SPRING 2004CENG 3521 Query Evaluation Chapters 12, 14.

SPRING 2004 CENG 352 1

Query Evaluation

Chapters 12, 14


Basic Steps in Query Processing1. Parsing and translation

2. Optimization

3. Evaluation


Basic Steps in Query Processing (Cont.)

• Parsing and translation– translate the query into its internal form. This is then

translated into relational algebra.– Parser checks syntax, verifies relations

• Evaluation– The query-execution engine takes a query-evaluation plan,

executes that plan, and returns the answers to the query.


Basic Steps: Optimization

• A relational algebra expression may have many equivalent expressions– E.g., balance2500(balance(account)) is equivalent to

balance(balance2500(account))

• Each relational algebra operation can be evaluated using one of several different algorithms– Correspondingly, a relational-algebra expression can be evaluated in

many ways.

• Annotated expression specifying detailed evaluation strategy is called an evaluation-plan.– e.g., can use an index on balance to find accounts with balance < 2500,

– or, can perform complete relation scan and discard accounts with balance 2500


Basic Steps: Optimization (Cont.)

• Query Optimization: Amongst all equivalent evaluation plans choose the one with lowest cost. – Cost is estimated using statistical information from the

database catalog• e.g. number of tuples in each relation, size of tuples, etc.

• We first study– How to measure query costs– Algorithms for evaluating relational algebra operations– How to combine algorithms for individual operations in order to

evaluate a complete expression

• Then– We study how to optimize queries, that is, how to find an

evaluation plan with lowest estimated cost


Measures of Query Cost• Cost is generally measured as total elapsed time for

answering query– Many factors contribute to time cost

• disk accesses, CPU, or even network communication

• Typically disk access is the predominant cost, and is also relatively easy to estimate. Measured by taking into account– Number of seeks * average-seek-cost

– Number of blocks read * average-block-transfer-cost

– Number of blocks written * average-block-transfer-cost


Some Common Techniques• Algorithms for evaluating relational operators use

some simple ideas extensively:– Indexing: Can use WHERE conditions to retrieve small

set of tuples (selections, joins)

– Iteration: Sometimes, faster to scan all tuples even if there is an index. (And sometimes, we can scan the data entries in an index instead of the table itself.)

– Partitioning: By using sorting or hashing, we can partition the input tuples and replace an expensive operation by similar operations on smaller inputs.


Statistics and Catalogs• Need information about the relations and indexes

involved. Catalogs typically contain at least:– # tuples (NTuples) and # pages (NPages) for each

relation.– # distinct key values (NKeys) and NPages for each index.– Index height, low/high key values (Low/High) for each

tree index.

• Catalogs updated periodically.– Updating whenever data changes is too expensive; lots of

approximation anyway, so slight inconsistency ok.

• More detailed information (e.g., histograms of the values in some field) are sometimes stored.


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.– Set-difference ( ) Tuples in reln. 1, but not in reln. 2.– Union ( ) Tuples in reln. 1 and in reln. 2.– Aggregation (SUM, MIN, etc.) and GROUP BY

• Since each op returns a relation, ops can be composed! After we cover the operations, we will discuss how to optimize queries formed by composing them.


Selection Operation• File scan – search algorithms that locate and retrieve

records that fulfill a selection condition.• Algorithm A1 (linear search). Scan each file block and

test all records to see whether they satisfy the selection condition.– Cost estimate (number of disk blocks scanned) = br

• br denotes number of blocks containing records from relation r

– If selection is on a key attribute, cost = (br /2) • stop on finding record

– Linear search can be applied regardless of • selection condition or• ordering of records in the file, or • availability of indices


Selection Operation (Cont.)• A2 (binary search). Applicable if selection is

an equality comparison on the attribute on which file is ordered. – Assume that the blocks of a relation are stored

contiguously

– Cost estimate (number of disk blocks to be scanned): log2(br) — cost of locating the first tuple by a binary

search on the blocks

• Plus number of blocks containing records that satisfy selection condition


Selections Using Indices• Index scan – search algorithms that use an index

– selection condition must be on search-key of index.• A3 (primary index on candidate key, equality). Retrieve a

single record that satisfies the corresponding equality condition – Cost = HTi + 1

• A4 (primary index on nonkey, equality) Retrieve multiple records. – Records will be on consecutive blocks – Cost = HTi + number of blocks containing retrieved records

• A5 (equality on search-key of secondary index).– Retrieve a single record if the search-key is a candidate key

• Cost = HTi + 1– Retrieve multiple records if search-key is not a candidate key

• Cost = HTi + number of records retrieved – Can be very expensive!

• each record may be on a different block – one block access for each retrieved record


Selections Involving Comparisons• Can implement selections of the form AV (r) or A V(r) by using

– a linear file scan or binary search,– or by using indices in the following ways:

• A6 (primary index, comparison). (Relation is sorted on A)– For A V(r) use index to find first tuple v and scan relation

sequentially from there– For AV (r) just scan relation sequentially till first tuple > v; do not use

index• A7 (secondary index, comparison).

– For A V(r) use index to find first index entry v and scan index sequentially from there, to find pointers to records.

– For AV (r) just scan leaf pages of index finding pointers to records, till first entry > v

– In either case, retrieve records that are pointed to• requires an I/O for each record• Linear file scan may be cheaper if many records are

to be fetched!


Implementation of Complex Selections• Conjunction: 1 2. . . n(r)

• A8 (conjunctive selection using one index). – Select a combination of i and algorithms A1 through A7 that results

in the least cost fori (r).– Test other conditions on tuple after fetching it into memory buffer.

• A9 (conjunctive selection using multiple-key index). – Use appropriate composite (multiple-key) index if available.

• A10 (conjunctive selection by intersection of identifiers). – Requires indices with record pointers. – Use corresponding index for each condition, and take intersection of

all the obtained sets of record pointers. – Then fetch records from file– If some conditions do not have appropriate indices, apply test in

memory.


Algorithms for Complex Selections• Disjunction:1 2 . . . n (r).

• A11 (disjunctive selection by union of identifiers). – Applicable if all conditions have available indices.

• Otherwise use linear scan.

– Use corresponding index for each condition, and take union of all the obtained sets of record pointers.

– Then fetch records from file

• Negation: (r)

– Use linear scan on file

– If very few records satisfy , and an index is applicable to • Find satisfying records using index and fetch from file


Schema for Examples

• Similar to old schema; rname added for variations.• Reserves:

– Each tuple is 40 bytes long, 100 tuples per page, 1000 pages.

• Sailors:– Each tuple is 50 bytes long, 80 tuples per page, 500

pages.

Sailors (sid: integer, sname: string, rating: integer, age: real)Reserves (sid: integer, bid: integer, day: dates, rname: string)


Simple Selections• With no index, unsorted: Must essentially scan the whole relation;

cost is 1000 I/Os (#pages in R).• With no index, sorted data: Utilize the sort order on rname by

doing a binary search to locate the first Joe. Cost is log2 1000 10 I/Os.

• With a B+ tree index on selection attribute: Use index to find qualifying data entries, then retrieve corresponding data records. Cost of finding the starting page is 2 or 3 I/Os; for a clustered index add one more I/O; for an unclustered index add one page per qualifying tuple.

• Hash index: 1 or 2 I/Os to retrieve the index pages. If 100 reservations by Joe then an additional 1-100 disk accesses depending how these records are distributed.

SELECT *FROM Reserves RWHERE R.rname = ‘Joe’


Using an Index for Selections

• Cost depends on #qualifying tuples, and clustering.• Assume we estimate roughly 10% of Reserves tuples

will be in result ( = 10,000 tuples, or 100 pages).– With a clustered index: cost is 100 I/Os + 1 or 2 disk

accesses for index.

– With an unclustered index: cost could be as high as 10,000 I/Os in the worst case. (might be cheaper to simply scan the entire relation)

SELECT *FROM Reserves RWHERE R.rname < ‘C%’


A Note on Complex Selections

• Selection conditions are first converted to conjunctive normal form (CNF):

(day<8/9/94 OR bid=5 OR sid=3 ) AND (rname=‘Paul’ OR bid=5 OR sid=3)

(day<8/9/94 AND rname=‘Paul’) OR bid=5 OR sid=3


Two Approaches to General Selections• Consider the selection condition:

day<8/9/94 AND bid=5 AND sid=3

• First approach: Find the most selective access path, retrieve tuples using it, and apply any remaining terms that don’t match the index:1. A B+ tree index on day can be used; then, bid=5 and sid=3 must be

checked for each retrieved tuple.2. Similarly, a hash index on <bid, sid> could be used; day<8/9/94 must

then be checked.

– Terms that match the index reduce the number of tuples retrieved; other terms are used to discard some retrieved tuples, but do not affect number of tuples/pages fetched.


Intersection of Rids• Second approach (if we have 2 or more matching indexes) :

– Get sets of rids of data records using each matching index.– Then intersect these sets of rids.– Retrieve the records and apply any remaining terms.

• For the given example condition:

– If we have a B+ tree index on day and an index on sid, we can retrieve rids of records satisfying day<8/9/94 using the first, rids of records satisfying sid=3 using the second, intersect, retrieve records and check bid=5.


The Projection Operation• To implement projection we have to do the

following:1. Remove unwanted attributes.2. Eliminate any duplicate tuples produced.

• The expensive part is removing duplicates.– SQL systems don’t remove duplicates unless the

keyword DISTINCT is specified in a query.

• There are two basic algorithms:1. Sorting Approach.2. Hashing Approach.


Approach based on sorting

• Modify Pass 1 of external sort to eliminate unwanted fields. If B buffer pages are available, runs of about 2B pages can be produced, but tuples in runs are smaller than input tuples. (Size ratio depends on # and size of fields that are dropped.)

• Modify merging passes to eliminate duplicates. Thus, number of result tuples smaller than input. (Difference depends on # of duplicates.)


Example

Cost: • In Pass 1, read original relation (1000 pages), write out

same number of smaller tuples.– Assume that each smaller tuple is 10 bytes long.– Thus cost is 250 pages.

• In merging passes, fewer tuples written out in each pass. – Assuming we have 20 buffer pages, the temporary relation can be

sorted in 2 passes.– In the first pass 250 pages are written out as 7 runs about 40 pages

long.– In the second pass we read the runs at a cost of 250 I/Os and

merge them.

• The total cost is 1500 I/Os.

SELECT DISTINCT R.sid, R.bidFROM Reserves R


Projection Based on Hashing• 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.

• Duplicate elimination phase: For each partition, read it and build an in-memory hash table, using hash fn h2 ( h1) on all fields, while discarding duplicates.– If partition does not fit in memory, can apply hash-based projection

algorithm recursively to this partition.

• Cost: For partitioning, read R, write out each tuple, but with fewer fields. This is read in next phase.– In our projection example this cost is 1000 + 2 * 250 = 1500 I/Os.


Discussion of Projection

• Sort-based approach is the standard; better handling of duplicate elimination and result is sorted.

• If an index on the relation contains all wanted attributes in its search key, can do index-only scan.– Apply projection techniques to data entries (much smaller!)

• If an ordered (i.e., tree) index contains all wanted attributes as prefix of search key, can do even better:– Retrieve data entries in order (index-only scan), discard unwanted

fields, compare adjacent tuples to check for duplicates.

SPRING 2004CENG 3521 Query Evaluation Chapters 12, 14.

Documents

measures of query cost

evaluation slide

queryevaluation plan

query optimization

balance balance

query evaluation chapters

transfercost slide

query processing