1. ICOM 5016– Intro. to Database Systems Lecture 11 - Buffer Management and RAID Dr. Manuel Rodríguez-Martínez Electrical and Computer Engineering Department

2. ICOM 5016Dr. Manuel Rodriguez Martinez2 Readings Read –Chapter 11

3. ICOM 5016Dr. Manuel Rodriguez Martinez3 Relational DBMS Architecture Disk Space Management Buffer Management File and Access Methods Relational Operators Query Optimizer Query Parser Client API Client DB Execution Engine Concurrency and Recovery

4. ICOM 5016Dr. Manuel Rodriguez Martinez4 Disk Space Managemet Disk Space Manager –DBMS module in charge of managing the disk space used to store relations –Duties Allocate space Write data Read data De-allocate space Disk Space Manager supplies a stream of data pages. –Minimal unit of I/O –Often the size of a block (sector, several sectors, or more)

5. ICOM 5016Dr. Manuel Rodriguez Martinez5 Disk Page Disk page is simply an array of bytes We impose the logic of an array of records! 123BobNY$1200 2178JilLA$9202 8273NedFL$2902 723AlPR$300 Disk Page Records Reading a Disk Page should be one I/O

6. ICOM 5016Dr. Manuel Rodriguez Martinez6 Disk arrangement option Suppose we need to create 10 GB of space to stored a database. Each page is 4 KB is size. –How to organize the disk to accomplish this. Cooked File –User the file system provide by OS –Create a file “mydb.dat” –Write to this file N pages of size 4KB N must be enough to reach the size of 10GB Page are full of bytes with zeros. –Have a hast table somewhere to store the information about this file “mydb.dat”. –Now you can start writing pages with actual data.

7. ICOM 5016Dr. Manuel Rodriguez Martinez7 Raw Disk Partition Don’t use the file provided by the DBMS Instead create a parition on the disk, but don’t format it with OS formats (e.g. FAT, FAT32, NTFS, LINUX) Make your own file system on the disk –Create a directory of pages –Need to implement all operation such as read, write, check, etc. –Need to implement you own files… –Faster and more efficient that OS files, but more complex.

8. ICOM 5016Dr. Manuel Rodriguez Martinez8 Buffer Management All Data Pages must be in memory in order to be accessed. Buffer Manager –deals with asking Disk Space Manager for pages from disk and store them into memory –Sends Disk Space Manager pages to be written to disk. Memory is faster that Disk –Keep as much data as possible in memory –If not enough space is available, need a policy to decide what pages to remove from memory.

9. ICOM 5016Dr. Manuel Rodriguez Martinez9 Buffer Pool Frame –Data strucuture that can hold a data page and control flags Buffer pool –Array of frames of size N. In C #define POOL_SIZE 100 #define PAGE_SIZE 4096 typedef struct frame { int pin_count; bool dirty; char page[PAGE_SIZE]; } frame; frame buffer_pool[POOL_SIZE];

10. ICOM 5016Dr. Manuel Rodriguez Martinez10 Buffer Pool DB Disk RAM Free Frame Disk Page

11. ICOM 5016Dr. Manuel Rodriguez Martinez11 Operational mode All requested data pages must first be placed into the buffer pool. pin_count is used to keep track of number of transactions that are using the page –0 means no body is using it dirty is used a flag (dirty bit) to indicate that a page has been modified since read from disk –Need to flush it to disk if the page is to be evicted from pool Page is an array of bytes where the actual is located –Need to interpret these bytes as the int, char, Date data types supported by SQL This is very complex and tricky!

12. ICOM 5016Dr. Manuel Rodriguez Martinez12 Buffer replacement If we need to bring a page from disk, we need to find a frame in the buffer to hold it Buffer pool keeps track on the number of frames in use –List of frames that are free If there is a free frame, we use it –Remove from list of free frame –Increment the pin_count –Store the data page into the byte array (page field) If the buffer is full, we need a policy to decide which page will be evicted

13. ICOM 5016Dr. Manuel Rodriguez Martinez13 Buffer replacement Algorithm Upon request of page X do –Look for page X in buffer pool –If found, return it –Else, determine if there is a free frame Y in the pool –If frame Y is found Increment its pin_count Read page in the frame’s byte array –Use a replacement policy to find a frame Z to replace Z must have pin_count == 0 –Increment the pin_count in Z –If dirty bit is set, write data currently in Z to disk –Read the new page into the byte array in Z

14. ICOM 5016Dr. Manuel Rodriguez Martinez14 Some issues Need to make sure pin_count is 0 –Nobody is using the frame Need to write the data to disk if dirty bit is zero This latter approach is called Lazy update –Write to disk only when you have to!!! –Careful, if power fails, you are in trouble. –DBMS need to periodically flush pages to disk Force write If no page is found with pin_count equal to 0, then either: –Wait until one is freed –Abort the transaction (insufficient resources)

15. ICOM 5016Dr. Manuel Rodriguez Martinez15 Buffer Replacement policies LRU – Least Recently Used –Evicts the page is the least recently used page in the pool. –Can be implemented by having a priority queue with the frame numbers. High Priority – not used frequently –Head of the queue is the LRU –Each time a page is used it must be removed from current queue position and put back at the end This queue need a method erase() that can erase stuff from the middle of the queue LRU is the most widely used policy for buffer replacement –Most cache managers also use it

16. ICOM 5016Dr. Manuel Rodriguez Martinez16 Other policies Most Recently Used –Evicts the page that was most recently accessed –Can be implemented with a priority queue FIFO –Pages are replaced in a strict First-In-First Out –Can be implemented with a FIFO List (queue in the strict sense) Random –Pick any page at random for replacement

17. ICOM 5016Dr. Manuel Rodriguez Martinez17 MTTF in terms of number of disks Suppose we have N disks in the array. Then the MTTF of the array is given by: This assumes any disk can fail with equal probability.

18. ICOM 5016Dr. Manuel Rodriguez Martinez18 MTTF in a disk array Suppose we have a single disk with a MTTF of 50,000 hrs (5.7 years). Then, if we build an array with 50 disks, then we have a MTTF for the array of 50,000/50 = 1000 hrs, or 42 days!, because any disk can fail at any given time with equal probability. –Disk failures are more common when disks are new (bad disk from factory) or old (wear due to usage). Morale of the story: More does not necessarily means better!

19. ICOM 5016Dr. Manuel Rodriguez Martinez19 Increasing MTTF with redundancy We can increase the MTTF in a disk array by storing some redundant information in the disk array. –This information can be used to recover from a disk failure. This information should be carefully selected so it can be used to reconstruct original data after a failure. What to store as redundant information? –Full data block –Parity bit for a set of bit locations across all the disks Where to store it? –Check disks – disks in the array used only for this purpose –All disks – spread redundant information on every disk in the array.

20. ICOM 5016Dr. Manuel Rodriguez Martinez20 Redundancy unit: Data Blocks One approach is to have a back-up copy of each data block in the array. This is called mirroring. Back up can be in: –another disk, or disk array –Tape (very slow …) Advantage: –Easy to recover from failure, just read the block from backup. Disadvantages: –Requires twice the storage space –Writes are more expensive Need to write data block to two different locations each time Snapshot writes are unfeasible (failures happen at any time!)

21. ICOM 5016Dr. Manuel Rodriguez Martinez21 Redundancy Unit: Parity bits Consider an array of N disks. Suppose k is the number of the k-th block in each disk. Each block consists of several kilobytes, and each byte is 8-bit. We can store redundant information about the i-th bit position in each data block. –Parity bit The parity bit gives the number of bits that were set to the value 1 in the group of corresponding bit locations of the data blocks. For example, if bit 1024 has a parity 0, then an even number of bits where set 1 at bit position 1024. Otherwise its value must be 1.

22. ICOM 5016Dr. Manuel Rodriguez Martinez22 Parity bits Consider bytes: –b1 = 00010001, b2 = 00111111, b3 = 00000011 If we take the XOR these bytes we get 00010001 00111111 00000011 00101101 - this byte has the parity for all bits in b1, b2, b3 Notice the following: –For bit position 0, the parity is 1,meaning an odd number of bits have value 1 for bit position 0. –For bit position 1, the parity is 0, meaning an even number of bits have value 1 for bit position 1

23. ICOM 5016Dr. Manuel Rodriguez Martinez23 Redundancy as blocks of parity bits For each corresponding data block compute and store a parity block that has the parity bits corresponding to the bit location in the data blocks. Disk Array Controller Bus Disk 0 Block 0 Disk 1 Block 0 Disk 2 Block 0 Disk 3 Block 0 Check Disk 0 Block 0

24. ICOM 5016Dr. Manuel Rodriguez Martinez24 Reliability groups in disk array Organize the disks in the array into groups called reliability groups. Each group has: –1 or more data disks, where the data blocks are stored –0 or more check disks where the blocks of parity bits are stored If a data disk fails, then the check disk(s) for its reliability group can be used to recover the data lost from that disk. There is a recovery algorithm that works for any failed disk m in the disk array. Can recover from up to 1 disk failure.

25. ICOM 5016Dr. Manuel Rodriguez Martinez25 Recovery algorithm Suppose we have an array of N disks, with M check disks (in this case there is one reliability group). Suppose disk p fails. We buy a replacement and then we can recover the data as follows. For each data block k on disk p: –Read data blocks k on every disk r, with r != p –Read parity block k from its check disk w –For each bit position i in block k of disk p: Count number of bits set 1 at bit i in each block coming from a disk other than p. Let this number be j If j is odd, and parity bit is 1 then bit position i is set to 0 If j is even, and parity bit is 0 then bit position i is set to 0 Else, bit position i is set to 1

26. ICOM 5016Dr. Manuel Rodriguez Martinez26 Recovery algorithm: Example Suppose we have an array of 5 disks, with 1 check disk (in this case there is one reliability group). Suppose disk 1 fails. We buy a replacement and then we can recover the data as follows. For each data block k on disk 1: –Read data blocks k on disks 0, 2, 3, 4, –Read parity block k on check disk 0 –For each bit position i in block k of disk 1: Count number of bits set 1 at bit I in each block k coming from disks 0, 2, 3, 4. Let this number be j If j is odd, and parity bit is 1 then bit position i is set to 0 If j is even, and parity bit is 0 then bit position i is set to 0 Else, bit position i is set to 1

27. ICOM 5016Dr. Manuel Rodriguez Martinez27 RAID Organization RAID: –Originally: Redundant Array of Inexpensive Disks –Now: Redundant Array of Independent Disks RAID organization combines the ideas of striping, redundancy as parity bits, and reliability groups. RAID system has one or more reliability groups –For simplicity we shall assume only one group … RAID systems can have various number of check disks for reliability groups, depending on the RAID level that is chosen for the system. Each RAID level represent a different tradeoff between storage requirements, write speed and recovery complexity.

28. ICOM 5016Dr. Manuel Rodriguez Martinez28 RAID Analysis Suppose we have a disk array with 4 data disks. Let’s analyze how many check disks we need to build a RAID with 1 reliability group of 4 data disks plus the check disks. Note: Effective space utilization is a measure of the amount of space in the disk array that is used to store data. It is given as a percentage by the formula:

29. ICOM 5016Dr. Manuel Rodriguez Martinez29 RAID Level 0 RAID Level 0: Non-redundant Uses data striping to distributed data blocks, and increase maximum disk bandwidth available. –Disk bandwidth refers to the aggregate rate of moving data from the disk array to the main memory. Ex. 200MB/sec Solution with lowest cost, but with little reliability. Write performance is the best since only 1 block is written in every write operation, and the cost is 1 I/O. Read performance is not the best, since a block can only be read from one site. Effective space utilization is 100%

30. ICOM 5016Dr. Manuel Rodriguez Martinez30 RAID Level 1 RAID Level 1: Mirrored Each data block is duplicated: –original copy + mirror copy No striping! Most expensive solution since it requires twice the space of the expected data set size. Every write involves two writes (original + copy) –cannot be done simultaneously to prevent double corruption –First write on data disk, then on copy at mirror disk Reads are fast since a block can be fetched from –Data disk –Mirror disk

31. ICOM 5016Dr. Manuel Rodriguez Martinez31 RAID Level 1 (cont…) In RAID Level 1, the data block can be fetched from the disk with least contention. Since we need to pair disks in groups of two (original + copy), the space utilization is 50%, independent on the amount of disks. RAID Level 1 is only good for small data workloads where the cost of mirroring is not an issue.

32. ICOM 5016Dr. Manuel Rodriguez Martinez32 RAID Level 0+1 RAID Level 0+1: Striping and Mirroring –Also called RAID Level 10 Combines mirroring and striping. Data is striped over the data disks. –Parallel I/O for high throughput (full disk array bandwidth) Each data disk is copied into a mirror disk Writes require 2 I/Os – (original disk + mirror disk) Blocks can be read from either original disk or mirror disk –Better performance since more parallelism can be achieved. –No need to wait for busy disk, just go to its mirror disk!

33. ICOM 5016Dr. Manuel Rodriguez Martinez33 RAID Level 1+0 (cont…) Space utilization is 50% (half data and half copies) RAID Level 1+0 is better than RAID 1 because of striping. RAID Level 1+0 is good for workloads with small data sets, where cost of mirroring is not an issue. Also good for workloads with high percentages of writes, since a write is always 2 I/Os to unloaded disks (specially the mirrors).

34. ICOM 5016Dr. Manuel Rodriguez Martinez34 RAID Level 2 RAID Level 2: Error-Correcting Codes Uses striping with a 1-bit striping unit. Hamming code for redundancy in C check disks. –Can indicate which disk failed –Make number of check disk grow logarithmically with respect to the number of data disks. (???) Read is expensive since to read 1 bit we need to read 1 physical data block, the one storing the bit. Therefore, to read 1 logical data block from the array we need to read multiple physical data blocks from each disk to get all the necessary bits.

35. ICOM 5016Dr. Manuel Rodriguez Martinez35 RAID Level 2 (Cont…) Since we are striping with 1-bit units, if we have an array with m data disks, then m reads for bits will require 1 block from each disk, for a total of m I/Os. Therefore, reading 1 logical data block from the RAID will require reading at least m blocks, and therefore the cost will be at least m I/Os. Level 2 is good for request of large contiguous data blocks since the system will fetch physical blocks that will have the require data. Level 2 is bad for request of small data since the I/Os will be wasted in fetching just a bits and throwing away the rest.

36. ICOM 5016Dr. Manuel Rodriguez Martinez36 RAID Level 2 (Cont…) Writes are expensive with Level 2 RAID. A write operation on N data disks involves: –Reading at least N data blocks into the memory. –Reading C check disks –Modifying the N data blocks with the new data. –Modifying C check disks to update hamming codes –Writing N + C blocks to the disk array. This is called a read-modify-write cycle. Level 2 has better space utilization than Level 1.

37. ICOM 5016Dr. Manuel Rodriguez Martinez37 RAID Level 3 Raid Level 3: Bit-Interleaved Parity Uses striping with a 1-bit striping unit. Does not uses Hamming codes, but simply computes bit parity. –Disk controller can tell which disk has failed. Only need 1 check disk to store parity bits of the data disks in the array. A RAID Level 3 system will have N disks, where N-1 are data disks, and one is the check disk.

38. ICOM 5016Dr. Manuel Rodriguez Martinez38 RAID Level 3 (Cont…) Reading or writing a logical data block in a RAID Level 3 involves reading at least N-1 data blocks from the array. Writing requires a read-modify-write cycle.

39. ICOM 5016Dr. Manuel Rodriguez Martinez39 RAID Level 4 RAID Level 4: Block-Interleaved Parity Uses striping with a 1-block striping unit. –Logical data block is the same as physical data block. Computes redundancy as parity bits, and has 1 check disk to store parity bits for all corresponding block in the array. Reads can be run in parallel –Works well for both large and small data requests. Writes require read-modify-write cycle but only involve: –Data disk for block being modified (target block k) –Check disk (parity block for block k)

40. ICOM 5016Dr. Manuel Rodriguez Martinez40 RAID Level 4 (Cont…) The parity block k is updated incrementally to avoid reading all data blocks k from all data disks. –Only need to read parity block k and block k to be modified –Parity is computed as follows: New parity block = ((Old block XOR New block) XOR Old parity block) In this way Read-modify-write cycle avoids reading the data block in each disk to compute the parity. Read-modify-write cycle only performs 4I/Os (2 reads and 2 writes of the target data block and parity block) Space utilization is the same as RAID Level 3.

41. ICOM 5016Dr. Manuel Rodriguez Martinez41 RAID Level 4 (Cont…) In RAID Level 3 and 4, the check disk is only used in writing operations. It does not help with the reads. Moreover, the check disk becomes a bottleneck since it must participate in every write operation.

42. ICOM 5016Dr. Manuel Rodriguez Martinez42 RAID Level 5 RAID Level 5: Block-Interleaved Distributed Parity Uses striping with a 1-block striping unit. Redundancy is stored as blocks of parity bits, but the parity blocks are distributed over all the disks in the array. –Every disk is both a data disk and a check disk. Best of both worlds: –Fast reads –Fast writes Reads are efficient since they can be run in parallel.

43. ICOM 5016Dr. Manuel Rodriguez Martinez43 RAID Level 5 (Cont…) Writes still involve a read-modify-write cycle But the cost of writing the parity block is lowered by scattering them over all the disks in the array. –Remove the contention at one check disk RAID Level 5 is a good general purpose system –Small reads –Large reads –Intensive writes Space utilization is equivalent to Level 3 and 4 since there is 1 disk worth of parity blocks in the system!

44. ICOM 5016Dr. Manuel Rodriguez Martinez44 RAID Level 6 RAID Level 6: P+Q Redundancy RAID Levels 2-4 only recover from 1 disk failure. In a large disk array, there is a high probability that two disk might fail simultaneously. RAID Level 6 provides recovery from 2 disk failures. Uses striping with 1-block striping unit Redundancy is stored as parity bits and Reed- Solomon codes. –Require two check disks for the data disks in the array

45. ICOM 5016Dr. Manuel Rodriguez Martinez45 RAID Level 6 (Cont…) Reads are like in RAID Level 5. Writes involve a read-modify-write cycle that involves 4 I/Os: –1 for data block –1 for parity block –2 for Reed-Solomon Codes