/dev/sda2 and /dev/sda3. The device is a lot slower than a single partition. Isn't md a pile of junk?
A: To have a RAID-0 device running a full speed, you must have partitions from different disks. Besides, putting the two halves of the mirror on the same disk fails to give you any protection whatsoever against disk failure.
A: It's not obvious that RAID-0 will always provide better performance; in fact, in some cases, it could make things worse. The ext2fs file system scatters files all over a partition, and it attempts to keep all of the blocks of a file contiguous, basically in an attempt to prevent fragmentation. Thus, ext2fs behaves "as if" there were a (variable-sized) stripe per file. If there are several disks concatenated into a single RAID-linear, this will result files being statistically distributed on each of the disks. Thus, at least for ext2fs, RAID-linear will behave a lot like RAID-0 with large stripe sizes. Conversely, RAID-0 with small stripe sizes can cause excessive disk activity leading to severely degraded performance if several large files are accessed simultaneously.In many cases, RAID-0 can be an obvious win. For example, imagine a large database file. Since ext2fs attempts to cluster together all of the blocks of a file, chances are good that it will end up on only one drive if RAID-linear is used, but will get chopped into lots of stripes if RAID-0 is used. Now imagine a number of (kernel) threads all trying to random access to this database. Under RAID-linear, all accesses would go to one disk, which would not be as efficient as the parallel accesses that RAID-0 entails.
A: To understand this, lets look at an example with three partitions; one that is 50MB, one 90MB and one 125MB. Lets call D0 the 50MB disk, D1 the 90MB disk and D2 the 125MB disk. When you start the device, the driver calculates 'strip zones'. In this case, it finds 3 zones, defined like this:Z0 : (D0/D1/D2) 3 x 50 = 150MB total in this zone Z1 : (D1/D2) 2 x 40 = 80MB total in this zone Z2 : (D2) 125-50-40 = 35MB total in this zone.You can see that the total size of the zones is the size of the virtual device, but, depending on the zone, the striping is different. Z2 is rather inefficient, since there's only one disk. Sinceext2fsand most other Unix file systems distribute files all over the disk, you have a 35/265 = 13% chance that a fill will end up on Z2, and not get any of the benefits of striping. (DOS tries to fill a disk from beginning to end, and thus, the oldest files would end up on Z0. However, this strategy leads to severe filesystem fragmentation, which is why no one besides DOS does it this way.)
md. Does it significantly increase the throughput? Is the performance really noticeable?
A: The answer depends on the configuration that you use.
- Linux MD RAID-0 and RAID-linear performance:
If the system is heavily loaded with lots of I/O, statistically, some of it will go to one disk, and some to the others. Thus, performance will improve over a single large disk. The actual improvement depends a lot on the actual data, stripe sizes, and other factors. In a system with low I/O usage, the performance is equal to that of a single disk.
- Linux MD RAID-1 (mirroring) read performance:
MD implements read balancing. That is, the RAID-1 code will alternate between each of the (two or more) disks in the mirror, making alternate reads to each. In a low-I/O situation, this won't change performance at all: you will have to wait for one disk to complete the read. But, with two disks in a high-I/O environment, this could as much as double the read performance, since reads can be issued to each of the disks in parallel. For N disks in the mirror, this could improve performance N-fold.
- Linux MD RAID-1 (mirroring) write performance:
Must wait for the write to occur to all of the disks in the mirror. This is because a copy of the data must be written to each of the disks in the mirror. Thus, performance will be roughly equal to the write performance to a single disk.
- Linux MD RAID-4/5 read performance:
Statistically, a given block can be on any one of a number of disk drives, and thus RAID-4/5 read performance is a lot like that for RAID-0. It will depend on the data, the stripe size, and the application. It will not be as good as the read performance of a mirrored array.
- Linux MD RAID-4/5 write performance:
This will in general be considerably slower than that for a single disk. This is because the parity must be written out to one drive as well as the data to another. However, in order to compute the new parity, the old parity and the old data must be read first. The old data, new data and old parity must all be XOR'ed together to determine the new parity: this requires considerable CPU cycles in addition to the numerous disk accesses.
A: Is the goal to maximize throughput, or to minimize latency? There is no easy answer, as there are many factors that affect performance:
- operating system - will one process/thread, or many be performing disk access?
- application - is it accessing data in a sequential fashion, or random access?
- file system - clusters files or spreads them out (the ext2fs clusters together the blocks of a file, and spreads out files)
- disk driver - number of blocks to read ahead (this is a tunable parameter)
- CEC hardware - one drive controller, or many?
- hd controller - able to queue multiple requests or not? Does it provide a cache?
- hard drive - buffer cache memory size -- is it big enough to handle the write sizes and rate you want?
- physical platters - blocks per cylinder -- accessing blocks on different cylinders will lead to seeks.
A: Since RAID-5 experiences an I/O load that is equally distributed across several drives, the best performance will be obtained when the RAID set is balanced by using identical drives, identical controllers, and the same (low) number of drives on each controller. Note, however, that using identical components will raise the probability of multiple simultaneous failures, for example due to a sudden jolt or drop, overheating, or a power surge during an electrical storm. Mixing brands and models helps reduce this risk.
A: When using the current (November 1997) RAID-4/5 implementation, it is strongly recommended that the file system be created withmke2fs -b 4096instead of the default 1024 byte filesystem block size.This is because the current RAID-5 implementation allocates one 4K memory page per disk block; if a disk block were just 1K in size, then 75% of the memory which RAID-5 is allocating for pending I/O would not be used. If the disk block size matches the memory page size, then the driver can (potentially) use all of the page. Thus, for a filesystem with a 4096 block size as opposed to a 1024 byte block size, the RAID driver will potentially queue 4 times as much pending I/O to the low level drivers without allocating additional memory.
Note: the above remarks do NOT apply to Software RAID-0/1/linear driver.
Note: the statements about 4K memory page size apply to the Intel x86 architecture. The page size on Alpha, Sparc, and other CPUS are different; I believe they're 8K on Alpha/Sparc (????). Adjust the above figures accordingly.
Note: if your file system has a lot of small files (files less than 10KBytes in size), a considerable fraction of the disk space might be wasted. This is because the file system allocates disk space in multiples of the block size. Allocating large blocks for small files clearly results in a waste of disk space: thus, you may want to stick to small block sizes, get a larger effective storage capacity, and not worry about the "wasted" memory due to the block-size/page-size mismatch.
Note: most ''typical'' systems do not have that many small files. That is, although there might be thousands of small files, this would lead to only some 10 to 100MB wasted space, which is probably an acceptable tradeoff for performance on a multi-gigabyte disk.
However, for news servers, there might be tens or hundreds of thousands of small files. In such cases, the smaller block size, and thus the improved storage capacity, may be more important than the more efficient I/O scheduling.
Note: there exists an experimental file system for Linux which packs small files and file chunks onto a single block. It apparently has some very positive performance implications when the average file size is much smaller than the block size.
Note: Future versions may implement schemes that obsolete the above discussion. However, this is difficult to implement, since dynamic run-time allocation can lead to dead-locks; the current implementation performs a static pre-allocation.
A: The chunk size is the amount of data contiguous on the virtual device that is also contiguous on the physical device. In this HOWTO, "chunk" and "stripe" refer to the same thing: what is commonly called the "stripe" in other RAID documentation is called the "chunk" in the MD man pages. Stripes or chunks apply only to RAID 0, 4 and 5, since stripes are not used in mirroring (RAID-1) and simple concatenation (RAID-linear). The stripe size affects both read and write latency (delay), throughput (bandwidth), and contention between independent operations (ability to simultaneously service overlapping I/O requests).Assuming the use of the ext2fs file system, and the current kernel policies about read-ahead, large stripe sizes are almost always better than small stripe sizes, and stripe sizes from about a fourth to a full disk cylinder in size may be best. To understand this claim, let us consider the effects of large stripes on small files, and small stripes on large files. The stripe size does not affect the read performance of small files: For an array of N drives, the file has a 1/N probability of being entirely within one stripe on any one of the drives. Thus, both the read latency and bandwidth will be comparable to that of a single drive. Assuming that the small files are statistically well distributed around the filesystem, (and, with the ext2fs file system, they should be), roughly N times more overlapping, concurrent reads should be possible without significant collision between them. Conversely, if very small stripes are used, and a large file is read sequentially, then a read will issued to all of the disks in the array. For a the read of a single large file, the latency will almost double, as the probability of a block being 3/4'ths of a revolution or farther away will increase. Note, however, the trade-off: the bandwidth could improve almost N-fold for reading a single, large file, as N drives can be reading simultaneously (that is, if read-ahead is used so that all of the disks are kept active). But there is another, counter-acting trade-off: if all of the drives are already busy reading one file, then attempting to read a second or third file at the same time will cause significant contention, ruining performance as the disk ladder algorithms lead to seeks all over the platter. Thus, large stripes will almost always lead to the best performance. The sole exception is the case where one is streaming a single, large file at a time, and one requires the top possible bandwidth, and one is also using a good read-ahead algorithm, in which case small stripes are desired.
Note that this HOWTO previously recommended small stripe sizes for news spools or other systems with lots of small files. This was bad advice, and here's why: news spools contain not only many small files, but also large summary files, as well as large directories. If the summary file is larger than the stripe size, reading it will cause many disks to be accessed, slowing things down as each disk performs a seek. Similarly, the current ext2fs file system searches directories in a linear, sequential fashion. Thus, to find a given file or inode, on average half of the directory will be read. If this directory is spread across several stripes (several disks), the directory read (e.g. due to the ls command) could get very slow. Thanks to Steven A. Reisman < > for this correction. Steve also adds:
I found that using a 256k stripe gives much better performance. I suspect that the optimum size would be the size of a disk cylinder (or maybe the size of the disk drive's sector cache). However, disks nowadays have recording zones with different sector counts (and sector caches vary among different disk models). There's no way to guarantee stripes won't cross a cylinder boundary.The tools accept the stripe size specified in KBytes. You'll want to specify a multiple of if the page size for your CPU (4KB on the x86).
mke2fs command:
mke2fs -b 4096 -R stride=nnn ...
What should the value of nnn be?
A: The-R strideflag is used to tell the file system about the size of the RAID stripes. Since only RAID-0,4 and 5 use stripes, and RAID-1 (mirroring) and RAID-linear do not, this flag is applicable only for RAID-0,4,5. Knowledge of the size of a stripe allowsmke2fsto allocate the block and inode bitmaps so that they don't all end up on the same physical drive. An unknown contributor wrote:I noticed last spring that one drive in a pair always had a larger I/O count, and tracked it down to the these meta-data blocks. Ted added theFor a 4KB block file system, with stripe size 256KB, one would use-R stride=option in response to my explanation and request for a workaround.-R stride=64.If you don't trust the
-Rflag, you can get a similar effect in a different way. Steven A. Reisman < > writes:Another consideration is the filesystem used on the RAID-0 device. The ext2 filesystem allocates 8192 blocks per group. Each group has its own set of inodes. If there are 2, 4 or 8 drives, these inodes cluster on the first disk. I've distributed the inodes across all drives by telling mke2fs to allocate only 7932 blocks per group.Some mke2fs pages do not describe the[-g blocks-per-group]flag used in this operation.
md commands in the startup scripts, so that everything will start automatically at boot time?
A: Rod Wilkens < > writes:What I did is put ``For raid-5, you will want to look at the return code formdadd -ar'' in the ``/etc/rc.d/rc.sysinit'' right after the kernel loads the modules, and before the ``fsck'' disk check. This way, you can put the ``/dev/md?'' device in the ``/etc/fstab''. Then I put the ``mdstop -a'' right after the ``umount -a'' unmounting the disks, in the ``/etc/rc.d/init.d/halt'' file.mdadd, and if it failed, do ato repair any damage.ckraid --fix /etc/raid5.conf
md0? This is for a news server, and I have 9 drives... Needless to say I need much more than two. Is this possible?
A: Yes. (describe how to do this)
A: Normally, Hardware RAID is considered superior to Software RAID, because hardware controllers often have a large cache, and can do a better job of scheduling operations in parallel. However, integrated Software RAID can (and does) gain certain advantages from being close to the operating system.For example, ... ummm. Opaque description of caching of reconstructed blocks in buffer cache elided ...
On a dual PPro SMP system, it has been reported that Software-RAID performance exceeds the performance of a well-known hardware-RAID board vendor by a factor of 2 to 5.
Software RAID is also a very interesting option for high-availability redundant server systems. In such a configuration, two CPU's are attached to one set or SCSI disks. If one server crashes or fails to respond, then the other server can
mdadd,mdrunandmountthe software RAID array, and take over operations. This sort of dual-ended operation is not always possible with many hardware RAID controllers, because of the state configuration that the hardware controllers maintain.
A: No, not unless the major version number changes. An MD version x.y.z consists of three sub-versions:x: Major version. y: Minor version. z: Patchlevel version.Version x1.y1.z1 of the RAID driver supports a RAID array with version x2.y2.z2 in case (x1 == x2) and (y1 >= y2). Different patchlevel (z) versions for the same (x.y) version are designed to be mostly compatible.The minor version number is increased whenever the RAID array layout is changed in a way which is incompatible with older versions of the driver. New versions of the driver will maintain compatibility with older RAID arrays.
The major version number will be increased if it will no longer make sense to support old RAID arrays in the new kernel code.
For RAID-1, it's not likely that the disk layout nor the superblock structure will change anytime soon. Most all Any optimization and new features (reconstruction, multithreaded tools, hot-plug, etc.) doesn't affect the physical layout.
mdstop /dev/md0 says that the device is busy.
A: There's a process that has a file open on/dev/md0, or/dev/md0is still mounted. Terminate the process orumount /dev/md0.
A: There is also a new utility callediotracein thelinux/iotracedirectory. It reads/proc/io-traceand analyses/plots it's output. If you feel your system's block IO performance is too low, just look at the iotrace output.
SPEED_LIMIT defined as 1024K/sec. What does this mean? Does this limit performance?
A:SPEED_LIMITis used to limit RAID reconstruction speed during automatic reconstruction. Basically, automatic reconstruction allows you toe2fsckandmountimmediately after an unclean shutdown, without first runningckraid. Automatic reconstruction is also used after a failed hard drive has been replaced.In order to avoid overwhelming the system while reconstruction is occurring, the reconstruction thread monitors the reconstruction speed and slows it down if its too fast. The 1M/sec limit was arbitrarily chosen as a reasonable rate which allows the reconstruction to finish reasonably rapidly, while creating only a light load on the system so that other processes are not interfered with.
A: Spindle synchronization is used to keep multiple hard drives spinning at exactly the same speed, so that their disk platters are always perfectly aligned. This is used by some hardware controllers to better organize disk writes. However, for software RAID, this information is not used, and spindle synchronization might even hurt performance.
A: Leonard N. Zubkoff replies: It is really fast, but you don't need to use MD to get striped swap. The kernel automatically stripes across equal priority swap spaces. For example, the following entries from/etc/fstabstripe swap space across five drives in three groups:/dev/sdg1 swap swap pri=3 /dev/sdk1 swap swap pri=3 /dev/sdd1 swap swap pri=3 /dev/sdh1 swap swap pri=3 /dev/sdl1 swap swap pri=3 /dev/sdg2 swap swap pri=2 /dev/sdk2 swap swap pri=2 /dev/sdd2 swap swap pri=2 /dev/sdh2 swap swap pri=2 /dev/sdl2 swap swap pri=2 /dev/sdg3 swap swap pri=1 /dev/sdk3 swap swap pri=1 /dev/sdd3 swap swap pri=1 /dev/sdh3 swap swap pri=1 /dev/sdl3 swap swap pri=1
A: In many cases, the answer is yes. Using several controllers to perform disk access in parallel will improve performance. However, the actual improvement depends on your actual configuration. For example, it has been reported (Vaughan Pratt, January 98) that a single 4.3GB Cheetah attached to an Adaptec 2940UW can achieve a rate of 14MB/sec (without using RAID). Installing two disks on one controller, and using a RAID-0 configuration results in a measured performance of 27 MB/sec.Note that the 2940UW controller is an "Ultra-Wide" SCSI controller, capable of a theoretical burst rate of 40MB/sec, and so the above measurements are not surprising. However, a slower controller attached to two fast disks would be the bottleneck. Note also, that most out-board SCSI enclosures (e.g. the kind with hot-pluggable trays) cannot be run at the 40MB/sec rate, due to cabling and electrical noise problems.
If you are designing a multiple controller system, remember that most disks and controllers typically run at 70-85% of their rated max speeds.
Note also that using one controller per disk can reduce the likelihood of system outage due to a controller or cable failure (In theory -- only if the device driver for the controller can gracefully handle a broken controller. Not all SCSI device drivers seem to be able to handle such a situation without panicking or otherwise locking up).