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Difference between revisions of "RAID5 Patches"

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   As ''handle_stripe()'' goes in logical block order, it
 
   As ''handle_stripe()'' goes in logical block order, it
 
   handles S0, then S8, and then again S0 and S8. After the first touch,  
 
   handles S0, then S8, and then again S0 and S8. After the first touch,  
   S0 is left with block #0 up-to-date, while #16 and P0 are not. Thus,
+
   S0 is left with block #0 up-to-date, while #16 and P0 are not. If the
   if the stripe is forced for completion, we would need to read block
+
   stripe is forced to completion, we would need to read block #16 or P0
  #16 or P0 to get a fully up-to-date stripe. Such reads hurt throughput
+
  to get a fully up-to-date stripe. Such reads hurt throughput, almost
   almost to death. If just a single process writes, then things are
+
   to death. If a single process writes, things are okay because no one
   OK, because nobody unplugs the queue, and there are no requests to
+
   unplugs the queue, and there are no requests to force completion of a
  force completion of a pending request. But, if there are more writers,
+
  pending request. If there are more writers and a queue unplug occurs
  then a queue unplug often occurs, and pending requests are often forced
+
  often, pending requests are often forced to completion. Consider that
  for completion. Take into account, that in reality, we use a large
+
  we use a large chunk size (128K, 256K, and even larger). In the end,  
  chuck size (128K, 256K and even larger). Hence, in the end, there
+
   there are many out-of-date stripes in the cache and many reads.
   are many out-of-date stripes in the cache and many reads.
 
  
 
; * memcpy() is a top consumer
 
; * memcpy() is a top consumer
   All requests go via internal cache, on dual-core, two-way Opteron.  
+
   All requests are handled via internal cache, on dual-core, two-way  
  It takes up to 30-33% of CPU doing 1 GB/s writes.
+
  Opteron. This takes up to 30-33% of CPU doing 1 GB/s writes.
  
 
; * Small requests
 
; * Small requests
 
   To fill I/O pipes and reach good throughput, we need very large
 
   To fill I/O pipes and reach good throughput, we need very large
   I/O requests. Lustre does this by using a bio subsystem on 2.6. But, as
+
   I/O requests. Lustre does this by using a bio subsystem on 2.6. As
   described above, RAID5 handles all blocks separately and issues separate
+
   described above, RAID5 handles all blocks separately, and issues separate
   I/O (bio) for every block. This is partially solved by an I/O scheduler
+
   I/O (bio) for each block. This is partially solved by an I/O scheduler
 
   that merges small requests into bigger ones. But, due to the nature of
 
   that merges small requests into bigger ones. But, due to the nature of
   the block subsystem, any process that wants I/O to get completed  
+
   the block subsystem, any process that wants I/O to be completed  
   ''unplugs'' the queue, and we can get many small requests in the pipe.
+
   ''unplugs'' the queue, causing many small requests in the pipe.
  
 
We have developed patches that address the described problems. You can find
 
We have developed patches that address the described problems. You can find

Revision as of 16:01, 14 May 2008

Notes about RAID5 Internals

Structures

In Linux, RAID5 handles all incoming requests by small units called stripes. A stripe is a set of blocks taken from all disks at the same position. A block is defined as a unit of PAGE_SIZE bytes.

For example, suppose you have 3 disks and have specified 8K chunksize. Internally, RAID5 will look like this:

S0 S8 S32 S40
Disk1 #0 #8 #32 #40
Disk2 #16 #24 #48 #56
Disk3 P0 P8 P32 P40

where:

  • Sn -- Number of internal stripe
  • #n -- An offset in sectors (512bytes)
  • Pn -- Parity for other blocks in the stripe (actually, it floats among disks)

As you can see, an 8K chunksize means 2 contiguous blocks.

Logic

make_request() goes through an incoming request, breaking it into blocks (PAGE_SIZE) and handling them separately. Given bio with bi_sector = 0 bi_size = 24K and the array described above, make_request() would handle #0,#8 and #16.

For every block, add_stripe_bio() and handle_stripe() are called.

add_stripe_bio() adds bio to a given stripe. Later, in handle_stripe(), we will be able to use bio and its data for serving requests.

handle_stripe() is a core of RAID5 (as discussed in the next section).

handle_stripe()

The routine works with a stripe. It checks what should be done, learns the current state of a stripe in the internal cache, decides what I/O is needed to satisfy user requests, and does recovery.

For example, if a user wants to write block #0 (8 sectors starting from sector 0), RAID5's responsibility is to store new data and update parity P0. There are a few possibilities here:

  1. Delay serving until the data for block #16 is ready -- probably the user will want to write #16 very soon?
  2. Read #16, make a new parity P0; write #0 and P0.
  3. Read P0, roll back the old #0 from P0 (so it looks like we did parity with #0) and re-compute parity with the new #0.

The first possibility looks like the best option because it does not require a very expensive read, but the problem is that user may need to write only #0 and not #16 in near future.

Also, the queue can get unplugged, meaning that the user wants all requests to complete. (Unfortunately, in the current block layer, there is no way to specify the exact request that the user is interested in, so any completion interest means immediate serving of the entire queue).

Problems

This is a short list of RAID5 problems that we encountered in the Thumper project:

* Order of handling is not good for large requests
  As handle_stripe() goes in logical block order, it
  handles S0, then S8, and then again S0 and S8. After the first touch, 
  S0 is left with block #0 up-to-date, while #16 and P0 are not. If the 
  stripe is forced to completion, we would need to read block #16 or P0
  to get a fully up-to-date stripe. Such reads hurt throughput, almost
  to death. If a single process writes, things are okay because no one
  unplugs the queue, and there are no requests to force completion of a
  pending request. If there are more writers and a queue unplug occurs
  often, pending requests are often forced to completion. Consider that
  we use a large chunk size (128K, 256K, and even larger). In the end, 
  there are many out-of-date stripes in the cache and many reads.
* memcpy() is a top consumer
  All requests are handled via internal cache, on dual-core, two-way 
  Opteron. This takes up to 30-33% of CPU doing 1 GB/s writes.
* Small requests
  To fill I/O pipes and reach good throughput, we need very large
  I/O requests. Lustre does this by using a bio subsystem on 2.6. As 
  described above, RAID5 handles all blocks separately, and issues separate
  I/O (bio) for each block. This is partially solved by an I/O scheduler
  that merges small requests into bigger ones. But, due to the nature of
  the block subsystem, any process that wants I/O to be completed 
  unplugs the queue, causing many small requests in the pipe.

We have developed patches that address the described problems. You can find them at ftp://ftp.clusterfs.com/pub/people/alex/raid5

Zero-copy Patch

In the current RAID5 implementation, there is a cache for each device. For each I/O instance, the RAID5 driver first updates the device cache, and then submits the read/write request to the real block device. The cache update operation consumes lots of CPU time to copy data to/from the device cache. For Lustre's bulk write, it is meaningless to update this cache space. This is the basis of the zero-copy patch for RAID5.

To avoid data copy, pages from bio are used directly to calculate the parity page and feed the I/O source to disks. These pages (bio pages from the filesystem layer) should not be modified when calculating the parity page. Otherwise, the wrong parity will be written to storage, which causes garbage data to be generated if the administrator tries to rebuild the array in the future.

If pages are being mapped, our solution is to lock the pages and then unmap them (to keep the pages from being modified). For this purpose, an additional page flag (PG_constant) has been introduced. Set this bit to let the RAID5 driver know that the page will not be modified during I/O. After I/O completes against this page, the bit will be cleared, so the page can be written again.