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Lecture 4: A Case for RAID (Part 2) Prof. Shahram Ghandeharizadeh Computer Science Department University of Southern California.

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Presentation on theme: "Lecture 4: A Case for RAID (Part 2) Prof. Shahram Ghandeharizadeh Computer Science Department University of Southern California."— Presentation transcript:

1 Lecture 4: A Case for RAID (Part 2) Prof. Shahram Ghandeharizadeh Computer Science Department University of Southern California

2 Smaller & Inexpensive Disks 25% annual reduction in size; 40% annual drop in price 1 inch in height, weighs 1 ounce (16 grams) 1 GB, Year 2008 IBM $125 Size of a refrigerator, 550 pounds (250 Kg) 1 GB, Year 1980 IBM $40,000

3 Inexpensive Disks Less than 9 Cents / Gigabyte of storage

4 Challenge: Managing Data is Expensive Cost of Managing Data is $100K/TB/Year : High availability: Down time is estimated at thousands of dollars per minute. Data loss results in lost productivity: 20 Megabytes of accounting data requires 21 days and costs $19K to reproduce. 50% of companies that lose their data due to a disaster never re- open; 90% go out of business in 2 years!

5 Challenge: Managing Data is Expensive Cost of Managing Data is $100K/TB/Year : High availability: Down time is estimated at thousands of dollars per minute. Data loss results in lost productivity: 20 Megabytes of accounting data requires 21 days and costs $19K to reproduce. 50% of companies that lose their data due to a disaster never re- open; 90% go out of business in 2 years! RAID

6 MTTF, MTBF, MTTR, AFR MTBF: Mean Time Between Failures Designed for repairable devices Number of hours since the system was started until its failure. MTTF: Mean Time To Failures Designed for non-repairable devices such as magnetic disk drives Disks of 2008 are more than 40 times more reliable than disks of MTTR: Mean Time To Repair Number of hours required to replace a disk drive, AND Reconstruct the data stored on the failed disk drive. AFR: Annualized Failure Rate Computed by assuming a temperature for the case (40 degrees centigrade), power-on-hours per year (say 8,760, 24x7), and 250 average motor start/stop cycles per year.

7 Focus on MTTF & MTTR MTTF: Mean Time To Failures Designed for non-repairable devices such as magnetic disk drives Disks of 2008 are more than 40 times more reliable than disks of MTTR: Mean Time To Repair Number of hours required to replace a disk drive, AND Reconstruct the data stored on the failed disk drive.

8 Assumptions MTTF of a disk is independent of other disks in a RAID. Assume: The MTTF of a disk is once every 100 years, and An array of 1000 such disks. The MTTF of any single disk in the array is once every 37 days.

9 RAID RAID organizes D disks into nG groups where each group consists of G disks and C parity disks. Example: D = 8 G = 4 C = 1 nG = 8/4 = 2 Disk 1Disk 2Disk 3Disk 4Parity 1Disk 5Disk 6Parity 2Disk 7Disk 8 Parity Group 1Parity Group 2

10 RAID RAID organizes D disks into nG groups where each group consists of G disks and C parity disks. Example: D = 8 G = 4 C = 1 nG = 8/4 = 2 Disk 1Disk 2Disk 3Disk 4Parity 1Disk 5Disk 6Parity 2Disk 7Disk 8 Parity Group 1Parity Group 2

11 RAID With 1 Group With G disks in a group and C check disks, a failure is encountered when: A disk in the group fails, AND A second disk fails before the failed disk of step 1 is repaired. MTTF of a group of disks with RAID is:

12 RAID With 1 Group (Cont…) Probability of another failure: MTTR includes the time required to: Replace the failed disk drive, Reconstruct the content of the failed disk. Performing step 2 in a lazy manner increases duration of MTTR. And the probability of another failure. What happens if we increase the number of data disks in a group?

13 RAID with nG Groups With nG groups, the Mean Time To Failure of the RAID is computed in a similar manner:

14 Review RAID 1 and 3 were presented in the previous lecture. Here is a quick review.

15 RAID 1: Disk Mirroring Contents of disks 1 and 2 are identical. Redundant paths keep data available in the presence of either a controller or disk failure. A write operation by a CPU is directed to both disks. A read operation is directed to one of the disks. Each disk might be reading different sectors simultaneously. Tandems architecture Controller 1Controller 2 CPU 1 Disk 1Disk 2

16 RAID 3: Small Blocks Reads Bit-interleaved. Bad news: Small reads of less than the group size, requires reading the whole group. E.g., read of one sector, requires read of 4 sectors. One parity group has the read rate identical to one disk Disk 1Disk 2Disk 3Disk 4Parity 0101

17 RAID 3: Small Block Reads Given a large number of disks, say D=12, enhance performance by constructing several parity groups, say 3. With G (4) disks per group and D (say 8), the number of read requests supported by RAID 3 when compared with one disks is the number of groups (2). Number of groups is D/G. Disk 1Disk 2Disk 3Disk 4Parity 1Disk 5Disk 6Parity 2Disk 7Disk 8 … Parity Group 1Parity Group 2

18 Any Questions?

19 A Few Questions? Assume one instance of RAID-1 organization. What are the values for: D G C nG

20 A Few Questions? Assume one instance of RAID-1 organization. What are the values for: D=1 G=1 C=1 nG=1

21 A Few Questions? Assume one instance of RAID-1 organization. What are the values for: D=1 G=1 C=1 nG=1 Is the availability characteristics of the following Level 3 RAID better than RAID 1? Disk 1Disk 2Disk 3Disk 4Parity 1 Parity Group

22 RAID 4 Enhances performance of small reads/writes/read-modify-write. How? Interleave data across disks at the granularity of a transfer unit. Minimum size is a sector. Parity block ECC1 is an exclusive or of the bits in blocks a, b, c, and d. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1

23 RAID 4 Small read retrieves its block from one disk. Now, 4 requests referencing blocks on different data disks may proceed in parallel. When compared with 1 disk, throughput of a D disk system is D times higher. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1

24 RAID 4: Failures (Cont…) If Disk 2 fails, a small read for Block b retrieves blocks a, c, d, and ECC 1 from disks 1, 3, 4, and Parity disks to compute the missing block. What is throughput relative to one disk now? Once Disk 2 is replaced with a new one, its content is constructed either eagerly or in a lazy manner. System cannot be too lazy because we want to minimize MTTR. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1

25 RAID 4: Failures (Cont…) If the Parity disk fails, read of data blocks may proceed as in normal mode of operation. Once the Parity disk is replaced, content of new Parity disk is constructed either eagerly or lazily. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1

26 RAID 4: Small Writes Performance of small writes is improved. To write Block b: Read the old Block b and old parity block ECC1, Compute the new parity using the old Block b, new Block b, and the old parity: New parity = (old block xor new block) xor old parity ECC1 A write requires 4 accesses: 2 reads and 2 writes. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1

27 RAID 4: Bottlenecks For writes, parity disk is a bottleneck. Two different writes to Block b and g must read ECC1 and ECC2 from the Parity disk. A queue will form on the Parity disk. Performance of small writes is same as RAID 3, D/2G. Disk 1Disk 2Disk 3Disk 4Parity Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock gBlock hECC 2

28 RAID 4: Summary

29 RAID 5: Resolve the Bottleneck Distribute data and check blocks across all disks. Disk 1Disk 2Disk 3Disk 4 Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock g Block h ECC 2 Disk 5 Block iBlock jECC 3Block kBlock l Block mECC 4Block n Block p Block o ECC 5Block qBlock r Block t Block s

30 RAID 5: Resolve the Bottleneck Write of Blocks a and j may proceed in parallel now. Disk 1Disk 2Disk 3Disk 4 Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock g Block h ECC 2 Disk 5 Block iBlock jECC 3Block kBlock l Block mECC 4Block n Block p Block o ECC 5Block qBlock r Block t Block s

31 RAID 5: Read Performance Check disks service read requests. With D disks broken into nG groups, number of parity disks is nG*C. nG = D/G. When compared with one disk, the throughput of a D disk system is D + CD/G times higher. Disk 1Disk 2Disk 3Disk 4 Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock g Block h ECC 2 Disk 5 Block iBlock jECC 3Block kBlock l

32 RAID 5: Write Performance For writes, read the referenced block and its parity block. Compute the new parity block. Write the new data block and its parity block. Continue to use the parity disk. With D disks broken into nG groups, number of parity disks is nG*C. nG = D/G. When compared with one disk, the throughput of a D disk system is D/4 + (CD/G)/4 times higher. Disk 1Disk 2Disk 3Disk 4 Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock g Block h ECC 2 Disk 5 Block iBlock jECC 3Block kBlock l

33 RAID 5: R-M-W Performance For R-M-W, read and write of the data block comes for free. the referenced block is already retrieved. Must perform one extra disk I/O to read they parity block. Compute the new parity block. Write the new data block and its parity block. Continue to use the parity disk. With D disks broken into nG groups, number of parity disks is nG*C. nG = D/G. When compared with one disk, the throughput of a D disk system is D/2 + (CD/G)/2 times higher. Disk 1Disk 2Disk 3Disk 4 Block aBlock bBlock cBlock dECC 1 Block eBlock fBlock g Block h ECC 2 Disk 5 Block iBlock jECC 3Block kBlock l

34 RAID 5: Summary

35 Significant improvement in the performance of small writes/R-M-W:

36 RAID Summary If your workload consists of small R-M-W operations, which RAID would you choose?


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