Chapter 2' Data Storage

1
Chapter 2. Data Storage
2
Outline

• Memory hierarchy
• Hardware Disks
• Access Times
• Example - Megatron 747
• Optimizations
• Disk failure
• RAIDs

3
Users
DBMSs
Operating Systems
Hardware - Data Storage
4
The Memory Hierarchy
DBMS
Programs, Main-memory DBMSs
Tertiary Storage
Main memory
Cache
5
Cache

• The cache is an integrated circuit or part of the
  processors chip
• Holding data or machine instructions
• Copy from main-memory
• If data being expelled from the cache has been
  modified, then the new value must be copied into
  the main memory.
• Typical performance
• Capacities up to a megabyte
• Access time 10 nanoseconds (10-8 seconds)
• Moving data bet. Cache and main memory 100
  nanoseconds (10-9 seconds)

6
Main Memory

• Everything that happens in the computer is
  resident in main memory
• Capacity around 100 Mbyte to 10 Gbyte
• Random access
• Typical access time is 10-100 nanoseconds

7
Virtual Memory

• Is a part of disk
• In a 32-bit address machine
• Virtual memory grows up to 232 bytes (4 Gbyte)
• Data is moved between disk and main memory in
  entire blocks, which are also called pages in
  main memory
• Main-memory database systems

8
Secondary Storage (1)

• Slower, more capacious than main memory
• Random access
• magnetic, optical, magneto-optical disks
• Disk read/write are done by moving a chuck of
  bytes called blocks (or pages)

9
Secondary Storage (2)

• Accessing a block 10-30 milliseconds
• Recently, one disk unit can store data ranging
  from 10 to 32 Gbytes
• A machine can have several disk units

10
Tertiary Storage (1)

• Have been developed to hold data volumes measured
  in terabytes
• Compared with secondary storage, it offers
• Higher read/write times
• Larger capacities and smaller cost per byte
• Not random access in general

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Tertiary Storage (2)

• Kinds of tertiary storage devices
• Ad-hoc tape storage
• Optical-disk juke boxes CD-ROMs
• Tape silo an automated version ad-hoc tape
  storage
• Capacities
• CD 2/3 Gbytes, 2.3 Gbytes
• Tapes 50 Gbytes
• Access time about 1000 times slower than
  secondary memory

12
Volatile and Nonvolatile

• Volatile vs. nonvolatile storage
• Flush memory
• A form of main memory
• Nonvolatile
• Becomes economical
• RAM disk
• A battery-backed main memory

13
Access Time vs. Capacity
14
Moores Law

• Gordon Moore observed that the followings double
  every 18 months
• The speed of processors, i.e., the number of
  instructions executed per second and the ratio of
  the speed to cost of a processor
• The cost of main memory per bit and the number of
  bits that can be put on one chip
• The cost of disk per bit and the number of bytes
  that a disk can hold
• Not applicable to
• Main memory access time, disk access time

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Disks
16
Disks A Top View

• Cylinder, Track, Sector, Gap
• Gaps often represents about 10 of the total
  tracks
• A entire section cannot be used if portion of it
  gets destroyed
• Typically a block consists of one or more
  sectors.

top view
17
The Disk Controller

• Controls one or more disk drives
• controlling the mechanical actuator
• selecting a surface or a sector on that surface
• Transferring bits via a data bus

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Disk Storage Characteristics (as of 1999)

• Rotation speed of the disk assembly
• 5400 RPM (one rotation every 11 milliseconds)
• Number of platters per unit
• Typical disk drive 5 platters (10 surfaces)
• Floppy/zip disk 1 platter (2 surfaces)
• Number of tracks per surface
• Have as many as 10,000 tracks
• 3.5 inch diskette 40 tracks
• Number of bytes per track
• Common disk 105 or more bytes
• 3.5 inch diskette 150K

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Megatron 747 Disk (1)

• Characteristics
• Have 4 platters (8 surfaces)
• 8192 (213) tracks per surface
• On average 256 (28) sectors per track
• 512 (29) bytes per sector

• Diameters of tracks
• outermost track is 3.5 inches
• innermost track is 1.5 inches
• Track consists of two parts
• gap 10
• data 90

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Megatron 747 Disk (2)

• The capacity of the disk
• 8 surfaces 8192 tracks 256 sectors 512
  bytes 8G bytes
• A single track on average
• 256 sectors 512 bytes 128K bytes 1 Mbits
• A cylinder is of 1 Mbytes on average
• If a block is 4096 bytes (212)
• A block uses 8 sectors ( 4096 bytes / 512 bytes)
• A track consists of 32 blocks ( 256 sectors / 8)

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Megatron 747 Disk (3)

• Each track in Megatron 747 has the different
  numbers of sectors
• outer 320 sectors
• middle 250 sectors
• inner 192 sectors
• The outermost track
• 1,801,800 bit / 9.9 ? 182,000 bpi
• The innermost track
• 47,880 bit / 4.2 ? 114,000 bpi

• If each track had the same number (i.e. 256) of
  sectors, then the density of bits around the
  tracks would be greater
• Length of the outermost track
• 0.9 3.5 ? ? 9.9 inch
• 1 megabit / 9.9 ? 100,000 bits per inch
• Length of the innermost track
• 0.9 1.5 ? ? 4.2 inch
• 1 megabit /4.2 ? 250,000 bits per inch

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The Latency of The Disk
block x in memory
I want block X
disk access time

• Disk access time
• seek time
• rotational delay
• transfer time
• others

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Seek Time

• The time to position the head assembly at the
  proper cylinder
• 0(zero) already to be at the proper cylinder
• Otherwise move to be at the proper cylinder

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Rotational latency Time

• The time for disk to rotate the first of the
  sectors containing the block
• One rotation takes 10 ms, so rotational latency
  on average 5 ms.

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Transfer Time/Other delays

• Transfer Time
• the time to read/writes the data on the
  appropriate disk surface
• 10 Mbytes per second
• Other delays (here, those are neglected)
• taken by the processor and disk controller
• due to contention for the disk controller
• other delays due to contention

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Modifying Blocks

• Not possible to modify a block on disk directly
• Sequence of procedures
• Read block (time rt)
• Modify in memory (time mt)
• Write block (time wt)
• Verify (time vt) if appropriate
• Total time
• rt mt wt vt

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Example 2.3 (1)

• Let us examine the time to read a 4096-byte block
  from the Megatron 747 disk
• Characteristic
• 4 platters (8 surfaces), 1 surface 8192 tracks
• 1 track 256 sectors, 1 sector 512 bytes
• Disk rotates at 3840 RPM, one rotation 1/64 of
  a second
• To move the head assembly
• 1ms (to start and stop) 1ms for every 500
  cylinders
• Heads move one track in 1.002 ms
• To move heads from innermost to outermost track
• 1 (8192 / 500) 17.4 ms

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Example 2.3 (2)

• Minimum time (the best case)
• No seek time, no rotational latency, only
  transfer time
• Note 1 track 256 sectors, 1 sector 512 bytes
• 4096 bytes / 512 bytes 8 sectors (including 7
  gap)
• gaps/sectors occupy 10/90 of track
• A track has 256 gaps and 256 sectors
• 36 7/256 324 8/256 11.109 degrees
• (11.109/360)/64 4.8e-4 seconds 0.5 ms

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Example 2.3 (3)

• Maximum time (the worst case)
• full seek time and rotational latency, plus
  transfer time
• full seek time 17.4 ms
• full rotational time 1/64 of a second 15.6 ms
• transfer time 0.5 ms
• 17.4 15.6 0.5 33.5 ms

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Example 2.3 (4)

• Average Time
• Transfer time 0.5 ms
• Average rotational time half of the full
  rotation 7.8 ms
• Average seek time
• average distance traveled 1/3 of the disk
  2730 cylinders
• 1 2730/500 6.5ms
• 0.5 7.8 6.5 14.8 ms

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RAM model vs. I/O model computation

• I/O model computation
• Dominance of I/O cost
• Remember, 105 - 106 in-memory operations take
  the same time as one disk I/O
• Should minimize the number of block accesses
• Data Structure vs. File Processing

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Using Secondary Storage Effectively

• In general database
• Whole databases are much too large to fit in main
  memory
• Key parts of databases are buffered in main
  memory
• Disk I/Os occur frequently
• Main memory sorts (such as Quick sort) are
  inadequate

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Merge Sort
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Two-Phase, Multiway Merge-Sort (1)

• Phase 1
• Sort main-memory-sized pieces of the data
• Fill all available main memory with blocks
• Sort the records in main memory
• Write the sorted records

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Two-Phase, Multiway Merge-Sort (2)

• Phase 2
• Merge all the sorted sublists into a single
  sorted list
• Find the smallest key among the first remaining
  elements of all the lists
• Move the smallest element to the first available
  position of the output block
• If output block is full, write it to disk and
  reinitialize the same buffer
• Repeat until all input blocks become exhausted.

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Main-memory Organization
37
Merge Sort Example (1)

• Assumption
• 10,000,000 tuples, 1 tuple 100 bytes
• So, 1 Gbyte data
• 50 Mbytes memory available
• 4096 byte blocks, so each block contains 40
  records
• Total of blocks 250,000
• of blocks in main memory 12,800 ( 50220 /
  212)
• Number of sublists
• 19 sublists (12,800 blocks) 1 sublists (6,800
  blocks)
• Each block read or write 15 ms

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Merge Sort Example (2)

• Computation
• First phase
• Read each of the 250,000 blocks once
• Write 250,000 new blocks
• Total time
• (250,000 15 ms) 2 7500 seconds 125
  minutes
• Second phase
• Similar with the first phase
• Total time 125 minutes

39
Improving the Access Time of Secondary Storage

• Place blocks on the same cylinder
• Divide the data among several small disks
• Mirroring disks
• Use a disk-scheduling algorithm
• Prefetch blocks to main memory in anticipation of
  their later use

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Organizing Data by Cylinders

• Use several adjacent cylinders
• Read all the blocks on a single track or on a
  cylinder consecutively
• Neglect all but the first seek time and the first
  rotational latency

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Example 2.9 (1)

• Recall examples 2.3 and 2.7
• Original data may be stored on consecutive
  cylinders
• Total of cylinders 1000 ( 1Gbytes / 1M bytes)
• Main memory can hold 50 cylinders (i.e. 50M)
• To read 50 cylinder data into main memory
• 6.5 ms for average seek time
• 49 ms for 49 one-cylinder seeks (1 ms each)
• 6.4 seconds for transfer of 12,800 blocks
• (12,800 0.5 ms) / 1000 6.4 seconds
• So, 6.5 49 6,400 6455.5 ms

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Example 2.9 (2)

• First phase
• Read
• ((6.5 ms 49 ms 6.4 seconds) 20 times)
  2.15 minutes
• Write The same as reading
• Total time 4.3 minutes
• Second phase
• Still takes about 125 minutes (WHY ?)

43
Using Multiple Disks in place of One

• Use several disks with their independent heads
• Transfer data at a higher rate
• Roughly speaking, total time could be divided by
  the number of disks

44
Example 2.10 (1)

• Replace one 747 by four 737s which have one
  platter and two surfaces
• Assumption
• Divide the given records among the four disks
• Occupy 1000 adjacent cylinders on each disk
• Fill ¼ of main memory each disk
• Recall previous examples
• Average seek time and rotational latency 0
• Number of full memory blocks 12,800
• ¼ memory size 3,200 blocks

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Example 2.10 (2)

• Computation
• First phase
• Transfer time 3200 0.5 ms 1.6 seconds
• Read (6.5 ms 49 ms 1.6 seconds) 20 33
  sec.
• Write similar with reading
• Total time about 1 minute

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Example 2.10 (3)

• Second phase
• Apply delicate techniques (?) to reduce disk I/O
  time
• Start comparisons among the 20 lists as soon as
  the first element of the block appears in main
  memory
• Use four output buffers
• Total time about 1 hours (?)

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Mirroring Disks

• Two or more disks hold identical copies of data
• Survive a head crash by either disk
• If we make n copies of a disk, we can read any n
  blocks in parallel.
• Using mirror disks does not speed up writing, but
  neither does it slow writing down (to some
  extent)

48
Scheduling Requests by the Elevator Algorithm

• Disk controller choose which of several requests
  to execute first, to increase throughput
• Elevator Algorithm
• Proceed in the same direction until the next
  cylinder with blocks to access is encountered
• When no requests ahead in direction of travel,
  reverse direction

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Example 2.11
Finishing times for block accesses using the
elevator algorithm
Finishing times for block accesses using the
first-come-first-served algorithm
Arrival times for six block-access requests
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Prefetching Data on Track- or Cylinder-sized
Chunks

• Can we predict the order in which blocks will be
  requested from disk ?
• For example,
• Devote two block buffers to each list when merged
  (when there is plenty of memory)
• When a buffer is exhausted, switch to the other
  buffer for the same list

51
Single Buffering

• Single buffering
• Read B1 ? Buffer
• Process Data in Buffer
• Read B2 ? Buffer
• Process Data in Buffer ...

• Computation
• P time to process/block
• R time to read in 1 block
• n of blocks
• Single buffer time n(PR)

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Single Buffering vs. Double Buffering

• Memory
• Disk

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Double Buffering

• Computation
• P processing time/block
• R IO time/block
• n of blocks
• Double buffering time R nP
• Single buffering time n(RP)

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Prefetching

• Combine prefetching with the cylinder-based
  strategy
• Store the sorted sublists on whole, consecutive
  cylinders
• Read whole tracks or cylinders whenever we need
  some records from a given list

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Example 2.14 (1)

• Consider the second phase of the sort
• Have in main memory two track-sized buffers
• A track 128KB
• Total space requirement 128KB 20 lists 2 5
  Mbyte
• Read all the blocks on 1000 cylinders (8000
  tracks)
• Computation
• average seek time 6.5 ms
• the time for disk to rotate once 15.6 ms
• total time (for reading) (6.5 15.6) 8000
  2.95 minutes

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Example 2.14 (2)

• Have in main memory two cylinder-sized buffers
  per sorted sublist
• 1 cylinder 8 tracks 128K 8 1M
• Use 40 buffers of a megabyte each
• 50 megabytes available main memory
• Need only do a seek once per cylinder
• Read all the block on 1000 cylinders (8000
  tracks)
• Total time (for reading)
• (6.5 8 15.6) 1000 cylinders) 2.19 minutes

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Block Size Selection

• Big block ? amortize I/O cost
• Big block ? read in more useless stuff and takes
  longer to read
• As memory prices drop, blocks get bigger

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Disk Failures

• Intermittent failure
• An attempt to read or write a sector is
  unsuccessful, but with repeated tries we are able
  to read or write successfully.
• Media decay
• A bit or bits are permanently corrupted, and the
  sector becomes unreadable.
• Write failure
• We can neither write successfully nor can we
  retrieve the previously written sector.
• Disk Crash
• When a disk becomes unreadable permanently

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Checksums (1)

• Each section has additional bits, called the
  checksum, to check reading or writing operations
• (w, s)
• w the data that is read
• s a status bit
• A simple form of checksum parity

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Checksums (2)

• Example 1 (even parity)
• The sequence of bits in a sector 01101000
• The parity bit is 1
• Data becomes 011010001
• Example 2 (even parity)
• The sequence of bits in a sector 11101110
• The parity bit is 0
• Data becomes 111011100

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Checksums (3)

• Possible that we cannot detect an error if more
  than one bit of the sector may be corrupted
• If we use n independent bits as a checksum, then
  the chance of missing an error is only 1/2n (WHY
  ?)

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Stable Storage (1)

• How to correct errors ?
• Stable storage is a technique for organizing a
  disk so that media decays or failed writes do not
  result in permanent loss.
• The general idea is that sectors are paired, and
  each pair represents one sector-contents X
• As the left (XL) and right (XR) copies

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Stable Storage (2)

• Writing policy
• Write the value of X into XL
• if status is good, write the value
• if status is bad, repeat writing
• If fails after a number of times, a media failure
  in the sector
• Repeat above scheme for XR
• Reading policy (to obtain the value of X)
• Read XL
• if status bad is returned, repeat reading
• if status good is returned, take that value as X
• If cant read XL , repeat above with XR

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Recovery from Disk Crashes

• Disk crash is fatal in mission-critical
  applications
• RAID (redundant arrays of independent disks)
• Here, we talk levels 5, 6, and 7
• These RAID schemes also handle failures discussed
  previously

65
The Failure Model of Disks

• Mean time to failure represents the length of
  time by which 50 of a population of disks will
  have failed catastrophically.
• For modern disks, it is about 10 years

Fraction surviving
Time
66
RAID Level 1

• To protect against data loss
• Use mirroring disks
• The only way data can be lost is if there is a
  second disk crash while the first crash is being
  repaired.

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How often will a data loss occur?

• Assume
• The process of replacing the failed disk
• take 3 hours, 1/8 day, 1/2920 year
• A failure rate of 5 per year
• Probability that the mirror disk will fail during
  copying
• (1/20) (1/2920) 1/58,400
• Mean time to a failure involving data loss
• One of the two disks will fail once in 5 years on
  the average
• 5 58,400 292,000 years

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RAID Level 4 (1)

• Use one redundant disks no matter how many data
  disks there are
• In the redundant disk, the ith block consists of
  parity checks for the ith blocks of all the data
  disks
• Use modulo-2 sum an even parity

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The Algebra of Modulo-2 Sums

• The commutative law
• x ? y y ? x
• The associative law
• x ? (y ? z) (x ? y) ? z
• The all-0 vector of the appropriate length is the
  identity for ?
• x ? O O ? x x
• ? is its own inverse
• x ? x O
• If x ? y z, y x ? z

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RAID Level 4 Reading (2)

• Read disks normally.
• We could read the redundant disk !
• Example
• read disk 2, 3, and 4, and get the contents of
  disk 1 using modulo-2 sum.

disk2 10101010 disk3 00111000 disk4
01100010 disk1 11110000
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RAID Level 4 Writing (3)

• When a block is written, we need to change the
  redundant disk
• Naïve approach
• N-1 reads of blocks not being rewritten
• One write of new block
• Rewrite new redundant disk
• In total, N1 disk I/Os
• There is a better way to do that !

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Writing Example (4)

• When disk 2 changes from 10101010 to 11001100

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RAID Level 4 Failure Recovery (5)

• Recomputing any missing data is simple, and does
  not depend on which disk (data or redundant) is
  failed.

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RAID Level 5

• We could treat each disk as the redundant disk
  for some of the blocks
• That is, do not have to treat one disk as the
  redundant disk and the others as data disks
• When there are n1 disks (disk 0 disk n)
• If (i mod n1) j, then we can treat the ith
  cylinder of disk j as redundant

75
Example 2.21 (1)

• How redundant blocks compute for 4 disks (n3)?
• Disk 0
• redundant for block 4, 8, 12,
• Disk 1
• redundant for block 1, 5, 9,
• Disk 2
• redundant for block 2, 6, 10,
• Disk 3
• redundant for block 3, 7, 11,

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Example 2.21 (2)

• The reading and writing load for each disk is the
  same
• If all blocks are equally likely to be written
• each disk has a 1/4 chance
• If not
• each disk has a 1/3 chance
• Each of four disks is involved in ½ of the writes
• 1/4 3/4 1/3 1/2

77
RAID Level 6 (1)

• To handle with any number of disk crashes data
  or redundant
• Here, focused on a simple example, where two
  simultaneous crashes are correctable and the
  strategy is based on a simple error-correcting
  code, Hamming code
• Consider a system with seven disks
• data disks disk 1-4
• redundant disks disk 5-7

78
RAID Level 6 (2)

• The relationship between data and redundant disks
• Note
• every possible column of three 0s and 1s,
  except for the all-0 column
• the columns for the redundant disk have a singe 1
• the columns for the data disks each have at least
  two 1s

79
RAID Level 6 (3)

• The disks with 1 in a row are treated as if they
  were the entire set of disks in a RAID level 4
  scheme.
• The bits of disk 5
• are the modulo-2 sum of bits of disk 1,2, and 3
• The bits of disk 6
• are the modulo-2 sum of bits of disk 1,2, and 4
• The bits of disk 7
• are the modulo-2 sum of bits of disk 1,3, and 4

80
RAID Level 6 Read/Write

• Reading Just read data from any data disk
  normally
• Writing
• Need to recalculate several redundant disks

81
A Writing Example (1)

• Writing
• Disk 2 is changed to be 0000111
• Corresponding redundant disks
• disk 5 and 6
• Using modulo-2 sum
• between old and new disk 2
• between modulo-2 sum of disk 2s and disk 5
• between modulo-2 sum of disk 2s and disk 6

82
A Writing Example (2)
83
RAID Level 6 Failure Recovery

• Assume that disk a and b fails simultaneously
• Find a row r in which the columns of a and b are
  different
• For example, a has 0 in row r, b has 1 in row r
• Compute the correct b by taking the modulo-2 sum
  of corresponding bits from all the disks other
  than b that have 1 in row r.
• Then, compute the correct a

84
A Recovery Example

• Pick the second row
• Disk 2
• modulo-2 sum of disks 1, 4, and 6
• 00001111
• Disk 5
• modulo-2 sum of disks 1, 2, and 3
• 11000111