* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download file_organize
Survey
Document related concepts
Transcript
Chapter 16-17 File Structures, Hashing, Indexing, and Physical Database Design 1 Oracle SQL*Loader • http://www.oracle.com/technetwork/database/enterpriseedition/sql-loader-overview-095816.html 2 Storage • Primary storage (main memory) – Can be operated on directly by computer CPU small, fast • Secondary storage – http://en.wikipedia.org/wiki/Hard_disk – Can not be operated on directly by computer CPU – Magnetic disks, optical disks, tapes, etc. – Larger capacities, inexpensive, slower than main memory 3 Table 16.1 Types of Storage with Capacity, Access Time, Max Bandwidth (Transfer Speed), and Commodity Cost Table 16.2 Specifications of Typical High-End Enterprise Disks from Seagate (a) Seagate Enterprise Performance 10 K HDD 1200 GB continued on next slide Storage capacity units • • • • Kilobytes – 1000 bytes Megabytes – 1 million bytes Gigabytes (Gbytes) – 1 billion bytes Terabytes – 1000 gigabytes 6 Memory Hierarchies and Storage Devices • Primary storage – Cache (static RAM)– most expensive, fast, used by CPU to speed up execution programs http://wombat.doc.ic.ac.uk/foldoc/foldoc.cgi?query=cache - Main memory (dynamic RAM) – work area for CPU 7 Secondary storage (Mass storage) – CD-ROM – Tapes – Disks Main memory database: entire database is stored in main memory 8 Figure 16.1 (a) A single-sided disk with read/write hardware. (b) A disk pack with read/write Figure 16.2 Different sector organizations on disk. (a) Sectors subtending a fixed angle. (b) Sectors maintaining a uniform recording density. Tracks The part of a disk which passes under one read/write head while the head is stationary. The number of tracks on a disk surface therefore corresponds to the number of different radial positions of the head(s). The collection of all tracks on all surfaces at a given radial position is known a cylinder and each track is divided into sectors. 11 Cylinder • The set of tracks on a multi-headed disk that may be accessed without head movement. That is, the collection of disk tracks which are the same distance from the spindle about which the disks rotate. 12 Sector • one sector lies within a continuous range of rotational angle of the disk 13 Data transfer between main memory and disks (in blocks) Hardware Address of a block – Surface number – Track number – Block number • Time requires – Seek time – Rotational delay time (latency) – Block transfer time 14 Table 16.2 (continued) Specifications of Typical HighEnd Enterprise Disks from Seagate (a) Seagate Enterprise Performance 10 K HDD - 1200 GB continued on next slide Table 16.2 (continued) Specifications of Typical High-End Enterprise Disks from Seagate (a) Seagate Enterprise Performance 10 K HDD - 1200 GB 17 Figure 16.3 Interleaved concurrency versus parallel execution. 18 Figure 16.4 Use of two buffers, A and B, for reading from disk. 19 Figure 16.5 Three record storage formats. (a) A fixed-length record with six fields and size of 71 bytes. (b) A record with two variable-length fields and three fixed-length fields. (c) A variablefield record with three types of separator characters. 20 Figure 16.6 Types of record organization. (a) Unspanned. (b) Spanned. Figure 16.7 Some blocks of an ordered (sequential) file of EMPLOYEE records with Name as the ordering key field. Table 16.3 Average Access Times for a File of b Blocks under Basic File Organizations File organization • Heap file (unordered file) place new records in no order at the end of the file • Sorted file ( sequential file) keeps the records ordered by the value of a particular file • Hashed file Uses hash function applied to a field (hash key) to determine a record’s placement on disk • B-trees, B+ trees – use tree structure 24 Hashing techniques Static hashing – hash address space is fixed Extendible hashing Linear hashing 25 Hashing algorithm 26 Hash Table (Wikipedia) http://en.wikipedia.org/wiki/Hash_table 27 28 Figure 16.8 Internal hashing data structures. (a) Array of M positions for use in internal hashing. (b) Collision resolution by chaining records. Figure 16.9 Matching bucket numbers to disk block addresses. Figure 16.10 Handling overflow for buckets by chaining. Figure 16.11 Structure of the extendible hashing scheme. 33 Figure 16.11 Structure of the extendible hashing scheme. Figure 16.12 Structure of the dynamic hashing scheme. Figure 16.13 Striping of data across multiple disks. (a) Bit-level striping across four disks. (b) Block-level striping across four disks. Figure 16.14 Some popular levels of RAID. (a) RAID level 1: Mirroring of data on two disks. (b) RAID level 5: Striping of data with distributed parity across four disks. 38 • A search tree of order p is a tree such that each node contains at most p - 1 search values and p pointers in the order < P1, K1, P2, K2, ..., Pq-1, Kq-1, Pq >, where q 1 p; each Pi is a pointer to a child node (or a null pointer); and each Ki is a search value from some ordered set of values. 39 40 41 B tree of order p 1. Each internal node in the B-tree is of the form <P1, <K1, Pr1> , P2, <K2, Pr2> , ..., <Kq-1,Prq-1> , Pq> where q 1 p. Each Pi is a tree pointer—a pointer to another node in the B-tree. Each Pri is a data pointer —a pointer to the record whose search key field value is equal to Ki (or to the data file block containing that record). 2. Within each node, K1 <K2 < ... < Kq-1. 3. For all search key field values X in the subtree pointed at by Pi (the ith subtree, see Figure 06.10a), we have: Ki-1 < X < Ki for 1 < i < q; X < Ki for i = 1; and Ki-1 < X for i = q. Each node has at most p tree pointers. Each node, except the root and leaf nodes, has at least (p/2) tree pointers. The root node has at least two tree pointers unless it is the only node in the tree. A node with q tree pointers, q 1 p, has q - 1 search key field values (and hence has q 1 data pointers). All leaf nodes are at the same level. Leaf nodes have the same structure as internal nodes except that all of their tree pointers Pi are null. 4. 5. 6. 7. 42 EXAMPLE 5: Suppose that the search field of Example 4 is a nonordering key field, and we construct a B-tree on this field. Assume that each node of the B-tree is 69 percent full. Each node, on the average, will have p * 0.69 = 23 * 0.69 or approximately 16 pointers and, hence, 15 search key field values. The average fan-out fo =16. We can start at the root and see how many values and pointers can exist, on the average, at each subsequent level: Root: Level 1: Level 2: Level 3: Level 4: 1 node 16 nodes 256 nodes 4096 nodes 65536 nodes 15 entries 240 entries 3840 entries 61,440 entries 983,040 entries 16 pointers 256 pointers 4096 pointers 65536 pointers 43 B+ Trees The structure of the internal nodes of a B+tree of order p is as follows: 1. Each internal node is of the form <P1, K1, P2, K2, ..., Pq-1, Kq-1, Pq> where q 1 p and each Pi is a tree pointer. 2. Within each internal node, K1 < K2 < ... <Kq-1. 3. For all search field values X in the subtree pointed at by Pi, we have Ki-1 < X 1 Ki for 1 < i < q; X 1 Ki for i = 1; and Ki-1 < X for i = q. 4. Each internal node has at most p tree pointers. 5. Each internal node, except the root, has at least (p/2) tree pointers. The root node has at least two tree pointers if it is an internal node. 6. An internal node with q pointers, q 1 p, has q - 1 search field values. 44 The structure of the leaf nodes of a B+-tree of order p (Figure 14.11b) is as follows: 1. Each leaf node is of the form <<K1, Pr1> , <K2, Pr2>, ..., <Kq-1, Prq-1>, Pnext> 2. 3. 4. 5. Where q 1 p, each Pri is a data pointer, and Pnext points to the next leaf node of the B+-tree. Within each leaf node, K1 < K2 < ... < Kq-1, q 1 p. Each Pri is a data pointer that points to the record whose search field value is Ki or to a file block containing the record (or to a block of record pointers that point to records whose search field value is Ki if the search field is not a key). Each leaf node has at least (p/2) values. All leaf nodes are at the same level. 45 46 47 EXAMPLE 7: Suppose that we construct a B+-tree on the field of Example 6. To calculate the approximate number of entries of the B+-tree, we assume that each node is 69 percent full. On the average, each internal node will have 34 * 0.69 or approximately 23 pointers, and hence 22 values. Each leaf node, on the average, will hold 0.69 * pleaf = 0.69 * 31 or approximately 21 data record pointers. A B+-tree will have the following average number of entries at each level: Root: Level 1: Level 2: Level 3: Level 4: 1 node 23 nodes 529 nodes 12,167nodes 279,841 nodes 22 entries 506 entries 11,638 entries 255,507entries 5,876,661entries 23 pointers 529 pointers 12,167 pointers 279,841 pointers 48 49 50 51 52