13.4.4
Error-Handling Capabilities of Stable
Storage
By: Bryant Panyarachun ID: 210
Types of Failures
• Media Failures
• If, after storing X in
sectors XL and XR, one of
them undergoes a media
failure, we can always
read X from the other,
unless they have both
failed.
• Probability of both failing
is extremely small.
• Write Failure
• As we write X there is a
system failure.
• Possible that X can be lost
in main memory, and
copy of X being written is
garbled.
• Ex. Half the sector has
new value of X while
other half stays the same.
Write failure: possible cases
Failure while writing XL
• XL will be “bad” but XR will
be “good”
• Thus, we can obtain the old
value of X. We may also
copy XR into XL to repair the
damage to XL.
• Unless there is a coincident
media failure at XR, which is
extremely unlikely.
Failure after writing XL
• We expect that XL will be
“good,” so we can read the
new value of X from XL.
• Since XR may not have the
correct value of X, we
should copy XL into XR.
13.4.5
Recovery from Disk Crashes
• Most serious mode of failure:
– Disk Crash or Head Crash.
– Data is permanently destroyed.
– No way to recover data unless was previously backed
up.
• Several schemes developed to reduce risk of data
loss
– Generally involve: redundancy, parity checks,
duplicated sectors.
– Commonly called RAID (redundancy arrays of
independent disks)
• Rate of disk crashes
– Measured by the mean time to failure
• time after 50% of population can be expected to fail and be
unrecoverable.
• Modern disks mean time to failure is about 10 years.
• If mean time to failure is n years, then in any given year,
1/nth of the surviving disks fail.
• Reality: disks tend to fail early or late.
– Mean time to disk crash not same as mean time to
data loss.
• There are a number of schemes to recover the data if a disk
crashes.
• Involves multiple disks; data disks and redundant disks.
• Either disks can be used to restore the other in case of a disk
crash.
13.4.6
Mirroring as a Redundancy Technique
Mirroring each disk
• Simplest scheme
• Referred to as RAID Level 1.
• Gives a mean time to memory loss that is
much greater than the mean time to disk
failure.
• Only way data can be lost is if there is a
second disk crash while the first crash is being
repaired.
Mirroring example
• Suppose each disk has a 10-year mean time to failure.
– Probability of failure in any given year is 10%.
• When disk fails, only need to replace it with a good
disk and copy the mirror disk to the new one.
• How often will both disks fail?
– Suppose process of replacing failed disk is 3 hours = 1/8 of
a day, or 1/2920 of a year.
– Since the average disks lasts 10 years, probability that
mirror disk will fail is (1/10) X (1/2920) or one in 29,200.
– If one disk fails every 10 years, then one of the 2 disks will
fail once in 5 years on average. One of 29,200 results in
data loss.
– Result mean time to failure involving data loss is 5 x 29,200
= 146,000 years.
13.4.7
Parity Blocks
• Mirroring uses same number of redundant
disks as data disks.
• Alternate approach: RAID 4
– Uses only 1 redundant disk regardless of number
of data disks.
– In the redundant disk, the ith block consists of
parity checks for the ith block of all data disks.
• The jth bits of all the ith blocks, including both the data
disks and the redundant disk, must have an even
number of 1’s among them. We choose the bit of the
redundant disk to make this condition true.
RAID 4
• In the redundant disk, choose bit j to be 1 if an
odd number of the data disks have 1 in that bit,
and we choose bit j of the redundant disk to be 0
if there are an even number of 1’s in that bit. This
calculation is called the modulo-2 sum.
Ex. Data disk: 1, 2, 3. Redundant disk: 4
Disk 1: 11110000
Disk 2: 10101010
Disk 3: 00111000
Disk 4: 01100010
Writing with RAID 4
• When writing a new block, we need to update
the corresponding block of the redundant disk.
– Naïve approach: read all corresponding data blocks of
all disks, calculate modulo-2 sum, and rewrite block of
redundant disk. Requires n + 1 disk I/O’s.
– Better to take modulo-2 sum of old a new data. Tells
us where there is a change in the number of 1’s
among the blocks.
– Any even number of 1’s changes to an odd number.
Changes are always by one; any even number of 1’s
changes to an odd number. If we change the same
positions of the redundant block, number of 1’s
becomes even again.
– Only requires 4 disk I/O’s
Writing with RAID 4
• 4 disk I/O’s
1. Read the old value of the data block being
changed.
2. Read the corresponding block of the redundant
disk.
3. Write the new data block.
4. Recalculate and write the block of the redundant
disk.
Writing with RAID 4 Example
Ex.
Disk 1: 11110000
Disk 2: 10101010
Disk 3: 00111000
Redundant disk: 01100010
• Want to change disk 2 to 11001100.
• Take modulo-2 sum of old and new values of
block on disk 2, to get 01100110.
–
–
01100110 shows that we must change positions, 2,
3, 6, 7 on block of redundant disk.
Replace redundant block with 00000100. (basically
modulo-2 sum of itself and 01100110.)
Failure Recovery
• If redundant disk crashes, replace new disk
and recompute the redundant blocks.
• If a data disk crashes, need to replace disk and
recompute data from the other disks.
• Since the number of 1’s among corresponding
bits of all disks is even, the bit in any position
is the modulo-2 sum of all the bits in the
corresponding positions of all the other disks.
Failure Recovery Example
• If the bit in question is 1, the number of corresponding
bits in the other disks that are 1 must be odd; their
modulo-2 sum is 1. If the bit in question is 0, then
there are an even number of 1’s among the
corresponding bits of the other disks; their modulo-2
sum is 0;
Disk 1: 11110000
Disk 2: ????????
Disk 3: 00111000
Disk 4: 01100010
• Taking the modulo-2 sum of each column gives us
10101010.