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ICOM 5016 – Introduction to Database Systems Lecture 13- File Structures Dr. Bienvenido Vélez Electrical and Computer Engineering Department Slides by.

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Presentation on theme: "ICOM 5016 – Introduction to Database Systems Lecture 13- File Structures Dr. Bienvenido Vélez Electrical and Computer Engineering Department Slides by."— Presentation transcript:

1 ICOM 5016 – Introduction to Database Systems Lecture 13- File Structures Dr. Bienvenido Vélez Electrical and Computer Engineering Department Slides by Dr. Manuel Rodríguez

2 ICOM 5016Dr. Manuel Rodriguez Martinez 2 Readings Read –New Book: Chapter 13

3 ICOM 5016Dr. Manuel Rodriguez Martinez 3 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 Martinez 4 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 Martinez 5 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

6 ICOM 5016Dr. Manuel Rodriguez Martinez 6 Buffer Management Buffer Manager supplies a stream of memory data pages. –Often the size of a block (sector, several sectors, or more) –Must provide in-memory access to more pages than physically fit in memory –Must implement a page replacement policy –Implements a cache of disk blocks

7 ICOM 5016Dr. Manuel Rodriguez Martinez 7 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

8 ICOM 5016Dr. Manuel Rodriguez Martinez 8 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

9 ICOM 5016Dr. Manuel Rodriguez Martinez 9 Storage arrangement options Suppose we need to create 10 GB of space to store a database. Each page is 4 KB is size. –How to organize the disk to accomplish this. Option #1: Cooked File System –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 has table somewhere to store the information about this file “mydb.dat”. –Now you can start writing pages with actual data.

10 ICOM 5016Dr. Manuel Rodriguez Martinez 10 Option #2: 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. –Provided by more advanced (and expensive) DBMS’s

11 Alternative DBMS Views of the Storage System Single File –Each database is stored as a single file –Example: SQLite –Pro: Easy to store and exchange databases –Cons: Concurrent/Transactional access very hard File System –Each database is stored as multiple files –Example: MySQL –Pro: Concurrent/Transactional access good, DBMS portable and simpler –Cons: DBMS must rely on OS for performance Block System (Raw Disk) –The DBMS implements its own storage system on raw disk partition –Example: Oracle 10g –Pro: RDBMS can achieve maximal performance –Cons: RDBMS complex/expensive

12 ICOM 5016Dr. Manuel Rodriguez Martinez 12 File and Access Methods Layer Buffer Manager provides a stream of pages But higher layers of DBMS need to see a stream of records A DBMS file layers provides this abstraction –File is a collection of records that belong to a relation R. For example: Relation students might be stored in DBMS internal file students.dat. This is internal to DBMS database!!! –File is made out of pages, and records are taken from pages File and Access Methods Layer implements various types of files to access the records –Access method – mechanism by which the records are extracted from the DBMS

13 ICOM 5016Dr. Manuel Rodriguez Martinez 13 File Types Heap File - Unordered collection of records –Records within a page a not ordered –Pages are not ordered –Simple to use and implement Sorted File – sorted collection or records –Within a page, records are ordered –Pages are ordered based on record contents –Efficient access to data, but expensive to maintain Index File – combines storage + data structure for fast access and lookups –Index entries – store value of attributes as search keys –Data entries – hold the data in the index file

14 ICOM 5016Dr. Manuel Rodriguez Martinez 14 Heap File 123BobNY$102 8387NedSJ$73 121JilNY$5595 81982TimMIA$4000 2381BillLA$500 4882AlSF$52303 9403NedNY$3333 1237PatWI$30 Page 0 Page 1 Page 2

15 ICOM 5016Dr. Manuel Rodriguez Martinez 15 Sorted File 121JilNY$5595 123BobNY$102 1237PatWI$30 2381BillLA$500 4882AlSF$52303 8387NedSJ$73 9403NedNY$3333 81982TimMIA$4000 Page 0 Page 1 Page 2

16 ICOM 5016Dr. Manuel Rodriguez Martinez 16 Index File 121JilNY$5595 123BobNY$102 1237PatWI$30 2381BillLA$500 4882AlSF$52303 8387NedSJ$73 9403NedNY$3333 81982TimMIA$4000 100 2000 9000 Index entry Data entries

17 ICOM 5016Dr. Manuel Rodriguez Martinez 17 Index files structure Index entries –Store search keys –Search key – a set of attributes in a tuple can be used to guide a search Ex. Student id –Search key do not necessarily have to be candidate keys For example: gpa can be a search key on relation: Students(sid, name, login, age, gpa) Data entries –Store the data records in the index file –Data record can have Actual tuples for the table on which index is defined Record identifier for tuples that match a given search key

18 ICOM 5016Dr. Manuel Rodriguez Martinez 18 Issues with Index files Index files for a relation R can occur in three forms: –Data entries store the actual data for relation R. Index file provides both indexing and storage. –Data entries store pairs : k – value for a search key. rid – rid of record having search key value k. Actual data record is stored somewhere else, perhaps on a heap file or another index file. –Data entries store pairs K – value for a search key Rid-list – list of rid for all records having search key value k Actual data record is stored somewhere else, perhaps on a heap file or another index file.

19 ICOM 5016Dr. Manuel Rodriguez Martinez 19 Operations on files Allocate file Scan operations –Grab each records one after one Can be used to step through all records Insert record –Adds a new record to the file Each record as a unique identifier called the record id (rid) Update record Find record with a given rid Delete record with a given rid De-allocate file

20 ICOM 5016Dr. Manuel Rodriguez Martinez 20 Implementing Heap Files Heap file links a collection of pages for a given relation R. Heap files are built on top of Buffer Manager. Each page has a page id –Often, we need to know the page size (e.g. 4KB) All pages for a given file have the same size. –Page id and page size can be used to compute an offset in a cooked file where the page is located. –In raw disk partition, page id should enable DBMS to find block in disk where the page is located.

21 ICOM 5016Dr. Manuel Rodriguez Martinez 21 Linked Implementation of Heap Files Header Page Data Page Data Page Data Page Data Page Data Page Linked List of pages With free space Linked List of full pages

22 ICOM 5016Dr. Manuel Rodriguez Martinez 22 Linked List of pages Each page has: –records –pointer to next page –pointer to previous page Pointer here means the integer with the page id of the next page. Header has two pointers –First page in the list of pages with free space –First page in the list of full pages Tradeoffs –Easy to use, good for fixed sized records –Complex to find space for variable length records need to iterate over list with space

23 ICOM 5016Dr. Manuel Rodriguez Martinez 23 Directory of pages Data Page 1 Data Page 2 Data Page 3 Data Page N...... header

24 ICOM 5016Dr. Manuel Rodriguez Martinez 24 Directory of pages Linked list of directory pages Directory page has –Pointer to a given page –Bit indicating if page is full or not –Alternatively, have amount of space that is available More complex to implement Makes it easier to find page with enough room to store a new record

25 ICOM 5016Dr. Manuel Rodriguez Martinez 25 Page formats Each page holds –records –Optional metadata for finding records within the page Page can be visualized as a collection of slots where records can be placed Each record has a record id in the form: – –page_id – id of the page where the record is located. –slot number – slot where the record is located.

26 ICOM 5016Dr. Manuel Rodriguez Martinez 26 Packed Fixed-Length Record... Slot 1 Slot 2 Slot 3 N Slot N Free Space Page header number of records

27 ICOM 5016Dr. Manuel Rodriguez Martinez 27 Unpacked Fixed-Length Record... Slot 1 Slot 2 Slot 3 N Slot N Free Space Page header number of slots 101 3 2 1 Slot bit vector

28 ICOM 5016Dr. Manuel Rodriguez Martinez 28 Variable-Length records rid=(i,N) Page i rid=(2,N) rid=(1,N)  12 …  15  24 N  Free Space N … 2 1 entries 12 bytes

29 ICOM 5016Dr. Manuel Rodriguez Martinez 29 Fixed-Length records for variable- length attributes Size of each record is determined by maximum size of the data type in each column F1F2F3F4 2624 Size in bytes Offset of F1: 0 Offset of F2: 2 Offset of F3: 8 Offset of F4: 10 Need to understand the schema and sizes to find a given column

30 ICOM 5016Dr. Manuel Rodriguez Martinez 30 Variable-length records and attributes Either 1.use a special symbol to separate fields 2.use a header to indicate offset of each field F1$F2$F3$F4 Option 1 F1F2F3F4 Option 2 Option 1 has the problem of determining a good $ Option 2 handles NULL easily


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