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  Disk Storage Devices  Files of Records  Operations on Files  Unordered Files  Ordered Files    Hashed Files  Dynamic and Extendible Hashing Techniques RAID Technology
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  Disk I/O is equivalent to one read or write operation of a single block  It is very expensive compared with what is likely to be done once the block gets in main memory  one random disk I/O ~ about 1,000,000 machine instructions in terms of time  Cost for computation that requires secondary storage is computed only by disk I/Os.
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  Preferred secondary storage device for high storage capacity and low cost.
 Data stored as magnetized areas on magnetic disk surfaces.
 A disk pack contains several magnetic disks connected to a rotating spindle.
  Disks are divided into concentric circular tracks on each disk surface.
 Track capacities vary typically from 4 to 50 Kbytes or more
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  A track is divided into smaller blocks or sectors  because it usually contains a large amount of information  The division of a track into sectors is hard-coded on the disk surface and cannot be changed.
 One type of sector organization calls a portion of a track that subtends a fixed angle at the center as a sector.
  A track is divided into blocks.
  The block size B is fixed for each system.
 Typical block sizes range from B=512 bytes to B=4096 bytes.
Whole blocks are transferred between disk and main memory for processing.
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   A read-write head moves to the track that contains the block to be transferred.
 Disk rotation moves the block under the read-write head for reading or writing.
A physical disk block (hardware) address consists of:    a cylinder number (imaginary collection of tracks of same radius from all recorded surfaces) the track number or surface number (within the cylinder) and block number (within track).
 Reading or writing a disk block is time consuming because of the seek time s and rotational delay (latency) rd.
  Double buffering can be used to speed up the transfer of contiguous disk blocks.
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Fill B
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 Data files decomposed into pages (blocks)   fixed size piece of contiguous information in the file sizes range from 512 bytes to several kilobytes  block is the smallest unit for transferring data between the main memory and the disk.  Address of a page (block):  ( cylinder# , track# (within cylinder) , sector# (within track )  
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  Track Sector One track 1 2 3
1 page/block = 4 Sectors 12
4 Gap
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   Page I/O --- one page I/O is the cost (or time needed) to transfer one page of data between the memory and the disk.  The cost of a (random) page I/O =  seek time + rotational delay + block transfer time    Seek time  time needed to position read/write head on correct track. Rotational delay ( latency )  time needed to rotate the beginning of page under read/write head. Block transfer time  time needed to transfer data in the page/block.
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    Average
rotational delay (rd)
 rd = ½ * (1/p) min = (60*1000)/(2*p) msec   OR rd = ½ * cost of 1 revolution   = ½ * (60*1000/p) msec where  p is speed of disk rotation (how many revolutions per minute - rpm) Example     Speed of disk rotatioon is p = 3600 rpm 60 revolutions/sec 1 rev. = 16.66 msec. (1 second = 1000 msec) rd = 8.33 ms
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Transfer rate (tr)
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tr = track size / cost of one revolution = track size / (60*1000/p) in msec
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Bulk transfer rate (btr)
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btr = (B/(B+G)) * tr bytes/msec
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Where B is the block size in bytes G is interblock gap size in bytes
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Block transfer time (btt)
  btt = B / tr btt = B / btr
not
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taking into acount G taking into acount G
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 Example:   Track size = 50 KB and p = 3600 rpm Block size B = 3KB = 3000 bytes  tr = (50*1000)/(60*1000/3600) = 3000 bytes/msec  btt = B / tr = 3000/3000 = 1 msec  
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 Average time for reading/writing
n
consecutive
pages that are in the same track or cylinder = s + rd +
n
* btt  Average time for reading/writing consecutively noncontigues pages/blocks that are in the same
n
cylinder = s +
n
* (rd + btt)  
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 A hard disk specifications:        4 platters, 8 Surfaces, 3.5 Inch diameter 2 13 = 8192 tracks/surface 2 8 = 256 sectors/track 2 9 = 512 bytes/sector Average seek time s = 25 ms Rotation rate rd = 3600 rpm = 60 rps 1 rev. = 16.66 msec    Transfer rate tr = 1 KB in 0.117 ms tr = 1 KB in 0.130 ms with gap 
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 What is the total capacity of this disk  8 GB (8*2 13 *2 8 *2 9 =2 33 )  How many bytes does one track hold?
 256 sectors/track*512 bytes/sector = 128KB  How many blocks per track?
 one block = 4096 bytes = 8 sectors (4096/512)  256/8 = 32 blocks/track  
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  How long does it take to access one block?
   One block = 4096 bytes  8 sectors = 4096/ 512 Rotation rate r  1 rev. = 16.66 msec.
Time to access 1 sector (s + r/2 + tr/(secters/KB)  25 + (16.66/2) + .117/2 = 33.3885 ms.
  time to access 1 block  time to access the first sector of the block + time to access the subsequent 7 sectors.
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 T = 25 + (16.66/2) + (0.117/2) * 1 + (0.13/2) *7 = 33.3885 + 0.455 ms = 33.8435ms
 Compare to one sector access time: 33.3885 ms  1 8 2 3 1 block = 8 Sectors
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 A buffer   is a contiguous reserved area in main memory available for storage of copies of disk blocks.
to speed up the processes.
 For a read command  the block from disk is copied into the buffer.
 For a write command  the contents of the buffer are copied into the disk.
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 RAM Block transfer Application Buffer  Record transfer
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 The database is stored as a collection of files.
 Each file is a sequence of records.  A record is a sequence of fields.
 Records are stored on disk blocks.  A file can have fixed-length records or variable- length records.
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 Fixed length records  Each record is of fixed length. Pad with spaces.   Variable length records   different records in the file have different sizes.
Arise in database systems in several ways:  different record types in a file.
 same record type with (variable-length fields, repeating field, or optional fields) 
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  
Blocking Factor
(bfr) - the number of records that can fit into a single block.  bfr = ⌊B/R⌋  B : Block size in bytes  R : Record size in bytes Example:    Record size R = 100 bytes Block Size B = 2,000 bytes Thus the blocking factor bfr = 2000/100 = 20   The number of blocks
b
records:  b = r/bfr blocks needed to store a file of
r
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    A block is the unit of data transfer between disk and memory.
Unspanned records :  A record is found in one and only one block.
 records do not span across block boundaries.  Used with fixed-length records having B  R Spanned records :  Records are allowed to span across block boundaries.
 Used with variable-length records having R  B   In variable-length records, either organization can be used.
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 A file header or file descriptor contains information about a file (e.g., the disk address, record format descriptions, etc.)  
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   Typical file operations include:  OPEN: Readies the file for access, and associates a pointer that will refer to a current file record at each point in time.
   FIND: Searches for the first file record that satisfies a certain condition, and makes it the current file record.
FINDNEXT: Searches for the next file record (from the current record) that satisfies a certain condition, and makes it the current file record.
READ: Reads the current file record into a program variable.
      INSERT: Inserts a new record into the file & makes it the current file record. DELETE: Removes the current file record from the file, usually by marking the record to indicate that it is no longer valid.
MODIFY: Changes the values of some fields of the current file record.
CLOSE: Terminates access to the file.
REORGANIZE: Reorganizes the file records.
 For example, the records marked deleted are physically removed from the file or a new organization of the file records is created.
READ_ORDERED: Read the file blocks in order of a specific field of the file.
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 The physical disk blocks that are allocated to hold the records of a file can be contiguous , linked , or indexed .
 In contiguous allocation, the file blocks are allocated to consecutive disk blocks.
 In linked allocation, each file block contains a pointer to the next file block.
 In indexed allocation, one or more index blocks contain pointers to the actual file blocks.
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Unordered/Heap/Pile File Organization
 a record can be placed anywhere in the file where there is space, or at the end   for full file scans or frequent updates Data unordered (unsorted) 
Ordered/Sorted/Sequential File Organization
 store records sorted in order, based on the value of the search key of each record  Need external sort or an index to keep sorted  
Hashing File Organization
 a hash function computed on some attribute of each record  the result specifies in which block of the file the record should be placed
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   Records are placed in the file in the order in which they are inserted. Such an organization is called a heap file.
 Insertion is at the end  takes constant time O(1) (very efficient)   Searching  requires a linear search (expensive) Deleting  requires a search, then delete  Select, Update and Delete  take b/2 time (linear time) in average  b is the number of blocks
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 For a file of unordered fixed-length records using unspanned blocks and contiguous allocation, it is straightforward to access any record by its position in the file.
 If the records are numbered 0,1,2, …, r-1 and   The records in each block are numbered 0,1,2, …, f-1 , where
f
is the blocking factor The the
i-th
 Block i/f record of the file is located in and in the  ( i mod f )-th record in that block  
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 A Heap file allows us to retrieve records:   by specifying the rid, or by scanning all records sequentially  Accessing a record by its position does not help locate a record based on a search condition.
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 
666666 MGT123 F1994 4.0
123456 CS305 S1996 4.0 page 0 987654 CS305 F1995 2.0
 717171 CS315 S1997 4.0
666666 EE101 S1998 3.0 page 1 765432 MAT123 S1996 2.0
515151 EE101 F1995 3.0
234567 CS305 S1999 4.0
page 2 878787 MGT123 S1996 3.0
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  Suitable for applications that require sequential processing of the entire file The records in the file are ordered by a search-key  
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 Some blocks of an ordered (sequential) file of EMPLOYEE records with NAME as the ordering key field.
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111111 MGT123 F1994 4.0
111111 CS305 S1996 4.0 page 0 123456 CS305 F1995 2.0
 123456 CS315 S1997 4.0
123456 EE101 S1998 3.0 page 1 232323 MAT123 S1996 2.0
234567 EE101 F1995 3.0
234567 CS305 S1999 4.0
page 2 313131 MGT123 S1996 3.0
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   Insertion is expensive    records must be inserted in the correct order  locate the position where the record is to be inserted  if there is free space insert there  if no free space insert the record in an overflow block  In either case, pointer chain must be updated Insert takes lg(b) plus the time to re-organize records. b is the number of blocks  Deletion  use pointer chains  Searching  very efficient (Binary search)  This requires lg(b) on the average
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 The following table shows the average access time to access a specific record for a given type of file  
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        Hashing for disk files is called External Hashing The file blocks are divided into M equal-sized buckets, numbered bucket 0 ,
bucket
1 , ..., bucket M-1 .
 Typically, a bucket corresponds to one (or a fixed number of) disk block.
One of the file fields is designated to be the hash key of the file.
The record with hash key value K is stored in bucket i, where i = h(K), and h is the hashing function.  i represents the address of the disk block in which the record is stored Search is very efficient on the hash key.
 A single-block access is needed (in most cases) to retrieve a record.
Collisions occur when a new record hashes to a bucket that is already full.
  An overflow file is kept for storing such records.
Overflow records that hash to each bucket can be linked together.
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 Hash table   The use of an array of records Simple hash function: h(K) = K mod M  
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   There are numerous methods for collision resolution, including the following:    Open addressing: Proceeding from the occupied position specified by the hash address, the program checks the subsequent positions in order until an unused (empty) position is found. Chaining: For this method, various overflow locations are kept, usually by extending the array with a number of overflow positions. In addition, a pointer field is added to each record location. A collision is resolved by placing the new record in an unused overflow location and setting the pointer of the occupied hash address location to the address of that overflow location. Multiple hashing: The program applies a second hash function if the first results in a collision. If another collision results, the program uses open addressing or applies a third hash function and then uses open addressing if necessary.
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     To reduce overflow records, a hash file is typically kept 70-80% full.
The hash function h should distribute the records uniformly among the buckets  Otherwise, search time will be increased because many overflow records will exist.
Main disadvantages of static external hashing:   Fixed number of buckets M is a problem if the number of records in the file grows or shrinks.
Ordered access on the hash key is quite inefficient (requires sorting the records).
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  Dynamic and Extendible Hashing Techniques  Hashing techniques are adapted to allow the dynamic growth and shrinking of the number of file records.
 These techniques include the following: dynamic hashing, extendible hashing, and linear hashing.  Both dynamic and extendible hashing use the binary representation of the hash value h(K) in order to access a directory.
  In dynamic hashing the directory is a binary tree.
In extendible hashing the directory is an array of size 2
d
where d is called the global depth. 
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  The directories can be stored on disk, and they expand or shrink dynamically.
 Directory entries point to the disk blocks that contain the stored records.
 An insertion in a disk block that is full causes the block to split into two blocks and the records are redistributed among the two blocks.
 The directory is updated appropriately.
 Dynamic and extendible hashing do not require an overflow area.   Linear hashing does require an overflow area but does not use a directory.
 Blocks are split in linear order as the file expands.
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