

Disk Storage,
Basic File Structures, and
Hashing

1


Introduction


In a computerized database, the data is stored on
computer storage medium, which includes:

Primary Storage
 can be processed directly by the CPU
 e.g., the main memory, cache
 fast, expensive, but of limited capacity

Secondary Storage
 cannot be processed directly by the CPU
 magnetic disks, optical disks, tapes
 slow, cost less, but have a large capacity.

2



Storage Hierarchy
Volatile
Cache
Primary
storage
Secondary
storage
unit price
Memory
Flash Memory
Magnetic Disk
speed
Non-volatile
Tertiary
storage
$$
Optical Disk
Magnetic Tape

3


Storage of Databases



For the following reasons, most databases are
stored permanently on secondary storage:

They are too large to fit entirely in main
memory

They must persist over long period of times, but
the main memory is a volatile storage

Secondary storage costs less
4


Secondary Storage


Magnetic-disk: cannot be directly processed by the CPU;
it must be brought to the main memory first.


Data is stored on spinning disk, and read/written
magnetically

Primary medium for the long-term storage of data;
typically stores entire database.

Non-volatile

slow access to data

large storage capacity (on the order of gigabytes)
5


Disk Storage Devices






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
6


Disk Storage Devices (contd.)




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 (faces) 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.
7


Disk Storage Devices (contd.)






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.
8


Physical Characteristics of Disks


9



Components of a Disk


The platters spin (say, 90rps).

The arm assembly is moved in or out to position a
head on a desired track.

Read-write head
 Positioned very close to the platter surface
(almost touching it)
 Reads or writes magnetically encoded
information.
 Only one head reads/writes at any one time.

Surface of platter divided into circular tracks
10


Physical Characteristics of Disks

Track
 an information storage circle on the surface of a
disk.
 Over 16,000 tracks per platter
 each track can store between 4KB and 50KB of
data.
 Each track is divided into sectors.




Tracks under heads make a cylinder (imaginary!)
Cylinder
 the tracks with the same diameter on all
surfaces of a disk pack.
 Cylinder i consists of i-th track of all the
platters
11


Physical Characteristics of Disks


Sector
 a part of a track with fixed size
 separated by fixed-size interblock gaps
 Typical sectors per track
 200 (on inner tracks) to 400 (on outer tracks)

12


Sectors


13




14



Disk I/O Model of Computation


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.
15



Pages and Blocks

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)
16




Pages and Blocks
Track
Gap
Sector
One track
1
3
2
4
...
1 page/block = 4 Sectors

17


Page I/O

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.




18


Page I/O


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
19



Page I/O

Transfer rate (tr)
 tr = track size / cost of one revolution

= track size / (60*1000/p) in msec

Bulk transfer rate (btr)
 btr = (B/(B+G)) * tr
bytes/msec
 Where B is the block size in bytes
 G is interblock gap size in bytes

Block transfer time (btt)
 btt = B / tr
not
 btt = B / btr
20

taking into acount G
taking into acount G



Page I/O


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
21



Page I/O

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 n
noncontigues pages/blocks that are in the same
cylinder = s + n * (rd + btt)
22



An Example



A hard disk specifications:
 4 platters, 8 Surfaces, 3.5 Inch diameter
 213 = 8192 tracks/surface
 28 = 256 sectors/track
 29 = 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
23


An Example





What is the total capacity of this disk
 8 GB (8*213*28*29=233)
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
24


An Example

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.
25



An Example

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
2
3
...
8
1 block = 8 Sectors

26



Buffering

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.
27



Accessing Data Through RAM Buffer 
RAM
Block transfer
Buffer
Application
block
Record
transfer

Page frames
28


Buffer Manager


Programs call on the buffer manager when they
need a block from disk.
 If the block is already in the buffer,
 the requesting program is given the address of the
block in main memory

If the block is not in the buffer,
 the buffer manager allocates space in the buffer for
the block, replacing (throwing out) some other
block, if required, to make space for the new block.
 The block that is thrown out is written back to disk
only if it was modified since the most recent time
that it was written to/fetched from the disk.

29


Buffer Manager

 Once space is allocated in the buffer, the buffer
manager reads the block from the disk to the buffer,
and passes the address of the block in main memory
to requester.

Buffer Replacement Policy:
 Frame is chosen for replacement by a
replacement policy:
 Least-recently-used (LRU), MRU, FIFO, etc.


Policy can have big impact on # of I/O’s;
depends on the access pattern.
30



File Organization

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 variablelength records.
31



File Organization


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)

32


File Organization


33


Fixed-Length Records

Insertion:
 Store record i starting from byte
n  (i – 1), where n is the size of
each record.

Deletion of record i:
 Packed format:

 move records i + 1, . . ., n to i, . . . ,
n–1
 OR
 move record n to i

Unpacked format (do not move
records, but)
 link all free records on a free list
 OR
 Use bitmap vector

34


Free Lists




Store the address of the first deleted record in the
file header.
Use this first record to store the address of the
second deleted record, and so on.
35


Page Formats: Fixed Length Records 

Record id = <page id, slot #>.
Slot 1
Slot 2
Slot 1
Slot 2
Free
Space
...
...
Slot N
Slot N
Slot M
N
PACKED

1 . . . 0 1 1M
number
of records
36
M ...
3 2 1
UNPACKED, BITMAP
number
of slots



Variable-Length Records Represenation 

Byte-String representation
 Attach an end-of-record () control character to
the end of each record
 Difficulty with deletion and growth

Slotted-page header contains:
 number of record entries
 location and size of each record
 end of free space in the block
37


Slotted Page Structure



Records can be moved around within a page to
keep them contiguous with no empty space
between them
 entry in the header must be updated.
 Pointers should not point directly to record instead they should point to the entry for the
record in header.
38


Fixed-Length Representation



Reserved Space
 can use fixed-length records of a known
maximum length
 unused space in shorter records filled with a
null or end-of-record symbol.
39


Fixed-Length Representation



List Representation by Pointers
 A variable-length record is represented by a list
of fixed-length records, chained together via
pointers.
 Can be used even if the maximum record length
is not known
40


Fixed-Length Representation


Disadvantage: space is wasted in all records except
the first in a a chain.
 Solution is to allow two kinds of block in file:
 Anchor block: contains the first records of chain
 Overflow block: contains records other than those
that are the first records of chairs.

41


Blocking Factor


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 needed to store a file of r
records:
 b = r/bfr blocks
42


Spanned & Unspanned Records



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.
43


Placing File Records on Disk



A file header or file descriptor contains information about
a file (e.g., the disk address, record format descriptions,
etc.)
44



Allocating File Blocks on Disk

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.
45




Organization of Records in Files


Heap 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)

Sorted/Ordered 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
46


Heap File Organization


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

47


Heap File Organization


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 record of the file is located in
 Block i/f and in the
 (i mod f)-th record in that block

48


Heap File Organization




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.
49


File Stored as a Heap File
666666
123456
987654
MGT123
CS305
CS305
F1994 4.0
S1996 4.0
F1995 2.0
717171
666666
765432
515151
CS315
EE101
MAT123
EE101
S1997 4.0
S1998 3.0
S1996 2.0
F1995 3.0
234567
CS305
S1999 4.0

page 0
page 1
page 2
878787

MGT123
S1996
50
3.0


Sequential File Organization




Suitable for applications that require sequential
processing of the entire file
The records in the file are ordered by a search-key
51


Files of Ordered Records



Some blocks of an
ordered (sequential)
file of EMPLOYEE
records with NAME
as the ordering key
field.
52


File Stored as a Sorted File
111111
111111
123456
MGT123
CS305
CS305
F1994 4.0
S1996 4.0
F1995 2.0
123456
123456
232323
234567
CS315
EE101
MAT123
EE101
S1997 4.0
S1998 3.0
S1996 2.0
F1995 3.0
234567
CS305
S1999 4.0

page 0
page 1
page 2
313131

MGT123
S1996
53
3.0


Sequential File Organization

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
54


Sequential File Organization


55



Hashing Techniques

A hash function maps the hash field of a record
into the address of the storage media in which
the record is stored.

Hashing provides very fast access to records,
where the search condition is an equality
condition on the hash field.

For internal files, hashing is implemented as a
hash table. The mapping that assigns each
element of the data a cell of the hash table is
called a hash function.
56




Hashing Techniques

Two records that yield the same hash value are
said to collide.

A good hash function must be easy to compute
and generate a low number of collisions.

The process of finding another position (for
colliding data) is called collision resolution.

There are several methods for collision
resolution, including open addressing,
chaining, and multiple hashing.
57




Hashing Techniques

Open addressing:
 Proceeding from the occupied position specified by
the hash function, check the subsequent positions in
order until an unused position is found.

Chaining:
 Associate an overflow area (or a linked list) to any
cell (hashing address) and then simply store the data
in this medium.

Multiple hashing:
 Apply a second hash function if the first results in a
collision.
 If another collision results, use open addressing, or
apply a third hash function, and then use open
addressing.
58




Hashing Techniques
59




Hashing Techniques


Hashing for disk files is called external hashing.

The target address space in external hashing is
made of buckets (which holds a disk block or a
cluster of contiguous blocks).

The collision problem is less severe, because as
many records as will fit in a bucket can hash to the
same bucket without causing collision problem.

A table maintained in the file header converts the
bucket number into the corresponding disk block
address.
60


Hashing Techniques



Matching bucket numbers to disk block addresses.
61



Hashing Techniques


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.
62


Hashed Files - Overflow handling


63



Hashing Techniques

The hashing scheme is called static hashing if a
fixed number of buckets is allocated.

Main disadvantage of static external hashing:
 The number of buckets must be chosen large
enough that can handle large files. That is, it is
difficult to expand or shrink the file
dynamically.

Three solutions to the above problem are:
 Dynamic hashing,
 Extendible hashing
 Linear hashing
64

