COSC 6385 Computer Architecture Storage Systems

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1 COSC 6385 Computer Architecture Storage Systems Edgar Gabriel Spring 2016 I/O problem Current processor performance: e.g. Pentium 4 3 GHz ~ 6GFLOPS Memory Bandwidth: 133 MHz * 4 * 64Bit ~ 4.26 GB/s Current network performance: Gigabit Ethernet: latency ~ 40 μs, bandwidth=125mb/s InfiniBand 4x: latency ~ 4 μs, bandwidth =1GB/s Disc performance: Latency: 7-12 ms Bandwidth: ~20MB/sec 60 MB/sec 1

2 Basic characteristics of storage devices Capacity: amount of data a device can store Transfer rate or bandwidth: amount of data at which a device can read/write in a certain amount of time Access time or latency: delay before the first byte is moved Prefix Abbreviation Base ten Base two kilo, kibi K, Ki 10^3 2^10=1024 Mega, mebi M, Mi 10^6 2^20 Giga, gibi G, Gi 10^9 2^30 Tera, tebi T, Ti 10^12 2^40 Peta, pebi P, Pi 10^15 2^50 Disk striping (I) Distribute a large file onto multiple disks Stripe factor: number of disks Stripe depth: size of each block 2

3 Disk striping Requirements for improving disk performance: Multiple physical disks Separate I/O channels to each disk Data transfer to all disks simultaneously Problem of simple disk striping: Minimum stripe depth (sector size) required for optimal disk performance since file size is limited, the number of disks which can be used in parallel is limited as well Loss of a single disk makes entire file useless Risk to loose a disk is proportional to the number of disks used RAID (Redundant Arrays of Independent Disks) Redundant arrays of independent disks (RAID) Central idea: replicate data over several disks such that no data is lost if a disk fails Several RAID levels defined RAID 0: disk striping without redundant storage ( JBOD = just a bunch of disks) No fault tolerance Good for high transfer rates Good for high request rates RAID 1: mirroring All data is replicated on two or more disks Does not improve write performance and just moderately the read performance 3

4 RAID level 2 RAID 2: Hamming codes Each group of data bits has several check bits appended to it forming Hamming code words Each bit of a Hamming code word is stored on a separate disk Very high additional costs: e.g. up to 50% additional capacity required Hardly used today since parity based codes faster and easier RAID level 3 Parity based protection: Based on exclusive OR (XOR) Reversible Example (data byte 1) XOR (data byte 2) (parity byte) Recovery (data byte 2) XOR (parity byte) (recovered data byte 1) 4

5 RAID level 3 (cont.) Data divided evenly into N subblocks (N = number of disks, typically 4 or 5) Computing parity bytes generates an additional subblock Subblocks written in parallel on N+1 disks For best performance data should be of size (N * sector size) Problems with RAID level 3: All disks are always participating in every operation => contention for applications with high access rates If data size is less than N*sector size, system has to read old subblocks to calculate the parity bytes RAID level 3 good for high transfer rates RAID level 4 Parity bytes for N disks calculated and stored parity bytes are stored on a separate disk Files are not necessarily distributed over N disks For read operations: Determine disks for the requested blocks Read data from these disks For write operations Retrieve the old data from the sector being overwritten Retrieve parity block from the parity disk Extract old data from the parity block using XOR operations Add the new data to the parity block using XOR Store new data Store new parity block Bottleneck: parity disk is involved in every operation 5

6 RAID level 5 Same as RAID 4, but parity blocks are distributed on different disks Block 1 Block 2 Block 3 Block 4 P(1,2,3,4) Block 5 Block 6 Block 7 P(5,6,7,8) Block 8 RAID level 6 Tolerates the loss of more than one disk Collection of several techniques E.g. P+Q parity: store parity bytes using two different algorithms and store the two parity blocks on different disks E.g. Two dimensional parity Parity disks 6

7 Is RAID level 1 + RAID level 0 RAID 1 mirroring RAID level 10 RAID 0 striping Also available: RAID 53 (RAID 0 + RAID 3) Dependability Module reliability measures : mean time to failure FIT: failures in time 1 FIT Often expressed as failures in 1,000,000,000 hours MTTR: mean time to repair MTBF: mean time between failures Module availability: MTBF MTTR M A MTTR (14) (15) (16) 7

8 Dependability - example Assume a disk subsystem with the following components and s: 10 disks, =1,000,000h 1 SCSI controller, =500,000h 1 power supply, =200,000h 1 fan, = 200,000h 1 SCSI cable, =1,000,000h What is the of the entire system? What is the probability, that the system fails within a 1 week period? Dependability example (II) Determine the sum of the failures in time of all components FIT system 10 1,000, , , ,000 1,000, ,000,000 1 FIT 23 1,000,000 43, system 500 system h 23,000 1,000,000,000 Probability that the system fails within a 1 week period: 1week P system 24*7 43,500 0, ,386% 8

9 Dependability example (III) What happens if we add a second power supply and we assume, that the MTTR of a power supply is 24 hours? Assumption: failures are not correlated of the pair of power supplies is the mean time to failure of the overall system divided by the probability, that the redundant unit fails before the primary unit has been replaced of the overall system: FIT system = 1/ power + 1/ power =2/ power system= 1/ FIT system = power /2 Dependability example (III) Probability, that 1 unit fails within MTTR: MTTR/ power pair MTTR 2 power / power ,000 2MTTR ,000,000 9

Dependability example (III) More generally, if power supply 1 has an of power_1 power supply 2 has an of power_2 FIT system = 1/ power_1 + 1/ power_2 system= 1/ FIT system pair = system /(MTTR/min(

10 Dependability example (III) More generally, if power supply 1 has an of power_1 power supply 2 has an of power_2 FIT system = 1/ power_1 + 1/ power_2 system= 1/ FIT system pair = system /(MTTR/min( power_1, power_2 )) Or if either power_1 or power_2 have been clearly declared to be the backup unit pair = system /(MTTR/ power_backup ) RAID 10 Reliability Assuming that the of a single hard drive is 250,000h and MTTR = 25h determine the overall of a single pair of hard drives (e.g. 1 and 1 ). FIT w/o = 1/250, /250,000 w/o = 250,000/2 = 125,000h w/ = w/o / ( MTTR/250,000) = 125,000/ ( 25/250,000) = 125,000/(1/10,000) = 125,000*10,000 = 1,250,000,000h What is the of 5 pair of hard drives as shown in the image above? FIT 5pairs = 5 * 1/1,250,000,000 5pairs = 1,250,000,000/5 = 250,000,000h 10

11 RAID 10 vs. RAID 5 Comparison to a RAID 5 configuration consisting of 5+1 hard drives Same data capacity in the overall configuration FIT 6disks w/o = 6 * 1/250,000 6disks w/o = 250,000/6 FIT 5 disks w/o = 5 * 1/250,000 5 disks w/o = 250,000/5 = 50,000 6 disks RAID5 w/ = 6dsisk w/o / ( MTTR/( 5 disks w/o ) = (250,000/6) / (25 / 50,000) = (250,000/6) / (1/2,000) = 250,000 * 2,000 / 6 = 250,000 * 1,000 / 3 = 250,000,000 / 3 RAID 10 has higher than RAID 5 Costs of RAID 5 lower than costs of RAID 10 ( 6 drives vs. 10 drives) 11

COSC 6385 Computer Architecture. Storage Systems

COSC 6385 Computer Architecture Storage Systems Spring 2012 I/O problem Current processor performance: e.g. Pentium 4 3 GHz ~ 6GFLOPS Memory Bandwidth: 133 MHz * 4 * 64Bit ~ 4.26 GB/s Current network performance: