Lecture 4: A Case for RAID (Part 2)Smaller & Inexpensive DisksInexpensive DisksChallenge: Managing Data is ExpensiveSlide 5MTTF, MTBF, MTTR, AFRFocus on MTTF & MTTRAssumptionsRAIDSlide 10RAID With 1 GroupRAID With 1 Group (Cont…)RAID with nG GroupsReviewRAID 1: Disk MirroringRAID 3: Small Blocks ReadsRAID 3: Small Block ReadsAny Questions?A Few Questions?Slide 20Slide 21RAID 4Slide 23RAID 4: Failures (Cont…)Slide 25RAID 4: Small WritesRAID 4: BottlenecksRAID 4: SummaryRAID 5: Resolve the BottleneckSlide 30RAID 5: Read PerformanceRAID 5: Write PerformanceRAID 5: R-M-W PerformanceRAID 5: SummarySlide 35RAID SummaryLecture 4: A Case for RAID (Part Lecture 4: A Case for RAID (Part 2)2)Prof. Shahram GhandeharizadehProf. Shahram GhandeharizadehComputer Science DepartmentComputer Science DepartmentUniversity of Southern CaliforniaUniversity of Southern CaliforniaSmaller & 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 Microdrive @ $125Size of a refrigerator, 550 pounds (250 Kg)1 GB, Year 1980 IBM 3380 @ $40,000Inexpensive DisksLess than 9 Cents / Gigabyte of storageChallenge: 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!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!RAIDMTTF, 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 1988.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.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 1988.MTTR: Mean Time To RepairNumber of hours required to replace a disk drive, ANDReconstruct the data stored on the failed disk drive.AssumptionsMTTF of a disk is independent of other disks in a RAID.Assume:1. The MTTF of a disk is once every 100 years, and2. An array of 1000 such disks.The MTTF of any single disk in the array is once every 37 days.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 = 2Disk 1 Disk 2 Disk 3 Disk 4 Parity 1 Disk 5 Disk 6 Parity 2Disk 7 Disk 8Parity Group 1 Parity Group 2RAID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 = 2Disk 1 Disk 2 Disk 3 Disk 4 Parity 1 Disk 5 Disk 6 Parity 2Disk 7 Disk 8Parity Group 1 Parity Group 2RAID With 1 GroupWith G disks in a group and C check disks, a failure is encountered when:1. A disk in the group fails, AND2. A second disk fails before the failed disk of step 1 is repaired.MTTF of a group of disks with RAID is:RAID With 1 Group (Cont…)Probability of another failure:MTTR includes the time required to:1. Replace the failed disk drive,2. 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?RAID with nG GroupsWith nG groups, the Mean Time To Failure of the RAID is computed in a similar manner:ReviewRAID 1 and 3 were presented in the previous lecture.Here is a quick review.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.Tandem’s architectureController 1 Controller 2CPU 1Disk 1 Disk 2RAID 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.0101111010101000000110100111101110110Disk 1 Disk 2 Disk 3 Disk 4 Parity01RAID 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 1 Disk 2 Disk 3 Disk 4 Parity 1 Disk 5 Disk 6 Parity 2Disk 7 Disk 8…Parity Group 1 Parity Group 2Any Questions?A Few Questions?Assume one instance of RAID-1 organization. What are the values for:DGCnGA Few Questions?Assume one instance of RAID-1 organization. What are the values for: D=1G=1C=1nG=1A 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 1 Disk 2 Disk 3 Disk 4 Parity 1Parity GroupRAID 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 1 Disk 2 Disk 3 Disk 4 ParityBlock a Block b Block c Block d ECC 1RAID 4Small read retrieves its block from one disk.Now, 4 requests referencing