User Perspective. Module III: System Perspective. Module III: Topics Covered. Module III Overview of Storage Structures, QP, and TM
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1 Module III Overview of Storage Structures, QP, and TM Sharma Chakravarthy UT Arlington base Management Systems: Sharma Chakravarthy Module I Requirements analysis EER modeling User Perspective EER to Relational mapping: Transform a real-world problem into a set of relations (or schema) Normalization of relations Specific DBMS independent Module II Create schema for a specific DBMS (e.g., Oracle) Understanding relational algebra Understanding SQL supported by all Relational DBMSs Understanding the relationship between RA and SQL How to connect to a DBMS, develop applications, and interfaces Slide 2 Module III: System Perspective This module provides an understanding of the other side system perspective! Is the architecture of a DBMS different from other applications? If so why? How is it different from a main-memory application? How does a DBMS support data sizes that are much larger (1000 to 10000) than its main memory size? What are the functional modules of a DBMS? How is SQL translated, optimized, and executed? How does a DBMS support hundreds and thousands of concurrent users? Hoe does a DBMS provide recovery from all types of failures (power, program, system, media)? Module III: Topics Covered Storage structures records, pages, and files (1 lecture) Index structures Tree-based and hashbased (2 lecture) NoSQL and cloud concepts new External sorting (1 lecture) may skip Query processing (1 lecture) Concurrency control overview (1.5 lecture) Recovery overview (1.5 lecture) Slide 3 Slide 4 1
2 DBMS Architecture Disks and Files Transaction Manager Lock Manager Concurrency control SQL commands from applications Query Optimization and Execution (parser, optimizer, Plan executor) Files and Access Methods Buffer Management Disk Space Management Log DB Recovery Manager System catalog, data files, Index files Slide 5 DBMS stores information on ( hard ) disks. This has major implications for DBMS design! READ: transfer data from disk to main memory (RAM). WRITE: transfer data from RAM to disk. Both are high-cost operations, relative to inmemory operations, so must be planned carefully! Unit of transfer (read/write) is a page (8K bytes) from disk Slide 6 Why Not Store Everything in Main Memory? Costs too much. $80 will buy you either 8 GB (4 GB 2 year ago) of RAM or 1 to 1.5 TB (250 GB 1 year ago) of disk today or 64GB of SSD (was $250 3 years ago) Main memory is volatile. We want data to be saved (persistent) between runs. (Obviously!) Typical storage hierarchy: Cache most expensive Main memory (RAM) for currently used data. Solid state disks (SSD) -- New!! (up to 512 GB) Disk for the main database (secondary storage). Other storage devices (flash card, usb stick, drum ) Tapes for archiving older versions of the data (tertiary storage). Accessing a Disk Disk is a mechanical device (as compared to SSD) Time to access (read/write) a disk block: seek time (moving arms to position disk head on track) rotational delay (waiting for block to rotate under head) transfer time (actually moving data to/from disk surface) Buffer size (2 MB typical, 8 MB, ) Seek time and rotational delay dominate. Seek time varies from about 1 to 20msec Rotational delay varies from 0 to 10msec Transfer rate is about 1 msec per 4KB page Key to lower I/O cost: reduce seek/rotation delays! Hardware vs. software solutions? DVD/CDROM and other devices least expensiveslide 7 Slide 8 2
3 Accessing a Disk Time to access (read/write) a disk block: Average seek time msec Average rotational delay msec transfer rate 13MB/sec Seek from one track to next 2.2 msec Max. seek time 15 msec Disk access takes about 10 msec whereas accessing memory location takes about 60 nano secs!! Memory is more than a Million times faster!! Slide 9 Impedance mismatch If you are studying DBMS, you have to understand this clearly CPU is way faster than disk access. In fact, it is a Million or more times faster Cpu accesses are in nano secs and disk accesses are still in milli secs! Because of this speed difference, it takes much longer to bring data from a disk So, cpu is waiting/idling most of the times for data (if not architected properly)! This is called impedance mismatch! Solution: BM and execute multiple Txs Disk Space Management Lowest layer of DBMS software manages space on disk. Higher levels call upon this layer to: allocate/de-allocate a page read/write a page Request for a sequence of pages must be satisfied by allocating the pages sequentially on disk! Higher levels don t need to know how this is done, or how free space is managed. Slide 11 Files of Records A relation is stored as a file of records (tuples) or block is OK when doing I/O, but higher levels of DBMS operate on records, and files of records. FILE: A collection of pages, each containing a collection of records. Must support: insert/delete/modify opeartions read a particular record (specified using record id) scan all records (possibly with some conditions on the records to be retrieved) Slide 12 3
4 Unordered (Heap) Files Heap (unordered) File Implemented as a List Simplest file structure contains records in no particular order. As a file grows and shrinks, disk pages are allocated and de-allocated. To support record level operations, we must: keep track of the pages in a file keep track of free space on pages keep track of the records on a page There are many alternatives for keeping track The header page id and Heap file name must be stored someplace. Each page contains 2 `disk pointers plus data. Disadvantage for variable length records! (all pages of this. Slide 13 will be on the free list) Slide 14 Header Full s s with Free Space Heap File Using a Directory Header DIRECTORY 1 2 N The entry for a page can include the number of free bytes on the page. The directory is a collection of pages; linked list implementation is just one alternative. Much smaller than linked list of all HF pages! Slide 15 Indexes A Heap file allows us to retrieve records: by specifying the rid, or by scanning all records sequentially (very slow) Sometimes, we want to retrieve records by specifying the values in one or more fields (associative retrieval), e.g., Find all students in the CS department Find all students with a gpa > 3 Indexes are file structures that enable us to answer such value-based queries efficiently. Slide 16 4
5 Range Searches ``Find all students with gpa > 3.0 If data is in contiguous sorted file, do binary search to find first such student, then scan to find others. Cost of binary search can be quite high. Simple idea: Create an `index file. k1 k2 kn Index File Binary Trees AVL Trees Indexes ISAM B Trees (height Balanced) B+ trees (also height balanced) N File * Can do binary search on (smaller) index file! Slide 17 Slide 18 B-tree structures. (a) A node in a B-tree with q 1 search values. (b) A B-tree. Values were inserted in the order 8, 5, 1, 7, 3, 12, 9, 6. The nodes of a B+-tree. (a) Internal node of a B+-tree with q 1 search values. (b) Leaf node of a B+-tree with q 1 search values and q 1 data pointers. Slide 19 Slide 20 5
6 2* 3* 5 13 Example B+ Tree Root 17 Entries < 17 Entries >= 17 5* 7* 8* 14* 16* 22* 24* 27* 29* 33* 34* 38* 39* Find 28*? 29*? All > 15* and < 30* Insert/delete: Find data entry in leaf, then change it. Need to adjust parent sometimes. And change sometimes bubbles up the tree Slide 21 B+ Trees in Practice Typical order: 100. Typical fill-factor: 67%. average fanout = 133 Typical capacities: Height 4: = 312,900,700 records Height 3: = 2,352,637 records Can often hold top levels in buffer pool: Level 1 = 1 page = 8 Kbytes Level 2 = 133 pages = 1 Mbyte Level 3 = 17,689 pages = 133 MBytes Slide 22 Overview of Query Optimization Module III Relational Query Optimization Sharma Chakravarthy UT Arlington sharma@cse.uta.edu base Management Systems: Sharma Chakravarthy Input: Sql query Output: Query Plan: Tree of Relational algebra operators, with choice of algorithm for each operator Main issues: For a given query, what plans are generated/considered? Algorithm to search plan space for cheapest (estimated) plan. How is the cost of a plan estimated? Using the cost formulas studied so far + assumptions Ideally: Want to find best plan. Practically: Avoid worst plans! We will study the System R approach. Slide 24 6
7 Why System R Optimizer Types of Query Evaluation Plans Most widely used currently; works well for < 10 joins. Cost estimation: Approximate art at best. Statistics, maintained in system catalogs, are used to estimate cost of operations and result sizes. Considers combination of CPU and I/O costs. Plan Space: Too large, must be pruned. Only the space of left-deep plans is considered. Left-deep plans allow output of each operator to be pipelined into the next operator without storing it in a temporary relation. Cartesian products avoided. Slide 25 Slide 26 Summary Query optimization involves Determining join order (cost based) Determining join algorithm for each join The above 2 are not separate steps; involves Search space exploration Pruning Constructing the final query evaluation plan Statistics are maintained by the system Uses histograms Module III Transaction Management Sharma Chakravarthy UT Arlington sharma@cse.uta.edu Slide 27 base Management Systems: Sharma Chakravarthy 7
8 TM has Two Distinct Components Concurrency Control the activity of coordinating the actions of processes that operate in parallel, access shared data, and therefore potentially interfere with each other Recovery the activity of ensuring that software and hardware failures do not corrupt or render persistent data inconsistent the database contains all the effects of committed transactions and none of the effects of uncommitted transactions Slide 29 Properties of transaction A transaction is a collection of operations with the following (ACID) properties: Atomicity: all or nothing property Consistency: A transaction transforms a consistent DB state to another consistent DB state. The actions taken by the Tx as a group do not violate any of the integrity constraints associated with the DB. Isolation: Even though transactions execute concurrently, it appears to each transaction T, that others executed either before T or after T, but not both. Durability: Once a transaction completes successfully (or commits), its changes to the DB state survive failures. Slide 30 Why Concurrency Control? Why Recovery? Concurrency control can increase processor utilization increase total transaction throughput can increase response time (slightly) for individual transactions. Why? Short transactions do not get delayed due to long running transactions The above are especially important in a DBMS where transactions access data from secondary storage devices (CPU is waiting for a disk read/write to complete!!) Slide 31 Needed to accommodate various kinds of failures logical errors (abort by the transaction) system errors (abort by the system deadlock, exceptions) system crashes: losing the contents of volatile storage (power failure) loss of non-volatile storage (or media failure) head crash Power failure?? Slide 32 8
9 Summary Concurrency control: synchronization of conflicting or interdependent actions across multiple transactions Anomalies due to reads and writes 2 phase locking Recovery Steal, no force WAL protocol Aries Recovery works in 3 phases: Analysis: Forward from checkpoint. Redo: Forward from oldest reclsn. Undo: Backward from end to first LSN of oldest Xact alive at crash. Slide 33 Thank You! 9
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