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SECONDARY STORAGE
MANAGEMENT
SECTIONS 13.1 – 13.3
Sanuja Dabade & Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
13.1 The Memory Hierarchy
13.1.1 The Memory Hierarchy
13.1.2 Transfer of Data Between Levels
13.1.3 Volatile and Nonvolatile Storage
13.1.4 Virtual Memory
13.2 Disks
13.2.1 Mechanics of Disks
13.2.2 The Disk Controller
13.2.3 Disk Access Characteristics
Presentation Outline (con’t)
13.3 Accelerating Access to Secondary
Storage
13.3.1 The I/O Model of Computation
13.3.2 Organizing Data by Cylinders
13.3.3 Using Multiple Disks
13.3.4 Mirroring Disks
13.3.5 Disk Scheduling and the Elevator
Algorithm
13.3.6 Prefetching and Large-Scale
Buffering
13.1.1 Memory Hierarchy
Several components for data storage having
different data capacities available
Cost per byte to store data also varies
Device with smallest capacity offer the fastest
speed with highest cost per bit
Memory Hierarchy Diagram
Programs,
DBMS
Main Memory DBMS’s
As Visual Memory
Tertiary Storage
Disk
Main Memory
Cache
File System
13.1.1 Memory Hierarchy
Cache
Lowest
level of the hierarchy
Data items are copies of certain locations of main
memory
Sometimes, values in cache are changed and
corresponding changes to main memory are delayed
Machine looks for instructions as well as data for those
instructions in the cache
Holds limited amount of data
13.1.1 Memory Hierarchy (con’t)
No need to update the data in main memory
immediately in a single processor computer
In multiple processors data is updated immediately
to main memory….called as write through
Main Memory
Everything happens in the computer i.e. instruction
execution, data manipulation, as working on
information that is resident in main memory
Main memories are random access….one can
obtain any byte in the same amount of time
Secondary storage
Is typically magnetic disk, transfer a single byte
between disk and main memory is around 10
milisecond
Used to store data and programs when they are not
being processed
More permanent than main memory, as data and
programs are retained when the power is turned
off
Tertiary Storage
Holds data volumes in terabytes
Used for databases much larger than what can be
stored on disk, but it has significantly higher
read/write times than secondary storage
13.1.2 Transfer of Data Between levels
Data moves between adjacent levels of the
hierarchy
At the secondary or tertiary levels accessing the
desired data or finding the desired place to store
the data takes a lot of time
Disk is organized into bocks
Entire blocks are moved to and from memory called
a buffer
13.1.2 Transfer of Data Between level
(cont’d)
A key technique for speeding up database
operations is to arrange the data so that when one
piece of data block is needed it is likely that other
data on the same block will be needed at the same
time
Same idea applies to other hierarchy levels
13.1.3 Volatile and Non Volatile
Storage
A volatile device forgets what data is stored on it
after power off
Non volatile holds data for longer period even
when device is turned off
All the secondary and tertiary devices are non
volatile and main memory is volatile
13.1.4 Virtual Memory
Typical software executes in virtual memory
Address space is typically 32 bit or 232 bytes or
4GB
Transfer between memory and disk is in terms of
blocks
Virtual memory is an artifact of the operating
system and its use of the machine’s hardware
13.2.1 Mechanism of Disk
Mechanisms of Disks
Use
of secondary storage is one of the important
characteristic of DBMS
Consists of 2 moving pieces of a disk
1.
disk assembly
2. head assembly
Disk
assembly consists of 1 or more platters
Platters rotate around a central spindle
Bits are stored on upper and lower surfaces of platters
13.2.1 Mechanism of Disk
Disk is organized into tracks
The track that are at fixed radius from center form
one cylinder
Tracks are organized into sectors
Tracks are the segments of circle separated by gap
13.2.2 Disk Controller
One or more disks are controlled by disk controllers
Disks controllers are capable of
Controlling
the mechanical actuator that moves the
head assembly
Selecting the sector from among all those in the cylinder
at which heads are positioned
Transferring bits between desired sector and main
memory
Possible buffering an entire track
13.2.3 Disk Access Characteristics
Accessing (reading/writing) a block requires 3 steps
Disk
controller positions the head assembly at the
cylinder containing the track on which the block is
located. It is a ‘seek time’
The disk controller waits while the first sector of the
block moves under the head. This is a ‘rotational
latency’
All the sectors and the gaps between them pass the
head, while disk controller reads or writes data in these
sectors. This is a ‘transfer time’
13.3 Accelerating Access to Secondary
Storage
Several approaches for more-efficiently accessing data
in secondary storage:
Place blocks that are together in the same cylinder.
Divide the data among multiple disks.
Mirror disks.
Use disk-scheduling algorithms.
Prefetch blocks into main memory.
Scheduling Latency – added delay in accessing data
caused by a disk scheduling algorithm.
Throughput – the number of disk accesses per second
that the system can accommodate.
13.3.1 The I/O Model of Computation
The number of block accesses (Disk I/O’s) is a good
time approximation for the algorithm.
This should be minimized.
Ex 13.3: You want to have an index on R to identify the
block on which the desired tuple appears, but not
where on the block it resides.
For Megatron 747 (M747) example, it takes 11ms to read
a 16k block.
A standard microprocessor can execute millions of
instruction in 11ms, making any delay in searching for the
desired tuple negligible.
13.3.2 Organizing Data by Cylinders
If we read all blocks on a single track or cylinder
consecutively, then we can neglect all but first seek time
and first rotational latency.
Ex 13.4: We request 1024 blocks of M747.
If data is randomly distributed, average latency is 10.76ms
by Ex 13.2, making total latency 11s.
If all blocks are consecutively stored on 1 cylinder:
6.46ms
+ 8.33ms * 16 = 139ms
(1 average seek)
(time per rotation)
(# rotations)
13.3.3 Using Multiple Disks
If we have n disks, read/write performance will
increase by a factor of n.
Striping – distributing a relation across multiple disks
following this pattern:
Data on disk R1: R1, R1+n, R1+2n,…
Data on disk R2: R2, R2+n, R2+2n,…
…
Data on disk Rn: Rn, Rn+n, Rn+2n, …
Ex 13.5: We request 1024 blocks with n = 4.
6.46ms + (8.33ms * (16/4)) = 39.8ms
(1 average seek)
(time per rotation)
(# rotations)
13.3.4 Mirroring Disks
Mirroring Disks – having 2 or more disks hold
identical copied of data.
Benefit 1: If n disks are mirrors of each other, the
system can survive a crash by n-1 disks.
Benefit 2: If we have n disks, read performance
increases by a factor of n.
Performance increases further by having the
controller select the disk which has its head closest
to desired data block for each read.
13.3.5 Disk Scheduling and the
Elevator Problem
Disk controller will run this algorithm to select which of
several requests to process first.
Pseudo code:
requests[] // array of all non-processed data requests
upon receiving new data request:
requests[].add(new request)
while(requests[] is not empty)
move head to next location
if(head location is at data in requests[])
retrieve data
remove data from requests[]
if(head reaches end)
reverse head direction
13.3.5 Disk Scheduling and the
Elevator Problem (con’t)
Events:
Head starting point
Request data at 8000
Request data at 24000
Request data at 56000
Get data at 8000
Request data at 16000
Get data at 24000
Request data at 64000
Get data at 56000
Request Data at 40000
Get data at 64000
Get data at 40000
Get data at 16000
64000
56000
48000
40000
32000
24000
16000
8000
Current
time
Current
0
time
13.6
26.9
34.2
45.5
56.8
4.3
10
20
30
data
time
8000..
4.3
24000..
13.6
56000..
26.9
64000..
34.2
40000..
45.5
16000..
56.8
13.3.5 Disk Scheduling and the
Elevator Problem (con’t)
Elevator
Algorithm
FIFO
Algorithm
data
time
data
time
8000..
4.3
8000..
4.3
24000..
13.6
24000..
13.6
56000..
26.9
56000..
26.9
64000..
34.2
16000..
42.2
40000..
45.5
64000..
59.5
16000..
56.8
40000..
70.8
13.3.6 Prefetching and Large-Scale
Buffering
If at the application level, we can predict the order
blocks will be requested, we can load them into
main memory before they are needed.
DISK FAILURES
Xiaqing He
ID: 204
Dr. Lin
Content
1)Focus on :
“How to recover from disk crashes”
common term RAID
“redundancy array of independent disks”
2)Several schemes to recover from disk crashes:
Mirroring—RAID level 1;
Parity checks--RAID 4;
Improvement--RAID 5;
RAID 6;
1) Mirroring
The simplest scheme to recovery from Disk Crashes,
is often referred to as RAID level 1
How does Mirror work?
-- making two or more copied of the data on
different disks
Benefit:
-- save data in case of one disk will fail;
-- divide data on several disks and let access to
several blocks at once
1) Mirroring (con’t)
For mirroring, when the data can be lost?
-- the only way data can be lost if there is a second (mirror/redundant) disk crash
while the first (data) disk crash is being repaired.
Possibility:
Suppose:
•
One disk: mean time to failure = 10 years;
•
One of the two disk: average of mean time to failure = 5 years;
•
The process of replacing the failed disk= 3 hours=1/2920 year;
So:
•
the possibility of the mirror disk will fail=1/10 * 1/2,920 =1/29,200;
•
The possibility of data loss by mirroring: 1/5 * 1/29,200 = 1/146,000
2)Parity Blocks
why changes?
-- disadvantages of Mirroring: uses so many redundant
disks
What’s new?
-- RAID level 4: uses only one redundant disk
How this one redundant disk works?
-- modulo-2 sum;
-- the jth bit of the redundant disk is the modulo-2 sum of
the jth bits of all the data disks.
Example
2)Parity Blocks(con’t)Example
Data disks:
Disk1: 11110000
Disk2: 10101010
Disk3: 00111000
Redundant disk:
Disk4: 01100010
2)RAID 4 (con’t)
Only one redundant disk
Reading
-- Similar with reading blocks from any disk;
Writing
1)change the data disk;
2)change the corresponding block of the redundant disk;
•
Why?
-- hold the parity checks for the corresponding blocks of all
the data disks
2)RAID 4 (con’t) _ writing
For a total N data disks:
1) naïve way:
•
read N data disks and compute the modulo-2 sum of the
corresponding blocks;
•
rewrite the redundant disk according to modulo-2 sum of
the data disks;
2) better way:
• Take modulo-2 sum of the old and new version of the data
block which was rewritten;
• Change the position of the redundant disk which was 1’s in
the modulo-2 sum;
2)RAID 4 (con’t)writingExample
•
•
•
•
•
•
Data disks:
Disk1: 11110000
Disk2: 10101010 01100110
Disk3: 00111000
to do:
Modulo-2 sum of the old and new version of disk 2: 11001100
So, we need to change the positions 1,2,5,6 of the redundant disk.
Redundant disk:
Disk4: 01100010 10101110
2)RAID 4 (con’t) _failure recovery
Redundant disk crash:
-- swap a new one and recomputed data from all the data disks;
One of Data disks crash:
-- swap a new one;
-- recomputed data from the other disks including data disks and redundant
disk;
How to recomputed? (same rule, that’s why there will be some
improvement)
-- take modulo-2 sum of all the corresponding bits of all the other disks
3) An Improvement: RAID 5
Why need a improvement?
-- Shortcoming of RAID level 4: suffers from a bottleneck defect (when updating data
disk need to read and write the redundant disk);
idea of RAID level 5 (RAID 5):
-- treat each disk as the redundant disk for some of the blocks;
Why it is feasible?
The rule of failure recovery for redundant disk and data disk is the same:
“take modulo-2 sum of all the corresponding bits of all the other disks”
So, there is no need to retreat one as redundant disk and others as data disks
3) RAID 5 (con’t)
How to recognize which blocks of each disk treat this disk as
redundant disk?
-- if there are n+1 disks which were labeled from 0 to
N, then we can treat the i cylinder of disk J as
redundant if J is the remainder when I is divided by
n+1;
th
Example;
3) RAID 5 (con’t)_example
•
•
•
•
•
•
N=3;
The first disk, labeled as 0 : 4,8,12…;
The second disk, labeled as 1 : 1,5,9…;
The third disk, labeled as 2 : 2,6,10…;
……….
Suppose all the 4 disks are equally likely to be written, for
one of the 4 disks, the possibility of being written:
1/4 + 3 /4 * 1/3 =1/2
If N=m => 1/m +(m-1)/m * 1/(m-1) = 2/m
4) Coping with multiple disk crashes
RAID 6
– deal with any number of disk crashes if using enough
redundant disks
Example
a system of seven disks ( four data disks_numer 1-4 and
3 redundant disks_ number 5-7);
• How to set up this 3*7 matrix ?
(why is 3? – there are 3 redundant disks)
1)every column values three 1’s and 0’s except for all three 0’s;
2) column of the redundant disk has single 1’s;
3) column of the data disk has at least two 1’s;
4) Coping with multiple disk crashes (con’t)
Reading:
read form the data disks and ignore the redundant
disk
Writing:
Change the data disk
change the corresponding bits of all the redundant
disks
4) Coping with multiple disk crashes (con’t)
•
•
•
•
In those system which has 4 data disks and 3 redundant disk, how they
can correct up to 2 disk crashes?
Suppose disk a and b failed:
find some row r (in 3*7 matrix)in which the column for a and b are
different (suppose a is 0’s and b is 1’s);
Compute the correct b by taking modulo-2 sum of the corresponding
bits from all the other disks other than b which have 1’s in row r;
After getting the correct b, Compute the correct a with all other disks
available;
Example
4) Coping with
(con’t)_example
multiple
disk
crashes
3*7 matrix
data disk
redundant
disk
disk number
1
2
3
4
5
6
7
1
1
1
0
1
0
0
1
1
0
1
0
1
0
1
0
1
1
0
0
1
4) Coping with multiple disk crashes
(con’t)_example
First block of all the disks
disk
1)
2)
3)
4)
5)
6)
7)
contents
11110000
10101010
00111000
01000001
01100010
00011011
10001001
4) Coping with multiple disk crashes
(con’t)_example
Two disks crashes;
disk
1)
2)
3)
4)
5)
6)
7)
contents
11110000
?????????
00111000
01000001
?????????
00011011
10001001
4) Coping with multiple disk crashes
(con’t)_example
In that 3*7 matrix, find in row 2, disk 2 and 5 have different value
and disk 2’s value is 1 and 5’s value is 0.
so: compute the first block of disk 2 by modulo-2 sum of all the
corresponding bits of disk 1,4,6;
then compute the first block of disk 2 by modulo-2 sum of all the
corresponding bits of disk 1,2,3;
1)
11110000
2)
????????? => 00001111
3)
00111000
4)
01000001
5)
????????? => 01100010
6)
00011011
7)
10001001
13.5 Arranging data on disk
Meghna Jain
ID-205
CS257
Prof: Dr. T.Y.Lin
Data elements are represented as records, which stores
in consecutive bytes in some disk block.
Basic layout techniques of storing data :
Fixed-Length Records
Allocation criteria - data should start at word boundary.
Fixed Length record header
1. A pointer to record schema.
2. The length of the record.
3. Timestamps to indicate last modified or last read.
Example
CREATE TABLE employee(
name CHAR(30) PRIMARY KEY,
address VARCHAR(255),
gender CHAR(1),
birthdate DATE
);
Data should start at word boundary and contain header and four fields
name, address, gender and birthdate.
Packing Fixed-Length Records into Blocks :
Simplest sort of record consists of fixed-length fields
Records are stored in the form of blocks on the disk and they
move into main memory when we need to update or access
them.
A block header is written first, and it is followed by series of
blocks.
Block header contains the following information :
Links to one or more blocks that are part of a network of
blocks.
Information about the role played by this block in such a
network.
Information about the relation, the tuples in this block
belong to.
A "directory" giving the offset of each record in the block.
Time stamp(s) to indicate time of the block's last
modification and/or access.
Example
Along with the header we can pack as many record as we
can
in one block as shown in the figure and remaining space will
be unused.
13.6 REPRESENTING BLOCK
AND RECORD ADDRESSES
Ramya Karri
CS257 Section 2
ID: 206
Introduction
Address of a block and Record
In
Main Memory
Address of the block is the virtual memory address of the first
byte
Address of the record within the block is the virtual memory
address of the first byte of the record
In
Secondary Memory: sequence of bytes describe the
location of the block in the overall system
Sequence of Bytes describe the location of the
block : the device Id for the disk, Cylinder number,
etc.
Addresses in Client-Server Systems
The addresses in address space are represented in two ways
Physical Addresses: byte strings that determine the place within
the secondary storage system where the record can be found.
Logical Addresses: arbitrary string of bytes of some fixed
length
Physical Address bits are used to indicate:
Host to which the storage is attached
Identifier for the disk
Number of the cylinder
Number of the track
Offset of the beginning of the record
ADDRESSES IN CLIENT-SERVER SYSTEMS (CONTD..)
Map Table relates logical addresses to
physical addresses.
Logical
Physical
Logical Address
Physical Address
Logical and Structured Addresses
Logical address and physical address
Gives more flexibility, when we
Move
the record around within the block
Move the record to another block
Gives us an option of deciding what to do
when a record is deleted?
Unused
Reco Reco Reco Reco
rd 4 rd 3 rd 2 rd 1
Offset table
Header
Pointer Swizzling
Translated from database address space to
the virtual address space
Important to learn about the management of
pointers
Every data item (block, record, etc.) has two
addresses:
database address: address on the disk
memory address, if the item is in virtual memory
POINTER SWIZZLING (CONTD…)
Translation Table: Maps database address to memory
address
Dbaddr
Mem-addr
Database address
Memory Address
All addressable items in the database have entries in
the map table, while only those items currently in
memory are mentioned in the translation table
POINTER SWIZZLING (CONTD…)
Pointer consists of the following two fields
Bit indicating the type of address
Database or memory address
Example 13.17
Disk
Memory
Swizzled
Block 1
Block 1
Unswizzled
Block 2
Example 13.7
Block 1 has a record with pointers to a second
record on the same block and to a record on
another block
If Block 1 is copied to the memory
The
first pointer which points within Block 1 can be
swizzled so it points directly to the memory
address of the target record
Since Block 2 is not in memory, we cannot swizzle
the second pointer
POINTER SWIZZLING (CONTD…)
Three types of swizzling
Automatic
Swizzling
As
soon as block is brought into memory, swizzle all
relevant pointers.
Swizzling
on Demand
Only
swizzle a pointer if and when it is actually
followed.
No Swizzling
Pointers
are not swizzled they are accesses using the
database address.
Programmer Control of Swizzling
Unswizzling
When
a block is moved from memory back to
disk, all pointers must go back to database (disk)
addresses
Use translation table again
Important to have an efficient data structure for
the translation table
Pinned records and Blocks
A block in memory is said to be pinned if it cannot
be written back to disk safely.
If block B1 has swizzled pointer to an item in block
B2, then B2 is pinned
Unpin
a block, we must unswizzle any pointers to it
Keep in the translation table the places in memory
holding swizzled pointers to that item
Unswizzle those pointers (use translation table to
replace the memory addresses with database (disk)
addresses
VARIABLE LENGTH DATA AND
RECORDS
Eswara Satya Pavan Rajesh Pinapala
CS 257
ID: 221
Topics
Records with Variable Length Fields
Records with Repeating Fields
Variable Format Records
Records that do not fit in a block
BLOBS
Example
name
0
297
addres
s
30
gender
286
birth date
287
Fig 1 : Movie star record with four
fields
Records with Variable Fields
An simple and effective way to represent
variable length records is as follows
Fixed length fields are Kept ahead of the
variable length fields
Record header contains
• Length of the record
• Pointers to the beginning of all variable
length fields except the first one.
Records with Variable Length Fields
header
information
record length
to address
gender birth date
name
address
Figure 2 : A Movie Star record with name and address
implemented as variable length character strings
Records with Repeating Fields
Records contains variable number of occurrences of a
field F
All occurrences of field F are grouped together and the
record
header contains a pointer to the first occurrence of field
F
L bytes are devoted to one instance of field F
Locating an occurrence of field F within the record
• Add to the offset for the field F which are the integer
multiples of L starting with 0 , L ,2L,3L and so on to
locate
•We stop upon reaching the offset of the field F.
Records with Repeating Fields
other header
information
record length
to address
to movie
pointers
name
address
pointers to movies
Figure 3 : A record with a repeating group of references to
Records with Repeating Fields
record header to name length of name
information
to address
length of
to movie
address
references number of
references
addres
s
name
Figure 4 : Storing variable-length fields separately from the
record
Records with Repeating Fields
A record contains a variable number oof
occurrences of a field
Advantage
Keeping the record itself fixed length allows record to be
searched more efficiently, minimizes the overhead in the
block headers, and allows records to be moved within or
among the blocks with minimum effort.
Disadvantage
Storing variable length components on another block
increases the number of disk I/O’s needed to examine all
components of a record.
Records with Repeating Fields
A compromise strategy is to allocate a fixed portion of
the record for the repeating fields
If the number of repeating fields is lesser than
allocated space, then there will be some unused
space
If the number of repeating fields is greater than
allocated space, then extra fields are stored in a
different location and
Pointer to that location and count of additional
occurrences is stored in the record
Variable Format Records
Records that do not have fixed schema
Variable format records are represented by sequence of
tagged fields
Each of the tagged fields consist of information
• Attribute or field name
• Type of the field
• Length of the field
• Value of the field
Why use tagged fields
• Information – Integration applications
• Records with a very flexible schema
Variable Format Records
code for name
code for string
type length
N
S 1
4
Clint
Eastwood
code for restaurant
owned
code for string type
length
R
S
1
6
Fig 5 : A record with tagged fields
Hog’s Breath Inn
Records that do not fit in a block
When the length of a record is greater than block size
,then
then record is divided and placed into two or more
blocks
Portion of the record in each block is referred to as a
RECORD FRAGMENT
Record with two or more fragments is called
SPANNED RECORD
Record that do not cross a block boundary is called
UNSPANNED RECORD
Spanned Records
Spanned records require the following extra
header
information
• A bit indicates whether it is fragment or not
• A bit indicates whether it is first or last
fragment of
a record
• Pointers to the next or previous fragment for
the
same record
Records that do not fit in a block
block header
record
header
record 1
block 1
record
2-a
record
2-b
record 3
block 2
Figure 6 : Storing spanned records across
blocks
BLOBS
Large binary objects are called BLOBS
e.g. : audio files, video files
Issues of BLOBS:
Storage of BLOBS
Retrieval of BLOBS
Record Modifications
Chapter 13
Section 13.8
Neha Samant
CS 257
(Section II) Id 222
83
Insertion
Insertion of records without order
Records can be placed in a block with empty space or in a new block.
Insertion of records in fixed order
Space available in the block
No space available in the block (outside the block)
Structured address
Pointer to a record from outside the block.
84
Insertion in fixed order
Space available within the block
Use of an offset table in the header of each block with pointers to the location of
each record in the block.
The records are slid within the block and the pointers in the offset table are
adjusted.
Offse
t
table
header
unuse
d
Record 4
Record 3
Record 2
Record 1
85
Insertion in fixed order
No space available within the block (outside the block)
Find space on a “nearby” block.
•
•
In case of no space available on a block, look at the following block in sorted
order of blocks.
If space is available in that block ,move the highest records of first block 1 to
block 2 and slide the records around on both blocks.
Create an overflow block
•
•
•
Records can be stored in overflow block.
Each block has place for a pointer to an overflow block in its header.
The overflow block can point to a second overflow block as shown below.
Block
B
Overflow
block for B
86
Deletion
Recover space after deletion
When using an offset table, the records can be slid around the block so there
will be an unused region in the center that can be recovered.
In case we cannot slide records, an available space list can be maintained in the
block header.
The list head goes in the block header and available regions hold the links in the
list.
87
Deletion
Use of tombstone
The tombstone is placed in a record in order to avoid pointers to the
deleted record to point to new records.
The tombstone is permanent until the entire database is
reconstructed.
If pointers go to fixed locations from which the location of the record is
found then we put the tombstone in that fixed location. (See examples)
Where a tombstone is placed depends on the nature of the record
pointers.
Map table is used to translate logical record address to physical
address.
88
Deletion
Use of tombstone
If we need to replace records by tombstones, place the bit that serves
as the tombstone at the beginning of the record.
This bit remains the record location and subsequent bytes can be
reused for another record
Record 1
Record 2
Record 1 can be replaced, but the tombstone remains, record 2 has
no tombstone and can be seen when we follow a pointer to it.
89
Update
Fixed Length update
No effect on storage system as it occupies same space as
before update.
Variable length update
Longer length
Short length
90
Update
Variable length update (longer length)
Stored on the same block:
Sliding records
Creation of overflow block.
Stored on another block
Move records around that block
Create a new block for storing variable length fields.
91
Update
Variable length update (Shorter length)
Same as deletion
Recover space
Consolidate space.
92
14.2
DATABASE SYSTEMS – The Complete Book
Presented By:
supervision of:BTREES &
Maciej Kicinski
INDEXES
Under the
BITMAP
Dr. T.Y.Lin
Structure
A balanced tree, meaning that all paths from the
leaf node have the same length.
There is a parameter n associated with each Btree
block. Each block will have space for n searchkeys
and n+1 pointers.
The root may have only 1 parameter, but all other
blocks most be at least half full.
Structure
● A typical
node >
● a typical interior
node would have
pointers pointing to
leaves with out
values
● a typical leaf would
have pointers point
to records
N search keys
N+1 pointers
Application
The search key of the Btree is the primary key for
the data file.
Data file is sorted by its primary key.
Data file is sorted by an attribute that is not a
key,and this attribute is the search key for the Btree.
Lookup
If at an interior node, choose the correct pointer to use. This is
done by comparing keys to search value.
If we are at a leaf, look among the keys there. If the ith key is K,
then the ith pointer will take us to the desired record
Insertion
When inserting, choose the correct leaf node to put
pointer to data.
If node is full, create a new node and split keys
between the two.
Recursively move up, if cannot create new pointer
to new node because full, create new node.
This would end with creating a new root node, if
the current root was full.
Deletion
Perform lookup to find node to delete and delete it.
If node is no longer half full, perform join on
adjacent node and recursively delete up, or key
move if that node is full and recursively change
pointer up.
Efficiency
Btrees allow lookup, insertion, and deletion of
records using very few disk I/Os.
Each level of a Btree would require one read. Then
you would follow the pointer of that to the next or
final read.
Efficiency
Three levels are sufficient for Btrees. Having each block have
255 pointers, 255^3 is about 16.6 million.
You can even reduce disk I/Os by keeping a level of a Btree in
main memory. Keeping the first block with 255 pointers
would reduce the reads to 2, and even possible to keep the
next 255 pointers in memory to reduce reads to 1.
CONCURRENCY CONTROL
18.1 – 18.2
Chiu Luk
CS257 Database Systems Principles
Spring 2009
Concurrency Control
Concurrency control in database management systems (DBMS)
ensures that database transactions are performed concurrently
without the concurrency violating the data integrity of a
database.
Executed transactions should follow the ACID rules. The DBMS
must guarantee that only serializable (unless Serializability is
intentionally relaxed), recoverable schedules are generated.
It also guarantees that no effect of committed transactions is lost,
and no effect of aborted (rolled back) transactions remains in
the related database.
Transaction ACID rules
Atomicity - Either the effects of all or none of its operations
remain when a transaction is completed - in other words, to
the outside world the transaction appears to be indivisible,
atomic.
Consistency - Every transaction must leave the database in
a consistent state.
Isolation - Transactions cannot interfere with each other.
Providing isolation is the main goal of concurrency control.
Durability - Successful transactions must persist through
crashes.
Serial and Serializable Schedules
In the field of databases, a schedule is a list of actions, (i.e. reading, writing,
aborting, committing), from a set of transactions.
In this example, Schedule D is the set of 3 transactions T1, T2, T3. The schedule
describes the actions of the transactions as seen by the DBMS. T1 Reads and
writes to object X, and then T2 Reads and writes to object Y, and finally T3 Reads
and writes to object Z. This is an example of a serial schedule, because the actions
of the 3 transactions are not interleaved.
Serial and Serializable Schedules
A schedule that is equivalent to a serial schedule has the serializability property.
In schedule E, the order in which the actions of the transactions are executed is not the same as in
D, but in the end, E gives the same result as D.
Serial Schedule TI
precedes T2
T1
T2
Read(A); A A+100
Write(A);
Read(B); B B+100;
Write(B);
A
25
B
25
125
125
Read(A);A A2;
Write(A);
Read(B);B B2;
Write(B);
250
250
250
250
Serial Schedule T2 precedes Tl
T1
T2
Read(A);A A2;
Write(A);
A
25
50
Read(B);B B2;
Write(B);
Read(A); A A+100
Write(A);
Read(B); B B+100;
Write(B);
B
25
50
150
150
150
150
serializable, but not serial, schedule
T1
T2
Read(A); A A+100
Write(A);
A
25
B
25
125
Read(A);A A2;
Write(A);
250
Read(B); B B+100;
Write(B);
125
Read(B);B B2;
Write(B);
r1(A); w1 (A): r2(A); w2(A); r1 (B); w1 (B); r2(B); w2(B);
250
250
250
nonserializable schedule
T1
T2
Read(A); A A+100
Write(A);
A
25
125
Read(A);A A2;
Write(A);
250
Read(B);B B2;
Write(B);
Read(B); B B+100;
Write(B);
B
25
50
250
150
150
schedule that is serializable only because of the detailed
behavior of the transactions
A
T1
T2’
25
Read(A); A A+100
Write(A);
125
Read(A);A A1;
Write(A);
125
Read(B);B B1;
Write(B);
Read(B); B B+100;
Write(B);
regardless of the consistent initial state: the final state will be consistent.
B
25
25
125
125
125
Non-Conflicting Actions
Two actions are non-conflicting if whenever they
occur consecutively in a schedule, swapping them
does not affect the final state produced by the
schedule. Otherwise, they are conflicting.
Conflicting Actions: General Rules
Two actions of the same transaction conflict:
r1(A)
w1(B)
Two actions over the same database element
conflict, if one of them is a write
r1(A)
w2(A)
w1(A) w2(A)
Conflict actions
Two or more actions are said to be in conflict if:
The actions belong to different transactions.
At least one of the actions is a write operation.
The actions access the same object (read or write).
The following set of actions is conflicting:
T1:R(X), T2:W(X), T3:W(X)
While the following sets of actions are not:
T1:R(X), T2:R(X), T3:R(X)
T1:R(X), T2:W(Y), T3:R(X)
Conflict Serializable
We may take any schedule and make as many
nonconflicting swaps as we wish.
With the goal of turning the schedule into a serial
schedule.
If we can do so, then the original schedule is
serializable, because its effect on the database
state remains the same as we perform each of the
nonconflicting
swaps.
Conflict Serializable
A schedule is said to be conflict-serializable when the schedule is conflict-equivalent to one or
more serial schedules.
Another definition for conflict-serializability is that a schedule is conflict-serializable if and only if
there exists an acyclic precedence graph/serializability graph for the schedule.
Which is conflict-equivalent to the serial schedule <T1,T2>, but not <T2,T1>.
Conflict equivalent / conflict-serializable
Let Ai and Aj are consecutive non-conflicting actions that
belongs to different transactions. We can swap Ai and Aj
without changing the result.
Two schedules are conflict equivalent if they can be turned one
into the other by a sequence of non-conflicting swaps of
adjacent actions.
We shall call a schedule conflict-serializable if it is conflictequivalent to a serial schedule.
conflict-serializable
T1
R(A)
W(A)
T2
R(A)
R(B)
W(A)
W(B)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(B)
T2
R(A)
W(A)
W(B)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(A)
T2
R(B)
W(B)
W(A)
R(B)
W(B)
conflict-serializable
T1
R(A)
W(A)
R(A)
W(B)
T2
Serial
Schedule
R(B)
W(A)
R(B)
W(B)
122
CONCURRENCY CONTROL
By Donavon Norwood
Ankit Patel
Aniket Mulye
INTRODUCTION
123
Enforcing serializability by locks
Locks
Locking scheduler
Two phase locking
Locking systems with several lock modes
Shared and exclusive locks
Compatibility matrices
Upgrading/updating locks
Incrementing locks
Locks
124
A lock table to help perform its job
It works like as follows :
A request from transaction
Scheduler checks in the lock table
Generates a serializable schedule of actions.
Consistency of transactions
125
Actions and locks must relate each other
Transactions
can only read & write only if has a lock
and has not released the lock.
Unlocking an element is compulsory.
Legality of schedules
No
two transactions can aquire the lock on same
element without the prior one releasing it.
Locking scheduler
126
Grants lock requests only if it is in a legal schedule.
Lock table stores the information about current locks on
the elements.
The locking scheduler (contd.)
127
A legal schedule of consistent transactions but
unfortunately it is not a serializable.
Locking schedule (contd.)
128
The locking scheduler delays requests that would
result in an illegal schedule.
Two-phase locking
129
Guarantees a legal schedule of consistent
transactions is conflict-serializable.
All lock requests proceed all unlock requests.
The growing phase:
Obtain
all the locks and no unlocks allowed.
The shrinking phase:
Release
all the locks and no locks allowed.
Working of Two-Phase locking
130
Assures serializability.
Two protocols for 2PL:
Strict
two phase locking : Transaction holds all its
exclusive locks till commit / abort.
Rigorous two phase locking : Transaction holds all
locks till commit / abort.
Possible to find a transaction Tj that has a 2PL and
a schedule S for Ti ( non 2PL ) and Tj that is not
conflict serializable.
Failure of 2PL.
131
2PL fails to provide security against deadlocks.
132
Locking Systems with Several Lock
Modes
Locking Scheme
Shared/Read
Lock ( For Reading)
Exclusive/Write Lock( For Writing)
Compatibility Matrices
Upgrading Locks
Update Locks
Increment Locks
Shared & Exclusive Locks
133
Consistency of Transactions
Cannot write without Exclusive Lock
Cannot read without holding some lock
This basically works on 2 principles
A read action can only proceed a shared or an exclusive
lock
A write lock can only proceed a exclusice lock
All locks need to be unlocked before commit
Shared and exclusive locks (cont.)
134
Two-phase locking of transactions
Must precede unlocking
Legality of Schedules
An element may be locked exclusively by one transaction or
by several in shared mode, but not both.
Compatibility Matrices
135
Has a row and column for each lock mode.
Rows
correspond to a lock held on an element by
another transaction
Columns correspond to mode of lock requested.
Example :
LOCK REQUESTED
S
X
LOCK
S
YES
NO
HOLD
X
NO
NO
Upgrading Locks
136
Suppose a transaction wants to read as well as
write :
It
requires a shared lock on the element
Performs the calculations on the element
And when its ready to write, It is granted a exclusive
lock.
Transactions with unpredicted read write locks can
use UPGRADING LOCKS.
Upgrading locks (cont.)
137
Indiscriminating use of upgrading produces a
deadlock.
Example : Both the transactions want to upgrade on
the same element
Update locks
138
Solves the deadlock occurring in upgrade lock
method.
A transaction in an update lock can read but cant
write.
Update lock can later be converted to exclusive
lock.
An update lock can only be given if the element has
shared locks.
Update locks (cont.)
139
An update lock is like a shared lock when you are
requesting it and is like a exclusive lock when you
have it.
Compatibility matrix :
S
X
U
S
YES
NO
YES
X
NO
NO
NO
U
NO
NO
NO
Increment Locks
140
Used for incrementing & decrementing stored
values.
It is widely used in banking transaction
E.g. - Transfer money from one bank to another,
Ticket selling transactions in which number seats are
decremented after each transaction.
Increment lock (cont.)
141
A increment lock does not enable read or write locks on
element.
Any number of transaction can hold increment lock on element
Shared and exclusive locks can not be granted if an increment
lock is granted on element
S
X
I
S
YES
NO
NO
X
NO
NO
NO
I
NO
NO
YES
Concurrency Control
142
Chapter 18
Section 18.5
Presented by
Khadke, Suvarna
CS 257
(Section II) Id 213
Overview
143
Assume knowledge of:
Lock
Two phase lock
Lock modes: shared, exclusive, update
A simple scheduler architecture based on
following principle :
Insert lock actions into the stream of reads, writes, and
other actions
Release locks when the transaction manager tells it that
the transaction will commit or abort
Scheduler That Inserts Lock Actions into
the transactions request stream
Scheduler That Inserts Lock Actions
145
If transaction is delayed, waiting for a lock, Scheduler
performs following actions
Part I: Takes the stream of requests generated by the
transaction & insert appropriate lock modes to db
operations (read, write, or update)
Part II: Take actions (a lock or db operation) from Part
I and executes it.
Determine the transaction (T) that action belongs and
status of T (delayed or not). If T is not delayed then
1. Database access action is transmitted to the database
and executed
Scheduler That Inserts Lock Actions
146
2.
3.
3.
If lock action is received by PartII, it checks the L Table
whether lock can be granted or not
i> Granted, the L Table is modified to include granted lock
ii>Not G. then update L Table about requested lock then
PartII delays transaction T
When a T = commits or aborts, PartI is notified by the
transaction manager and releases all locks.
If any transactions are waiting for locks PartI notifies
PartII.
Part II when notified about the lock on some DB element,
determines next transaction T’ to get lock to continue.
The Lock Table
147
A relation that associates database elements with
locking information about that element
Implemented with a hash table using database
elements as the hash key
Size is proportional to the number of lock elements
only, not to the size of the entire database
DB
element A
Lock
informatio
n for A
Lock Table Entries Structure
148
Some Sort of
information found in
Lock Table entry
1>Group modes
-S: only shared locks are
held
-X: one exclusive lock and
no other locks
- U: one update lock and
one or more shared locks
2>wait : one transaction
waiting for a lock on A
3>A list : T currently
hold locks on A or
Waiting for lock on A
Handling Lock Requests
149
Suppose transaction T requests a lock on A
If there is no lock table entry for A, then there are
no locks on A, so create the entry and grant the lock
request
If the lock table entry for A exists, use the group
mode to guide the decision about the lock request
Handling Lock Requests
150
If group mode is U (update) or X (exclusive)
No other lock can be granted
Deny the lock request by T
Place an entry on the list saying T requests a lock
And Wait? = ‘yes’
If group mode is S (shared)
Another shared or update lock can be granted
Grant request for an S or U lock
Create entry for T on the list with Wait? = ‘no’
Change group mode to U if the new lock is an update lock
Handling Unlock Requests
151
Now suppose transaction T unlocks A
Delete T’s entry on the list for A
If T’s lock is not the same as the group mode, no
need to change group mode
Otherwise check entire list for new group mode
S: GM(S) or nothing
U: GM(S) or nothing
X: nothing
Handling Unlock Requests
152
If the value of waiting is “yes" need to grant one or more locks using following
approaches
First-Come-First-Served:
Grant the lock to the longest waiting request.
No starvation (waiting forever for lock)
Priority to Shared Locks:
Grant all S locks waiting, then one U lock.
Grant X lock if no others waiting
Priority to Upgrading:
If there is a U lock waiting to upgrade to an X lock, grant that first.
Concurrency Control
153
Managing Hierarchies of Database Elements (18.6)
Presented by
Ronak Shah
(214)
March 9, 2009
Managing Hierarchies of Database
Elements
Two problems that arise with locks when there is
a tree structure to the data are:
When the tree structure is a hierarchy of
lockable elements
Determine how locks are granted for both large elements
(relations) and smaller elements (blocks containing tuples
or individual tuples)
When the data itself is organized as a tree (Btree indexes)
This will be discussed in the next section
Locks with Multiple Granularity
A database element can be a relation, block or a
tuple
Different systems use different database elements
to determine the size of the lock
Thus some may require small database elements
such as tuples or blocks and others may require
large elements such as relations
Example of Multiple Granularity
Locks
Consider a database for a bank
Choosing relations as database elements means we would
have one lock for an entire relation
If we were dealing with a relation having account balances,
this kind of lock would be very inflexible and thus provide
very little concurrency
Why? Because balance transactions require exclusive locks
and this would mean only one transaction occurs for one
account at any time
But as each account is independent of others we could
perform transactions on different accounts simultaneously
…(contd.)
Thus it makes sense to have block element for the lock so
that two accounts on different blocks can be updated
simultaneously
Another example is that of a document
With similar arguments as above, we see that it is better
to have large element (a complete document) as the lock
in this case
Warning (Intention) Locks
These are required to manage locks at different
granularities
In the bank example, if the a shared lock is obtained for the
relation while there are exclusive locks on individual tuples,
unserializable behavior occurs
The rules for managing locks on hierarchy of
database elements constitute the warning protocol
Database Elements Organized in
Hierarchy
Rules of Warning Protocol
These involve both ordinary (S and X) and warning
(IS and IX) locks
The rules are:
Begin at the root of hierarchy
Request the S/X lock if we are at the desired element
If the desired element id further down the hierarchy, place a
warning lock (IS if S and IX if X)
When the warning lock is granted, we proceed to the child
node and repeat the above steps until desired node is
reached
Compatibility Matrix for Shared,
Exclusive and Intention Locks
IS
IX
S
X
IS
Yes
Yes
Yes
No
IX
Yes
Yes
No
No
S
Yes
No
Yes
No
X
No
No
No
No
• The above matrix applies only to locks held by
other transactions
Group Modes of Intention Locks
An element can request S and IX locks at the same
time if they are in the same transaction (to read
entire element and then modify sub elements)
This can be considered as another lock mode, SIX,
having restrictions of both the locks i.e. No for all
except IS
SIX serves as the group mode
Example
Consider a transaction T1 as follows
Select * from table where attribute1 = ‘abc’
Here, IS lock is first acquired on the entire relation; then
moving to individual tuples (attribute = ‘abc’), S lock in
acquired on each of them
Consider another transaction T2
Update table set attribute2 = ‘def’ where attribute1 = ‘ghi’
Here, it requires an IX lock on relation and since T1’s IS lock
is compatible, IX is granted
On
reaching the desired tuple (ghi), as there is no lock,
it gets X too
If T2 was updating the same tuple as T1, it would have
to wait until T1 released its S lock
Phantoms and Handling Insertions
Correctly
This arises when transactions create new sub
elements of lockable elements
Since we can lock only existing elements the new
elements fail to be locked
The problem created in this situation is explained in
the following example
Example
Consider a transaction T3
Select
sum(length) from table where attribute1 = ‘abc’
This calculates the total length of all tuples having
attribute1
Thus, T3 acquires IS for relation and S for targeted
tuples
Now, if another transaction T4 inserts a new tuple
having attribute1 = ‘abc’, the result of T3 becomes
incorrect
Example (…contd.)
This is not a concurrency problem since the serial
order (T3, T4) is maintained
But if both T3 and T4 were to write an element X, it
could lead to unserializable behavior
r3(t1);r3(t2);w4(t3);w4(X);w3(L);w3(X)
r3 and w3 are read and write operations by T3 and w4
are the write operations by T4 and L is the total length
calculated by T3 (t1 + t2)
At the end, we have result of T3 as sum of lengths of t1 and
t2 and X has value written by T3
This is not right; if value of X is considered to be that written
by T3 then for the schedule to be serializable, the sum of
lengths of t1, t2 and t3 should be considered
Example (…contd.)
Else if the sum is total length of t1 and t2 then for the
schedule to be serializable, X should have value written by
T4
This problem arises since the relation has a phantom
tuple (the new inserted tuple), which should have
been locked but wasn’t since it didn’t exist at the
time locks were taken
The occurrence of phantoms can be avoided if all
insertion and deletion transactions are treated as
write operations on the whole relation
CONCURRENCY
CONTROL
SECTION 18.7
THE TREE PROTOCOL
By :
Saloni Tamotia (215)
BASICS
B-Trees
- Tree data structure that keeps data sorted
- allow searches, insertion, and deletion
- commonly used in database and file systems
Lock
- Enforce limits on access to resources
- way of enforcing concurrency control
Lock Granularity
- Level and type of information that lock
protects.
TREE PROTOCOL
Kind of graph-based protocol
Alternate to Two-Phased
Locking (2PL)
Nodes of the tree DO NOT
form a hierarchy based on
containment
Way to get to the node is
through its parent
Example: B-Tree
ADVANTAGES OF
TREE PROTOCOL
Unlocking takes less time as
compared to 2PL
Freedom from deadlocks
18.7.1 MOTIVATION FOR
TREE-BASED LOCKING
Consider B-Tree Index, treating individual
nodes as lockable database elements.
Concurrent use of B-Tree is not possible
with standard set of locks and 2PL.
Therefore, a protocol is needed which can
assure serializability by allowing access to
the elements all the way at the bottom of the
tree even if the 2PL is violated.
18.7.1 MOTIVATION FOR
TREE-BASED LOCKING (cont.)
Reason for : “Concurrent use of B-Tree is not
possible with standard set of locks and 2PL.”
every transaction must begin with locking
the root node
2PL transactions can not unlock the root
until all the required locks are acquired.
18.7.2 ACCESSING TREE
STRUCTURED DATA
Assumptions:
Only
one kind of lock
Consistent transactions
Legal schedules
No 2PL requirement on transaction
18.7.2 RULES FOR ACCESSING
TREE STRUCTURED DATA
RULES:
First lock can be at any node.
Subsequent locks may only be acquired if parent
node has a lock.
Nodes may be unlocked any time.
No relocking of the nodes even if the node’s
parent is still locked
18.7.3 WHY TREE
PROTOCOL WORKS?
Tree protocol implies a serial order on
transactions in the schedule.
Order of precedence:
Ti < s Tj
If Ti locks the root before Tj, then Ti locks
every node in common with Tj before Tj.
ORDER OF PRECEDENCE
CONCURRENCY
CONTROL
SECTION 18.8
Timestamps
By :
Rupinder Singh (216)
What is Timestamping?
Scheduler assign each transaction T a unique
number, it’s timestamp TS(T). In other word,
timestamp is a id for transaction
Timestamps must be issued in ascending order,
at the time when a transaction first notifies the
scheduler that it is beginning.
Timestamp TS(T)
Two methods of generating Timestamps.
Use
the value of system, clock as the timestamp.
Use a logical counter that is incremented after a
new timestamp has been assigned.
Scheduler maintains a table of currently active
transactions and their timestamps irrespective
of the method used
Timestamps for database element X
and commit bit
RT(X):- The read time of X, which is the highest
timestamp of transaction that has read X.
WT(X):- The write time of X, which is the highest
timestamp of transaction that has write X.
C(X):- The commit bit for X, which is true if and only if
the most recent transaction to write X has already
committed.
Physically Unrealizable Behavior
Read too late:
A transaction U that started after transaction T, but
wrote a value for X before T reads X.
U writes X
T reads X
T start
U start
Physically Unrealizable Behavior
Write too late
A transaction U that started after T, but read X
before T got a chance to write X.
U reads X
T writes X
T start
U start
Figure: Transaction T tries to write too late
Dirty Read
It is possible that after T reads the value of X written
by U, transaction U will abort.
U writes X
T reads X
U start
T start
U aborts
T could perform a dirty read if it reads X when shown
Rules for Timestamps-Based
scheduling
1. Scheduler receives a request rT(X)
a) If TS(T) ≥ WT(X), the read is physically realizable.
1. If C(X) is true, grant the request, if TS(T) > RT(X), set RT(X) :=
TS(T); otherwise do not change RT(X).
2. If C(X) is false, delay T until C(X) becomes true or transaction
that wrote X aborts.
b) If TS(T) < WT(X), the read is physically unrealizable.
Rollback T.
Rules for Timestamps-Based
scheduling (Cont.)
2. Scheduler receives a request WT(X).
a) if TS(T) ≥ RT(X) and TS(T) ≥ WT(X), write is physically
realizable and must be performed.
1. Write the new value for X,
2. Set WT(X) := TS(T), and
3. Set C(X) := false.
b) if TS(T) ≥ RT(X) but TS(T) < WT(X), then the write is physically
realizable, but there is already a later values in X.
a. If C(X) is true, then the previous writers of X is
committed, and ignore the write by T.
b. If C(X) is false, we must delay T.
c) if TS(T) < RT(X), then the write is physically unrealizable, and T
must be rolled back.
Rules for Timestamps-Based
scheduling (Cont.)
3. Scheduler receives a request to commit T. It must find all the
database elements X written by T and set C(X) := true. If any
transactions are waiting for X to be committed, these
transactions are allowed to proceed.
4. Scheduler receives a request to abort T or decides to rollback
T, then any transaction that was waiting on an element X that T
wrote must repeat its attempt to read or write.
Multiversion Timestamps
Multiversion schemes keep old versions of data item
to increase concurrency.
Each successful write results in the creation of a new
version of the data item written.
Use timestamps to label versions.
When a read(X) operation is issued, select an
appropriate version of X based on the timestamp of
the transaction, and return the value of the selected
version.
Timestamps and Locking
Generally, timestamping performs better than locking
in situations where:
Most transactions are read-only.
It is rare that concurrent transaction will try to read
and write the same element.
In high-conflict situation, locking performs better than
timestamps
At a Glance
Introduction
Validation based scheduling
Validation based Scheduler
Expected exceptions
Validation rules
Example
Comparisons
Summary
Introduction
What is optimistic concurrency control?
(assumes no unserializable behavior will occur)
Timestamp- based scheduling and
Validation-based scheduling
(allows T to access data without locks)
Validation based scheduling
Scheduler keeps a record of what the active
transactions are doing.
Executes in 3 phases
1.
2.
3.
Read- reads from RS( ), computes local address
Validate- compares read and write sets
Write- writes from WS( )
Validation based Scheduler
Contains an assumed serial order of transactions.
Maintains three sets:
START(
): set of T’s started but not completed
validation.
VAL( ): set of T’s validated but not finished the writing
phase.
FIN( ): set of T’s that have finished.
Expected exceptions
1. Suppose there is a transaction U, such that:
U is in VAL or FIN; that is, U has validated,
FIN(U)>START(T); that is, U did not finish before T started
RS(T) ∩WS(T) ≠φ; let it contain database element X.
2. Suppose there is transaction U, such that:
• U is in VAL; U has successfully validated.
•FIN(U)>VAL(T); U did not finish before T entered its validation phase.
•WS(T) ∩ WS(U) ≠φ; let x be in both write sets.
Validation rules
Check that RS(T) ∩ WS(U)= φ for any previously
validated U that did not finish before T has started
i.e. FIN(U)>START(T).
Check that WS(T) ∩ WS(U)= φ for any previously
validated U that did not finish before T is validated
i.e. FIN(U)>VAL(T)
Example
Solution
Validation of U:
Nothing to check
Validation of T:
WS(U) ∩ RS(T)= {D} ∩{A,B}=φ
WS(U) ∩ WS(T)= {D}∩ {A,C}=φ
Validation of V:
RS(V) ∩ WS(T)= {B}∩{A,C}=φ
WS(V) ∩ WS(T)={D,E}∩ {A,C}=φ
RS(V) ∩ WS(U)={B} ∩{D}=φ
Validation of W:
RS(W) ∩ WS(T)= {A,D}∩{A,C}={A}
WS(W) ∩ WS(V)= {A,D}∩{D,E}={D}
WS(W) ∩ WS(V)= {A,C}∩{D,E}=φ
(W is not validated)
Comparison
Concurrency
control
Mechanisms
Storage Utilization
Delays
Locks
Space in the lock table is
proportional to the number of
database elements locked.
Delays transactions but
avoids rollbacks
Timestamps
Space is needed for read and
write times with every
database element, neither or
not it is currently accessed.
Do not delay the
transactions but cause
them to rollback unless
Interface is low
Validation
Space is used for timestamps
and read or write sets for each
currently active transaction,
plus a few more transactions
that finished after some
currently active transaction
began.
Do not delay the
transactions but cause
them to rollback unless
interface is low
Summary
Concurrency control by validation
The three phases
Validation Rules
Comparison
21.1 INTRODUCTION TO
INFORMATION INTEGRATION
CS257 Fan Yang
Need for Information Integration
All the data in the world could put in a single
database (ideal database system)
In the real world (impossible for a single database):
databases are created independently
hard to design a database to support future use
University Database
Registrar: to record student and grade
Bursar: to record tuition payments by students
Human Resources Department: to record employees
Other department….
Inconvenient
Record grades for students who pay tuition
Want to swim in SJSU aquatic center for free in
summer vacation?
(all the cases above cannot achieve the function by
a single database)
Solution: one database
How to integrate
Start over
build one database: contains all the legacy
databases; rewrite all the applications
result: painful
Build a layer of abstraction (middleware)
on top of all the legacy databases
this layer is often defined by a collection of classes
BUT…
Heterogeneity Problem
What is Heterogeneity Problem
Aardvark Automobile Co.
1000 dealers has 1000 databases
to find a model at another dealer
can we use this command:
SELECT * FROM CARS
WHERE MODEL=“A6”;
Type of Heterogeneity
Communication Heterogeneity
Query-Language Heterogeneity
Schema Heterogeneity
Data type difference
Value Heterogeneity
Semantic Heterogeneity
Conclusion
One database system is perfect, but impossible
Independent database is inconvenient
Integrate database
1. start over
2. middleware
heterogeneity problem
CHAPTER 21.2
MODES OF INFORMATION
INTEGRATION
ID: 219
Name: Qun Yu
Class: CS257 219 Spring 2009
Instructor: Dr. T.Y.Lin
Content Index
21.2 Modes of Information Integration
21.2.1 Federated Database Systems
21.2.2 Data Warehouses
21.2.3 Mediators
Federations
The simplest architecture for integrating several
DBs
One to one connections between all pairs of
DBs
n DBs talk to each other, n(n-1) wrappers are
needed
Good when communications between DBs are
limited
Wrapper
Wrapper : a software translates incoming queries
and outgoing answers. In a result, it allows
information sources to conform to some shared
schema.
Federations Diagram
DB1
DB2
2 Wrappers
2 Wrappers
2 Wrappers
2 Wrappers
2 Wrappers
2 Wrappers
DB3
DB4
A federated collection of 4 DBs needs 12 components to translate queries
from one to another.
Example
Car dealers want to share their inventory. Each dealer queries the
other’s DB to find the needed car.
Dealer-1’s DB relation: NeededCars(model,color,autoTrans)
Dealer-2’s DB relation: Auto(Serial, model, color)
Options(serial,option)
wrapper
Dealer-1’s DB
wrapper
Dealer-2’s DB
Example…
For(each tuple(:m,:c,:a) in NeededCars){
if(:a=TRUE){/* automatic transmission wanted */
SELECT serial
FROM Autos, Options
WHERE Autos.serial = Options.serial AND Options.option = ‘autoTrans’
AND Autos.model = :m AND Autos.color =:c;
}
Else{/* automatic transmission not wanted */
SELECT serial
FROM Auto
WHERE Autos.model = :m AND
Autos.color = :c AND
NOT EXISTS( SELECT * FROM Options WHERE serial = Autos.serial
AND option=‘autoTrans’);
}
}
Dealer 1 queries Dealer 2 for needed cars
Data Warehouse
Sources are translated from their local schema to
a global schema and copied to a central DB.
User transparent: user uses Data Warehouse just
like an ordinary DB
Warehouse Diagram
User
query
result
Warehouse
Combiner
Extractor
Extractor
Source 1
Source 2
Example
Construct a data warehouse from sources DB of 2 car dealers:
Dealer-1’s schema: Cars(serialNo, model,color,autoTrans,cdPlayer,…)
Dealer-2’s schema: Auto(serial,model,color)
Options(serial,option)
Warehouse’s schema:
AutoWhse(serialNo,model,color,autoTrans,dealer)
Extractor --- Query to extract data from Dealer-1’s data:
INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)
SELECT serialNo,model,color,autoTrans,’dealer1’
From Cars;
Example
Extractor --- Query to extract data from Dealer-2’s data:
INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)
SELECT serialNo,model,color,’yes’,’dealer2’
FROM Autos,Options
WHERE Autos.serial=Options.serial AND
option=‘autoTrans’;
INSERT INTO AutosWhse(serialNo, model, color, autoTans, dealer)
SELECT serialNo,model,color,’no’,’dealer2’
FROM Autos
WHERE NOT EXISTS ( SELECT * FROM serial =Autos.serial
AND option = ‘autoTrans’);
Construct Data Warehouse
There are mainly 3 ways to constructing
the data in the warehouse:
1)
Periodically reconstructed from the current data in the
sources, once a night or at even longer intervals.
Advantages:
simple algorithms.
Disadvantages:
1) need to shut down the warehouse;
2) data can become out of date.
Construct Data Warehouse
2)
Updated periodically based on the changes(i.e. each
night) of the sources.
Advantages:
involve smaller amounts of data. (important when warehouse is
large and needs to be modified in a short period)
Disadvantages:
1) the process to calculate changes to the warehouse is complex.
2) data can become out of date.
Construct Data Warehouse
3)
Changed immediately, in response to each change or a
small set of changes at one or more of the sources.
Advantages:
data won’t become out of date.
Disadvantages:
requires too much communication, therefore, it is
generally too expensive.
(practical for warehouses whose underlying sources changes
slowly.)
Mediators
Virtual warehouse, which supports a virtual view
or a collection of views, such as a collection of
classes, that integrates several sources.
Mediator doesn’t store any data.
Mediators’ tasks:
1)receive user’s query,
2)send queries to wrappers,
3)combine results from wrappers,
4)send the final result to user.
A Mediator diagram
Result
User query
Mediator
Query
Result
Result
Wrapper
Query
Result
Source 1
Query
Wrapper
Query
Result
Source 2
Example
Same data sources as the example of data warehouse, the mediator
Integrates the same two dealers’ source into a view with schema:
AutoMed(serialNo,model,color,autoTrans,dealer)
When the user have a query:
SELECT sericalNo, model
FROM AkutoMed
Where color=‘red’
Example
In this simple case, the mediator forwards the same query to each
Of the two wrappers.
Wrapper1: Cars(serialNo, model, color, autoTrans, cdPlayer, …)
SELECT serialNo,model
FROM cars
WHERE color = ‘red’;
Wrapper2: Autos(serial,model,color); Options(serial,option)
SELECT serial, model
FROM Autos
WHERE color=‘red’;
The mediator needs to interprets serial into serialNo, and then
returns the union of these sets of data to user.
Example
There may be different options for the mediator to forward user query,
for example, the user queries if there are a specific model&color car
(i.e. “Gobi”, “blue”).
The mediator decides 2nd query is needed or not based on the result of
1st query. That is, If dealer-1 has the specific car, the mediator doesn’t
have to query dealer-2.
CHAPTER 21 INFORMATION
INTEGRATION
21.3 WRAPPERS IN MEDIATOR-BASED
SYSTEMS
Presented by: Kai Zhu
Professor: Dr. T.Y. Lin
Class ID: 220
Introduction
Templates for Query patterns
Wrapper Generator
Filter
Wrappers in Mediator-based
Systems
More complicated than that in most data warehouse
system.
Able to accept a variety of queries from the mediator
and translate them to the terms of the source.
Communicate the result to the mediator.
How to design a wrapper?
Classify the possible queries that the mediator can
ask into templates, which are queries with parameters
that represent constants.
Templates for Query Patterns:
Use notation T=>S to express the idea
that the template T is turned by the
wrapper into the source query S.
Example 1
Dealer 1
Cars (serialNo, model, color, autoTrans,
navi,…)
For use by a mediator with schema
AutoMed (serialNo, model, color, autoTrans,
dealer)
Outline of the code would be
SELECT *
FROM AutosMed
WHERE color = ’$c’;
=>
SELECT serialNo, model, color, autoTrans, ’dealer1’
FROM Cars
WHERE color=’$c’;
(Template T => Source query S)
Wrapper Generators
The wrapper generator creates a table holds the
various query patterns contained in the templates.
The source queries that are associated with each.
A driver is used in each wrapper, the task of the
driver is to:
Accept a query from the mediator.
Search the table for a template that matches the
query.
The source query is sent to the source, again using a
“plug-in” communication mechanism.
The response is processed by the wrapper.
Filter
Have a wrapper filter to supporting more queries.
It is possible to filter the returned tuples at the
wrapper and pass only the desired tuples to the
mediator
Example 2
SELECT *
FROM AutosMed
WHERE model = ’$m’ AND color = ’$c’;
=>
SELECT serialNo, model, color, autoTrans, ’dealer1’
FROM Cars
WHERE model = ’$m’ AND color=’$c’;
Now we suppose the only template we have is color. However
the wrapper is asked by the Mediator to find “blue Gobi
model car.”
Solution:
1. Use template with $c=‘blue’ find all blue cars
and store them in a temporary relation:
TemAutos (serialNo, model, color, autoTrans,
dealer)
2.The wrapper then return to the mediator the
desired set of automobiles by excuting the local query:
SELECT*
FROM TemAutos
WHERE model= ’Gobi’;
INFORMATION
INTEGRATION
SECTIONS 21.4 – 21.5
Sanuja Dabade & Eilbroun Benjamin
CS 257 – Dr. TY Lin
Presentation Outline
21.4 Capability Based Optimization
21.4.1The Problem of Limited Source Capabilities
21.4.2 A notation for Describing Source
Capabilities
21.4.3 Capability-Based Query-Plan Selection
21.4.4 Adding Cost-Based Optimization
21.5 Optimizing Mediator Queries
21.5.1 Simplified Adornment Notation
21.5.2 Obtaining Answers for Subgoals
21.5.3 The Chain Algorithm
21.5.4 Incorporating Union Views at the Mediator
21.4 Capability Based Optimization
Introduction
Typical
DBMS estimates the cost of each query plan
and picks what it believes to be the best
Mediator – has knowledge of how long its sources will
take to answer
Optimization of mediator queries cannot rely on cost
measure alone to select a query plan
Optimization by mediator follows capability based
optimization
21.4.1 The Problem of Limited Source
Capabilities
Many sources have only Web Based interfaces
Web sources usually allow querying through a
query form
E.g. Amazon.com interface allows us to query about
books in many different ways.
But we cannot ask questions that are too general
E.g.
Select * from books;
21.4.1 The Problem of Limited Source
Capabilities (con’t)
Reasons why a source may limit the ways in which
queries can be asked
Earliest
database did not use relational DBMS that
supports SQL queries
Indexes on large database may make certain queries
feasible, while others are too expensive to execute
Security reasons
E.g.
Medical database may answer queries about averages,
but won’t disclose details of a particular patient's
information
21.4.2 A Notation for Describing
Source Capabilities
For relational data, the legal forms of queries are
described by adornments
Adornments – Sequences of codes that represent the
requirements for the attributes of the relation, in their
standard order
f(free) – attribute can be specified or not
b(bound) – must specify a value for an attribute but any
value is allowed
u(unspecified) – not permitted to specify a value for a
attribute
21.4.2 A notation for Describing
Source Capabilities….(cont’d)
c[S](choice from set S) means that a value must be specified
and value must be from finite set S.
o[S](optional from set S) means either do not specify a value
or we specify a value from finite set S
A prime (f’) specifies that an attribute is not a part of the
output of the query
A capabilities specification is a set of adornments
A query must match one of the adornments in its capabilities
specification
21.4.2 A notation for Describing
Source Capabilities….(cont’d)
E.g. Dealer 1 is a source of data in the form:
Cars (serialNo, model, color, autoTrans, navi)
The adornment for this query form is b’uuuu
21.4.3 Capability-Based Query-Plan
Selection
Given a query at the mediator, a capability based
query optimizer first considers what queries it can ask
at the sources to help answer the query
The process is repeated until:
Enough
queries are asked at the sources to resolve all the
conditions of the mediator query and therefore query is
answered. Such a plan is called feasible.
We can construct no more valid forms of source queries, yet
still cannot answer the mediator query. It has been an
impossible query.
21.4.3 Capability-Based Query-Plan
Selection (cont’d)
The simplest form of mediator query where we
need to apply the above strategy is join relations
E.g we have sources for dealer 2
Autos(serial,
model, color)
Options(serial, option)
Suppose
that ubf is the sole adornment for Auto and
Options have two adornments, bu and uc[autoTrans, navi]
Query is – find the serial numbers and colors of Gobi
models with a navigation system
21.4.4 Adding Cost-Based
Optimization
Mediator’s Query optimizer is not done when the
capabilities of the sources are examined
Making an intelligent, cost based query
optimization requires that the mediator knows a
great deal about the costs of queries involved
Sources are independent of the mediator, so it is
difficult to estimate the cost
21.5 Optimizing Mediator Queries
Chain algorithm – a greed algorithm that finds a
way to answer the query by sending a sequence of
requests to its sources.
Will
always find a solution assuming at least one
solution exists.
The solution may not be optimal.
21.5.1 Simplified Adornment Notation
A query at the mediator is limited to b (bound) and
f (free) adornments.
We use the following convention for describing
adornments:
nameadornments(attributes)
where:
name
is the name of the relation
the number of adornments = the number of attributes
21.5.2 Obtaining Answers for
Subgoals
Rules for subgoals and sources:
Suppose
we have the following subgoal:
Rx1x2…xn(a1, a2, …, an),
and source adornments for R are: y1y2…yn.
If yi
is b or c[S], then xi = b.
If xi = f, then yi is not output restricted.
The
adornment on the subgoal matches the adornment
at the source:
If yi
is f, u, or o[S] and xi is either b or f.
21.5.3 The Chain Algorithm
Maintains 2 types of information:
An
adornment for each subgoal.
A relation X that is the join of the relations for all the
subgoals that have been resolved.
Initially, the adornment for a subgoal is b iff the
mediator query provides a constant binding for the
corresponding argument of that subgoal.
Initially, X is a relation over no attributes, containing
just an empty tuple.
21.5.3 The Chain Algorithm (con’t)
1.
First, initialize adornments of subgoals and X.
Then, repeatedly select a subgoal that can be
resolved. Let Rα(a1, a2, …, an) be the subgoal:
Wherever α has a b, we shall find the argument in R
is a constant, or a variable in the schema of R.
Project X onto its variables that appear in R.
21.5.3 The Chain Algorithm (con’t)
2.
For each tuple t in the project of X, issue a query to the
source as follows
If a component of β is b, then the corresponding component
of α is b, and we can use the corresponding component of t
for source query.
If a component of β is c[S], and the corresponding component
of t is in S, then the corresponding component of α is b, and
we can use the corresponding component of t for the source
query.
If a component of β is f, and the corresponding component of
α is b, provide a constant value for source query.
21.5.3 The Chain Algorithm (con’t)
3.
If a component of β is u, then provide no binding for this
component in the source query.
If a component of β is o[S], and the corresponding
component of α is f, then treat it as if it was a f.
If a component of β is o[S], and the corresponding
component of α is b, then treat it as if it was c[S].
Every variable among a1, a2, …, an is now bound. For
each remaining unresolved subgoal, change its
adornment so any position holding one of these
variables is b.
21.5.3 The Chain Algorithm (con’t)
4.
5.
Replace X with X πs(R), where S is all of the
variables among: a1, a2, …, an.
Project out of X all components that correspond to
variables that do not appear
in the head or in any
α
unresolved subgoal.
If every subgoal is resolved, then X is the answer.
If every subgoal is not resolved, then the algorithm
fails.
21.5.3 The Chain Algorithm Example
Mediator query:
Q:
Answer(c) ← Rbf(1,a) AND Sff(a,b) AND Tff(b,c)
Example:
Relation
Data
Adornment
R
S
T
w
x
x
y
y
z
1
2
2
4
4
6
1
3
3
5
5
7
1
4
5
8
bf
c’[2,3,5]f
bu
21.5.3 The Chain Algorithm Example
(con’t)
Initially, the adornments on the subgoals are the
same as Q, and X contains an empty tuple.
S
and T cannot be resolved because they each have ff
adornments, but the sources have either a b or c.
R(1,a) can be resolved because its adornments are
matched by the source’s adornments.
Send R(w,x) with w=1 to get the tables on the
previous page.
21.5.3 The Chain Algorithm Example
(con’t)
Project the subgoal’s relation onto its second
component, since only the second component of
R(1,a) is a variable.
a
2
3
4
This is joined with X, resulting in X equaling this
relation.
Change adornment on S from ff to bf.
21.5.3 The Chain Algorithm Example
(con’t)
Now we resolve Sbf(a,b):
Project
X onto a, resulting in X.
Now, search S for tuples with attribute a equivalent to
attribute a in X.
a
b
2
4
3
5
Join this relation with X, and remove a because it
doesn’t appear in the head nor any unresolved
b
subgoal:
4
5
21.5.3 The Chain Algorithm Example
(con’t)
Now we resolve Tbf(b,c):
b
c
4
6
5
7
5
8
Join this relation with X and project onto the c
attribute to get the relation for the head.
Solution is {(6), (7), (8)}.
21.5.4 Incorporating Union Views at
the Mediator
This implementation of the Chain Algorithm does not
consider that several sources can contribute tuples to
a relation.
If specific sources have tuples to contribute that
other sources may not have, it adds complexity.
To resolve this, we can consult all sources, or make
best efforts to return all the answers.
21.5.4 Incorporating Union Views at
the Mediator (con’t)
Consulting All Sources
We
can only resolve a subgoal when each source for its
relation has an adornment matched by the current
adornment of the subgoal.
Less practical because it makes queries harder to
answer and impossible if any source is down.
Best Efforts
We
need only 1 source with a matching adornment to
resolve a subgoal.
Need to modify chain algorithm to revisit each subgoal
when that subgoal has new bound requirements.
LOCAL-AS-VIEW MEDIATORS
Priya Gangaraju(Class Id:203)
Local-as-View Mediators.
In a LAV mediator, global predicates defined are
not views of the source data.
for each source, expressions are defined, involving
the global predicates that describe the tuples that
the source is able to produce.
Queries are answered at the mediator by
discovering all possible ways to construct the query
using the views provided by the source.
Motivation for LAV Mediators
Sometimes the the relationship between what the
mediator should provide and what the sources
provide is more subtle.
For example, consider the predicate Par(c, p)
meaning that p is a parent of c which represents the
set of all child parent facts that could ever exist.
The sources will provide information about whatever
child-parent facts they know.
Motivation(contd..)
There can be sources which may provide childgrandparent facts but not child- parent facts at all.
This source can never be used to answer the childparent query under GAV mediators.
LAV mediators are that a certain source provides
grand parent facts.
They help discover how and when to use that source
in a given query.
Terminology for LAV Mediation
The queries at the mediator and the queries that
describe the source will be single Datalog rules.
A query that is a single Datalog rule is often called
a conjunctive query.
The global predicates of the LAV mediator are used
as the subgoals of mediator queries.
There are conjunctive queries that define views.
Contd..
Their heads each have a unique view predicate that
is the name of a view.
Each view definition has a body consisting of global
predicates and is associated with a particular
source.
It is assumed that each view can be constructed with
an all-free adornment.
Example..
Consider global predicate Par(c, p) meaning that p
is a parent of c.
One source produces parent facts. Its view is
defined by the conjunctive queryV1(c, p) Par(c, p)
Another source produces some grand parents facts.
Then its conjunctive query will be –
V2(c, g) Par(c, p) AND Par(p, g)
Example contd..
The query at the mediator will ask for great-grand
parent facts that can be obtained from the sources.
The mediator query is –
Q(w, z) Par(w, x) AND Par(x, y) AND Par(y, z)
One solution can be using the parent predicate(V1)
directly three times.
Q(w, z) V1(w, x) AND V1 (x, y) AND V1(y, z)
Example contd..
Another solution can be to use V1(parent facts) and
V2(grandparent facts).
Q(w, z) V1(w, x) AND V2(x, z)
Or
Q(w, z) V2(w, y) AND V1(y, z)
Expanding Solutions.
Consider a query Q, a solution S that has a body
whose subgoals are views and each view V is
defined by a conjunctive query with that view as the
head.
The body of V’s conjunctive query can be
substituted for a subgoal in S that uses the
predicate V to have a body consisting of only
global predicates.
Expansion Algorithm
A solution S has a subgoal V(a1, a2,…an) where ai’s
can be any variables or constants.
The view V can be of the form
V(b1, b2,….bn) B
Where B represents the entire body.
V(a1, a2, … an) can be replaced in solution S by a
version of body B that has all the subgoals of B with
variables possibly altered.
Expansion Algorithm contd..
1.
2.
3.
The rules for altering the variables of B are:
First identify the local variables B, variables that
appear in the body but not in the head.
If there are any local variables of B that appear in B
or in S, replace each one by a distinct new variable
that appears nowhere in the rule for V or in S.
In the body B, replace each bi by ai for
i=
1,2…n.
Example.
Consider the view definitions,
V1(c, p) Par(c, p)
V2(c, g) Par(c, p) AND Par(p, g)
One of the proposed solutions S is
Q(w, z) V1(w, x) AND V2(x, z)
The first subgoal with predicate V1 in the solution
can be expanded as Par(w, x) as there are no local
variables.
Example Contd.
The V2 subgoal has a local variable p which
doesn’t appear in S nor it has been used as a local
variable in another substitution. So p can be left as
it is.
Only x and z are to be substituted for variables c
and g.
The Solution S now will be
Q(w, z) Par(w, x) AND Par(x, p) AND Par(p,z)
Containment of Conjunctive Queries
1.
2.
A containment mapping from Q to E is a function т
from the variables of Q to the variables and
constants of E, such that:
If x is the ith argument of the head of Q, then
т(x) is the ith argument of the head of E.
Add to т the rule that т(c)=c for any constant c. If
P(x1,x2,… xn) is a subgoal of Q, then P(т(x1),
т(x2),… т(xn)) is a subgoal of E.
Example.
Consider two Conjunctive queries:
Q1: H(x, y) A(x, z) and B(z, y)
Q2: H(a, b) A(a, c) AND B(d, b) AND A(a, d)
When we apply the substitution,
Т(x) = a, Т(y) = b, Т(z) = d, the head of Q1 becomes
H(a, b) which is the head of Q2.
So,there is a containment mapping from Q1 to Q2.
Example contd..
The first subgoal of Q1 becomes A(a, d) which is the
third subgoal of Q2.
The second subgoal of Q1 becomes the second
subgoal of Q2.
There is also a containment mapping from Q2 to Q1
so the two conjunctive queries are equivalent.
Why the Containment-Mapping Test
Works
Suppose there is a containment mapping т from Q1 to
Q2.
When Q2 is applied to the database, we look for
substitutions σ for all the variables of Q2.
The substitution for the head becomes a tuple t that is
returned by Q2.
If we compose т and then σ, we have a mapping from
the variables of Q1 to tuples of the database that
produces the same tuple t for the head of Q1.
Finding Solutions to a Mediator Query
There can be infinite number of solutions built from the
views using any number of subgoals and variables.
LMSS Theorem can limit the search which states that
•
If a query Q has n subgoals, then any answer produced by
any solution is also produced by a solution that has at most n
subgoals.
If the conjunctive query that defines a view V has in its
body a predicate P that doesn’t appear in the body of
the mediator query, then we need not consider any
solution that uses V.
Example.
Recall the query
Q1: Q(w, z) Par(w, x) AND Par(x, y) AND
Par(y, z)
This query has three subgoals, so we don’t have to
look at solutions with more than three subgoals.
Why the LMSS Theorem Holds
Suppose we have a query Q with n subgoals and
there is a solution S with more than n subgoals.
The expansion E of S must be contained in Query
Q, which means that there is a containment
mapping from Q to E.
We remove from S all subgoals whose expansion
was not the target of one of Q’s subgoals under the
containment mapping.
Contd..
We would have a new conjunctive query S’ with at
most n subgoals.
If E’ is the expansion of S’ then, E’ is a subset of Q.
S is a subset of S’ as there is an identity mapping.
Thus S need not be among the solutions to query Q.
The Query Compiler
16.1 Parsing and Preprocessing
Meghna Jain(205)
Dr. T. Y. Lin
Presentation Outline
16.1 Parsing and Preprocessing
16.1.1 Syntax Analysis and Parse Tree
16.1.2 A Grammar for Simple Subset of SQL
16.1.3 The Preprocessor
16.1.4 Processing Queries Involving Views
Query compilation is divided
into three steps
1. Parsing: Parse SQL query into parser tree.
2. Logical query plan: Transforms parse tree into
expression tree of relational algebra.
3.Physical query plan:
Transforms logical query plan
into physical query plan.
. Operation performed
. Order of operation
. Algorithm used
. The way in which stored data is obtained and passed from
one
operation to another.
Query
Parser
Preprocessor
Logical Query plan
generator
Query rewrite
Preferred logical
query plan
Form a query to a logical query
plan
Syntax Analysis and Parse
Tree
Parser takes the sql query and convert it to
parse tree. Nodes of parse tree:
1. Atoms: known as Lexical elements such as
key words, constants, parentheses, operators,
and other schema elements.
2. Syntactic categories: Subparts that plays a
similar role in a query as <Query> ,
<Condition>
Grammar for Simple Subset of SQL
<Query> ::= <SFW>
<Query> ::= (<Query>)
<SFW> ::= SELECT <SelList> FROM <FromList> WHERE <Condition>
<SelList> ::= <Attribute>,<SelList>
<SelList> ::= <Attribute>
<FromList> ::= <Relation>, <FromList>
<FromList> ::= <Relation>
<Condition> ::= <Condition> AND <Condition>
<Condition> ::= <Tuple> IN <Query>
<Condition> ::= <Attribute> = <Attribute>
<Condition> ::= <Attribute> LIKE <Pattern>
<Tuple> ::= <Attribute>
Atoms(constants), <syntactic categories>(variable),
::= (can be expressed/defined as)
Query and Parse Tree
StarsIn(title,year,starName)
MovieStar(name,address,gender,birthdate)
Query:
Give titles of movies that have at least one star born in 1960
SELECT title FROM StarsIn WHERE starName IN
(
SELECT name FROM MovieStar WHERE
birthdate LIKE '%1960%'
);
Another query equivalent
SELECT title
FROM StarsIn, MovieStar
WHERE starName = name AND
birthdate LIKE '%1960%' ;
Parse Tree
<Query>
<SFW>
SELECT <SelList> FROM
<Attribute>
<FromList>
WHERE
<RelName> , <FromList>
title
StarsIn
<Condition>
starName
=
AND
<RelName>
MovieStar
<Attribute>
<Condition>
<Attribute>
name
<Query>
<Condition>
<Attribute> LIKE <Pattern>
birthdate
‘%1960’
The Preprocessor
Functions of Preprocessor
1. Checks relation uses : Every relation mentioned in FROMclause must be a relation or a view in current schema.
2. Check and resolve attribute uses: Every attribute mentioned in SELECT
or WHERE clause must be an attribute of same relation in the current
scope.
3. Check types: All attributes must be of a type appropriate to their uses.
StarsIn(title,year,starName)
MovieStar(name,address,gender,birthdate)
Query:
Give titles of movies that have at least one star born in 1960
SELECT title FROM StarsIn WHERE starName IN
(
SELECT name FROM MovieStar WHERE
birthdate LIKE '%1960%'
);
Preprocessing Queries Involving Views
When an operand in a query is a virtual view, the preprocessor
needs to replace the operand by a piece of parse tree that
represents how the view is constructed from base table.
Base Table: Movies( title, year, length, genre, studioname,
producerC#)
View definition : CREATE VIEW ParamountMovies AS
SELECT title, year FROM movies
WHERE studioName = 'Paramount';
Example based on view:
SELECT title FROM ParamountMovies WHERE year = 1979;
16.2 ALGEBRAIC LAWS FOR
IMPROVING QUERY PLANS
Ramya Karri
ID: 206
Optimizing the Logical Query Plan
The translation rules converting a parse tree to a logical query
tree do not always produce the best logical query tree.
It is often possible to optimize the logical query tree by
applying relational algebra laws to convert the original tree
into a more efficient logical query tree.
Optimizing a logical query tree using relational algebra laws
is called heuristic optimization
Relational Algebra Laws
These laws often involve the properties of:
commutativity - operator can be applied to operands independent
of order.
E.g.
A + B = B + A - The “+” operator is commutative.
associativity - operator is independent of operand grouping.
E.g.
A + (B + C) = (A + B) + C - The “+” operator is
associative.
Associative and Commutative
Operators
The relational algebra operators of cross-product (×), join
(⋈), union, and intersection are all associative and
commutative.
Commutative
Associative
R X S=S X R
(R X S) X T = S X (R X T)
R⋈S=S⋈R
(R ⋈ S) ⋈ T= S ⋈ (R ⋈ T)
RS=SR
(R S) T = S (R T)
R ∩S =S∩ R
(R ∩ S) ∩ T = S ∩ (R ∩ T)
Laws Involving Selection
Complex selections involving AND or OR can be broken into two or
more selections: (splitting laws)
σC1 AND C2 (R) = σC1( σC2 (R))
σC1 OR C2 (R) = ( σC1 (R) ) S ( σC2 (R) )
Example
R={a,a,b,b,b,c}
p1 satisfied by a,b, p2 satisfied by b,c
σp1vp2 (R) = {a,a,b,b,b,c}
σp1(R) = {a,a,b,b,b}
σp2(R) = {b,b,b,c}
σp1 (R) U σp2 (R) = {a,a,b,b,b,c}
Laws Involving Selection (Contd..)
Selection is pushed through both arguments for
union:
σC(R S) = σC(R) σC(S)
Selection is pushed to the first argument and
optionally the second for difference:
σC(R - S) = σC(R) - S
σC(R - S) = σC(R) - σC(S)
Laws Involving Selection (Contd..)
All other operators require selection to be pushed to only
one of the arguments.
For joins, may not be able to push selection to both if
argument does not have attributes selection requires.
σC(R × S) = σC(R) × S
σC(R ∩ S) = σC(R) ∩ S
σC(R ⋈ S) = σC(R) ⋈ S
σC(R ⋈D S) = σC(R) ⋈D S
Laws Involving Selection (Contd..)
Example
Consider relations R(a,b) and S(b,c) and the
expression
σ (a=1 OR a=3) AND b<c (R ⋈S)
σ a=1 OR a=3(σ b<c (R ⋈S))
σ a=1 OR a=3(R ⋈ σ b<c (S))
σ a=1 OR a=3(R) ⋈ σ b<c (S)
Laws Involving Projection
Like selections, it is also possible to push projections
down the logical query tree. However, the
performance is less than selections because
projections just reduce the number of attributes
other than number of tuples.
Laws Involving Projection
Laws for pushing projections with joins:
πL(R × S) = πL(πM(R) × πN(S))
πL(R ⋈ S) = πL((πM(R) ⋈ πN(S))
πL(R ⋈D S) = πL((πM(R) ⋈D πN(S))
Laws Involving Projection
Laws for pushing projections with set operations.
Projection can be performed entirely before union.
πL(R UB S) = πL(R) UB πL(S)
Projection can be pushed below selection as long as we
also keep all attributes needed for the selection (M = L
attr(C)).
πL ( σC (R)) = πL( σC (πM(R)))
Laws Involving Join
We have previously seen these important rules
about joins:
1. Joins are commutative and associative.
2. Selection can be distributed into joins.
3. Projection can be distributed into joins.
Laws Involving Duplicate Elimination
The duplicate elimination operator (δ) can be
pushed through many operators.
R has two copies of tuples t, S has one copy of t,
δ (RUS)=one copy of t
δ (R) U δ (S)=two copies of t
Laws Involving Duplicate Elimination
Laws for pushing duplicate elimination operator (δ):
δ(R × S) = δ(R) × δ(S)
δ(R S) = δ(R)
δ(S)
δ(R D S) = δ(R)
D δ(S)
δ( σC(R) = σC(δ(R))
Laws Involving Duplicate Elimination
The duplicate elimination operator (δ) can also be
pushed through bag intersection, but not across
union, difference, or projection in general.
δ(R ∩ S) = δ(R) ∩ δ(S)
Laws Involving Grouping
The grouping operator (γ) laws depend on the aggregate operators
used.
There is one general rule, however, that grouping subsumes duplicate
elimination:
δ(γL(R)) = γL(R)
The reason is that some aggregate functions are unaffected by
duplicates (MIN and MAX) while other functions are (SUM, COUNT,
and AVG).
Section 16.3
DATABASE SYSTEMS – The Complete Book
Presented By:
Deepti Kundu
Under the supervision of:
THE QUERY
COMPILER
Dr. T.Y.Lin
Topics to be covered
From Parse to Logical Query Plans
Conversion
to Relational Algebra
Removing Subqueries From Conditions
Improving the Logical Query Plan
Grouping Associative/ Commutative Operators
16.3 From Parse to Logical Query Plans ►
Review
Query
Parser
Section 16.1
Preprocessor
Logical query
plan generator
Section 16.3
Query Rewriter
Preferred logical query plan
Two steps to turn Parse tree into Preferred
Logical Query Plan
Replace the nodes of the parse tree, in appropriate groups, by
an operator or operators of relational algebra.
Take the relational algebra expression and turn it into an
expression that we expect can be converted to the most
efficient physical query plan.
Reference Relations
StarsIn (movieTitle, movieYear, starName)
MovieStar (name, address, gender, birthdate)
Conversion to Relational Algebra
If we have a <Query> with a <Condition> that has no
subqueries, then we may replace the entire construct –
the select-list, from-list, and condition – by a relationalalgebra expression.
The relational-algebra expression consists of the
following from bottom to top:
The products of all the relations mentioned in the
<FromList>, which Is the argument of:
A selection σC, where C is the <Condition> expression in
the construct being replaced, which in turn is the argument
of:
A projection πL , where L is the list of attributes in the
<SelList>
A query : Example
SELECT movieTitle
FROM Starsin, MovieStar
WHERE starName = name AND
birthdate LIKE ‘%1960’;
SELECT movieTitle
FROM Starsin, MovieStar
WHERE starName = name AND
birthdate LIKE ‘%1960’;
Translation to an algebraic expression tree
Removing Subqueries From Conditions
For parse trees with a <Condition> that has a
subquery
Intermediate operator – two argument selection
It is intermediate in between the syntactic
categories of the parse tree and the relationalalgebra operators that apply to relations.
Using a two-argument σ
πmovieTitle
σ
<Condition>
StarsIn
<Tuple>
<Attribute>
starName
IN
πname
σ birthdate LIKE ‘%1960'
MovieStar
Two argument selection with condition involving
IN
Now say we have, two arguments – some relation and the
second argument is a <Condition> of the form t IN S.
‘t’ – tuple composed of some attributes of R
‘S’ – uncorrelated subquery
Steps to be followed:
1.
2.
3.
Replace the <Condition> by the tree that is the expression for S ( δ is
used to remove duplicates)
Replace the two-argument selection by a one-argument selection σC.
Give σC an argument that is the product of R and S.
Two argument selection with condition involving
IN
σ
R
σC
<Condition>
t
IN
X
S
R
δ
S
The effect
Improving the Logical Query Plan
Algebraic laws to improve logical query plans:
Selections
can be pushed down the expression tree as
far as they can go.
Similarly, projections can be pushed down the tree, or
new projections can be added.
Duplicate eliminations can sometimes be removed, or
moved to a more convenient position in the tree.
Certain selections can be combined with a product
below to turn the pair of operations into an equijoin.
Grouping Associative/ Commutative Operators
An operator that is associative and commutative
operators may be though of as having any number of
operands.
Reorder these operands so that the multiway join is
executed as sequence of binary joins.
Its more time consuming to execute them in the order
suggested by parse tree.
For each portion of subtree that consists of nodes with
the same associative and commutative operator (natural
join, union, and intersection), we group the nodes with
these operators into a single node with many children.
The effect of query rewriting
Π movieTitle
Starname = name
StarsIn
σbirthdate LIKE ‘%1960’
MovieStar
Final step in producing logical query plan
=>
R
U
U
U
R
S
T
V
W
U
U
S
T
V
W
An Example to summarize
“find movies where the average age of the stars was at most
40 when the movie was made”
SELECT distinct m1.movieTitle, m1,movieYear
FROM StarsIn m1
WHERE m1.movieYear – 40 <= (
SELECT AVG (birthdate)
FROM StartsIn m2, MovieStar s
WHERE m2.starName = s.name AND
m1.movieTitle = m2.movieTitle AND
m1.movieYear = m2.movieyear
);
SELECT distinct m1.movieTitle, m1,movieYear
FROM StarsIn m1
WHERE m1.movieYear – 40 <= (
SELECT AVG (birthdate)
FROM StartsIn m2, MovieStar s
WHERE m2.starName = s.name AND
m1.movieTitle = m2.movieTitle AND
m1.movieYear = m2.movieyear );
Selections combined with a product to turn the
pair of operations into an equijoin…
Condition pushed up the expression tree…
`
Selections combined…
(16.4)
DATABASE SYSTEMS – The Complete Book
THE QUERY COMPILER
Presented By:
Under the supervision of:
Maciej Kicinski
Dr. T.Y.Lin
Topics to be covered
From Parse to Logical Query Plans
Conversion to Relational Algebra
Removing Subqueries From Conditions
Improving the Logical Query Plan
Grouping Associative/ Commutative Operators
Estimating the Cost of Operation
Estimating Sizes of Intermediate Relations
Estimating the Size of a Projection
Estimating the Size of a Selection
Estimating the Size of a Join
Estimating Sizes for Other Operations
Estimating the Cost of Operations
After getting to the logical query plan, we turn it into
physical plan.
Consider all the possible physical plan and estimate their
costs – this evaluation is known as cost-based
enumeration.
The one with least estimated cost is the one selected to be
passed to the query-execution engine.
Selection for each physical plan
We select for each physical plan:
An order and grouping for associative-and-commutative
operations like joins, unions, and intersections.
An algorithm for each operator in the logical plan, for
instance, deciding whether a nested-loop join or hash-join
should be used.
Additional operators – scanning, sorting etc. – that are
needed for the physical plan but that were not present
explicitly in the logical plan.
The way in which the arguments are passed from on
operator to the next.
Estimating Sizes of Intermediate Relations
1.
2.
3.
Give accurate estimates.
Easy to compute.
Logically consistent; the size estimate for an
intermediate relation should not depend on how
that relation is computed.
Estimating the Size of a Projection
We should treat a classical, duplicate-eliminating
projection as a bag-projection.
The size of the result can be computed exactly.
There may be reduction in size (due to eliminated
components) or increase in size (due to new
components created as combination of attributes).
Estimating the Size of a Selection
While performing selection, we may reduce the
number of tuples but the sizes of tuple remain same.
Size can be computed as:
S = σ A=c (R)
Where A is an attribute of R and c is a constant
The recommended estimate is
T(S) = T(R)/ V(R,A)
Estimating Sizes of Other Operations
Union
Intersection
Difference
Duplicate Elimination
Grouping and Aggregation
CHOOSING AN ORDER FOR JOINS
Chapter 16.6 by:
Chiu Luk
ID: 210
Introduction
This section focuses on critical problem in cost-based
optimization:
Selecting
order for natural join of three or more
relations
Compared to other binary operations, joins take
more time and therefore need effective
optimization techniques
Introduction
Significance of Left and Right Join
Arguments
The argument relations in joins determine the cost of
the join
The left argument of the join is
Called
the build relation
Assumed to be smaller
Stored in main-memory
Significance of Left and Right Join
Arguments
The right argument of the join is
Called
the probe relation
Read a block at a time
Its tuples are matched with those of build relation
The join algorithms which distinguish between the
arguments are:
One-pass
join
Nested-loop join
Index join
Join Trees
Order of arguments is important for joining two
relations
Left argument, since stored in main-memory, should
be smaller
With two relations only two choices of join tree
there are n! ways to order the arguments and
therefore n! join trees, where n is the no. of relations
Join Trees
Order of arguments is important for joining two
relations
Left argument, since stored in main-memory, should
be smaller
With two relations only two choices of join tree
With more than two relations, there are n! ways to
order the arguments and therefore n! join trees,
where n is the no. of relations
Join Trees
Total # of tree shapes T(n) for n relations given by
recurrence:
T(1) = 1
T(2) = 1
T(3) = 2
T(4) = 5 … etc
Left-Deep Join Trees
Consider 4 relations. Different ways to join them are
as follows
Continue.
In fig (a) all the right children are leaves. This is a
left-deep tree
In fig (c) all the left children are leaves. This is a
right-deep tree
Fig (b) is a bushy tree
Considering left-deep trees is advantageous for
deciding join orders
Join order
Join order selection
A1
A2
A3
..
Left deep join trees
An
An
Ai
Dynamic
Best
programming
plan computed for each subset of relations
Best plan (A1, .., An) = min cost plan of(
Best plan(A2, .., An)
A1
Best plan(A1, A3, .., An)
A2
….
Best plan(A1, .., An-1))
An
Dynamic Programming to Select a Join
Order and Grouping
Three choices to pick an order for the join of many relations
are:
Consider all of the relations
Consider a subset
Use a heuristic o pick one
Dynamic programming is used either to consider all or a subset
Construct a table of costs based on relation size
Remember only the minimum entry which will required to
proceed
Dynamic Programming to Select a Join
Order and Grouping
Dynamic Programming to Select a Join
Order and Grouping
Dynamic Programming to Select a Join
Order and Grouping
Dynamic Programming to Select a Join
Order and Grouping
A Greedy Algorithm for Selecting a
Join Order
It is expensive to use an exhaustive method
Better approach is to use a join-order heuristic for
the query optimization
Greedy algorithm is an example of that
Make
one decision at a time about order of join and
never backtrack on the decisions once made
COMPLETING THE PHYSICAL-QUERYPLAN AND CHAPTER 16 SUMMARY
(16.7-16.8)
CS257 Spring 2009
Professor Tsau Lin
Student: Suntorn Sae-Eung
Donavon Norwood
Outline
370
16.7 Completing the Physical-Query-Plan
I. Choosing a Selection Method
II. Choosing a Join Method
III. Pipelining Versus Materialization
IV. Pipelining Unary Operations
V. Pipelining Binary Operations
VI. Notation for Physical Query Plan
VII. Ordering the Physical Operations
16.8 Summary of Chapter 16
Before complete Physical-Query-Plan
371
A query previously has been
Parsed
and Preprocessed (16.1)
Converted to Logical Query Plans (16.3)
Estimated the Costs of Operations (16.4)
Determined costs by Cost-Based Plan Selection (16.5)
Weighed costs of join operations by choosing an Order
for Joins
16.7 Completing the Physical-Query-Plan
372
3 topics related to turning LP into a complete
physical plan
1.
2.
3.
Choosing of physical implementations such as
Selection and Join methods
Decisions regarding to intermediate results
(Materialized or Pipelined)
Notation for physical-query-plan operators
I. Choosing a Selection Method (A)
373
Algorithms for each selection operators
1. Can we use an created index on an attribute?
If yes, index-scan. Otherwise table-scan)
2. After retrieve all condition-satisfied tuples in (1), then
filter them with the rest selection conditions
Choosing a Selection Method(A) (cont.)
374
Recall Cost of query = # disk I/O’s
How costs for various plans are estimated from σC(R) operation
1. Cost of table-scan algorithm
a)
b)
B(R)
T(R)
if R is clustered
if R is not clustered
2. Cost of a plan picking an equality term (e.g. a = 10) w/ index-scan
a)
b)
B(R) / V(R, a)
T(R) / V(R, a)
clustering index
nonclustering index
3. Cost of a plan picking an inequality term (e.g. b < 20) w/ index-scan
a)
b)
B(R) / 3
T(R) / 3
clustering index
nonclustering index
Example
375
Selection: σx=1 AND y=2 AND z<5 (R)
- Where parameters of R(x, y, z) are :
T(R)=5000,
V(R,x)=100, and
-
-
B(R)=200,
V(R, y)=500
Relation R is clustered
x, y have nonclustering indexes, only index on z is
clustering.
Example (cont.)
376
Selection options:
1.
2.
3.
4.
Table-scan filter x, y, z. Cost is B(R) = 200 since R is
clustered.
Use index on x =1 filter on y, z. Cost is 50 since
T(R) / V(R, x) is (5000/100) = 50 tuples, index is not
clustering.
Use index on y =2 filter on x, z. Cost is 10 since
T(R) / V(R, y) is (5000/500) = 10 tuples using
nonclustering index.
Index-scan on clustering index w/ z < 5 filter x ,y.
Cost is about B(R)/3 = 67
Example (cont.)
377
Costs
option 1 = 200
option 2 = 50
option 3 = 10
option 4 = 67
The lowest Cost is option 3.
Therefore, the preferred physical plan
1.
2.
retrieves all tuples with y = 2
then filters for the rest two conditions (x, z).
II. Choosing a Join Method
378
Determine costs associated with each join algorithms:
1. One-pass join, and nested-loop join devotes enough buffer
to joining
2. Sort-join is preferred when attributes are pre-sorted or two
or more join on the same attribute such as
(R(a, b) S(a, c)) T(a, d)
- where sorting R and S on a will produce result of R S to
be sorted on a and used directly in next join
Choosing a Join Method (cont.)
379
3. Index-join for a join with high chance of using index
created on the join attribute such as R(a, b)
S(b,
c)
4. Hashing join is the best choice for unsorted or nonindexing relations which needs multipass join.
III. Pipelining Versus Materialization
380
Materialization (naïve way)
store (intermediate) result of each operations on disk
Pipelining (more efficient way)
Interleave the execution of several operations, the tuples produced by
one operation are passed directly to the operations that used it
store (intermediate) result of each operations on buffer, on main
memory
IV. Pipelining Unary Operations
381
Unary = a-tuple-at-a-time or full relation
selection and projection are the best candidates
for pipelining.
In buf
Unary
operation
Out buf
Unary
operation
Out buf
R
In buf
M-1 buffers
Pipelining Unary Operations (cont.)
Pipelining Unary Operations are implemented by
iterators
382
V. Pipelining Binary Operations
383
Binary operations : , , - , , x
The results of binary operations can also be
pipelined.
Use one buffer to pass result to its consumer, one
block at a time.
The extended example shows tradeoffs and
opportunities
Example
384
Consider physical query plan for the expression
(R(w, x) S(x, y)) U(y, z)
Assumption
R occupies 5,000 blocks, S and U each 10,000 blocks.
The intermediate result R S occupies k blocks for some k.
Both joins will be implemented as hash-joins, either onepass or two-pass depending on k
There are 101 buffers available.
Example (cont.)
385
First consider join
R S, neither relations
fits in buffers
Needs two-pass
hash-join to partition
R into 100 buckets
(maximum possible) each bucket has 50 blocks
The 2nd pass hash-join uses 51 buffers, leaving the
rest 50 buffers for joining result of R S with U.
Example (cont.)
386
Case 1: suppose k 49, the result of
occupies at most 49 blocks.
Steps
1.
2.
3.
4.
R
S
Pipeline in R S into 49 buffers
Organize them for lookup as a hash table
Use one buffer left to read each block of U in turn
Execute the second join as one-pass join.
Example (cont.)
387
The total number of I/O’s
is 55,000
45,000 for two-pass hash
join of R and S
10,000 to read U for onepass hash join of
(R S) U.
Example (cont.)
388
1.
2.
3.
Case 2: suppose k > 49 but < 5,000, we can still
pipeline, but need another strategy which
intermediate results join with U in a 50-bucket, twopass hash-join. Steps are:
Before start on R S, we hash U into 50 buckets of 200
blocks each.
Perform two-pass hash join of R and U using 51 buffers as
case 1, and placing results in 50 remaining buffers to form
50 buckets for the join of R S with U.
Finally, join R S with U bucket by bucket.
Example (cont.)
389
The number of disk I/O’s is:
20,000
to read U and write its tuples into buckets
45,000 for two-pass hash-join R S
k to write out the buckets of R S
k+10,000 to read the buckets of R S and U in the
final join
The total cost is 75,000+2k.
Example (cont.)
390
Compare Increasing I/O’s between case 1 and
case 2
k
49 (case 1)
Disk
k
I/O’s is 55,000
> 50 5000 (case 2)
k=50
, I/O’s is 75,000+(2*50) = 75,100
k=51 , I/O’s is 75,000+(2*51) = 75,102
k=52 , I/O’s is 75,000+(2*52) = 75,104
Notice: I/O’s discretely grows as k increases from 49 50.
Example (cont.)
391
1.
2.
Case 3: k > 5,000, we cannot perform two-pass
join in 50 buffers available if result of R S is
pipelined. Steps are
Compute R S using two-pass join and store the
result on disk.
Join result on (1) with U, using two-pass join.
Example (cont.)
392
The number of disk I/O’s is:
45,000
for two-pass hash-join R and S
k
to store R S on disk
30,000 + k for two-pass join of U in R S
The total cost is 75,000+4k.
Example (cont.)
393
In summary, costs of physical plan as function of R
S size.
VI. Notation for Physical Query Plans
394
Several types of operators:
1.
2.
3.
4.
Operators for leaves
(Physical) operators for Selection
(Physical) Sorts Operators
Other Relational-Algebra Operations
In practice, each DBMS uses its own internal
notation for physical query plan.
Notation for Physical Query Plans (cont.)
395
1.
Operator for leaves
A leaf operand is replaced in LQP tree
TableScan(R) : read all blocks
SortScan(R, L) : read in order according to L
IndexScan(R, C): scan index attribute A by
condition C of form Aθc.
IndexScan(R, A) : scan index attribute R.A. This
behaves like TableScan but more efficient if R is not
clustered.
Notation for Physical Query Plans (cont.)
396
2.
(Physical) operators for Selection
Logical operator σC(R) is often combined with
access methods.
If σC(R) is replaced by Filter(C), and there is no
index on R or an attribute on condition C
Use TableScan or SortScan(R, L) to access R
If condition C Aθc AND D for condition D, and
there is an index on R.A, then we may
Use operator IndexScan(R, Aθc) to access R and
Use Filter(D) in place of the selection σC(R)
Notation for Physical Query Plans (cont.)
397
3.
(Physical) Sort Operators
Sorting can occur any point in physical plan, which
use a notation SortScan(R, L).
It is common to use an explicit operator Sort(L) to
sort relation.
Can apply at the top of physical-query-plan tree if
the result needs to be sorted with ORDER BY clause
(г).
Notation for Physical Query Plans (cont.)
398
4.
Other Relational-Algebra Operations
Descriptive text definitions and signs to elaborate
Operations performed e.g. Join or grouping.
Necessary parameters e.g. theta-join or list of elements
in a grouping.
A general strategy for the algorithm e.g. sort-based,
hashed based, or index-based.
A decision about number of passed to be used e.g. onepass, two-pass or multipass.
An anticipated number of buffers the operations will
required.
Notation for Physical Query Plans (cont.)
399
Example of a physical-query-plan
A physical-query-plan in example 16.36 for the case k >
5000
TableScan
Two-pass hash join
Materialize (double line)
Store operator
Notation for Physical Query Plans (cont.)
400
Another example
A physical-query-plan in example 16.36 for the case k <
49
TableScan
(2) Two-pass hash join
Pipelining
Different buffers needs
Store operator
Notation for Physical Query Plans (cont.)
401
A physical-query-plan in example 16.35
Use Index on condition y = 2 first
Filter with the rest condition later on.
402
VII. Ordering of Physical
Operations
The PQP is represented as a tree structure
implied order of operations.
Still, the order of evaluation of interior nodes may
not always be clear.
Iterators are used in pipeline manner
Overlapped time of various nodes will make
“ordering” no sense.
Ordering of Physical Operations (cont.)
403
3 rules summarize the ordering of events in a PQP
tree:
1.
Break the tree into sub-trees at each edge that
represent materialization.
2.
Order the execution of the subtree (two direction)
3.
Execute one subtree at a time.
Bottom-top
Left-to-right
All nodes of each sub-tree are executed
simultaneously.
COMPILATION OF QUERIES
Compilation means turning a query into a physical
query plan, which can be implemented by query
engine.
Steps of query compilation :
Parsing
Semantic
checking
Selection of the preferred logical query plan
Generating the best physical plan
404
THE PARSER
405
The first step of SQL query processing.
Generates a parse tree
Nodes in the parse tree corresponds to the SQL
constructs
Similar to the compiler of a programming language
VIEW EXPANSION
406
A very critical part of query compilation.
Expands the view references in the query tree to
the actual view.
Provides opportunities for the query optimization.
SEMANTIC CHECKING
Checks the semantics of a SQL query.
Examines a parse tree.
Checks :
Attributes
Relation
names
Types
407
Resolves attribute references.
CONVERSION TO A LOGICAL QUERY
PLAN
408
Converts a semantically parsed tree to a algebraic
expression.
Conversion is straightforward but sub queries need
to be optimized.
Two argument selection approach can be used.
ALGEBRAIC TRANSFORMATION
409
Transform a logical query plan to an actual plan by using
algebraic transformations.
The laws used for this transformation :
Commutative and associative laws
Laws involving selection
Pushing selection
Laws involving projection
Laws about joins and products
Laws involving duplicate eliminations
Laws involving grouping and aggregation
ESTIMATING SIZES OF RELATIONS
True running time is taken into consideration when
selecting the best logical plan.
Two factors the affects the most in estimating the
sizes of relation :
Size
of relations ( No. of tuples )
No. of distinct values for each attribute of each relation
410
Histograms are used by some systems.
COST BASED OPTIMIZING
Best physical query plan represents the least costly
plan.
Factors that decide the cost of a query plan :
Order
and grouping operations like joins, unions and
intersections.
Nested loop and the hash loop joins used.
Scanning and sorting operations.
Storing intermediate results.
411
PLAN ENUMERATION STRATEGIES
Common approaches for searching the space for
best physical plan .
Dynamic
programming : Tabularizing the best plan for
each sub expression
Selinger style programming : sort-order the results as a
part of table
Greedy approaches : Making a series of locally
optimal decisions
Branch-and-bound : Starts with enumerating the worst
plans and reach the best plan
412
LEFT-DEEP JOIN TREES
413
Left – Deep Join Trees are the binary trees with a
single spine down the left edge and with leaves as
right children.
This strategy reduces the number of plans to be
considered for the best physical plan.
Restrict the search to Left – Deep Join Trees when
picking a grouping and order for the join of several
relations.
PHYSICAL PLANS FOR SELECTION
414
Breaking a selection into an index-scan of relation,
followed by a filter operation.
The filter then examines the tuples retrieved by the
index-scan.
Allows only those to pass which meet the portions of
selection condition.
PIPELINING VERSUS MATERIALIZING
415
This flow of data between the operators can be controlled to
implement “ Pipelining “ .
The intermediate results should be removed from main memory
to save space for other operators.
This techniques can implemented by using “ materialization “ .
Both the pipelining and the materialization should be
considered by the physical query plan generator.
An operator always consumes the result of other operator and
is passed through the main memory.
Query Execution
416
Chapter 15
Section 15.1
Presented by
Khadke, Suvarna
CS 257
(Section II) Id 213
Agenda
417
Query Processor and major parts of Query
processor
Physical-Query-Plan Operators
Scanning Tables
Basic approaches to locate the tuples of a
relation R
Sorting While Scanning Tables
Computation Model for Physical Operator
I/O Cost for Scan Operators
Iterators
What is a Query Processor
418
Group of components of a DBMS that converts
a user queries and data-modification
commands into a sequence of database
operations
It also executes those operations
Must supply detail regarding how the query is
to be executed
Major parts of Query processor
419
Query Execution:
The algorithms
that manipulate
the data of the
database.
Focus on the
operations of
extended
relational
algebra.
Outline of Query Compilation
420
Query compilation
Parsing : A parse tree for the
query is constructed
Query Rewrite : The parse tree is
converted to an initial query plan
and transformed into logical query
plan (less time)
Physical Plan Generation :
Logical Q Plan is converted into
physical query plan by selecting
algorithms and order of execution
of these operator.
Physical-Query-Plan Operators
421
Physical operators are implementations of the
operator of relational algebra.
They can also be use in non relational algebra
operators like “scan” which scans tables, that is,
bring each tuple of some relation into main
memory
Scanning Tables
422
One of the basic thing we can do in a Physical
query plan is to read the entire contents of a
relation R.
Variation of this operator involves simple
predicate, read only those tuples of the relation R
that satisfy the predicate.
Scanning Tables
423
Basic approaches to locate the tuples of a relation
R
Table
Scan
Relation R is stored in secondary memory with its
tuples arranged in blocks
get the blocks one by one
Index-Scan
If there is an index on any attribute of Relation R,
we can use this index to get all the tuples of
Relation R
Sorting While Scanning Tables
424
Number of reasons to sort a relation
Query could include an ORDER BY clause,
requiring that a relation be sorted.
Algorithms to implement relational algebra
operations requires one or both arguments to be
sorted relations.
Physical-query-plan operator sort-scan takes a
relation R, attributes on which the sort is to be
made, and produces R in that sorted order
Computation Model for Physical
Operator
425
Physical-Plan Operator which is essential for good
Query Processor should be selected wisely.
For “cost” of each operator is estimated by
number of disk I/O’s for an operation.
The total cost of operation depends on the size of
the answer, and includes the final write back cost
to the total cost of the query.
Parameters for Measuring Costs
426
Parameters that affect the performance of a query
Buffer space availability in the main memory
at the time of execution of the query
Size of input and the size of the output
generated
The size of memory block on the disk and the
size in the main memory also affects the
performance
Parameters for Measuring Costs
427
B: The number of blocks are needed to hold all
tuples of relation R.
Also denoted as B(R)
T:The number of tuples in relationR.
Also denoted as T(R)
V: The number of distinct values that appear in a
column of a relation R
V(R, a)- is the number of distinct values of column
for a in relation R
I/O Cost for Scan Operators
428
If relation R is clustered, then the number of
disk I/O for the table-scan operator is = ~B
disk I/O’s
If relation R is not clustered, then the number of
required disk I/O generally is much higher
A index on a relation R occupies many fewer
than B(R) blocks
That means a scan of the entire relation R
which takes at least B disk I/O’s will require
more I/O’s than the entire index
Iterators for Implementation of Physical
Operators
429
Many physical operators can be implemented as
an Iterator.
Three methods forming the iterator for an
operation are:
1. Open( ) :
This method starts the process of getting tuples
It initializes any data structures needed to
perform the operation
Iterators for Implementation of Physical
Operators
430
2. GetNext( ):
Returns the next tuple in the result
If there are no more tuples to return, GetNext
returns a special value NotFound
3. Close( ) :
Ends the iteration after all tuples
It calls Close on any arguments of the
operator
Query Execution
431
One-Pass Algorithms for Database Operations (15.2)
Presented by
Ronak Shah
(214)
April 22, 2009
Introduction
432
The choice of an algorithm for each operator is an essential
part of the process of transforming a logical query plan into a
physical query plan.
Main classes of Algorithms:
Sorting-based methods
Hash-based methods
Index-based methods
Division based on degree difficulty and cost:
1-pass algorithms
2-pass algorithms
3 or more pass algorithms
One-Pass Algorithm Methods
433
Tuple-at-a-time, unary operations: (selection & projection)
Full-relation, unary operations
Full-relation, binary operations (set & bag versions of union)
434
One-Pass Algorithms for Tuple-ata-Time Operations
Tuple-at-a-time operations are selection and projection
read the blocks of R one at a time into an input buffer
perform the operation on each tuple
move the selected tuples or the projected tuples to the output
buffer
The disk I/O requirement for this process depends only on how
the argument relation R is provided.
If R is initially on disk, then the cost is whatever it takes to
perform a table-scan or index-scan of R.
435
A selection or projection being
performed on a relation R
436
One-Pass Algorithms for Unary, fillRelation Operations
Duplicate Elimination
To eliminate duplicates, we can read each block of R one at a
time, but for each tuple we need to make a decision as to
whether:
1.
2.
It is the first time we have seen this tuple, in which case we
copy it to the output, or
We have seen the tuple before, in which case we must not
output this tuple.
One memory buffer holds one block of R's tuples, and the
remaining M - 1 buffers can be used to hold a single copy of
every tuple.
437
Managing memory for a one-pass
duplicate-elimination
Duplicate Elimination
438
When a new tuple from R is considered, we compare it with all
tuples seen so far
if it is not equal: we copy both to the output and add it to the inmemory list of tuples we have seen.
if there are n tuples in main memory: each new tuple takes
processor time proportional to n, so the complete operation takes
processor time proportional to n2.
We need a main-memory structure that allows each of the
operations:
Add a new tuple, and
Tell whether a given tuple is already there
Duplicate Elimination (…contd.)
439
The different structures that can be used for such main memory
structures are:
Hash table
Balanced binary search tree
440
One-Pass Algorithms for Unary, fillRelation Operations
Grouping
The grouping operation gives us zero or more grouping attributes
and presumably one or more aggregated attributes
If we create in main memory one entry for each group then we
can scan the tuples of R, one block at a time.
The entry for a group consists of values for the grouping
attributes and an accumulated value or values for each
aggregation.
Grouping
441
The accumulated value is:
For MIN(a) or MAX(a) aggregate, record minimum
/maximum value, respectively.
For any COUNT aggregation, add 1 for each tuple of group.
For SUM(a), add value of attribute a to the accumulated sum
for its group.
AVG(a) is a hard case. We must maintain 2 accumulations:
count of no. of tuples in the group & sum of a-values of these
tuples. Each is computed as we would for a COUNT & SUM
aggregation, respectively. After all tuples of R are seen, take
quotient of sum & count to obtain average.
442
One-Pass Algorithms for Binary
Operations
Binary operations include:
Union
Intersection
Difference
Product
Join
Set Union
443
We read S into M - 1 buffers of main memory and build a
search structure where the search key is the entire tuple.
All these tuples are also copied to the output.
Read each block of R into the Mth buffer, one at a time.
For each tuple t of R, see if t is in S, and if not, we copy t to
the output. If t is also in S, we skip t.
Set Intersection
444
Read S into M - 1 buffers and build a search structure with
full tuples as the search key.
Read each block of R, and for each tuple t of R, see if t is
also in S. If so, copy t to the output, and if not, ignore t.
Set Difference
445
Read S into M - 1 buffers and build a search structure with
full tuples as the search key.
To compute R -s S, read each block of R and examine each
tuple t on that block. If t is in S, then ignore t; if it is not in S
then copy t to the output.
To compute S -s R, read the blocks of R and examine each
tuple t in turn. If t is in S, then delete t from the copy of S in
main memory, while if t is not in S do nothing.
After considering each tuple of R, copy to the output those
tuples of S that remain.
Bag Intersection
446
Read S into M - 1 buffers.
Multiple copies of a tuple t are not stored individually.
Rather store 1 copy of t & associate with it a count equal to
no. of times t occurs.
Next, read each block of R, & for each tuple t of R see
whether t occurs in S. If not ignore t; it cannot appear in the
intersection. If t appears in S, & count associated with t is
(+)ve, then output t & decrement count by 1. If t appears in
S, but count has reached 0, then do not output t; we have
already produced as many copies of t in output as there
were copies in S.
Bag Difference
447
To compute S -B R, read tuples of S into main memory &
count no. of occurrences of each distinct tuple.
Then read R; check each tuple t to see whether t occurs in S,
and if so, decrement its associated count. At the end, copy
to output each tuple in main memory whose count is positive,
& no. of times we copy it equals that count.
To compute R -B S, read tuples of S into main memory &
count no. of occurrences of distinct tuples.
Bag Difference (…contd.)
448
Think of a tuple t with a count of c as c reasons not to copy t to
the output as we read tuples of R.
Read a tuple t of R; check if t occurs in S. If not, then copy t to
the output. If t does occur in S, then we look at current count c
associated with t. If c = 0, then copy t to output. If c > 0, do
not copy t to output, but decrement c by 1.
Product
449
Read S into M - 1 buffers of main memory
Then read each block of R, and for each tuple t of R
concatenate t with each tuple of S in main memory.
Output each concatenated tuple as it is formed.
This algorithm may take a considerable amount of
processor time per tuple of R, because each such tuple must
be matched with M - 1 blocks full of tuples. However,
output size is also large, & time/output tuple is small.
Natural Join
450
Convention: R(X, Y) is being joined with S(Y, Z), where Y
represents all the attributes that R and S have in common, X is
all attributes of R that are not in the schema of S, & Z is all
attributes of S that are not in the schema of R. Assume that S is
the smaller relation.
To compute the natural join, do the following:
1. Read all tuples of S & form them into a main-memory
search structure.
Hash table or balanced tree are good e.g. of such
structures. Use M - 1 blocks of memory for this purpose.
Natural Join (…contd.)
451
2.
Read each block of R into 1 remaining main-memory
buffer.
For each tuple t of R, find tuples of S that agree with t on
all attributes of Y, using the search structure.
For each matching tuple of S, form a tuple by joining it
with t, & move resulting tuple to output.
QUERY EXECUTION
15.3
Nested-Loop Joins
By:
Saloni Tamotia (215)
Introduction to Nested-Loop Joins
Used for relations of any side.
Not necessary that relation fits in main memory
Uses “One-and-a-half ” pass method in which for each
variation:
One argument read just once.
Other argument read repeatedly.
Two kinds:
Tuple-Based Nested Loop Join
Block-Based Nested Loop Join
ADVANTAGES OF NESTED-LOOP
JOIN
Fits in the iterator framework.
Allows us to avoid storing
intermediate relation on disk.
Tuple-Based Nested-Loop Join
Simplest variation of the
nested-loop join
Loop ranges over
individual tuples
Tuple-Based Nested-Loop Join
Algorithm to compute the Join R(X,Y) | | S(Y,Z)
FOR each tuple s in S DO
FOR each tuple r in R DO
IF r and s join to make tuple t THEN
output t
R and S are two Relations with r and s as tuples.
carelessness in buffering of blocks causes the use of
T(R)T(S) disk I/O’s
IMPROVEMENT & MODIFICATION
To decrease the cost
Method 1: Use algorithm for Index-Based joins
We find tuple of R that matches given tuple of S
We need not to read entire relation R
Method 2: Use algorithm for Block-Based joins
Tuples of R & S are divided into blocks
Uses enough memory to store blocks in order to reduce
the number of disk I/O’s.
Block-Based Nested-Loop Join Algorithm
Access to arguments is organized by block.
While reading tuples of inner relation we use
less number of I/O’s disk.
Using enough space in main memory to store tuples
of relation of the outer loop.
Allows to join each tuple of the inner relation
with as many tuples as possible.
Block-Based Nested-Loop Join Algorithm
ALGORITHM:
FOR each chunk of M-1 blocks of S DO
FOR each block b of R DO
FOR each tuple t of b DO
find the tuples of S in memory that join with t
output the join of t with each of these tuples
Block-Based Nested-Loop Join Algorithm
Assumptions:
B(S)
≤ B(R)
B(S) > M
This means that the neither relation fits in the
entire main memory.
Analysis of Nested-Loop Join
Number of disk I/O’s:
[B(S)/(M-1)]*(M-1 +B(R))
or
B(S) + [B(S)B(R)/(M-1)]
or approximately B(S)*B(R)/M
TWO-PASS ALGORITHMS
BASED ON SORTING
SECTION 15.4
Rupinder Singh
Two-Pass Algorithms Based on
Sorting
Two-pass Algorithms: where data from the
operand relations is read into main memory,
processed in some way, written out to disk again,
and then reread from disk to complete the
operation
Basic idea
Step 1: Read M blocks of R into main memory.
Step 2:Sort these M blocks in main memory, using
an efficient, main-memory sorting algorithm. so
we expect that the time to sort will not exceed the
disk 1/0 time for step (1).
Step 3: Write the sorted list into M blocks of disk.
Duplicate Elimination Using Sorting
δ(R)
First sort the tuples of R in sublists
Then use the available main memory to hold one
block from each sorted sublist
Then we repeatedly copy one to the output and
ignore all tuples identical to it.
The total cost of this algorithm is 3B(R)
This algorithm requires only √B(R)blocks of main
memory, rather than B(R) blocks(one-pass
algorithm).
Example
Suppose that tuples are integers, and only two
tuples fit on a block. Also, M = 3 and the relation R
consists of 17 tuples:
2,5,2,1,2,2,4,5,4,3,4,2,1,5,2,1,3
After first-pass
Sublists
Elements
R1
1,2,2,2,2,5
R2
2,3,4,4,4,5
R3
1,1,2,3,5
Example
Second pass
Sublist
In memory
Waiting on disk
R1
1,2
2,2, 2,5
R2
2,3
4,4, 4,5
R3
1,1
2,3,5
Sublist
In memory
Waiting on disk
R1
2
2,2, 2,5
R2
2,3
4,4, 4,5
After processing tuple 1
2,3
Output: 1 R3
Continue the same process with next tuple.
5
Grouping and Aggregation Using
Sorting γ(R)
Two-pass algorithm for grouping and aggregation is quite
similar to the previous algorithm.
Step 1:Read the tuples of R into memory, M blocks at a
time. Sort each M blocks, using the grouping attributes of L
as the sort key. Write each sorted sublist to disk.
Step 2:Use one main-memory buffer for each sublist, and
initially load the first block of each sublist into its buffer.
Step 3:Repeatedly find the least value of the sort key
(grouping attributes) present among the first available
tuples in the buffers.
This algorithm takes 3B(R) disk 1/0's, and will work as long
as B(R) < M².
A Sort-Based Union Algorithm
For bag-union one-pass algorithm is used.
For set-union
Step 1:Repeatedly bring M blocks of R into main memory, sort their
tuples, and write the resulting sorted sublist back to disk.
Step 2:Do the same for S, to create sorted sublists for relation S.
Step 3:Use one main-memory buffer for each sublist of R and S.
Initialize each with the first block from the corresponding sublist.
Step 4:Repeatedly find the first remaining tuple t among all the
buffers. Copy t to the output. and remove from the buffers all copies
of t (if R and S are sets there should be at most two copies)
This algorithm takes 3(B(R)+B(S)) disk 1/0's, and will work as long as
B(R)+B(S) < M².
Sort-Based Intersection and
Difference
For both set version and bag version, the algorithm is same
as that of set-union except that the way we handle the
copies of a tuple t at the fronts of the sorted sublists.
For set intersection, output t if it appears in both R and S.
For bag intersection, output t the minimum of the number
of times it appears in R and in S.
For set difference, R-S, output t if and only if it appears in R
but not in S.
For bag difference, R-S, output t the number of times it
appears in R minus the number of times it appears in S.
A Simple Sort-Based Join Algorithm
When taking a join, the number of tuples from the
two relations that share a common value of the
join attribute(s), and therefore need to be in main
memory simultaneously, can exceed what fits in
memory
To avoid facing this situation, are can try to reduce
main-memory use for other aspects of the
algorithm, and thus make available a large number
of buffers to hold the tuples with a given joinattribute value
A Simple Sort-Based Join Algorithm
Given relations R(X, Y) and S(Y, Z) to join, and given M blocks of main
memory for buffers.
Step 1:Sort R and S, using a two-phase, multiway merge sort, with Y as the
sort key.
Step 2:Merge the sorted R and S. The following steps are done repeatedly:
Find the least value y of the join attributes Y that is currently at the front of the
blocks for R and S.
If y does not appear at the front of the other relation, then remove the
tuple(s) with sort key y.
Otherwise, identify all the tuples from both relations having sort key y.
Output all the tuples that can be formed by joining tuples from R and S with a
common Y-value y.
If either relation has no more unconsidered tuples in main memory.,reload the
buffer for that relation.
A Simple Sort-Based Join Algorithm
The simple sort-join uses 5(B(R) + B(S)) disk I/0's.
It requires B(R) ≤ M² and B(S) ≤ M² to work.
A More Efficient Sort-Based Join
If we do not have to worry about very large numbers of
tuples with a common value for the join attribute(s), then
we can save two disk 1/0's per block by combining the
second phase of the sorts with the join itself
To compute R(X, Y) ►◄ S(Y, Z) using M main-memory
buffers
Create sorted sublists of size M, using Y as the sort key, for both
R and S.
Bring the first block of each sublist into a buffer
Repeatedly find the least Y-value y among the first available
tuples of all the sublists. Identify all the tuples of both relations
that have Y-value y. Output the join of all tuples from R with all
tuples from S that share this common Y-value
A More Efficient Sort-Based Join
The number of disk I/O’s is 3(B(R) + B(S))
It requires B(R) + B(S) ≤ M² to work
Summary of Sort-Based Algorithms
Operators
Approximate
M required
Disk I/O
γ,δ
√B
3B
U,∩,−
√(B(R) + B(S))
3(B(R) + B(S))
►◄
√(max(B(R),B(S)))
5(B(R) + B(S))
►◄(more efficient)
√(B(R) + B(S))
3(B(R) + B(S))
By
Swathi Vegesna
QUERY EXECUTION
15.5 TWO-PASS ALGORITHMS BASED ON
HASHING
At a glimpse
Introduction
Partitioning Relations by Hashing
Algorithm for Duplicate Elimination
Grouping and Aggregation
Union, Intersection, and Difference
Hash-Join Algorithm
Sort based Vs Hash based
Summary
Introduction
Hashing is done if the data is too big to store in main
memory buffers.
Hash
all the tuples of the argument(s) using an
appropriate hash key.
For all the common operations, there is a way to select
the hash key so all the tuples that need to be
considered together when we perform the operation
have the same hash value.
This reduces the size of the operand(s) by a factor
equal to the number of buckets.
Partitioning Relations by Hashing
Algorithm:
initialize M-1 buckets using M-1 empty buffers;
FOR each block b of relation R DO BEGIN
read block b into the Mth buffer;
FOR each tuple t in b DO BEGIN
IF the buffer for bucket h(t) has no room for t THEN
BEGIN
copy the buffer t o disk;
initialize a new empty block in that buffer;
END;
copy t to the buffer for bucket h(t);
END ;
END ;
FOR each bucket DO
IF the buffer for this bucket is not empty THEN
write the buffer to disk;
Duplicate Elimination
For the operation δ(R) hash R to M-1 Buckets.
(Note that two copies of the same tuple t will hash to the same
bucket)
Do duplicate elimination on each bucket Ri independently,
using one-pass algorithm
The result is the union of δ(Ri), where Ri is the portion of R that
hashes to the ith bucket
Requirements
Number of disk I/O's: 3*B(R)
B(R) < M(M-1), only then the two-pass, hash-based
algorithm will work
In order for this to work, we need:
hash
function h evenly distributes the tuples among the
buckets
each bucket Ri fits in main memory (to allow the onepass algorithm)
i.e., B(R) ≤ M2
Grouping and Aggregation
Hash all the tuples of relation R to M-1 buckets, using a hash
function that depends only on the grouping attributes
(Note: all tuples in the same group end up in the same bucket)
Use the one-pass algorithm to process each bucket
independently
Uses 3*B(R) disk I/O's, requires B(R) ≤ M2
Union, Intersection, and Difference
For binary operation we use the same hash function
to hash tuples of both arguments.
R U S we hash both R and S to M-1
R ∩ S we hash both R and S to 2(M-1)
R-S we hash both R and S to 2(M-1)
Requires 3(B(R)+B(S)) disk I/O’s.
Two pass hash based algorithm requires
min(B(R)+B(S))≤ M2
Hash-Join Algorithm
Use same hash function for both relations; hash function should
depend only on the join attributes
Hash R to M-1 buckets R1, R2, …, RM-1
Hash S to M-1 buckets S1, S2, …, SM-1
Do one-pass join of Ri and Si, for all i
3*(B(R) + B(S)) disk I/O's; min(B(R),B(S)) ≤ M2
Sort based Vs Hash based
For binary operations, hash-based only limits size
to min of arguments, not sum
Sort-based can produce output in sorted order,
which can be helpful
Hash-based depends on buckets being of equal
size
Sort-based algorithms can experience reduced
rotational latency or seek time
Summary
Partitioning Relations by Hashing
Algorithm for Duplicate Elimination
Grouping and Aggregation
Union, Intersection, and Difference
Hash-Join Algorithm
Sort based Vs Hash based
Index-Based Algorithms
488
Chapter 15
Section 15.6
Presented by
Fan Yang
CS 257
Class ID218
Clustering and Nonclustering
Indexes
489
Clustered Relation: Tuples are packed into roughly
as few blocks as can possibly hold those tuples
Clustering indexes: Indexes on attributes that all the
tuples with a fixed value for the search key of this
index appear on roughly as few blocks as can hold
them
490
Clustering and Nonclustering
Indexes
A relation that isn’t clustered cannot have a
clustering index
A clustered relation can have nonclustering indexes
Index-Based Selection
491
For a selection σC(R), suppose C is of the form a=v,
where a is an attribute
For clustering index R.a:
the number of disk I/O’s will be B(R)/V(R,a)
Index-Based Selection
492
The actual number may be higher:
1. index is not kept entirely in main memory
2. they spread over more blocks
3. may not be packed as tightly as possible into
blocks
Example
493
B(R)=1000, T(R)=20,000 number of I/O’s required:
1. clustered, not index
1000
2. not clustered, not index
20,000
3. If V(R,a)=100, index is clustering
10
4. If V(R,a)=10, index is nonclustering 2,000
Joining by Using an Index
494
Natural join R(X, Y) S S(Y, Z)
Number of I/O’s to get R
Clustered: B(R)
Not clustered: T(R)
Number of I/O’s to get tuple t of S
Clustered: T(R)B(S)/V(S,Y)
Not clustered: T(R)T(S)/V(S,Y)
Example
495
R(X,Y): 1000 blocks S(Y,Z)=500 blocks
Assume 10 tuples in each block,
so T(R)=10,000 and T(S)=5000
V(S,Y)=100
If R is clustered, and there is a clustering index on Y
for S
the number of I/O’s for R is: 1000
the number of I/O’s for S
is10,000*500/100=50,000
Joins Using a Sorted Index
496
Natural join R(X, Y) S (Y, Z) with index on Y for
either R or S
Extreme case: Zig-zag join
Example:
relation R(X,Y) and R(Y,Z) with index on Y for both
relations
search keys (Y-value) for R: 1,3,4,4,5,6
search keys (Y-value) for S: 2,2,4,6,7,8
CHAPTER 15.7
BUFFER MANAGEMENT
ID: 219
Name: Qun Yu
Class: CS257 219 Spring 2009
Instructor: Dr. T.Y.Lin
What does a buffer manager do?
Assume there are M of main-memory buffers needed for
the operators on relations to store needed data.
In practice:
1) rarely allocated in advance
2) the value of M may vary depending on system
conditions
Therefore, buffer manager is used to allow processes
to get the memory they need, while minimizing the
delay and unclassifiable requests.
The role of the buffer manager
Read/Writes
Requests
Buffers
Buffer
manager
Figure 1: The role of the buffer manager : responds to requests for
main-memory access to disk blocks
15.7.1 Buffer Management Architecture
Two broad architectures for a buffer manager:
1) The buffer manager controls main memory directly.
• Relational DBMS
2) The buffer manager allocates buffers in virtual memory,
allowing the OS to decide how to use buffers.
• “main-memory” DBMS
• “object-oriented” DBMS
Buffer Pool
Key setting for the Buffer manager to be efficient:
The buffer manager should limit the number of buffers in
use so that they fit in the available main memory, i.e.
Don’t exceed available space.
The number of buffers is a parameter set when the DBMS is
initialized.
No matter which architecture of buffering is used, we simply
assume that there is a fixed-size buffer pool, a set of
buffers available to queries and other database actions.
Buffer Pool
Page Requests from Higher Levels
BUFFER POOL
disk page
free frame
MAIN MEMORY
DISK
DB
choice of frame dictated
by replacement policy
Data must be in RAM for DBMS to operate on it!
Buffer Manager hides the fact that not all data is in RAM.
15.7.2 Buffer Management Strategies
Buffer-replacement strategies:
When a buffer is needed for a newly requested
block and the buffer pool is full, what block to
throw out the buffer pool?
Buffer-replacement strategy -- LRU
Least-Recently Used (LRU):
To throw out the block that has not been read or written
for the longest time.
• Requires more maintenance but it is effective.
• Update the time table for every access.
• Least-Recently Used blocks are usually less likely to
be accessed sooner than other blocks.
Buffer-replacement strategy -- FIFO
First-In-First-Out (FIFO):
The buffer that has been occupied the longest by the
same block is emptied and used for the new block.
• Requires less maintenance but it can make more
mistakes.
• Keep only the loading time
• The oldest block doesn’t mean it is less likely to be
accessed.
Example: the root block of a B-tree index
Buffer-replacement strategy – “Clock”
The “Clock” Algorithm (“Second Chance”)
Think of the 8 buffers as arranged in a circle, shown as
Figure 3
Flag 0 and 1:
buffers with a 0 flag are ok to sent their contents back
to disk, i.e. ok to be replaced
buffers with a 1 flag are not ok to be replaced
Buffer-replacement strategy – “Clock”
0
0
1
0
the buffer with
a 0 flag will
be replaced
0
0
1
1
Start point to
search a 0 flag
The flag will
be set to 0
By next time the hand
reaches it, if the content of
this buffer is not accessed,
i.e. flag=0, this buffer will
be replaced.
That’s “Second Chance”.
Figure 3: the clock algorithm
Buffer-replacement strategy -- Clock
a buffer’s flag set to 1 when:
a block is read into a buffer
the contents of the buffer is accessed
a buffer’s flag set to 0 when:
the buffer manager needs a buffer for a new block, it
looks for the first 0 it can find, rotating clockwise. If it
passes 1’s, it sets them to 0.
System Control helps Buffer-replacement strategy
System Control
The components of a DBMS can give advice to the buffer
manager in order to avoid some of the mistakes that would
occur with a strict policy such as LRU,FIFO or Clock.
For example:
A “pinned” block means it can’t be moved to disk without
first modifying certain other blocks that point to it.
In FIFO, use “pinned” to force root of a B-tree to remain in
memory at all times.
15.7.3 The Relationship Between Physical
Operator Selection and Buffer Management
Problem:
Physical Operator expected certain number of
buffers M for execution.
However, the buffer manager may not be able
to guarantee these M buffers are available.
15.7.3 The Relationship Between Physical
Operator Selection and Buffer Management
Questions:
Can the algorithm adapt to changes of M, the
number of main-memory buffers available?
When available buffers are less than M, and some
blocks have to be put in disk instead of in memory.
How the buffer-replacement strategy impact the
performance (i.e. the number of additional I/O’s)?
Example
FOR each chunk of M-1 blocks of S DO BEGIN
read these blocks into main-memory buffers;
organize their tuples into a search structure whose
search key is the common attributes of R and S;
FOR each block b of R DO BEGIN
read b into main memory;
FOR each tuple t of b DO BEGIN
find the tuples of S in main memory that
join with t ;
output the join of t with each of these tuples;
END ;
END ;
END ;
Figure 15.8: The nested-loop join algorithm
Example
The outer loop number (M-1) depends on the average
number of buffers are available at each iteration.
The outer loop use M-1 buffers and 1 is reserved for a block
of R, the relation of the inner loop.
If we pin the M-1 blocks we use for S on one iteration of the
outer loop, we shall not lose their buffers during the round.
Also, more buffers may become available and then we could
keep more than one block of R in memory.
Will these extra buffers improve the running time?
Example
CASE1: NO
Buffer-replacement strategy: LRU
Buffers for R: k
We read each block of R in order into buffers.
By end of the iteration of the outer loop, the last k blocks of R
are in buffers.
However, next iteration will start from the beginning of R
again.
Therefore, the k buffers for R will need to be replaced.
Example
CASE 2: YES
Buffer-replacement strategy: LRU
Buffers for R: k
We read the blocks of R in an order that alternates:
firstlast and then lastfirst.
In this way, we save k disk I/Os on each iteration of the outer
loop except the first iteration.
Other Algorithms and M buffers
Other Algorithms also are impact by M and the
buffer-replacement strategy.
Sort-based algorithm
If M shrinks, we can change the size of a
sublist.
Unexpected result: too many sublists to
allocate each sublist a buffer.
Hash-based algorithm
If M shrinks, we can reduce the number of
buckets, as long as the buckets still can fit in M
buffers.
CHAPTER 15
QUERY EXECUTION
15.8 ALGORITHMS USING MORE THAN
TWO PASSES
Presented by: Kai Zhu
Professor: Dr. T.Y. Lin
Class ID: 220
Introduction
Why we use more than 2 passes
Multi-pass Sort-based Algorithms
Conclusion
Why we use more than two passes:
Two passes are usually enough, however, for the
largest relation, we use as many passes as necessary.
Multi-pass Sort-based Algorithms
Suppose we have M main-memory buffers
available to sort a relation R, which we assume is
stored clustered.
Then we do the following:
BASIS:
If R fits in M blocks (i.e., B(R)<=M)
1. Read R into main memory.
2. Sort it using any main-memory
sorting algorithm.
3. Write the sorted relation to disk.
INDUCTION:
If R does not fit into main memory.
1. Partition the blocks holding R into
M groups,
which we shall call R1, R2, R3…
2. Recursively sort Ri for each
i=1,2,3…M.
3. Merge the M sorted sublists.
If we are not merely sorting R, but performing a
unary operation such as δ or γ on R.
We can modify the above so that at the final merge
we perform the operation on the tuples at the front
of the sorted sublists.
That is:
For a δ, output one copy of each distinct tuple, and
skip over copies of the tuple.
For a γ, sort on the grouping attributes only, and
combine the tuples with a given value of these
grouping attributes.
Conclusion
The two pass algorithms based on sorting or hashing
have natural recursive analogs that take three or
more passes and will work for larger amounts of
data.