INFO20003 Database Systems
INFO20003 Database Systems 1© University of Melbourne 2018
INFO20003 Database Systems
Lecture 11
Query Processing Part I
Semester 2 2018, Week 6
Dr Renata Borovica-Gajic
INFO20003 Database Systems 2© University of Melbourne 2018
Remember this? Components of a DBMS
DBMS
Index files
Heap files
Database
Concurrency
control module
File and access
methods mgr.
Buffer pool mgr.
Disk space mgr.
Storage module Concurrency
control module
Transaction
mgr.
Lock mgr.
Crash recovery
module
Log mgr.
Query processing module
Parser/
Compiler
Optimizer Executor
This is one of several possible architectures;
each system has its own slight variations.
TODAY &
Next time
Will briefly
touch upon …
INFO20003 Database Systems 3© University of Melbourne 2018
Coverage
• Query Processing Overview
• Selections
• Projections
Readings: Chapter 12 and 14, Ramakrishnan & Gehrke, Database Systems
INFO20003 Database Systems 4© University of Melbourne 2018
Query processing overview
• Some database operations are EXPENSIVE
• DBMSs can greatly improve performance by being ‘smart’
– e.g., can speed up 1,000,000x over naïve approach
• Main weapons are:
1. clever implementation techniques for operators
2. exploiting ‘equivalencies’ of relational operators
3. using cost models to choose among alternatives
INFO20003 Database Systems 5© University of Melbourne 2018
Query processing workflow
Query Parser
Query Optimizer
Plan
Generator
Plan Cost
Estimator
Query Plan Evaluator
Catalog Manager
Usually there is a
heuristics-based
rewriting step before
the cost-based steps.
Schema Statistics
Select *
From Blah B
Where B.blah = “foo”
Query
Next week
INFO20003 Database Systems 6© University of Melbourne 2018
Relational Operations
• We will consider how to implement:
–Selection () Selects a subset of rows from relation
–Projection () Deletes unwanted columns from relation
–Join ( ) Allows us to combine two relations
• Operators can be then be composed creating query plans
INFO20003 Database Systems 7© University of Melbourne 2018
Query Processing
• Query Processing Overview
• Selections
• Projections
Readings: Chapter 14, Ramakrishnan & Gehrke, Database Systems
INFO20003 Database Systems 8© University of Melbourne 2018
Schema for Examples
• Sailors (S):
–Each tuple is 50 bytes long, 80 tuples per page, 500 pages
–N = NPages(S) = 500, pS=NTuplesPerPage(S) = 80
–NTuples(S) = 500*80 = 40000
• Reserves (R):
–Each tuple is 40 bytes long, 100 tuples per page, 1000 pages
–M= NPages(R) = 1000, pR=NTuplesPerPage(R) =100
–NTuples(R) = 100000
Sailors (sid: integer, sname: string, rating: integer, age: real)
Reserves (sid: integer, bid: integer, day: dates, rname: string)
INFO20003 Database Systems 9© University of Melbourne 2018
Simple Selections
• Of the form
• Example:
• The best way to perform a selection depends on:
1. available indexes/access paths
2. expected size of the result (number of tuples and/or
number of pages)
SELECT *
FROM Reserves R
WHERE R.rname LIKE ‘C%’
R attr valueop R. ( )
INFO20003 Database Systems 10© University of Melbourne 2018
Estimate result size (reduction factor)
• Size of result approximated as:
size of relation * ς(reduction factors)
• Reduction factor is usually called selectivity. It estimates
what portion of the relation will qualify for the given
predicate, i.e. satisfy the given condition.
This is estimated by the optimizer (will be taught next week)
E.g. 30% of records qualify, or 5% of records qualify
INFO20003 Database Systems 11© University of Melbourne 2018
Alternatives for Simple Selections
1. With no index, unsorted:
–Must scan the whole relation, i.e. perform Heap Scan
–Cost = Number of Pages of Relation, i.e. NPages(R)
–Example: Reserves cost(R)= 1000 IO (1000 pages)
2. With no index, but file is sorted:
–cost = binary search cost + number of pages containing results
–Cost = log2(NPages(R)) + (RF*NPages(R))
–Example: Reserves cost(R)= 10 I/O + (RF*NPages(R))
3. With an index on selection attribute:
–Use index to find qualifying data entries,
–Then retrieve corresponding data records
–Discussed next….
INFO20003 Database Systems 12© University of Melbourne 2018
Index Clustering: Review
Clustered vs. unclustered
Data entries
(Index File)
(Data file)
Data Records
Data entries
Data Records
CLUSTERED UNCLUSTERED
INFO20003 Database Systems 13© University of Melbourne 2018
Using an Index for Selections
• Cost depends on the number of qualifying tuples
• Clustering is important when calculating the total cost
• Steps to perform:
1. Find qualifying data entries:
– Go through the index: height typically small, 2-4 I/O in case of B+tree, 1.2 I/O
in case of hash index (negligible if many records retrieved)
– Once data entries are reached, go through data entries one by one and look
up corresponding data records (in the data file)
2. Retrieve data records (in the data file)
• Cost:
1. Clustered index:
Cost = (NPages(I) + NPages(R))*RF
2. Unclustered index:
Cost = (NPages(I) + NTuples(R))*RF
INFO20003 Database Systems 14© University of Melbourne 2018
Our example
• Example: Let’s say that 10% of Reserves tuples qualify, and let’s
say that index occupies 50 pages
• RF = 10% = 0.1, NPages(I) = 50, NPages(R) = 1000
• Cost:
1. Clustered index:
Cost = (NPages(I) + NPages(R))*RF
Cost = (50+ 1000)*0.1 = 105 (I/O)
2. Unclustered index:
Cost = (NPages(I) + NTuples(R))*RF
Cost = (50+ 100000)*0.1 = 10005 (I/O)
3. Heap Scan:
Cost = NPages(R) = 1000 (I/O)
Cheapest access path
INFO20003 Database Systems 15© University of Melbourne 2018
General Selection Conditions
• Typically queries have multiple predicates (conditions)
• Example: day<8/9/94 AND rname=‘Paul’ AND bid=5 AND sid=3
• A B-tree index matches (a combination of) predicates that
involve only attributes in a prefix of the search key
–Index on
(a)
–Index on
to answer b=3
–This implies that only reduction factors of predicates that are
part of the prefix will be used to determine the cost (they are
called matching predicates (or primary conjucts))
INFO20003 Database Systems 16© University of Melbourne 2018
Selections approach
1. Find the cheapest access path
– An index or file scan with the least estimated page I/O
2. Retrieve tuples using it
– Predicates that match this index reduce the number of
tuples retrieved (and impact the cost)
3. Apply the predicates that don’t match the index (if any) later
on
– These predicates are used to discard some retrieved tuples, but
do not affect number of tuples/pages fetched (nor the total cost)
– In this case selection over other predicates is said to be done
“on-the-fly”
INFO20003 Database Systems 17© University of Melbourne 2018
Cheapest Access Path: Example
• Example: day < 8/9/94 AND bid=5 AND sid=3
• A B+ tree index on day can be used;
–RF = RF(day)
–Then, bid=5 and sid=3 must be checked for each retrieved
tuple on the fly
• Similarly, a hash index on
–ς𝑅𝐹 = RF(bid)*RF(sid)
–Then, day<8/9/94 must be checked on the fly
• How about a B+tree on
• How about a B+tree on
• How about a Hash index on
INFO20003 Database Systems 18© University of Melbourne 2018
Query Processing
• Overview
• Selections
• Projections
Readings: Chapter 14, Ramakrishnan & Gehrke, Database Systems
INFO20003 Database Systems 19© University of Melbourne 2018
• Issue with projection is removing duplicates
• Projection can be done based on hashing or sorting
The Projection Operation
SELECT DISTINCT R.sid, R.bid
FROM Reserves R
INFO20003 Database Systems 20© University of Melbourne 2018
• Basic approach is to use sorting
–1. Scan R, extract only the needed attributes
–2. Sort the result set (typically using external merge sort)
–3. Remove adjacent duplicates
The Projection Operation
12,10
12,10
11,80
12,75
13,20
13,20
13,75
INFO20003 Database Systems 21© University of Melbourne 2018
External Merge Sort
• If data does not fit in memory do several passes
• Sort runs: Make each B pages sorted (called runs)
• Merge runs: Make multiple passes to merge runs
–Pass 2: Produce runs of length B(B-1) pages
–Pass 3: Produce runs of length B(B-1)2 pages
–…
–Pass P: Produce runs of length B(B-1)P pages
B Memory buffers
INPUT 1
INPUT B-1
OUTPUT
DiskDisk
INPUT 2
. . . . . .. . .
Readings: Chapter 13, Ramakrishnan & Gehrke, Database Systems
We will let you know
how many passes there are
INFO20003 Database Systems 22© University of Melbourne 2018
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records,
sorting on a single attribute (just showing the attribute value)
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
Pass 2
Main Memory
INFO20003 Database Systems 23© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
1,1
Pass 2
Main Memory
# buffer pages in memory B = 4, each page 2 records
External Merge Sort: Example
INFO20003 Database Systems 24© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
1,1 2,3
Pass 2
Main Memory
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 25© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
1,1 2,3
Pass 2
Main Memory
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 26© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
1 2,3
Pass 2
Main Memory
1
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 27© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
2,3
Pass 2
Main Memory
1
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 28© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
2,3 2,3
Pass 2
Main Memory
1
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 29© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
2,3
2,3 2,3
Pass 2
Main Memory
1
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 30© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
3
2,3 2,3
Pass 2
Main Memory
1,2
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 31© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
3
3 2,3
Pass 2
Main Memory
1,2
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 32© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
3
3 3
Pass 2
Main Memory
1,2
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 33© University of Melbourne 2018
Input file
4-page runs
3,4 6,2 9,4 8,7 5,6 3,1 9,2
2,3
5,66,7
4,4
8,9
1,1
2,3
6,1
6,9
8,2 3,4 5,5
5,5
6,8
2,3
6,3
3,4
Pass 1
3
3 3
Pass 2
Main Memory
1,2
External Merge Sort: Example
# buffer pages in memory B = 4, each page 2 records
INFO20003 Database Systems 34© University of Melbourne 2018
The Projection Operation Cost
• Sorting with external sort:
–1. Scan R, extract only the needed attributes
–2. Sort the result set using EXTERNAL SORT
–3. Remove adjacent duplicates
Cost = ReadTable +
WriteProjectedPages +
SortingCost +
ReadProjectedPages
WriteProjectedPages = NPages(R)* PF
PF: Projection Factor says how much are we projecting, ratio with respect to
all attributes (e.g. keeping ¼ of attributes, or 10% of all attributes)
SortingCost = 2*NumPasses*ReadProjectedPages
Read the entire table and keep only projected attributes
Write pages with projected attributes to disk
Sort pages with projected attributes with external sort
Read sorted projected pages to discard adjacent
duplicates
Every time we read and write
INFO20003 Database Systems 35© University of Melbourne 2018
Our example
• Example: Let’s say that we project ¼ of all attributes, and let’s say
that we have 20 pages in memory
• PF = 1/4 = 0.25, NPages(R) = 1000
• With 20 memory pages we can sort in 2 passes
Cost = ReadTable +
WriteProjectedPages +
SortingCost +
ReadProjectedPages
= 1000 + 0.25 * 1000 + 2*2*250 + 250 = 2500 (I/O)
INFO20003 Database Systems 37© University of Melbourne 2018
Projection based on Hashing
• Hashing-based projection
–1. Scan R, extract only the needed attributes
–2. Hash data into buckets
•Apply hash function h1 to choose one of B output buffers
–3. Remove adjacent duplicates from a bucket
•2 tuples from different partitions guaranteed to be distinct
INFO20003 Database Systems 38© University of Melbourne 2018
Projection Based on Hashing
B main memory buffersDisk
Original
Relation Buckets
2
INPUT
1
hash
function
h1
B-1
. . .
INFO20003 Database Systems 39© University of Melbourne 2018
1. Partition data into B partitions with h1 hash function
2. Load each partition, hash it with another hash function (h2) and eliminate
duplicates
Projection based on External Hashing
B main memory buffers DiskDisk
Original
Relation OUTPUT
2
INPUT
1
hash
function
h1
B-1
Partitions
1
2
B-1
. . .
INFO20003 Database Systems 40© University of Melbourne 2018
1. Partitioning phase:
–Read R using one input buffer
–For each tuple:
•Discard unwanted fields
•Apply hash function h1 to choose one of B-1 output buffers
–Result is B-1 partitions (of tuples with no unwanted fields)
•2 tuples from different partitions guaranteed to be distinct
2. Duplicate elimination phase:
–For each partition
•Read it and build an in-memory hash table
–using hash function h2 (<> h1) on all fields
•while discarding duplicates
–If partition does not fit in memory
•Apply hash-based projection algorithm recursively to this
partition (we will not do this…)
Projection based on External Hashing
INFO20003 Database Systems 41© University of Melbourne 2018
Our example:
Cost = ReadTable +
WriteProjectedPages +
ReadProjectedPages
= 1000 + 0.25 * 1000 + 250 = 1500 (I/O)
Cost of External Hashing
Cost = ReadTable +
WriteProjectedPages +
ReadProjectedPages
Read the entire table and project attributes
Write projected pages into corresponding partitions
Read partitions one by one, create another hash
table and discard duplicates within a bucket
INFO20003 Database Systems 42© University of Melbourne 2018
What’s examinable
• Understand the logic behind relational operators
• Learn alternatives for selections and projections (for now)
–Be able to calculate the cost of alternatives
• Important for Assignment 3 as well
INFO20003 Database Systems 43© University of Melbourne 2018
Next Lecture
– –
• Query Processing Part II
‒ Join alternatives