate for each unit, and a read unit that reads out the data stored in a distributed manner in a plurality of storage units in units of one or a plurality of data blocks for each predetermined number of storage units Is to function.

The program according to the present invention uses the distance information, which is an index of the interaction between the storage devices of the plurality of storage devices, to configure a redundant group in which the storage devices having a small interaction are grouped and data is stored. The data blocks are divided into a plurality of data blocks of the same data amount, the same number of blocks are distributed and arranged in the plurality of storage devices in units of the data blocks, and the data blocks of the same content are duplicated in each storage device in the redundancy group. The computer is made to function as a data storage means for storing and a reading means for reading out data stored in a distributed manner in a plurality of storage devices in units of one or a plurality of data blocks for each redundancy group.

The program according to the present invention uses the distance information, which is an index of the interaction between the storage devices of the plurality of storage devices, to configure a redundant group in which the storage devices having a small interaction are put together, and data is stored. A data storage unit that divides the data into a plurality of data blocks of the same data amount and stores the data blocks of the same content in duplicate in each storage device in the redundant group when the data blocks are distributed and stored in the plurality of storage devices. It makes a computer function as.

[0036]

BEST MODE FOR CARRYING OUT THE INVENTION An embodiment of the present invention will be described below. Embodiment 1. 1 is a diagram showing a configuration example of a data storage device according to a first embodiment of the present invention. In the figure, 1
Are connected to the n node machines 3-1 to 3-1 via the network 2.
A host machine (data storage device) connected to 3-n, which is a computer device that executes a program that realizes data storage according to the first embodiment. Reference numeral 2 denotes a network connecting the host machine 1 and the node machine, and a communication infrastructure such as the Internet or an intranet can be considered. 3-1 to 3-n are node machines, which are computer devices having disks as secondary storage devices. The host machine 1, the network 2, and the node machines 3-1 to 3-n form one system as a whole. Here, the medium for connecting the host machine 1 and the node machines 3-1 to 3-n is not limited to the network 2, and may be connected using a bus to form a parallel system. Further, as the node machine 3, I
An I / O control device or SCSI may be used.

FIG. 2 is a diagram showing the configuration of the host machine shown in FIG. In the figure, reference numeral 4 denotes a main storage device, which includes an operating system 5, a data transfer control unit (reading unit) 6, and a file block management unit (data storage unit).
Software for implementing functions such as 7, a communication control unit 8 and a node machine management unit (data storage unit) 9 is stored. The operating system 5 stored in the main storage device 4 controls the entire hardware of the host machine 1. Reference numeral 10 represents a central processing unit (data storage means,
The reading means) executes each software in the main storage device 4. A network adapter 11 controls connection of the host machine 1 to the network 2.
Reference numeral 12a is a system bus that connects the main storage device 4 and the central processing unit 10, and 12b is a network adapter 1
1 and the disk controller 14, and a PCI bus for connecting other devices. A bus coupler 13 couples the system bus 12a and the PCI bus 12b. A disk controller 14 controls the disk devices (storage devices) 15 and 16.

Here, the data transfer control unit 6, the file block management unit 7, the communication control unit 8, and the node machine management unit 9 have been described as an example realized by software, but they may be firmware. You may comprise as the hardware which implement | achieves the said function. Also, a combination of software, firmware and hardware may be used.

FIG. 3 is a diagram showing the relationship of each software of the data storage device according to the first embodiment of the present invention. In the figure, 7a is a file block management table stored in the main storage device 4, and stores information for specifying a read block which is a file block transferred from each disk at the time of data transfer. A node machine management table 9a stores information related to management of node machines.
The node machines 3-1 to 3-n execute software that realizes the functions of the operating system 17, the communication control unit 18, and the disk access control unit 19, and
A plurality of disk devices (storage devices) 20 as the next storage device
Shall have.

FIG. 4 is a diagram showing the structure of the node machine management table in FIG. In the figure, reference numeral 21 is node machine management information, which is one of the pieces of information constituting the node machine management table 9a, and the node machine number 21a and IP address 21 are used as information for identifying each node machine.
b and port number 21c are set. The node machine management unit 9 acquires the disk information 22 that identifies the disk connected to each node machine based on this node machine management information 21, and sets the disk number 22a and node machine number 22b in the disk information 22. . Reference numeral 22 is disk information for specifying a disk connected to the node machine, which is composed of a disk number 22a for specifying the disk, a node machine number for specifying each node machine and a redundancy group number 22c for specifying the redundancy group. Reference numeral 23 is layer information, which is lower layer information 23a that identifies the layer relationship of the disks connected to each node machine.
And upper layer information 23b. Reference numeral 24 is distance information, which is an index indicating the possibility of inaccessibility between two node machines at the same time. That is, the distance information 24 is
It can be said that this is an index showing the interaction between the disks.

Next, an example of setting the hierarchy information 23 at the time of initial setting will be described. FIG. 5 is a diagram showing an example of a hierarchical relationship of disks constituting the data storage device according to the first embodiment. In the figure, # D1 to # D8 are disks constituting a data storage device. Also, reference numerals 25-1 and 25-2
Disks surrounded by boxes 5-2, 26-1, and 26-2 represent disks connected to each node machine. Further, the disks enclosed by the frames 25 and 26 are connected to the node machines existing on the same floor. Further, 27 represents the entire system that constitutes the data storage device according to the first embodiment. In this way, what represents the entire system is called a root. A frame for classifying disks, such as a node machine or a floor, is called a frame. That is, in the hierarchical relationship of the disks, node machine frame 25-1, node machine frame 25-
2. Use the floor frame 25.

The disks belonging to the node machine frame 25-1 are # D1 and # D2, and the disks belonging to the floor frame 25 are # D1 to # D4. The disks belonging to the route 27 are all disks in the system and are # D1 to # D8. That is, there is a hierarchical relationship between such frames. In the example of FIG. 5, the disks belonging to the node machine frame 25-1 are # D1 and # D2, and the disks belonging to the floor frame 25 are the node machine frames 25-1 and 25-.
It is a disc belonging to 2. The discs belonging to the route 27 are also discs belonging to the floor frames 25 and 26.

More specifically, with respect to the disk # D1, the upper layer is the node machine frame 25-1 to which the disk # D1 belongs.
The upper layer of -1 is the floor frame 25 to which the node machine frame 25-1 belongs, and the upper layer of the floor frame 25 is the root 27.

In the example of FIG. 4, the lower layer information 23a is the disk # D1 and the upper layer information 23b is the node machine frame 25-1. Similarly, if the disk # D2 is the lower layer information 23a, its upper layer information 23b.
Becomes the node machine frame 25-1. Hereinafter, all the other disks are set as the lower layer information 23a and the upper layer information 23b is set.

When the node machine frame 25-1 is the lower layer information 23a, the upper layer information 23b is the floor frame 25. Similarly, when the node machine frame 25-2 is the lower layer information 23a, the upper layer information 23b is the floor frame 25. Hereinafter, all other node machine frames are used as lower layer information, and the upper layer information 23b is set.

Further, when the floor frames 25 and 26 are the lower layer information 23a, the upper layer information 23b is the route 27.

The node machine management unit 9 sets the above hierarchical relationship as the hierarchy information 23 in the node machine management table 9a. By using this layer information 23, the layer relationship from each disk to the root is determined.

Here, in the data storage device of the present invention, the frame for classifying the disks may be set freely. In FIG. 5, the frame of the node machine and the floor are set, but a larger frame such as a region or a smaller frame such as a bus may be added arbitrarily. That is, the hierarchy information is 1
It is possible to arbitrarily set from the hierarchy to the N hierarchy.

FIG. 6 is a diagram showing another example of the hierarchical relationship of the disks constituting the data storage device according to the first embodiment.
As shown in FIG. 6, there may be disks and node machines belonging to a plurality of frames in the same layer. That is, the node machine frame 26-1 is the floor frame 2
5 and 26, and if the node machine frame 26-1 is the lower layer information 23a, both the floor frames 25 and 26 are set in the upper layer information 23b.

Generally, disks belonging to the same frame in a lower hierarchy are likely to be inaccessible at the same time when a system failure occurs. For example, as shown in FIG. 5, when a disaster occurs on the floor corresponding to the floor frame 25, it becomes impossible to access the disks of the disks # D1, # D2, # D3 and # D4. Also, node machine frame 2
Even when the node machine corresponding to 5-1 fails, access to the disks # D1 and # D2 becomes impossible.

That is, the disks # D1 and # D2 classified into a lower hierarchy like the node machine frame 25-1.
Is disabled at the same time even if a failure occurs in a higher-level frame such as the floor frame 25 to which the node machine frame 25-1 belongs. On the contrary, disk #D
With a combination of 1 and disk # D8, it is unlikely that two disks will be inaccessible at the same time.

The host machine 1 is located at the root of the hierarchical structure.
By placing, the hierarchical structure viewed from the host machine 1 can be constructed. For example, consider a case where the host machine 1 is installed on the floor corresponding to the floor frame 25 in the system having the hierarchical structure shown in FIG.

FIG. 7 is a diagram showing the hierarchical structure of the system viewed from the host machine arranged as described above. In the figure, # HD1 and # HD2 are disks connected to the host machine 1 and belong to the host machine frame 29.
Reference numeral 28 is a host machine hierarchy corresponding to the highest hierarchy of the hierarchical structure, and 29 is a host machine frame indicating that the disk is connected to the host machine 1.

As shown in FIG. 7, when a hierarchical structure is built around the host machine 1 installed on the floor corresponding to the floor frame 25, the node machine frames 25-1,
The frame 25-2 and the floor frame 26 are classified as frames in the same layer with the host machine layer 28 as an upper layer. In this way, by considering the hierarchical structure of the system, it is possible to build the hierarchical structure of the system centered on the system installed at the root of the hierarchical structure.

Next, an example of setting the distance information 24 at the time of initial setting will be described. Here, the distance information is set to have a small value for disks that are likely to be inaccessible at the same time and a large value for disks that are less likely to be inaccessible at the same time.

A procedure for setting the distance information 24 using the layer information 23 will be described. When the layer information 23 is used, if the two discs belong to the same frame in the lower layer, the distance information 24 becomes small. Conversely, if the lower layer does not belong to the same frame but the higher layer belongs to the same frame, the distance information 24 becomes large.

That is, from the relationship between the layer information 23 and the distance information 24, two discs having a small distance information 24 between discs are likely to be inaccessible at the same time, and conversely, two discs having a large distance information between discs. The disk can be determined to be unlikely to be inaccessible at the same time.

Here, when the host machine 1 is placed at the root as shown in FIG. 7 when the layer information 23 is set, the distance information 24 between the disks is determined from the relationship between the layer information 23 and the distance information 24. It can be determined that the two small disks are highly inaccessible at the same time when viewed from the host machine 1. On the other hand, conversely, two disks with large distance information 24 between the disks are
From the viewpoint, it is judged that there is a low possibility that access will be disabled at the same time. Also, a disk connected to the host machine 1,
Distance information for disks installed outside the host machine 2
4 takes the maximum value in the total distance information. Therefore, from the viewpoint of the host machine 1, the possibility of being inaccessible at the same time is the lowest.

Next, the operation will be described. FIG. 8 is a flow chart showing the operation of the data storage device according to the first embodiment. Along with this flow chart, the distance information 24 is obtained by using the hierarchy information 23.
The operation of setting will be described. Further, as a specific example, a case of setting the distance information 24 between the disc # D1 and the disc # D3 in FIG. 5 will also be described.

First, the node machine management unit 9 checks the hierarchy information 23 of the entire system and acquires the maximum number of layers of the system (step ST1). Here, the maximum number of layers is the number of layers up to the root of the disk having the largest number of layers up to the root. This value is used as the maximum value of the distance information 24. In the example of FIG. 5, disks # D1 to # D8
The number of layers from the root to the root is all four.

Next, the node machine management unit 9 judges whether or not the setting of the distance information 24 has been completed for all combinations of disks (step ST2). At this time, if the setting is not completed for all the combinations of the disks, the node machine management unit 9 selects two disks for which the setting of the distance information 24 between the disks is not completed, and the distance information is selected. The maximum value (the number of layers in the hierarchy) is set to the value of 24 (step ST3). On the other hand, if the setting is completed for all the combinations of disks, the node machine management unit 9 ends the process. In the example of FIG. 5, since the setting of the distance information 24 between the disks # D1 and # D3 has not been completed, the maximum number of tiers 4 is set as the value.

For the two disks for which the maximum value (the number of layers of the hierarchy) is set in the distance information 24 in step ST3, the node machine management section 9 checks the hierarchy information 23 from each disk to the root (step ST4). In the example of FIG. 5, the node machine management unit 9 causes the disk # D1 to
To root, and layer information from disk # D3 to root is obtained. This allows disk # D1
As the hierarchical information, the disk # D1 → node machine frame 25-1 → floor frame 25 → root is obtained. Further, as the layer information of the disk # D3, the disk # D3 → node machine frame 25-2 → floor frame 25 → route is obtained.

Next, the hierarchy information 2 checked in step ST4
3, the node machine management unit 9 traces the layer information 23 from the root toward the disk, and determines whether or not each layer information 23 of the two selected disks has a common part (step ST5). . More specifically, the node machine management unit 9 first compares the layer information 23 of the highest layer of the two selected disks. Here, when the layer information 23 is different, the setting of the distance information 24 of the two discs is ended, and the process returns to step ST2. In the example of FIG. 5, first, the highest level information among the layer information 23 of the disks # D1 and # D3 is compared.

On the other hand, when the tier information 23 is the same, the node machine management section 9 determines the distance information 24 of the two disks.
Is decreased by 1 (step ST6). In the example of FIG. 5, the highest information among the layer information 23 is the disk #D.
It is determined that both 1 and # D3 are equal in the route. As a result, the value of the distance information 24 is subtracted from 4 by 1 to become 3.

Subsequently, the node machine management unit 9 further acquires each layer information 23 of the two disks while tracing the layer information 23 from the root toward the disk (step ST7), and returns to the processing of step ST5. It is determined whether or not each layer information 23 has a common part. Figure 5
In the above example, the layer information 23 of the disks # D1 and # D3 is 1
When they are traced in the disc direction and compared again, these are common to the portions belonging to the floor frame 25. For this reason,
The value of the distance information 24 of the disks # D1 and # D3 is 2.

Further, the node machine management unit 9 repeats the above-mentioned operation until the common part in the hierarchy information 23 of the two disks disappears, and repeats the procedure of lowering the hierarchy information 23 in the disk direction. To set. In the example of FIG. 5, if one layer information is traced in the disk direction and recomparison is performed, the layer information 23 of the disks # D1 and # D3 differs between the node machines 25-1 and 25-2. As a result, the value of the distance information 24 between the disc # D1 and the disc # D3 is set to 2.

By applying the above operation to all the disks, the distance information 24 is obtained based on the layer information 23.
Can be set.

Further, the disks # D5 and # D6 shown in FIG.
Consider the case where the distance information 24 of the disk existing over a plurality of frames is set as described above. in this case,
In step ST4 of FIG. 8, when obtaining the layer information 23 from the disks # D5, # D6 to the root, the layer information 23 of all combinations is obtained. That is, the layer information 23 from the disk # D5 to the root shown in FIG.
5 → node machine frame 26-1 → floor frame 2
5 → root and disk # D5 → node machine frame 26-1 → floor frame 26 → root.

Here, when setting the distance information 24 between the disk # D5 and another disk, the value of the distance information 24 is increased by using the above-mentioned two types of hierarchical information from the disk # D5 to the root. Use one. For example, when the distance information 24 between the disc # D1 and the disc # D5 is obtained, the value of the distance information 24 becomes 2 when the layer information 23 of the disc # D5 that passes through the floor frame 25 is used. On the other hand, if the one passing through the floor frame 26 is used, the value of the distance information 24 becomes 3. Therefore, the value of the distance information 24 between the disc # D1 and the disc # D5 is set to 3.

Next, another method of setting the distance information 24 will be described. Here, without using the hierarchy information 23,
A method of calculating the operating rate based on the disk failure rate and setting it as the distance information 24 will be described. FIG. 9 is a diagram showing the configuration of the data storage device according to the first embodiment and the failure rate in the hierarchical relationship of the disks. In the figure, 3
0-1 and 30-2 are node machines, which are routers 31
-1 is connected to the host machine 32, and the node machines 30-3 and 30-4 are connected to the host machine 32 via the router 31-2. Here, the host machine 32 is assumed to have the same configuration as the host machine 1 in FIG. The disks # D1 and # D2 are connected to the node machine 30-1. B1 is disk #
The failure rate between D1 and the node machine 30-1 is shown, and B2 shows the failure rate between the disk # D2 and the node machine 30-1. Similarly, disks # D3 to #
The failure rates B3 to B8 with the node machine connected to D8 are also set.

B9 is the failure rate between the node machine 30-1 and the router 31-1. Similarly, failure rates B9 to B12 with the routers connected to the node machines 30-2, 30-3, and 30-4 are also set. Further, B13 is a router 31-1 and a host machine 3
2 is the failure rate, and B14 is the failure rate between the router 31-2 and the host machine 32.

The node machine management unit 9 uses the host machine 3
The operating rate is calculated assuming that the access path from 2 to either of the two disks is operating, and the operating rate is set as the distance information 24 of the two disks. Here, the operating rate is defined by operating rate = (1−failure rate). Therefore, the disk # D1 and the node machine 30-
The operating rate between 1 and 1 is (1-B1). Further, the operation rate when there is only one route from the router 31-1 to the disk # D1 is (1-B1) (1-B9). Further, the probability that the disks # D1 and # D2 will simultaneously fail is B1 * B2. Therefore, the node machine 30
The operating rate when the access path from -1 to either of the disks # D1 and # D2 is operating is (1-B1 * B
2).

Next, a case where the distance information 24 of the disks # D1 and # D5 shown in FIG. 9 is set will be described. At this time, the operating rates from the host machine 32 to the disk # D1 are (1-B1) (1-B9) (1-B13).
On the other hand, the operating rates from the host machine 32 to the disk # D5 are (1-B5) (1-B11) (1-B14).

Here, from the host machine 32, the disk #
If the operation rate in the state where the access route to either D1 or # D5 is established is calculated, {1- {1- (1-B
1) (1-B9) (1-B13)} {1- (1-B5)
(1-B11) (1-B14)}}. This value is set as the distance information 24 between the disks # D1 and # D5.

Further, as the next example, the disks # D1, #
A case of setting the distance information 24 of D3 will be described.
At this time, the disks # D1 and # D3 take different routes to the router 31-1, and a common route from the router 31-1 to the host machine 32. In this example, first, as viewed from the router 31-1, the disk # D1,
The operation rate in the state where the access route to either # D3 is established is calculated. Referring to FIG. 9, this operating rate is {1- {1- (1-B1) (1-B9)} {1-
(1-B3) (1-B10)}}.

Using this value and the operating rate from the router 31-1 to the host machine 32, the operating rate in the state where the access path from the host machine 32 to either of the disks # D1 and # D3 is established is obtained. And {1- {1- (1-
B1) (1-B9)} {1- (1-B3) (1-B1
0)}} (1-B13). This value is disk #D
It is set as the distance information 24 of 1 and # D3.

Similarly, the distance information 24 can be set by obtaining the failure rates of all the disks.

FIG. 10 is a diagram showing another configuration of the data storage device according to the first embodiment and the failure rate in the hierarchical relationship of the disks. As shown in FIG. 10, some systems may have a plurality of paths from the disk to the host machine. Here, in FIG. 10, the route between the node machine 30-3 and the router 31-1 is added to the configuration shown in FIG. 9, and the failure rate is B15. Also, the node machine 3
The route between 0-2 and router 31-2 is also added, and the failure rate B
Sixteen. Even in such a case, the operating rate from the host machine 32 to the two disks can be calculated, and the operating rate can be set as the distance information 24 of the two disks.

Next, the creation of a redundant group at the time of initial setting will be described. Here, the redundant group is a set of disks in which the same data is stored, and is a combination of disks that are unlikely to be inaccessible at the same time. First, as an example of creating a redundant group, a method of combining disks based on the distance information 24 will be described.

In this case, the node machine management unit 9 determines the number of disks in each redundancy group according to the data multiplicity. That is, in a system having an M (M; integer of 2 or more) multiplex data redundancy configuration, M disks are combined as one redundancy group.

More specifically, when the node machine management unit 9 creates a certain redundant group combination, the distance information 24 between the disks in the redundant group is created.
Find the maximum value of. This value is also obtained in each redundancy group that is arbitrarily combined, and the value that minimizes the distance information 24 is obtained. Then, the node machine management unit 9 obtains a value that minimizes the distance information 24 for all combinations of redundant groups, and selects a combination of redundant groups having the largest value among these.

For example, consider a case where a combination of redundancy groups with a multiplicity of 2 is created with the configuration shown in FIG. In this case, the node machine management unit 9 sends {# D1, # D5},
{# D2, # D6}, {# D3, # D7}, {# D4
When a combination of redundant groups such as # D8} is selected, the maximum value of the distance information 24 is 3 in each redundant group. Therefore, the minimum value among them is 3. In the system having the structure shown in FIG. 5, this minimum value is the maximum value among all the combinations of redundancy groups. Therefore,
The above combination is selected as the combination of the redundancy groups in the multiplicity 2 of this system.

Similarly, by increasing the number of disks in each redundancy group to M, it is possible to construct an M multiplex redundant configuration.

Further, by providing at least one combination of disks having the distance information 24 of N or more in all the redundancy groups, a system which operates even when a failure occurs in the N-1 or lower hierarchy is constructed. be able to.

Furthermore, the data multiplicity may be changed for each redundancy group. That is, it is possible to have a redundant configuration in which the data multiplicity of a certain redundancy group is M-multiplexing and the data multiplicity of another redundancy group is N-multiplexing.

In the above, the example in which the distance information 24 obtained from the hierarchy information 23 is used when creating the redundant group has been shown. However, the distance between the disks obtained based on the physical distance between the disks. May be used to find the combination of redundancy groups.

Here, a procedure for creating a redundancy group using the distance between disks obtained based on the physical distance will be described. The node machine management unit 9 has 1
When a combination of two redundancy groups is created, a value that maximizes the physical distance between the disks in the redundancy group is calculated. This value is also obtained in each redundant group that is arbitrarily combined, and the minimum value is obtained.
Then, the node machine management unit 9 obtains this minimum value for all combinations of redundancy groups, and selects the combination of redundancy groups having the largest value among these.

FIG. 11 is a diagram showing the physical positional relationship of the disks installed on the same floor. For example, the disc # D1 and the disc # D2 are discs installed at positions close to each other on the floor, and the disc # D3 and the disc # D7.
Indicates that the disk is located far away on the floor. The node machine management unit 9 is, for example, as shown in FIG.
For a set of disks arranged on a plane as shown in (3), a redundancy group is created by using the distance between the disks obtained based on the physical position information between the disks.

Consider a case where a redundant group of multiplex 2 is created under the conditions shown in FIG. In this case, the node machine management unit 9 causes the {# D1, # D8}, {# D2, # D7},
When a combination of redundant groups such as {# D3, # D5}, {# D4, # D6} is selected, the maximum distance value is obtained in each redundant group. Next, the node machine management unit 9
Is the value of the distance between the disks # D4 and # D6 when the minimum value among them is calculated. This value is the maximum value among the values obtained for the combination of all redundancy groups. For this reason, the created redundancy group combination has a multiplicity of 2 in the system having the configuration shown in FIG.
Will be selected as a redundancy group in.

Similarly, by increasing the number of disks in each redundancy group to M, it is possible to construct an M-multiplexed data redundancy configuration.

Further, in the example of FIG. 11, it is possible to change the data multiplicity for each redundancy group. That is, it is possible to have a redundant configuration in which the data multiplicity of a certain redundancy group is M-multiplexing and the data multiplicity of another redundancy group is N-multiplexing.

Next, the data writing operation to each disk of the data storage device according to the first embodiment will be described.
FIG. 12 is a flowchart showing the data write operation of the data storage device according to the first embodiment, and an example of the write operation will be described with reference to this figure. A data write command from the central processing unit 10 is issued from the upper application to the file block management unit 7 (step ST1a). As a result, the file block management unit 7
Inquires of the node machine management unit 9 to acquire the redundancy group number 22c (step ST2a).

Next, the file block management section 7 divides the write data into blocks (step ST3).
a). Subsequently, the file block management unit 7 determines whether or not writing of all data blocks in the redundancy group has been completed (step ST4a). At this time, if the writing of all the data blocks is completed, the data writing operation is ended. On the other hand, if writing of all data blocks has not been completed, the process proceeds to step ST5a.

In step ST5a, the file block management section 7 refers to the redundancy group number 22c,
The redundancy group to be written is selected so that the blocks are evenly arranged in the redundancy group. After that, the file block management unit 7 sequentially writes the same block to the disks in the same redundancy group from the head of the disk and executes the same for all the blocks (step S
T6a).

Here, in the actual data writing process to the node machines 3-1 to 3-n, the file block management unit 7 uses the communication control unit 8, the operating system 5, the network 2, the operating system 17, and the communication control unit 18. It is executed by issuing a disk write command to the disk access control unit 19 via the.

FIG. 13 is a diagram showing a storage diagram of blocks written in each disk in the write flow shown in FIG. In the figure, reference numerals 33-1, 33-2, 33-3,
Disks surrounded by a frame 33-4 represent the same redundancy group. As shown in FIG. 13, data is stored in the disks in each redundancy group in an equal amount of data.

Further, although FIG. 13 shows an example in which the data multiplicity is 2, by setting the number of disks in the redundant group to M, it is possible to construct an M-multiplexed redundant configuration.

Note that hash distribution may be used as the method of determining the group in which the data block is written, or round robin may be used.

Next, read block number 7 of each disk
The operation of setting a-2 will be described. FIG. 14 shows FIG.
It is a figure which shows the internal structure of the file block management table inside. In the figure, 7a-1 is a disc number for identifying each disc. 7a-2 is a read block number, which identifies a read block which is a file block transferred from each disk at the time of data transfer.

The file block management unit 7 updates the file block management table 7a when writing data. Here, the file block management unit 7 predetermines a file block to be transferred from each disk as a read block at the time of data transfer, and holds information identifying this as a read block number 7a-2 in the file block management table 7a. To do.

FIG. 15 is a diagram showing a read block number set at the time of writing data by the data storage device of the first embodiment. As shown in FIG. 15, the file block management unit 7 includes redundancy groups 33-1, 33-2, 33-.
The same redundant groups 33-1, 33-2, 33-3, 33- are used to transfer blocks from 3, 33-4.
The read block number 7a-2 is determined so that duplicated blocks are not accessed in each disk belonging to No. 4. For example, for the disk # D1 belonging to the redundancy group 33-1, the file block management unit 7 sets the code 3
Information for identifying a data block surrounded by a frame with 4-1 is set as a read block number 7a-2. At this time, for the disk # D5, the information specifying the data block enclosed by the frame with the reference numeral 34-2, which does not overlap with the disk # D1, is the read block number 7a-2.
Is specified as. Hereinafter, other redundant groups 33-2 to
The same applies to 33-4.

For a redundant group having an M multiplex structure, assuming that the total number of blocks of a disk in the redundant group is N, each disk sets N / M consecutive blocks as a read block.

Next, the operation during data transfer will be described. FIG. 16 is a flowchart showing the data transfer operation of the data storage device according to the first embodiment. First, the upper application issues a data transfer instruction to the data transfer control unit 6 (step ST1b). As a result, the data transfer control unit 6 passes the transfer data information to the file block management unit 7.

The file block management section 7 refers to the file block management table 7a to search for a disk whose transfer block is the read block number 7a-2 (step ST2b). After that, the file block management unit 7 determines the disk number 7a of the corresponding disk.
-1 and the read block number 7a-2 are returned to the data transfer control unit 6.

Next, the data transfer control unit 6 issues a block transfer instruction to the disk access control unit 19 based on the disk number 7a-1 and the read block number 7a-2 returned from the file block management unit 7. Yes (step ST3b).

The operation of the data storage device according to the first embodiment when a disk fails will be described. When a disk failure is detected, the disk number of the failed disk is passed to the node machine management unit 9. Subsequently, the node machine management unit 9 sends the disk number 22a of the failed disk and the redundancy group number 22c to which the failed disk belongs to the file block management unit 7. As a result, the file block management unit 7 uses the redundancy group number 22c received from the node machine management unit 9 as a basis, and reads block numbers 7a of the disks belonging to the redundancy group.
Rewrite -2.

The rewriting operation of the read block number by the file block management unit 7 will be described in detail below. FIG. 17 is a diagram showing the read block numbers set by the data storage device according to the first embodiment when a disk fails. The example of FIG. 17 indicates that the disks # D7 and # D8 have failed. First, since the disk # D7 has failed, the file block management unit 7
The read block number 7a-2 in the disk # D2 belonging to the same redundant group as 7 is rewritten. In performing this rewriting, the file block management unit 7
Sets the read block number 7a-2 so that the load can be evenly applied to the disks operable in the redundancy group. That is, in the redundant group 33-2, the only operable disk is the disk # D2. Therefore, in the disk # D2, all data in the redundancy group is set as the read block number 7a-2. Then disk #
Since D7 is an inoperable disc, the read block number 7a-2 is set to empty. Similarly, for the disks # D3 and # D8 belonging to the redundancy group 33-3,
The read block number 7a-2 is rewritten.

Next, an example of operation at the time of adding disks will be described. At the time of adding a disk, the node machine management unit 9 performs information registration work of the added disk. At this time, the node machine management unit 9 uses the node machine management table 9a.
A setting field relating to the additional disk is added to the disk information 22 of No. 2, and the disk number 22a is allocated and the node machine number 22b is set. Furthermore, classify the additional disks into one of the redundancy groups, and set the redundancy group number 2
Set 2c. Subsequently, the node machine management unit 9 sets the distance information 24 between the additional disk and another disk.

When the information registration work for the additional disk is completed, the node machine management section 9 transfers the redundant group number 22c of the additional disk to the file block management section 9. The file block management unit 9 rewrites the read block number 7a-2 so that the disk load of the redundant group to which the additional disk belongs is equalized.

As described above, according to the first embodiment, data is divided into a plurality of data blocks of the same data amount, and the same number of blocks are distributed and arranged in the plurality of disk devices in units of the data blocks. At the same time, data blocks of the same content are stored redundantly for each predetermined number of disk devices (redundancy groups), and the data stored by being distributed to a plurality of disk devices is stored in units of one or more data blocks for each redundancy group. Since it is read, the data is stored so that the data amount of each disk device becomes equal, and the data transfer amount from each disk device at the time of data transfer becomes equal, so that the decrease in the throughput of the entire system is minimized. Can be suppressed to Also, by storing each data block sequentially for each redundancy group, no special calculation is required when deciding from which disk in the redundancy group the data will be transferred when performing the data transfer process. It is possible to operate at high speed in the process of sequentially transferring all the data from the head of the disc.

Further, according to the first embodiment, the distance information between the disks is defined from the hierarchical relation of the system, and the disks are grouped based on the distance information.
It is possible to cope with a failure in a specific hierarchy of the system.

Furthermore, according to the first embodiment, it is possible to construct a redundant configuration centering on a specific part of the system.

Further, according to the first embodiment, since the redundancy group of the disks is created and the data multiplicity is set in the redundancy group unit, the data multiplicity is changed for the specific disk and the specific redundancy group. Can be appropriately performed.

Furthermore, according to the first embodiment, by adding a disk in the redundancy group and increasing the data multiplicity, the system reliability can be improved and the data transfer speed can be increased. be able to.

Further, according to the first embodiment, in the parallel system, the throughput of the parallel system as a whole can be improved by adding a new disk to the redundant group having the disk having the poor data transfer performance. .

Embodiment 2. The second embodiment differs from the first embodiment in the procedure of creating a redundant group and the procedure of writing a data block to each redundant group. For example, when the data multiplicity is M, the number of disks in which one block is written is M, and duplicated blocks of a certain disk are not all held by the same M disks, but 2 * (M-1). It is distributed and stored in the book disk.

The basic hardware configuration of the data storage device according to the second embodiment is the same as that of the first embodiment. Further, the redundant group is also created by combining disks having large distance information in the same manner as in the first embodiment. Here, the difference from the first embodiment is that
There is a redundancy group for each disk. That is, in the second embodiment, when N disks exist, N redundancy groups exist. In the configuration of the redundant group, the disk in which the original data is written is called the original disk, and the disk in which the copy of the original disk is written is called the duplicate disk.

Next, the operation will be described. FIG. 18 is a flow chart showing the operation of the data storage device according to the second embodiment, and the redundant group creating operation with the data multiplicity of 2 will be described using this example and the example of the hierarchical relationship of the disks shown in FIG. First, the node machine management unit 9 determines whether or not the allocation of redundant groups has been completed for all disks (step ST1c). At this time, if the allocation of the redundant group to all the disks has been completed, the processing ends. On the other hand, if the assignment of the redundant groups to all the disks has not been completed, the node machine management unit 9 selects an arbitrary disk for which the assignment of the redundant groups has not been completed as one original disk (step ST2c). Here, in the example of FIG. 5, it is assumed that the disk # D1 is selected.

Next, in the case of the data multiplicity M, the node machine management unit 9 sets the redundant groups of the original disks until the sum of the numbers of the original disks and the duplicate disks becomes 2 * (M-1) +1. create. Specifically, the node machine management unit 9
Always creates a redundant group of original disks, and always determines whether the number of duplicate disks is 2 * (M-1) (step ST3c). At this time, if the number of duplicate disks is 2 * (M-1), the process returns to step ST1c and the above-described operation is repeated. On the other hand, if the number of duplicate disks is not 2 * (M-1), the process proceeds to step ST4c. In the example of FIG. 5, since the data multiplicity is 2, the node machine management unit 9 creates the redundant group of the disk # D1 until the sum of the original disk and the duplicate disk becomes three.

At step ST4c, the node machine management section 9 selects the disk having the largest distance information from the original disk. In the example of FIG. 5, the disks having the largest distance information with the disk # D1 are the disks # D5 to # D8.

Here, the node machine management unit 9 determines whether or not there are a plurality of disks selected as disks having the largest distance information from the original disk (step S).
T5c). At this time, if the number of selected disks is one, the node machine management unit 9 proceeds to step ST15c to set the disk as a redundant group of the original disk, and also sets the original disk in the redundant group of the selected disk. to add.

On the other hand, when there are a plurality of selected disks, the node machine management section 9 reselects the disk having the largest distance information from the duplicate disk of the original disk (step ST6c). In the example of FIG. 5, the current disk # D1
Since there is no duplicate disk, the disk remains unchanged after being reselected in step ST6c.
# D8 is selected.

Subsequently, the node machine management section 9 determines whether or not there are a plurality of disks selected again in step ST6c (step ST7c). At this time, if the number of selected disks is one, the node machine management unit 9 proceeds to the process of step ST15c. On the other hand, when there are a plurality of selected disks, the node machine management unit 9 selects the disk with the smallest number of allocated duplicate disks among the selected disks (step ST8c). In the example of FIG. 5, since the duplicate disks are not assigned to the disks # D5 to # D8, the number of assigned duplicate disks is 0, and the disks # D5 to # D8 are selected again.

Thereafter, the node machine management section 9 determines whether or not there are a plurality of disks selected in step ST8c (step ST9c). At this time, if the number of selected disks is one, the node machine management unit 9
Advances to the process of step ST15c. On the other hand, when there are a plurality of selected disks, the node machine management unit 9
Selects the upper layer having the smallest allocation of duplicate disks in the upper layer of the selected disks, and selects the disks included in the upper layer (step ST10c).
In the example of FIG. 5, the number of duplicate disks allocated to the node machine frame 26-1 and the node machine frame 26-2, which are the upper layers of the disks # D5 to # D8, is compared.

Here, the number of duplicate disks of the node machine frame 26-1 is the number of duplicate disks assigned to the disks # D5 and # D6. The number of duplicate disks of the node machine frame 26-2 is disk #D.
This is the number of duplicate disks allocated to # 7 and # D8.
In the example of FIG. 5, the duplicate disks are not assigned to the disks # D5 to # D8. Therefore, the number of duplicate disks assigned to the node machine frames 26-1 and 26-2 is 0. Therefore, step ST1
Also in 0c, the disks # D5 to # D8 are selected.

Subsequently, the node machine management section 9 determines whether or not there are a plurality of disks selected in step ST10c (step ST11c). At this time, if the number of selected disks is one, the node machine management unit 9 proceeds to the process of step ST15c.

On the other hand, when there are a plurality of selected disks, the node machine management unit 9 raises the hierarchy by which the disk should be selected by one and judges whether or not there is an upper hierarchy by which the disk should be selected ( Step ST12c). At this time, if there is an upper layer for which a disk should be selected, the process returns to step ST10c and the node machine management unit 9
Compares the allocated number of duplicate disks in the upper layer. In the example of FIG. 5, the upper layer of the node machine frames 26-1 and 26-2 is the floor frame 2.
Equal to six. Therefore, the number of duplicated disks allocated in the upper layer is equal, and the disk # D5
~ # D8 is selected.

After that, if it is determined in step ST11c that a plurality of disks are present, the node machine management unit 9 proceeds to step ST12c to further raise the upper layer by one, and allocates a duplicate disk in the upper layer. Compare the number of items. This operation is repeated until it is determined in step ST12c that there is no upper layer.

If it is determined in step ST12c that there is no upper layer from which a disk should be selected, the node machine management section 9 proceeds to step ST13c and determines whether there are a plurality of disks selected in step ST10c. Determine whether or not. At this time, if the number of selected disks is one, the node machine management unit 9 proceeds to step ST.
It progresses to the processing of 15c. On the other hand, when there are a plurality of selected disks, the node machine management unit 9 selects an arbitrary disk from the plurality of selected disks (step ST14c). After this, the process proceeds to step ST15c.

In the example of FIG. 5, the number of duplicate disks is the same up to the highest layer. Therefore, step ST14c
At, any disc is selected from the discs # D5 to # D8. Here, when the disk # D5 is selected, the disk # D5 is set as the redundant group of the disk # D1 in step ST15c, and accordingly, the disk # D5 is set as the redundant group of the disk # D5.
1 is set.

FIG. 19 is a diagram showing a correspondence table of redundant groups created for the hierarchical relationships in FIG. Here, the redundancy group corresponding to the data multiplicity of 2 is the one shown in FIG.
It is created by the flow of.

Next, the operation of setting a redundancy group in a system having a disk redundancy configuration in which the data multiplicity is M (M; an integer of 2 or more) will be described. First, according to the flow of FIG. 18 described above, a redundant group having a disk redundancy configuration with a multiplicity of 2 is set for all the original disks. Next, while increasing the data multiplicity, the number of redundant groups is increased for all the original disks.

For example, in the case of a disk redundancy configuration with a data multiplicity of 3, the redundancy group of disk # D1 is a redundancy group of disks # D5 and # D7 which are members of the redundancy group of disk # D1 with a data multiplicity of 2. Is newly added as a redundant group of the disk # D1.

That is, in the data multiplicity 2, the redundant group of the disk # D5 is the disks # D1 and # D4. The redundant group of the disk # D7 is the disks # D1 and # D3. Therefore, in the system having the data multiplicity of 3, the redundancy group of the disk # D1 is {# D3, # D4, # D5, # D7}.

Similarly, when a redundant group is determined for another source disk, a redundant group with a multiplicity of 3 shown in FIG. 19 is set. A redundant group can be set in a system having a disk redundancy configuration with a multiplicity of M by increasing the number of redundant groups of all the original disks by the same procedure each time the data multiplicity is increased.

Also in the second embodiment, the data is written so that the data amount becomes even in the redundant group unit.

Next, the data write operation in the second embodiment will be described. FIG. 20 is a flow chart showing the data write operation by the data storage device of the second embodiment, and the data write operation for the disk having the redundant group configuration shown in FIG. 19 will be described with reference to this figure. First, the file block management unit 7 determines the order of the redundant groups for writing data (step S).
T1d). More specifically, the file block management unit 7 first selects an arbitrary original disk. For example,
It is assumed that the disk # D1 is selected in the system having the configuration of FIG. 5 and the redundancy group setting in the case of the multiplicity of 2 in FIG. Next, the file block management unit 7 selects one having the largest distance information in the redundancy group of the disk # D1. In the above system, the redundancy group of disk # D1 is disk # D5,
# D7, and both have the same distance information. For this reason,
Either may be selected. Here, the disk # D5 is selected as an example.

Then, the file block management section 7 selects the one having the largest distance information in the redundancy group of the disk # D5. That is, one of the disks # D1 and # D4 belonging to the redundancy group of the disk # D5 is selected. Here, since the disk # D1 has already been selected, the disk # D4 is selected. Hereinafter, the same operation is repeated for all the disks. This allows
The file block management unit 7 writes the data block to the disk # D1, the disk # D5, and the disk #D.
4, disk # D8, disk # D2, disk #D
6, disc # D3, and disc # D7 are decided in this order.

Next, the file block management section 7 divides the write data into blocks (step ST2).
d). Subsequently, the file block management unit 7 determines whether or not writing of all data blocks in the redundancy group has been completed (step ST3d). At this time, if the writing of all the data blocks is completed, the data writing operation is ended. On the other hand, if the writing of all the data blocks is not completed, the process proceeds to step ST4d.

In step ST4d, the file block management section 7 first writes a data block to the original disk in the writing order determined as described above. After this, the file block management unit 7
If the data multiplicity is M (M; integer of 2 or more),
A data block that is the same as the data block written to the original disk is stored in the redundancy group (M
-1) Write on the duplicate disks. More specifically, the file block management unit 7 excludes the original disk in the redundancy group while determining whether or not the data block has been written to the (M-1) duplicate disks, and 2 * (M-1). A duplicate disk is selected from the individual disks and a data block is written (step ST5).
d). At this time, if the data blocks have been written to the (M-1) duplicate disks, the process returns to step ST2d and the above-described operation is repeated.

On the other hand, if the data block is not written to the (M-1) duplicated disks, the file block management unit 7 selects the disk having the largest distance information from the original disk, excluding the disk to which the block has already been written. Select (step ST6d).

After that, the file block management unit 7
It is determined whether or not there are a plurality of disks selected in step ST6d (step ST7d). At this time, if a plurality of selected disks do not exist, the file block management unit 7 proceeds to the processing of step ST11d to write the data block to the selected disk, and returns to the processing of step ST5d.

On the other hand, when there are a plurality of selected disks, the file block management section 7 selects the disk with the smallest number of written data blocks (the number of written blocks) among the selected disks (step ST8).
d). That is, the disks are selected so that blocks are written evenly on each disk.

Subsequently, the file block management section 7 determines whether or not there are a plurality of disks selected in step ST8d (step ST9d). At this time, if there are not a plurality of selected disks, the file block management unit 7 proceeds to the process of step ST11d. on the other hand,
When there are a plurality of selected disks, the file block management unit 7 selects one of the plurality of disks selected in step ST8d (step ST10d).

After that, the file block management unit 7
In step ST11d, the data block is written to the disc selected in step ST10d. By repeating this operation, the duplicate data block is written to the (M-1) duplicate disks.

FIG. 21 is a diagram showing a data block storage example of a system in which a redundancy group setting corresponding to the multiplicity of 2 in FIG. 19 is made. Here, the data block storage in FIG. 21 is performed by the flow of FIG. 20 described above.

The file block management section 7 updates the file block management table 7a when writing data and sets the read block number 7a-2. At this time, the block written as the original disk is set as the read block of each disk.

FIG. 22 is a diagram showing a case where a read block is set in the configuration of FIG. In the figure, it is shown that the data block surrounded by a circle symbol is set as a read block. For example, for disk # D1 {B
1, B9, B17} is set to the read block. As shown in FIG. 22, the data blocks B1 to B24 forming the original data are assigned as the read blocks of the respective disks.

Further, the operation at the time of data transfer by the data storage device of the second embodiment is performed by the flow of FIG. 16 as in the first embodiment. Therefore, duplicated description will be omitted.

Next, the operation at the time of a disk failure in the second embodiment will be described. First, when a disk failure is detected, the node machine management table 9a is corrected by the node machine management 9 as in the first embodiment. After that, the file block management unit 7
The file block management table 7a is modified. The correction operation of the file block management table 7a will be described in detail.

FIG. 23 is a diagram showing the read block numbers set when the disk fails in the configuration of FIG. In the figure, the vertical line attached to the disk # D2 indicates that the disk # D2 has failed. Further, as in the case of FIG. 22, the data block surrounded by the circle symbol is the read block. In FIG. 23, disk # D2
After the file block management table 7a is corrected at the time of failure, the block to be the read block is surrounded by a circle symbol. For example, before the failure of the disk # D2, the read block of the disk # D8 is {B as shown in FIG.
6, B14, B22}. However, after the failure of the disk # D2, the read blocks of the disk # D8 are {B5, B6, B13, B2 as shown in FIG.
1}.

FIG. 24 is a flowchart showing the operation of the data storage device according to the second embodiment when a disk fails, which will be described in detail with reference to this figure, FIG. 22 and FIG. First,
The file block management unit 7 modifies the file block management table 7a when the node machine management unit 9
The total number of target disks and the disk number of the failed disk are acquired from (step ST1e). In the example of FIG. 23, the disk number of the failed disk is # D2, and the total number of disks is 7.

Next, the file block management section 7 refers to the file block management table 7a and acquires the read block number 7a-2 of the failed disk (step ST2e). In the example of FIG. 23, the file block management unit 7 determines that the read block number of the failed disk # D2 is {B5,
B13, B21} is acquired.

Subsequently, the file block management unit 7 determines whether or not all the data blocks forming the original data are assigned as the read blocks of any disk (step ST3e). At this time, if all the blocks are assigned to any of the disks, the file block management unit 7 ends the process.

On the other hand, if there is an unallocated block, the file block management section 7 selects a disk having a copy of the read block that is not allocated to any disk (step ST4e). Specifically, the file block management unit 7 selects a disk that stores a copy of the read block number 7a-2 of the failed disk. In the example of FIG. 23, disk #D
The disk storing the copy of the read block number 7a-2 of No. 2 is disk # D8. Therefore, the file block management unit 7 selects the disk # D8 and reassigns the read block number 7a-2.

Next, the file block management section 7 obtains the number of read blocks to be newly assigned to the disk selected in step ST4e (step ST5e). Here, the number of blocks to be newly allocated is calculated by the following equation (1). New_B = ceil (Rem_B / (Rem_D + 1)) (1) where New_B is the number of blocks to be newly allocated, Rem_B is the number of blocks whose allocation has not been decided, and Rem_D is the number of disks to which read blocks have not been reallocated. Shows. Also, ceil () means rounding up after the decimal point.

Here, among the blocks forming the original data, in order to evenly allocate the blocks which are not currently assigned as read blocks to any of the disks to the other disks, at least the number of blocks of the selected disk must be set. You must ask if you should newly allocate the block. The above formula (1) is a formula for determining the number of blocks to be newly allocated.

In the example of FIG. 23, for disk # D8, New_B = ceil (3 / (7-1 + 1)) = 1
Therefore, it is necessary to newly allocate one read block. That is, one read block is newly added to the three read blocks currently allocated, and the total number of read blocks after the reallocation becomes four.

Next, the file block management section 7 selects all the read blocks that are not currently assigned to any disk, and selects these read blocks and step ST.
From the read blocks currently assigned to the disk selected in 4e, blocks corresponding to the number of read blocks to be newly assigned in step ST5e are selected (step ST6e). In the example of FIG. 23, the disk #
As a read block newly set to D8, {B5, B13, B21} which is not currently assigned to any disk as a read block, and {B6, B1 which is currently assigned to a disk # D8 as a read block.
4, B22}, the number of blocks to be newly allocated,
That is, one read block is selected.

Therefore, the read block newly set on the disk # D8 is {B5, B6, B13, B21}.
Becomes Based on this information, the file block management unit 7 rewrites the read block number 7a-2 of the file block management table 7a (step ST7).
e).

After that, the file block management unit 7
Return to the processing of step ST3e, and similarly perform disk #D.
For 4 and # D5, new read block numbers are obtained and the file block management table 7a is rewritten.

By the above operation, the read block of each disk is reset as shown in FIG. 23 after the failure of the disk # D2.

In the example of FIG. 23, the data multiplicity is 2
However, the same procedure can be used to achieve the data multiplicity M.

Further, the operation of the data storage device according to the second embodiment when a failure occurs in the upper hierarchy will be described. FIG. 25 is a diagram showing an example of disk storage in which data is distributed in the system having the configuration of FIG. 5 based on the procedure of the second embodiment.
1 shows a situation in which a failure occurs in -1 and data access to disks # D1 and # D2 is not possible.

As shown in FIG. 25, even when failures occur in a plurality of disks at the same time, the read block number 7a-2 is rearranged by the flow shown in FIG. 24 so that the load of each disk is evenly distributed. Has been done.

In FIG. 25, the data multiplicity is 2, but the data multiplicity M can be realized by the same procedure.

As described above, according to the second embodiment, when data cannot be read from any of the disks, the data block group to be read from the disk is distributed to the data block groups having the same contents. Since the read blocks of the failed disk are set evenly on non-failed disks because the disks are read evenly from other disks that are stored in parallel, the specific disk becomes a bottleneck at the time of the disk failure and becomes parallel. The problem that the throughput of the entire system is reduced does not occur. Also, the same effect can be obtained when a failure occurs in a plurality of disks.

Further, according to the second embodiment, the distance information is defined between the disks, and at the same time, a redundant group is made up of disks which are less likely to fail and a copy of the disks is created, so that a failure in the upper layer is caused. It is possible to provide a parallel system in which data can be accessed even when it occurs.

In the second embodiment, the disks of the host machine may be included in the group and the data may be arranged in the host machine.

Third embodiment. The third embodiment is different from the second embodiment in data storage. That is,
In the third embodiment, each disk is divided into two partitions to write data.

FIG. 26 is a diagram showing an example of block storage of the data storage device according to the third embodiment of the present invention. In the figure, each part separated by a boundary line attached to a cylindrical shape representing one disk indicates a partition of the disk. Thus, in the third embodiment,
The block set in the read block (data surrounded by a circle in the figure) and the duplicated data of another disk are stored in different partitions.

Here, when writing the data block to each disk, the file block management unit 7 according to the third embodiment alternately writes the block set as the read block and the duplicated data of another disk to different partitions. . The other operations are the same as those in the above embodiment.

As described above, according to the third embodiment, when the blocks are accessed sequentially and the data is transferred, the data transferred from each disk are arranged in a specific partition.
Blocks can be transferred from each disk at high speed, and block data transfer performance is not significantly deteriorated. Moreover, the number of seeks can be reduced for sequential access of a large amount of data.

In the example of FIG. 26, the data multiplicity is 2
However, the case of the data multiplicity M can be similarly realized.

Further, in the third embodiment, the disks of the host machine may be grouped and the data may be arranged in the host machine.

In the first to third embodiments described above,
The reliability level may be set for each redundancy group based on the distance information. That is, when setting the redundancy group, disks constituting these are selected so that they are classified into a bus level, a machine level, a regional level, etc. and have different reliability levels. By doing so, it is possible to deal with the failure with different reliability levels in each redundancy group.

[0177]

As described above, according to the present invention, data is divided into a plurality of data blocks of the same data amount, and the same number of blocks are distributed and arranged in a plurality of storage devices in units of the data blocks. Since the data blocks of the same content are redundantly stored for each of the predetermined number of storage devices, and the data stored in a distributed manner in the plurality of storage devices are read in units of one or more data blocks for each of the predetermined number of storage devices. Since the data is stored so that the data amount of each storage device becomes equal, and the data transfer amount from each storage device at the time of data transfer becomes equal, the decrease in the throughput of the entire device is minimized. The effect is that you can. Also, by storing each data block sequentially for each redundancy group, no special calculation is required when deciding from which storage device in the redundancy group the data should be transferred when performing the data transfer process. Therefore, there is an effect that it is possible to operate at high speed in the process of sequentially transferring all the data from the head of the storage device.

According to the present invention, the distance information, which is an index of the interaction between the storage devices of the plurality of storage devices, is used to form a redundant group in which the storage devices having a small interaction are put together, and a plurality of data are stored. When storing the data blocks in different storage devices, the data is divided into a plurality of data blocks of the same data amount, and the data blocks of the same content are stored in duplicate in each storage device in the redundancy group. Since the data distributed and stored in the device is read in units of one or a plurality of data blocks for each redundancy group, the data is stored so that the data amount of each storage device is equalized, and each storage device at the time of data transfer is read. Since the amount of data transferred from the device becomes even, there is an effect that a decrease in throughput of the entire device can be suppressed to a minimum.
Further, there is an effect that it is possible to provide a highly reliable data storage device that flexibly copes with a failure due to a specific interaction between the storage devices.

According to the present invention, since the data blocks are stored in a plurality of storage devices by round robin, there is an effect that the data can be stored so that the data amount of each storage device becomes equal.

According to the present invention, since the storage device in which the data block is to be stored is determined by using the hash distribution,
There is an effect that the data can be stored so that the data amount of each storage device becomes equal.

According to the present invention, since a plurality of data blocks read at once from each storage device are arranged adjacent to each other in each storage area, the data blocks can be transferred from each storage device at high speed. The advantage is that the transfer performance of the data block is less deteriorated. Further, there is an effect that the number of seeks can be reduced for the sequential access of a large amount of data.

According to the present invention, when data cannot be read from any of the storage devices, a plurality of data blocks to be read from the storage device are stored in a distributed manner with the data block groups having the same contents as those stored therein. Since the data blocks that should be read from the failed storage device are read evenly from the non-failed storage device, the specific storage device becomes a bottleneck at the time of the failure, and the throughput of the entire device decreases. This has the effect of suppressing the problem of

According to the present invention, since the distance information is obtained based on the hierarchical structure formed by a plurality of storage devices, there is an effect that it is possible to flexibly cope with a failure in a specific hierarchy of the storage device. There is also an effect that a redundant configuration can be built around a specific part of the device.

According to the present invention, since the distance information is obtained based on the position information of each storage device, there is an effect that it is possible to flexibly cope with a failure caused by the positional relationship of the storage devices.

According to the present invention, the distance information is obtained based on the failure rate determined according to the connection path between the storage devices, so that it is possible to flexibly cope with the failure caused by the connection relation of the storage devices. There is an effect.

According to the present invention, since the multiplicity, which is the degree to which data blocks having the same contents are overlapped, is set for each redundancy group, the reliability of the device can be improved and the data transfer can be speeded up. There is an effect that can be.

According to the present invention, since the reliability level is set for each redundancy group based on the distance information, there is an effect that a failure can be dealt with at a different reliability level for each redundancy group.

[Brief description of drawings]

FIG. 1 is a diagram showing a configuration example of a data storage device according to a first embodiment of the present invention.

FIG. 2 is a diagram showing a configuration of a host machine in FIG.

FIG. 3 is a diagram showing the relationship of each software of the data storage device according to the first embodiment of the present invention.

FIG. 4 is a diagram showing a configuration of a node machine management table in FIG.

FIG. 5 is a diagram showing an example of a hierarchical relationship of disks constituting the data storage device according to the first embodiment.

FIG. 6 is a diagram showing another example of the hierarchical relationship of disks constituting the data storage device according to the first embodiment.

FIG. 7 is a diagram showing a hierarchical structure of the system viewed from the host machine.

FIG. 8 is a flowchart showing the operation of the data storage device according to the first embodiment.

FIG. 9 is a diagram showing a configuration of the data storage device according to the first embodiment and a failure rate in a hierarchical relationship of disks.

FIG. 10 is a diagram showing another configuration of the data storage device according to the first embodiment and a failure rate in a hierarchical relationship of disks.

FIG. 11 is a diagram showing a physical positional relationship of disks installed on the same floor.

FIG. 12 is a flowchart showing a data write operation of the data storage device according to the first embodiment.

FIG. 13 is a diagram showing a storage diagram of blocks written in each disc in the write flow shown in FIG. 12;

FIG. 14 is a diagram showing an internal structure of a file block management table in FIG.

FIG. 15 is a diagram showing a read block number set at the time of writing data by the data storage device of the first embodiment.

FIG. 16 is a flowchart showing a data transfer operation of the data storage device according to the first embodiment.

FIG. 17 is a diagram showing a read block number set when a disk fails by the data storage device according to the first embodiment.

FIG. 18 is a flowchart showing the operation of the data storage device according to the second embodiment.

19 is a diagram showing a redundancy group correspondence table created for the hierarchical relationship in FIG.

FIG. 20 is a flowchart showing a data write operation by the data storage device according to the second embodiment.

FIG. 21 is a diagram showing an example of data block storage in a system in which a redundant group setting corresponding to the multiplicity of 2 in FIG. 19 is made.

22 is a diagram showing a case where a read block is set in the configuration of FIG.

FIG. 23 is a diagram showing a read block number set when a disk fails in the configuration of FIG. 21.

FIG. 24 is a flowchart showing an operation when a disk fails in the data storage device according to the second embodiment.

FIG. 25 is a diagram showing an example of disk storage in which data is distributed in the system having the configuration of FIG. 5 based on the procedure of the second embodiment.

FIG. 26 is a diagram showing an example of block storage of a data storage device according to a third embodiment of the present invention.

FIG. 27 is a diagram showing an example of data storage using mirroring.

FIG. 28 is a diagram showing an example of data storage by a conventional disk array.

[Explanation of symbols]

1 host machine (data storage device), 2 network, 3-1 to 3-n node machine, 4 main storage device, 5
Operating system, 6 data transfer control unit (reading unit), 7 file block management unit (data storage unit), 7a file block management table,
7a-1 disk number, 7a-2 read block number, 8 communication control unit, 9 node machine management unit (data storage means), 9a node machine management table, 10
Central processing unit (data storage means, reading means),
11 network adapter, 12a system bus,
12b PCI bus, 13 bus coupling device, 14 disk controller, 15, 16 disk device (storage device), 17 operating system, 18 communication control unit, 19 disk access control unit, 20 disk device (storage device), 21 node machine management Information, 21
a Node machine number, 21b IP address, 21c
Port number, 22 disk information, 22a disk number, 22b node machine number, 23 layer information, 23
a lower layer information, 23b upper layer information, 24 distance information, 25,26 floor frame, 25-1,25
2, 26, 26-1, 26-2 node machine frame, 27 whole system, 28 host machine hierarchy, 29
Host machine frame, 30-1, 30-2, 30-
3, 30-4 node machine, 31-1, 31-2 router, 32 host machine, 33-1, 33-2, 33
-3, 33-4 Redundancy group.

   ─────────────────────────────────────────────────── ─── Continued front page    (72) Inventor Mitsunori Gun             2-3 2-3 Marunouchi, Chiyoda-ku, Tokyo             Inside Ryo Electric Co., Ltd. F-term (reference) 5B018 GA04 HA04 MA14                 5B065 BA01 CA12 CC08 CH13 CH18                       EA31

Claims (17)

[Claims] 1. Data is divided into a plurality of data blocks of the same data amount, and the same number of blocks are distributed and arranged in a plurality of storage devices in units of the data blocks.
Data storage means for redundantly storing the same data block for each predetermined number of the storage devices, and one or a plurality of data stored in a distributed manner in the plurality of storage devices. And a data storage device having a reading means for reading the data block by unit. 2. A redundant group in which the storage devices having a small interaction are grouped together by using distance information which is an index of the interaction between the storage devices of the plurality of storage devices, and data is stored in the plurality of storage devices. A data storage unit that divides the data into a plurality of data blocks of the same data amount and stores the data blocks of the same content in duplicate in each storage device in the redundancy group when the data blocks are distributed and stored in the devices. A data storage device comprising: a reading unit that reads out data stored in a distributed manner in the plurality of storage devices in units of one or a plurality of data blocks for each of the redundancy groups. 3. A redundant group in which the storage devices having a small interaction are grouped together by using distance information, which is an index of interaction between the storage devices of the plurality of storage devices, and data is stored in the plurality of storage devices. A data storage device that divides the data into a plurality of data blocks of the same data amount and stores the data blocks of the same content in duplicate in each storage device in the redundancy group when the data blocks are distributed and stored in the devices. 4. The data storage device according to claim 1, wherein the data storage means stores the data block in a round robin manner in a plurality of storage devices. 5. The data storage device according to claim 1, wherein the data storage means determines the storage device in which the data block should be stored by using hash distribution. 6. The data storage according to claim 1, wherein the data storage means arranges a plurality of data blocks read at a time from each storage device adjacently in each storage area. apparatus. 7. A plurality of reading units store a data block group to be read from the storage device in a distributed manner when the storage device cannot read data in any of the storage devices. 3. Evenly reading from the storage device of claim 1.
The described data storage device. 8. The data storage device according to claim 3, wherein the distance information is obtained based on a hierarchical structure formed by a plurality of storage devices. 9. The data storage device according to claim 3, wherein distance information is obtained based on position information in which each storage device is arranged. 10. The data storage device according to claim 3, wherein the distance information is obtained based on a failure rate determined according to a connection path between the storage devices. 11. The data storage device according to claim 3, wherein a multiplicity, which is a degree of overlapping data blocks having the same content, is set for each redundancy group. 12. The reliability level is set for each redundancy group based on the distance information.
The described data storage device. 13. The data is divided into a plurality of data blocks of the same data amount, and the same number of blocks are distributed and arranged in a plurality of storage devices in units of the data blocks, and the same contents are stored for each predetermined number of the storage devices. A data storing step of storing the data blocks in duplicate, and a reading step of reading the data stored in a distributed manner in the plurality of storage devices in units of one or a plurality of data blocks for each of the predetermined number of storage devices. Prepared data management method. 14. In the data storing step, a distance group, which is an index of the interaction between the storage devices of the plurality of storage devices, is used to form a redundant group in which the storage devices having a small interaction are put together. When the data is distributed and stored in the plurality of storage devices, the data is divided into a plurality of data blocks of the same data amount, and the storage device in the redundancy group is duplicated with the same data block. 14. The data management method according to claim 13, wherein the data is stored. 15. The data is divided into a plurality of data blocks of the same amount of data, the same number of blocks are distributed and arranged in a plurality of storage devices in units of the data blocks, and the same contents are stored for each of a predetermined number of the storage devices. A data storage means for storing the data blocks in duplicate, a computer as a read means for reading out the data stored dispersedly in the plurality of storage devices in units of one or a plurality of data blocks for each of the predetermined number of storage devices. A program to make it work. 16. A redundant group in which the storage devices having a small interaction are grouped together by using distance information, which is an index of interaction between the storage devices of a plurality of storage devices, and data is stored in the same amount of data. Divided into a plurality of data blocks, and the same number of blocks are distributed and arranged in the plurality of storage devices in units of the data blocks, and the data blocks having the same content are redundantly stored in the storage devices in the redundancy group. A program for causing a computer to function as a reading unit that reads out data stored in a distributed manner in the plurality of storage devices in units of one or a plurality of data blocks for each of the redundancy groups. 17. A redundant group in which the storage devices having a small interaction are grouped together by using distance information, which is an index of interaction between the storage devices of the plurality of storage devices, and data is stored in the plurality of storage devices. As a data storage unit that divides the data into a plurality of data blocks of the same data amount and stores the data blocks of the same content in duplicate in each storage device in the redundant group when the data blocks are distributed and stored in the devices. A program for operating a computer.