Lecture 4 A Case for RAID Part 2 Prof Shahram Ghandeharizadeh Computer Science Department University of Southern California Smaller Inexpensive Disks 25 annual reduction in size 40 annual drop in price 1 GB Year 1980 1 GB Year 2008 IBM 3380 40 000 IBM Microdrive 125 Size of a refrigerator 550 pounds 250 Kg 1 inch in height weighs 1 ounce 16 grams Inexpensive Disks Less than 9 Cents Gigabyte of storage 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 Challenge Managing Data is Expensive RAID 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 MTTF MTBF MTTR AFR MTBF Mean Time Between Failures 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 Designed for repairable devices Number of hours since the system was started until its failure 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 2 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 RAID RAID organizes D disks into nG groups where each group consists of G disks and C parity disks Example Disk 1 D 8 G 4 C 1 nG 8 4 2 Disk 2 Disk 3 Disk 4 Parity 1 Parity Group 1 Disk 5 Disk 6 Disk 7 Disk 8 Parity 2 Parity Group 2 RAID RAID organizes D disks into nG groups where each group consists of G disks and C parity disks Example Disk 1 D 8 G 4 C 1 nG 8 4 2 Disk 2 Disk 3 Disk 4 Parity 1 Parity Group 1 Disk 5 Disk 6 Disk 7 Disk 8 Parity 2 Parity Group 2 RAID With 1 Group With G disks in a group and C check disks a failure is encountered when 1 2 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 RAID With 1 Group Cont Probability of another failure MTTR includes the time required to 1 2 Performing step 2 in a lazy manner increases duration of MTTR Replace the failed disk drive Reconstruct the content of the failed disk 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 architecture CPU 1 Controller 1 Controller 2 Disk 1 Disk 2 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 01011110101010000001101001111 Disk 1 0 1 Disk 2 1 1 Disk 3 0 1 Disk 4 1 0 Parity 0 1 RAID 3 Small Block Reads Disk 1 Given a large number of disks say D 12 enhance performance by constructing several parity groups say 3 Disk 2 Disk 3 Disk 4 Parity 1 Parity Group 1 Disk 5 Disk 6 Disk 7 Disk 8 Parity 2 Parity Group 2 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 Any Questions A Few Questions Assume one instance of RAID 1 organization What are the values for D G C nG A Few Questions Assume one instance of RAID 1 organization What are the values for D 1 G 1 C 1 nG 1 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 1 Disk 2 Disk 3 Disk 4 Parity Group Parity 1 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 Disk 1 Disk 2 Disk 3 Disk 4 Block a Block b Block c Block d Parity ECC 1 Parity block ECC1 is an exclusive or of the bits in blocks a b c and d RAID 4 Small read retrieves its block from one disk Disk 1 Disk 2 Disk 3 Disk 4 Block a Block b Block c Block d Parity ECC 1 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 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 Disk 1 Disk 2 Disk 3 Disk 4 Block a Block b Block c Block d Parity ECC 1 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 …