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INFO20003 Database Systems 1© University of Melbourne INFO20003 Database Systems Lecture 11 Query Processing Part I Week 6 Dr Renata Borovica-Gajic INFO20003 Database Systems 2© University of Melbourne 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 Coverage • Query Processing Overview • Selections • Projections Readings: Chapter 12 and 14, Ramakrishnan & Gehrke, Database Systems INFO20003 Database Systems 4© University of Melbourne 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 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 Relational Operations • We will consider how to implement: –Selection (σ) Selects a subset of rows from relation –Projection (π) 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 Query Processing • Query Processing Overview • Selections • Projections Readings: Chapter 14, Ramakrishnan & Gehrke, Database Systems INFO20003 Database Systems 8© University of Melbourne 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 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.BID > 20; σ R attr valueop R. ( ) INFO20003 Database Systems 10© University of Melbourne Estimate result size (reduction factor) • Size of result approximated as: size of relation * ∏ (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 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 Index Clustering: Review Clustered vs. unclustered Data entries (Index File) (Data file) Data Records Data entries Data RecordsCLUSTERED UNCLUSTERED INFO20003 Database Systems 13© University of Melbourne 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 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, NTuplesPerPage(R) = 100 • 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 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
matches predicates on: (a,b,c), (a,b) and (a) –Index on
matches a=5 AND b=3, but will not used 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 conjuncts)) INFO20003 Database Systems 16© University of Melbourne 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 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
could be used; –∏ = RF(bid)*RF(sid) –Then, day<8/9/94 must be checked on the fly • How about a B+tree on
? (Y/N) • How about a B+tree on
? (Y/N) • How about a Hash index on
? (Y/N) INFO20003 Database Systems 18© University of Melbourne Query Processing • Overview • Selections • Projections Readings: Chapter 14, Ramakrishnan & Gehrke, Database Systems INFO20003 Database Systems 19© University of Melbourne • 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 • 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 Projection Based on Hashing B main memory buffersDisk Original Relation Buckets 2INPUT 1 hashfunction h1 B-1 . . . INFO20003 Database Systems 39© University of Melbourne 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 2INPUT 1 hashfunction h1 B-1 Partitions 1 2 B-1 . . . INFO20003 Database Systems 40© University of Melbourne 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 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 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 Next Lecture - - • Query Processing Part II ‒ Join alternatives
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