Understanding Striping, Redundancy, and RAID Levels in Database Management Systems, Slides of Introduction to Database Management Systems

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Database Management Systems Design

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The problem with Striping

• Striping has the advantage of speeding up disk access time.
• But the use of a disk array decreases the reliability of the storage system because more disks mean more possible points of failure.
• Mean-time-to-failure (MTTF)
  • Mean time to have the disk fail and lose its data
• MTTF is inversely proportional to the number of components in used by the system. - The more we have the more likely they will fall apart!

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MTTF in a disk array

• Suppose we have a single disk with a MTTF of 50,000 hrs (5.7 years).
• Then, if we build an array with 50 disks, then we have a MTTF for the array of 50,000/50 = 1000 hrs, or 42 days!, because any disk can fail at any given time with equal probability. - Disk failures are more common when disks are new (bad disk from factory) or old (wear due to usage).
• Morale of the story: More does not necessarily means better!

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Increasing MTTF with redundancy

• We can increase the MTTF in a disk array by storing some redundant information in the disk array. - This information can be used to recover from a disk failure.
• This information should be carefully selected so it can be used to reconstruct original data after a failure.
• What to store as redundant information?
  • Full data block
  • Parity bit for a set of bit locations across all the disks
• Where to store it?
  • Check disks – disks in the array used only for this purpose
  • All disks – spread redundant information on every disk in the array.

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Redundancy Unit: Parity bits

• Consider an array of N disks. Suppose k is the number of the k-th block in each disk. Each block consists of several kilobytes, and each byte is 8-bit.
• We can store redundant information about the i-th bit position in each data block. - Parity bit
• The parity bit gives the number of bits that were set to the value 1 in the group of corresponding bit locations of the data blocks.
• For example, if bit 1024 has a parity 0, then an even number of bits where set 1 at bit position 1024. Otherwise its value must be 1.

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Parity bits

• Consider bytes:
  • b1 = 00010001, b2 = 00111111, b3 = 00000011
• If we take the XOR these bytes we get 00010001 00111111 00000011 00101101 - this byte has the parity for all bits in b1, b2, b
• Notice the following:
  • For bit position 0, the parity is 1,meaning an odd number of bits have value 1 for bit position 0.
  • For bit position 1, the parity is 0, meaning an even number of bits have value 1 for bit position 1

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Reliability groups in disk array

• Organize the disks in the array into groups called reliability groups.
• Each group has:
  • 1 or more data disks , where the data blocks are stored
  • 0 or more check disks where the blocks of parity bits are stored
• If a data disk fails, then the check disk(s) for its reliability group can be used to recover the data lost from that disk.
• There is a recovery algorithm that works for any failed disk m in the disk array.
• Can recover from up to 1 disk failure.

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Recovery algorithm

• Suppose we have an array of N disks, with M check disks (in this case there is one reliability group).
• Suppose disk p fails. We buy a replacement and then we can recover the data as follows.
• For each data block k on disk p:
  • Read data blocks k on every disk r, with r != p
  • Read parity block k from its check disk w
  • For each bit position i in block k of disk p:
    • Count number of bits set 1 at bit i in each block coming from a disk other than p. Let this number be j
    • If j is odd, and parity bit is 1 then bit position i is set to 0 Docsity.com

RAID Organization

• RAID:

• Originally: Redundant Array of Inexpensive Disks
• Now: Redundant Array of Independent Disks
• RAID organization combines the ideas of striping, redundancy as parity bits, and reliability groups.
• RAID system has one or more reliability groups
• For simplicity we shall assume only one group …
• RAID systems can have various number of check disks for reliability groups, depending on the RAID level that is chosen for the Docsity.com

RAID Analysis

• Suppose we have a disk array with 4 data disks.
• Let’s analyze how many check disks we need to build a RAID with 1 reliability group of 4 data disks plus the check disks.
• Note: Effective space utilization is a measure of the amount of space in the disk array that is used to store data. It is given as a percentage by the formula:

Total Disksinthe Array

Effective SpaceUtilization = Data Disks

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RAID Level 1

• RAID Level 1: Mirrored
• Each data block is duplicated:
  • original copy + mirror copy
• No striping!
• Most expensive solution since it requires twice the space of the expected data set size.
• Every write involves two writes (original + copy) - cannot be done simultaneously to prevent double corruption - First write on data disk, then on copy at mirrorDocsity.com

RAID Level 1 (cont…)

• In RAID Level 1, the data block can be fetched from the disk with least contention.
• Since we need to pair disks in groups of two (original + copy), the space utilization is 50%, independent on the amount of disks.
• RAID Level 1 is only good for small data workloads where the cost of mirroring is not an issue.

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RAID Level 1+0 (cont…)

• Space utilization is 50% (half data and half copies)
• RAID Level 1+0 is better than RAID 1 because of striping.
• RAID Level 1+0 is good for workloads with small data sets, where cost of mirroring is not an issue.
• Also good for workloads with high percentages of writes, since a write is always 2Docsity.com

RAID Level 2

• RAID Level 2: Error-Correcting Codes
• Uses striping with a 1-bit striping unit.
• Hamming code for redundancy in C check disks. - Can indicate which disk failed - Make number of check disk grow logarithmically with respect to the number of data disks. (???)
• Read is expensive since to read 1 bit we need to read 1 physical data block, the one storing the bit. Docsity.com