Storage and Fault Tolerance

Question 1

What does RAID stand for?

Answer 1

Redundant Array of Independent Disks

Question 2

What is a RAID?

Answer 2

• Several disks play the role of one
• Each disk detects its own errors using codes in each sector

Question 3

What should a RAID do?

Answer 3

• Provide better performance
• Normal read/write even when we have a bad sector or a whole disk fails.

Question 4

What is RAID0?

Answer 4

Uses “striping” (track 0 on first disk, track 1 on second, etc) for increased throughput

Question 5

What is RAID0’s performance?

Answer 5

• Nx data throughput (N is number of disks)
• less queuing delay (latency)

Question 6

MTTF?

Answer 6

Mean Time to Failure

Question 7

MTTDL

Answer 7

Mean Time to Data Loss

Question 8

Failure rate

Answer 8

f = failures per disk per second

Question 9

Single disk MTTF

Answer 9

1/f

Question 10

Single disk MTTDL

Answer 10

Single disk MTTF

Question 11

What is the MTTF of RAID0 with N disks?

Answer 11

MTTF_N = MTTDL_N = MTTD_1 / N

Question 12

What is RAID1?

Answer 12

Uses “mirroring” (same data on multiple disks) for reliability.

Question 13

What is the throughput of a RAID0 with N disks?

Answer 13

N x the throughput of one disk

Question 14

What is throughput of RAID1?

Answer 14

Write: same as 1 disk
Read: N x throughput of one disk

Question 15

Reliability of RAID1?

Answer 15

RAID1 tolerates any faults that affect one disk. Has ECC on each disk sector, so it knows where the error is and on which disk. It can then use the mirror copy on the other disk instead.

Question 16

MTTR

Answer 16

Mean Time to Repair

Question 17

What is the MTTDL for RAID1 if we don’t repair a damaged disk?

Answer 17

MTTDL = (MTTF_1/N) + MTTF_1

Question 18

What is the MTTDL for RAID1 if we DO repair a damaged disk?

Answer 18

MTTDL = (MTTF_1/N)*(MTTF_1/MTTR)

i.e., MUCH MUCH MUCH better than not repairing.

Question 19

What is RAID4?

Answer 19

Uses “block-interleaved parity” to both improve performance and reliability.

For N disks, N-1 have data striped just like in RAID0 and 1 has the parity bits calculated from the other disks using bitwise XOR.

Question 20

What is the throughput of RAID4?

Answer 20

Reads: (N-1) * throughput of one disk
Writes: 1/2 the throughput of one disk!

When we write in RAID4, we need to write to a disk, read from the parity disk, and write to a parity disk. The parity read can happen in parallel with the main write, so overall it takes twice as long.

THIS is why we need RAID5.

Question 21

MTTF of RAID4

Answer 21

(MTTF_1/N) * (MTTF_1/(N-1)*MTTR_1)

But what’s the N-1 about? This is because it takes MTTF_1/N time for the first failure. We plan to repair before the second failure, which takes MTTF_1/(N-1) * 1/MTTR_1.

Question 22

How do we compute new parity on RAID4 write?

Answer 22

First we XOR the new data and old data for the block we’re writing to (which finds all changes). Then we XOR that with the old parity. The result is stored as the new parity.

Because we only have one parity disk, that creates a bottleneck on writes. Hence the need for RAID5.

Question 23

What is RAID5?

Answer 23

DISTRIBUTED block-interleaved parity. Similar to RAID4, but the parity stripes are distributed among all disks (the first might be on disk 4, the next on disk 1, the next on 2, etc).

Question 24

What is the throughput of RAID5?

Answer 24

Where N is the total number of disks…

Reads: N * throughput of one disk (because now we can actually read from all N disks at once!)
Writes: N/4 * throughput of one disk! (because we still need 4 total accesses per write, but they’re distributed over all N disks)

Question 25

What is the reliability of RAID5?

Answer 25

Same as RAID4! We can always recover from the loss of one disk. If we lose parity, we can still read/write from the other disks. If we lose one data disk, we can reconstruct data from parity.

So it’s just as reliable, without bottleneck on writes.

MTTF is the same as RAID4.

Question 26

MTTF of RAID5

Answer 26

Same as RAID4!

(MTTF_1/N(N-1)) * (MTTF_1/MTTR_1)

But what’s the N-1 about? This is because it takes MTTF_1/N time for the first failure. We plan to repair before the second failure, which takes MTTF_1/(N-1) * 1/MTTR_1.

Question 27

What is RAID6?

Answer 27

• Two parity blocks per group.
• Can work when two stripes per group have failed
• One true parity block
• The other “parity” block is a different kind of check block
• When one disk fails, use parity
• When two fail, solve some equations using both parity and “parity” blocks to recover data

Question 28

When should you use RAID6?

Question 29

RAID5 vs RAID6

Answer 29

• RAID6 has 2x the overhead (what is the “overhead”?)
• More write overhead: when we write, we need to read and write the data block plus BOTH check blocks
• RAID6 is only useful when the chance of a second disk failing during our MTTR is actually high. This is unusual.

Question 30

Is RAID6 overkill?

Answer 30

RAID6 seems like overkill when you assumes disk failures are independent. Under RAID5, the likelihood of a second independent failure happening during the repair period is very low. So why use RAID6?

Because disk failures are NOT necessarily independent. For example, say you remove the wrong disk when trying to replace the failed one: you now have a second, correlated failure. RAID6 would’ve been nice.