Chapter 2 Data Storage

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1 Chapter 2

2 22 CHAPTER 2. DATA STORAGE

3 2.1. THE MEMORY HIERARCHY 23

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6 26 CHAPTER 2. DATA STORAGE main memory, yet is essentially random-access, with relatively small differences

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11 Figure 2.4: A typical disk

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13 Figure 2.6: Schematic of a simple computer system Rotation Speed of the Disk Assembly RPM, i.e., one rotation every 11 milliseconds, is common, although higher and lower speeds are found. Number of Platters per Unit. A typical disk drive has about five platters and therefore ten surfaces. However, the common diskette ("floppy" disk) and "zip" disk have a single platter with two surfaces, and disk drives with

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15 2.2. DISKS 35 b) The sectors containing the block move under the disk head as the entire disk assembly rotates. The time taken between the moment at which the command to read a block is issued and the time that the contents of the block appear in main memory is called the latency of the disk. It can be broken into the following components: 1. The time taken by the processor and disk controller to process the request, usually a fraction of a millisecond, which we shall neglect. We shall also neglect time due to contention for the disk controller (some other process might be reading or writing the disk at the same time) and other delays due to contention, such as for the bus. 2. The time to position the head assembly at the proper cylinder. This time, called seek time, can be 0 if the heads happen already to be at the proper cylinder. If not, then the heads require some minimum time to start moving and stop again, plus additional time that is roughly proportional to the distance traveled. Typical minimum times, the time to start, move by one track, and stop, are a few milliseconds, while maximum times to

16 36 CHAPTER 2. DATA STORAGE

17 2.2. DISKS 37

18 38 CHAPTER 2. DATA STORAGE

19 2.2. DISKS 39 proper sector(s) to rotate under the head, but, instead of reading the data

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21 2.3. USING SECONDARY STORAGE EFFECTIVELY 41 of computation is often called the "RAM model" or random-access model of computation. However, when implementing a DBMS, one must assume that the data does not fit into main memory. One must therefore take into account the use of secondary, and perhaps even tertiary storage in designing efficient

22 42 CHAPTER 2. DATA STORAGE Example 2.4 : Suppose our database has a relation R and a query asks for the tuple of R that has a certain key value k. As we shall see, it is quite desirable

23 2.3. USING SECONDARY STORAGE EFFECTIVELY 43 fastest. Moieover, we would use a strategy where we sort only the key fields with attached pointers to the full iccoids. Only when the keys and their pointers were in sorted order, would we use the pointeis to bring every record to its proper position. Unfortunately, these ideas do not work very well when secondary memory

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26 Figure 2.11: Main-memory organization for multiway merging queues" 6 that takes time proportional to the logarithm of the number of sublists to find the smallest element. 2. Move the smallest element to the first available position of the output block.

27 2.3. USING SECONDARY STORAGE EFFECTIVELY 47 How Big Should Blocks Be?

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29 2.4. IMPROVING THE ACCESS TIME OF SECONDARY STORAGE 49 * c) The size of blocks is doubled, to 8192 bytes (again, as throughout this exercise, all other parameteis are unchanged). d) The size of available main memory is doubled to 100 megabytes.

30 50 CHAPTER 2. DATA STORAGE assumption may be appropriate for a system that is executing a large number

31 2.4. IMPROVING THE ACCESS TIME OF SECONDARY STORAGE Organizing Data by Cylinders Since seek time represents about half the average time to access a block, there

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33 2.4. IMPROVING THE ACCESS TIME OF SECONDARY STORAGE 53 random, data-dependent way. If the core algorithm of phase 2 the selection

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35 2.4. IMPROVING THE ACCESS TIME OF SECONDARY STORAGE 55

36 56 CHAPTER 2. DATA STORAGE Cylinder First time of Request available = 1000 " ,

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38 58 CHAPTER 2. DATA STORAGE Waiting for the Last of Two Blocks Suppose there are two blocks at random positions around a cylinder. Let Xi and X2 be the positions, in fractions of the full circle, so these are

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41 2.4. IMPROVING THE ACCESS TIME OF SECONDARY STORAGE 61 Cylinder First time of Request available

42 62 CHAPTER 2. DATA STORAGE

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44 64 CHAPTER 2. DATA STORAGE

45 2.5. DISK FAILURES 65

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47 2.6. RECOVERY FROM DISK CRASHES Exercises for Section 2.5 Exercise 2.5.1: Compute the parity bit for the following bit sequences: * a) b) c)

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52 72 CHAPTER 2. DATA STORAGE on disk 2, to get That tells us we must change positions 2, 3, 6, and

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55 2.6. RECOVERY FROM DISK CRASHES 75 We shall see shortly that the particular choice of bits in this matrix gives us a simple rule by which we can recover from two simultaneous disk crashes. Reading We may read data from any data disk normally. The redundant disks can be ignored. Writing The idea is similar to the writing strategy outlined in Section 2.6.4, but now several redundant disks may be involved. To write a block of some data disk, we compute the modulo-2 sum of the new and old versions of that block. These bits are then added, in a modulo-2 sum, to the corresponding blocks of all those redundant disks that have 1 in a row in which the written disk also has 1.

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57 2.6. RECOVERY FROM DISK CRASHES 77 Additional Observations About RAID Level 6

58 78 CHAPTER 2. DATA STORAGE Disk Contents 1) ) ) ) )???????? 6) )

59 2.6. RECOVERY FROM DISK CRASHES 79 Error-Correcting Codes and RAID Level 6 There is a broad theory that guides our selection of a suitable matrix, like that of Fig. 2.17, to determine the content of redundant disks. A code of length n is a set of bit-vectors (called code words) of length n. The Hamming distance between two code words is the number of positions in which they differ, and the minimum distance of a code is the binallest Hamming distance of any two different rode words. If C is any code of length n, we can require that the corresponding bits on n disks have one of the sequences that are members of the code. As a very simple example, if we are using a disk and its minor, then n = 2,

60 80 CHAPTER 2. DATA STORAGE a) Express this situation by giving a parity check matrix analogous to Fig !! b) It is possible to recover from some but not all situations where two disks fail at the same time. Determine for which pairs it is possible to recover and for which pairs it is not. *! Exercise : Suppose we have eight data disks numbered 1 through 8, and three redundant disks: 9, 10, and 11. Disk 9 is a parity check on disks

61 2.7. SUMMARY OF CHAPTER Disk

62 82 CHAPTER 2. DATA STORAGE +

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