Advanced Database System

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Advanced Database System

• CS 641
• Lecture 4
• Jan 24th 2008

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Accelerating access to secondary storage

• Place blocks that are accessed together on the
  same cylinder.
• Divide the data among several smaller disks
  rather than a large one.
• Mirror a disk
• Use a disk scheduling algorithm
• Prefetch blocks to main memory

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Organizing data by cylinders

• Objective reduce seek time
• Method analyze application behavior, put data
  that is likely to be accessed together on a
  single cylinder or adjacent cylinders
• Thus, if read all the blocks on a single track or
  on a cylinder consecutively, we can neglect all
  but the first seek time and the first rotational
  latency.

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Example

• Megatron 747
• Avg transfer time ,seek time and rotational
  latency are 0.25ms, 6.46ms and 4.17ms
  respectively.
• Sorting for 10,000,000 records need 74mins in
  TPMMS.
• Each cylinder stores 8M bytes (512 blocks)
• We store data on 100,000/512196 cylinders.
• We must read 100M/8M13 different cylinders to
  fill main memory once.

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Example (Cont.)

• The total time to fill main memory once
• 6.46ms for one avg seek
• 12 ms for 12 one-cylinder seeks
• 1.60s for 6400 blocks (0.25ms per block)
• We need to fill main memory 16 times. Thats
  about 1.61625.6s.
• Both read and write operations take 225.6s51.2s
• However, this mechanism can not help phase 2
  because we have to read/write one block each
  time. Still need 37mins.(?)

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Using multiple disks

• For single disk, only one head can read/write
  data at a certain time.
• With multiple disks, we can read/write data onto
  different disks at the same time.

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Example

• Replace Megatron 747 with 4 Megatron 737.
• Divide the given records among 4 disks. Thus the
  time to fill 100M main memory is 1.6/40.4s
• Entire phase 1 takes 51.2/412.8s
• In phase 2, to take advantage of 4 disks, TPMMS
  need to be modified
• Comparison starts as soon as the first element of
  the new block appears in the main memory
• The speedup is about in the 2-3 times range.

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Mirroring disks

• Have two or more disks hold identical copies of
  data.
• It is used for reliability, however, it can also
  be use to speed up access to data.
• For example, if we have four copies, system can
  guarantee to retrieve 4 blocks at once. (read in
  parallel)
• It can speed up read, but not write.

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Disk scheduling

• Useful to reduce the access latency for many
  small processes that each access a few blocks.
• Definition In disk I/O, seek time and rotational
  latency are very important. Since all disk
  requests are linked in queues, reduce them
  causing the system to speed up. Disk Scheduling
  Algorithms are used to reduce the total seek time
  and rotational latency of all requests.

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Shortest Seek Time First (SSTF)
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Elevator
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Example
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Elevator
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FCFS
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Comments

• Elevator algorithm can further improve the
  throughput when the average number of requests
  waiting for the disk increases.

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Example

• Assume megatron 747, 16384 cylinders.
• 1000 I/O requests
• avg 10.254.175.42ms
• compare to previous random access (10.88ms)
• 1000 I/Os finish in 5.42s, avg delay to satisfy
  a request is half of it 2.71s
• 32768 I/O requests
• avg 1/20.25(1/2)(2/3)8.333.53ms
• 32768 I/Os finish in 116s, avg delay to satisfy
  a request is 58s.

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Prefetching (double buffering)

• Load the blocks into main memory before they are
  needed.
• Advantage better schedule disk I/Os.
• We can gain speedup in block access.

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Example

• In previous example, if two blocks are given to
  each sorted sublist (in phase 2), in case the
  records in one block are used up, we can switch
  to another one without delay.
• However, if those 100,000 blocks are random
  accessed, we did not get much benefits. To solve
  that
• Store the sorted sublists on whole.
• Read whole tracks or who cylinder.

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Another Example

• In previous example, each sublist has 2
  track-sized buffers, need 16M main memory.
• Time to read the whole track 6.468.3314.79ms.
• We have 196 cylinders (3136 tracks)
• The time to read 313614.79ms46.4s
• 2 cylinder-sized (16 tracks) buffers, need 256M
  main memory
• Time to read the whole cylinder
  6.46168.33140ms.
• The time to read 196140ms27.44s

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Two different applications

• A) regular situation, where blocks can be read
  and written in a sequence that can be predicted
  in advance. Only one process using the disk
  (TPMMS phase 1)
• B) a collection of short processes, execute in
  parallel, share the same disk, and can not be
  predicted in advance. (TPMMS phase 2)

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Cylinder-based organization

• advantage excellent for A)
• disadvantage no help for B)

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Multiple disks

• Advantage increasing read/write access rates for
  both applications
• Problems read/write to the same disk can not be
  satisfied at the same time.
• Disadvantage cost

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Mirroring

• Advantage increasing read rates for both
  applications improve fault tolerance for both
  applications
• Disadvantage pay for two or more disks but get
  the capacity of one.

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Disk scheduling (Elevator algorithm)

• Advantage reduce the average time to read/write
  when the accesses to blocks are unpredictable.
• Problem it is more efficient for a large amount
  of pending I/O requests and the average delay for
  the processes are high

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Prefetching

• Advantage speeds up access when the needed
  blocks are known but the timing of requests is
  data-dependent.
• Disadvantage requires extra main memory buffers.
  No help when accesses are random.

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Exercises 11.5.2

• Suppose use two Megatron 747 disks as mirrors.
  However, instead of allowing reads of any blocks
  from either disk, we keep the head of the first
  disk in the inner half of the cylinder, and the
  head of the second disk in the outer half of the
  cylinders. Assuming read requests are on random
  tracks
• What is the average rate at which this system can
  read blocks?
• How does this rate compare to no restriction?
• What disadvantages do you foresee for this system?

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Disk failure

• Intermittent failure an attempt to read/write a
  sector fails, but repeated tries succeed.
• Media decay a bit or bits are permanently
  corrupted, thus the corresponding sector is
  damaged.
• Write failure attempt to write a sector is fail
• Disk crash the entire disk becomes unreadable,
  suddenly or permanently.

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Techniques

• Parity check detect intermittent failures
• Stable storage prevent media decay/write failure
  damages
• RAID coping with disk crashes.
• Basic idea using additional storage to keep
  redundant information.

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A useful model

• Disk sectors are ordinarily stored with some
  redundant bits.
• For read a pair(w,s) is returned, where w is the
  data in the sector that is read, and s is a
  status bit that tells whether or not the read was
  successful.

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Checksum

• Additional bits are set depending on the values
  of the data bits stored in that sector.
• A simple form is based on the parity of all the
  bits in the sector.
• If an odd number of 1s among the bits, it is odd
  parity, or their parity bit is 1.
• If an even number of 1s among the bits, it is
  even parity, or their parity bit is 0.
• Thus, the number of 1s among a collection of
  bits and their parity bit is always even.

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Example

• 01101000, the parity bit is 1, will be stored as
  011010001
• 11101110, the parity bit is 0, will be stored as
  111011100.
• If one of these bits fail, it can be detected, if
  more than one bits corrupted, 50 the error can
  not be detected.
• More parity bits can be used to detect more
  errors.
• In general, if n independent bits are used as a
  checksum, the chance of missing an error is only
  1/2n

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Stable storage

• Problem checksum can detect error, but can not
  fix the error. For critical applications, such as
  banks, airlines, this is not enough.
• Stable storage pair the sectors, and each pair
  representing one sector-contents X.
• We call these two copies as XL and XR
• If the read returns (w, good) for either XL or
  XR, then w is the true value of X.

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Writing policy

• Write the value of X into XL. Check that the
  value has status good. Otherwise, repeat write,
  if still not succeed for several times, a media
  failure is detected, find a spare sector for XL
• Repeat for XR

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Reading policy

• To obtain the value of X, read XL. If status
  bad, repeat several times. If eventually status
  becomes good, take the value for X
• If can not read XL, repeat for XR

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Error-handling capabilities

• Media failure
• Only both of them fail, then we can not get the
  value for X otherwise, we can read X.
• Write failure if a system failure (power
  outrage) occurs when write X, failure can be
  detected
• The failure occurred as we were writing XL,
  status of XL becomes bad, but XR has the old
  value for X
• The failure occurred after we wrote XL, XL status
  is good, write XR with the value in XL

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Exercises

• 11.6.1

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Recovery from disk crashes

• The most common strategy is RAID, Redundant
  Arrays of Independent Disks.

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The failure model for disks

• Mean time to failure (MTTF) the length of time
  by which 50 of a population of disks will have
  failed catastrophically.
• For modern disks 10 years.

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A survival rate curve for disks
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Mechanism

• One or more disks that hold the data (data disks)
  and,
• adding one or more disks that hold information
  that is completely determined by the contents of
  the data disks (redundant disks)
• In case a disk crash of either a data disk or a
  redundant disk, the other disks can be used to
  restore the failed disk, so no permanent
  information loss.

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Mirroring (RAID 1)

• Mirror each data disk with a redundant disk.
• Data loss occurs only when both disks fail
  simultaneously.
• Example
• A disk has 10 probability to fail each year
• Replace a new one takes 3 hours,(1/2920 of a
  year).
• The probability the mirror disk fails is 1/29200.
• One of two disks will failure once in 5 years on
  average
• The mean time to data loss is 1/5(1/29200)1/1460
  00 or once per 146,000 years.

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Parity blocks (RAID 4)

• Mirroring approach use as many redundant disks as
  data disks.
• RAID 4 only use 1 redundant disk no matter how
  many data disks there are.
• Mechanism assume all the disks are identical, in
  the redundant disk, the ith block consists of
  parity checks for the ith blocks of all the data
  disks.

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Example

• 3 data disks 1,2,3. A redundant disk 4.
• Disk 1 11110000
• Disk 2 10101010
• Disk 3 00111000
• Then on redundant disk
• Disk 4 01100010

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Reading example

• Suppose we are reading a block from disk 1, and
  another request comes in to read a different
  block of the same disk.
• Ordinary, we have to wait for the first request
  to finish.
• However, we can computer it by reading blocks
  from other disks.

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Reading example (Cont.)

• Disk 2 10101010
• Disk 3 00111000
• Disk 4 01100010
• We can get
• Disk 1 11110000

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Writing

• When write a new block of a data disk, we need to
  change the corresponding block of the redundant
  disk as well.
• Assume n data disks, a naive approach needs n1
  disk I/Os.
• A better approach only check old and new version
  of the data block to be written.
• Read the old value of the block to be changed
• Read the corresponding block of the redundant
  disk
• Write the new data block
• Recalculate and write the block of the redundant
  disk.

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Writing example

• Disk 1 11110000
• Disk 2 10101010
• Disk 3 00111000
• Disk 4 01100010
• Now, change the block in disk 2 to 11001100
• Thus, the values in positions 2,3,6 and 7 have
  been changed
• The block on redundant disk will do the modulo-2
  sum of 0110110 to 01100010, and get 00000100.
• 4 disk I/Os are needed.

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Failure recovery

• If redundant disk fails, replace with a new disk,
  and recompute the redundant blocks
• If one of the data disk fails, replace with a new
  disk, and recompute its data from other disks.
• Rule the bit in any position is the modulo-2 sum
  of all the bits in the corresponding positions of
  all the other disks.

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Example

• If disk 2 fails
• Disk 1 11110000
• Disk 2 ????????
• Disk 3 00111000
• Disk 4 01100010
• Take the modulo-2 sum of each column
• Disk 2 10101010

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An improvement RAID 5

• Problem for RAID 4
• if n data disks, the number of disk writes to
  the redundant disk will be n times the average
  number of writes to any one data disk.
• To solve it, RAID 5 treat each disk as the
  redundant disk for some of the blocks.

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Example
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Example

• N4 (4 disks)
• average access to each disk is
• 1/43/41/31/2 of the writes.

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Coping with multiple disk crashes (RAID 6)

• RAID 4 and 5 can not deal with multiple errors.
• To deal with two simultaneous crashes, hamming
  code (error-correcting code) is used.
• In the following discussion
• 4 data disks (1,2,3,4) and 3 redundant disks
  (5,6,7)

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Redundancy pattern
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Rules

• The bits of disk 5 are the modulo-2 sum of the
  corresponding bits of disks 1,2, and 3
• The bits of disk 6 are the modulo-2 sum of the
  corresponding bits of disks 1,2, and 4
• The bits of disk 7 are the modulo-2 sum of the
  corresponding bits of disks 1,3, and 4

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Reading

• We can read data from any data disk normally. The
  redundant disk can be ignored.

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Writing

• The idea is similar, but now several redundant
  disks may be involved.

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Example

• Disk 1 11110000
• Disk 2 10101010
• Disk 3 00111000
• Disk 4 01000001
• Disk 5 01100010
• Disk 6 00011011
• Disk 7 10001001
• Rewrite disk 2 to be 00001111, positions 1,3,6,8
  changes.
• Since disk 5 and 6 calculate the bits involve
  disk 2, we must modify them accordingly.
• Disk 5 11000111
• Disk 6 10111110

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Failure recovery (2 simultaneous disk crashes)

• Disk 1 11110000
• Disk 2 ????????
• Disk 3 00111000
• Disk 4 01000001
• Disk 5 ????????
• Disk 6 10111110
• Disk 7 10001001
• Since disk 6 are calculated with disk 1,2,4. we
  can recover the data in disk 2 with disk 1,4,6.
• Disk 2 00001111
• Disk 5 is calculated with disk 1,2,3.
• Disk 5 11000111

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Exercises

• 11.7.1
• 11.7.4 a)
• 11.7.5 a)
• 11.7.6
• 11.7.8 a)

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Summary

• Memory Hierarchy
• Tertiary storage
• Disk/secondary storage
• Blocks, sectors, cylinders, etc.
• Disk access time
• TPMMS
• Speed up disk access
• Disk failure model
• Checksum
• Stable storage
• RAID