Relational Query Optimization

1
Relational Query Optimization

• Chapter 15 Ramakrishnan Gehrke
• (Sections 15.1-15.6)

2
What you will learn from this lecture

• Cost-based query optimization (System R)
• Plan space considered
• How cost is estimated
• Selectivity estimation using histograms.

3
Highlights of System R Optimizer

• Impact
• Most widely used currently works well for lt 10
  joins.
• Cost estimation Approximations used.
• Statistics, maintained in system catalogs, used
  to estimate cost of operations and result sizes.
• System R considers combination of CPU and I/O
  costs.
• Plan Space Too large, must be pruned.
• Only the space of left-deep plans is considered.
• Left-deep plans allow output of each operator to
  be pipelined into the next operator without
  storing it in a temporary relation.
• Cartesian products avoided.

4
Schema for Examples
Ratings(uid integer, sid integer, time date,
rating integer) Songs(sid integer, sname
string, genre string, year date)

• Ratings
• Each tuple is 40 bytes long, 100 tuples per
  page, 1000 pages.
• Songs
• Each tuple is 50 bytes long, 80 tuples per page,
  500 pages.

5
Query Blocks Units of Optimization
SELECT S.sname FROM Songs S WHERE S.sid IN
(SELECT R.sid FROM Ratings R
WHERE R.ratinggt7 AND R.time gt S.year)

• A SQL query is parsed into a collection of query
  blocks, and these are optimized one block at a
  time.
• Nested blocks are usually treated as calls to a
  subroutine, made once per outer tuple. (This is
  an over-simplification, but serves for now.)

Outer block
Nested block

• For each block, the plans considered are
• All available access methods, for each reln in
  FROM clause.
• All left-deep join trees (i.e., all ways to
  join the relations one-at-a-time, with the inner
  reln in the FROM clause, considering all reln
  permutations and join methods.)
• i.e., we pipeline what is being computed.

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Cost Estimation (Recap 1)

• For each plan considered, must estimate cost
• Must estimate cost of each operation in plan
  tree.
• Depends on input cardinalities.
• Weve already discussed how to estimate the cost
  of operations (sequential scan, index scan,
  joins, etc.)
• Must estimate size of result for each operation
  in tree!
• Use information about the input relations.
• For selections and joins, assume independence of
  predicates.
• Well discuss the System R cost estimation
  approach.
• Inexact, but works ok in practice.
• More sophisticated techniques known now.

7
Statistics and Catalogs (Recap 2)

• 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.

8
Size Estimation and Reduction Factors (Recap 3)
SELECT attribute list FROM relation list WHERE
term1 AND ... AND termk

• Consider a query block
• Maximum tuples in result is the product of the
  cardinalities of relations in the FROM clause.
• Reduction factor (RF) associated with each term
  reflects the impact of the term in reducing
  result size. Result cardinality Max tuples
  product of all RFs.
• Implicit assumption that terms are independent!
• Term colvalue has RF 1/NKeys(I), given index I
  on col
• Term col1col2 has RF 1/MAX(NKeys(I1), NKeys(I2))
• Term colgtvalue has RF (High(I)-value)/(High(I)-Low
  (I) 1)

9
Histograms

• Example Songs(sid, sname, genre, year). Suppose
  recorded years ?1994,2008 and there are 45
  tuples.
• Suppose the year values occur in the following
  pattern

year frequency
94 2
95 3
96 3
97 1
98 2
99 1
00 3
01 8
02 4
03 2
04 0
05 1
06 2
07 4
08 9

• For year gt 07, what does our
• uniformity assumption tell us?
• RF 1/15, so 45 x 1/15 3
• tuples qualify. But in reality, 9
• tuples do!

frequency table

• Can we improve estimation
• by maintaining small amount of
• additional info.?
• Idea repeat what we did for the
• interval 94,08 to subintervals!

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Histogram Generation Use
Can slice up the frequency table using different
schemes. Each scheme leads to a type of
histogram.
Suppose we want to create a Histogram with 5
buckets.
age frequency
94 2
95 3
96 3
97 1
98 2
99 1
00 3
01 8
02 4
03 2
04 0
05 1
06 2
07 4
08 9
Bucket (Bin)
Split up the range into subintervals of equal
width.
Make uniformity assumption inside each bucket.
e.g., 95 is estimated to occur 8/3 2.67
times (approx).
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Equiwidth Histograms

• Equiwidth split 94,08 into 5 buckets (i.e.,
  bins) of equal width sum up frequencies
  bin-wide.

Frequency ? Ntuples. Note this is already much
closer to actual distribution. Whats our
current estimate for year gt 07? 1/3 x 15 5
tuples. Compare with actual data. Compare with
uniformity assumption. Uniformity assumption
histogram with 1 bucket!
bucket frequency
94-96 8
97-99 4
00-02 15
03-05 3
06-08 15
Equiwidth histogram With 5 buckets (bins).
12
Equidepth Histogram Generation
Can slice up the frequency table using different
schemes. Each scheme leads to a type of
histogram.
Want to make a 5 bucket Histogram again.
year frequency
94 2
95 3
96 3
97 1
98 2
99 1
00 3
01 8
02 4
03 2
04 0
05 1
06 2
07 4
08 9
Split up the range into subintervals such that
each contains about the same total
occurrences, i.e., 45/5 9 in our example.
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Equidepth Histograms

• Equidepth

• Estimate for year gt 07 is 9.
• Even better than equiwidth HG.
• Happens to be exact in this case.
• What is the maximum error in
• estimate for EW HG and ED HG
• in the present example?
• What are the main differences
• between equiwidth and
• equidepth HGs?

bucket frequency
94-97 9
98-00 6
01-02 12
03-07 9
08 9
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A slightly different slicing

• Some of the bars have been moved. The slicing
  cannot be perfect in general.

year frequency
94 2
95 3
96 3
97 1
98 2
99 1
00 3
01 8
02 4
03 2
04 0
05 1
06 2
07 4
08 9
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Equidepth Histograms

• Equidepth

• How does this HG compare with
• the previous ED HG?

bucket frequency
94-98 11
99-01 12
02-05 7
06-07 6
08 9
First ED HG.
bucket frequency
94-97 9
98-00 6
01-02 12
03-07 9
08 9
Which HG would you put your money on?
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Histogram Use

• How do you calculate RF using a histogram?
• Straightforward for Attrval.
• What about Attr lt val? (we use 1st EDHG below.)
• E.g. year gt 98
• tuplesuniform 10/15 x 45 30. RFuniform
  30/45 10/15 2/3.
• tuplesewhg 1 x 4/3 33 34.33. RFewhg
  34.33/45.
• tuplesedhg 2 x 6/3 30 34. RFedhg 34/45.
• Actual (tuples with year gt 98) 34. Actual RF
  34/45.
• In general, can define optimal histograms (given
  buckets) to minimize expected error.
• Interesting algorithmic problem with lots of
  research literature.
• HGs useful for selectivity estimation for
  multi-dimensional queries.

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Relational Algebra Equivalences

• Allow us to choose different join orders and to
  push selections and projections ahead of joins.
• Selections
  (Cascade)

(Commute)

• Projections

(Cascade)
when a1 ? a2 ? ? an.
(Associative)

• Joins

R (S T) (R S) T
(Commutative)
(R S) (S R)
R (S T) (T R) S

• Show that

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More Equivalences

• A projection commutes with a selection that only
  uses attributes retained by the projection.
• Selection between attributes of the two arguments
  of a cross-product converts cross-product to a
  join.
• A selection on just attributes of R commutes with
  R S. (i.e., (R S)
  (R) S )
• Similarly, if a projection follows a join R
  S, we can push it by retaining only attributes
  of R (and S) that are needed for the join or are
  kept by the projection.

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Some more equivalences

• Join vs. Group-by
• ?_R.sid max(time)(Songs Ratings) ?
• ?_R.sid, max(time) ( ?_R.sid
  max(time)(Ratings) Songs)
• ? ?_R.sid max(time)(Ratings).
• provided the RIC Ratings.sid ? Songs.sid holds
  (needed only for second equivalence).
• Does the equivalence hold for other aggregate
  functions e.g., what about avg(rating)? or
  count(time)?

R.sidS.sid
R.sidS.sid
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One more on Group-By and Join.

• ??_A agg(D)r(A,B) s(B,C,D) ??
• ?_A,agg(D)(r(A,B) ? ?_Bagg(D)(s(B,C,D))).
  
• Does this equivalence always hold? What
  constraints, if any, are needed for it to hold?
  If the constraint is relaxed, what then?

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Enumeration of Alternative Plans

• There are two main cases
• Single-relation plans
• Multiple-relation plans
• For queries over a single relation, queries
  consist of a combination of selects, projects,
  and aggregate ops
• Each available access path (file i.e., table
  scan/index scan/index probe) is considered, and
  the one with the least estimated cost is chosen.
• The different operations are essentially carried
  out together (e.g., if an index is used for a
  selection, projection is done for each retrieved
  tuple, and the resulting tuples are pipelined
  into the aggregate computation).

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Cost Estimates for Single-Relation Plans

• Index I on primary key matches equality
  selection
• Cost is Height(I)1 for a B tree, about 1.2 for
  hash index.
• Additional cost if alternative 2 or 3 is used.
• Clustered index I matching one or more selects
• (NPages(I)NPages(R)) product of RFs of
  matching selects. this is mainly for secondary
  keys.
• Non-clustered index I matching one or more
  selects
• (NPages(I)NTuples(R)) product of RFs of
  matching selects.
• Sequential scan of file
• NPages(R).
• Note Typically, no duplicate elimination on
  projections! (Exception Done on answers if user
  says DISTINCT.)

secondary key
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Example
SELECT S.sid FROM Songs S WHERE
S.genrefriendship

• If we have an index on genre
• (1/NKeys(I)) NTuples(R) (1/10) 40000 tuples
  retrieved. assuming, distinct genres 10.
• Suppose index size Npages(I) 50 pages.
• Clustered index (1/NKeys(I))
  (NPages(I)NPages(R)) (1/10) (50500) pages
  are retrieved. (This is the cost.)
• Unclustered index (1/NKeys(I))
  (NPages(I)NTuples(R)) (1/10) (5040000)
  pages are retrieved.
• Doing a file scan
• We retrieve all file pages (500).

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An Example with Aggregation

• SELECT year, COUNT()
• FROM Songs
• WHERE yeargt2000 AND genrefriendship
• GROUP BY year
• HAVING COUNT DISTINCT (sname) gt 2
• PROJ_year, count(sname) (
• HAVING_count disinct (sname)gt2(
• GROUP BY_year count distinct(sname) (
• PROJ_year,sname(
  SEL_yeargt2000genrefriendship
• (Songs))))).

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Plan w/o indexes

• Scan Songs file apply SEL/PROJ on the fly.
• Write out result into temp.
• Sort by year (for GROUP BY) aggregate during
  merge do HAVING on the fly.
• Whats the cost?
• 500 I/Os for scan
• RF for yeargt2000 is 0.5 (approx.) RF for
  genrefriendship is 0.1 (used default value).
• RF 0.5 x 0.1 PROJ reduces size by 0.5 (if all
  fields have the same size). Temp. size 500 x
  0.5 x 0.1 x 0.5 13 pages (approx), so 13 I/Os
  for writing it.
• 3 x 13 for sorting it. can be less if we have
  sufficient buffer space.
• Total 552 I/Os.

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Plans w/ indexes.

• Pick one (whatever QOzer estimates is best) index
  among those available probe the index and verify
  unmatched selection conditions as pages are read
  in.
• Use multiple indexes do a rid intersection then
  fetch resulting pages.
• If GROUP-BY attrs are a prefix of a Btree index,
  use the index to retrieve tuples in sort order
  (simplifies GROUP-BY). Rest of the ops done on
  each tuple, then each group.
• Suppose there is a dense index that includes all
  attrs mentioned in query. Then only need to scan
  the indexs data entries. If index matches some
  of the SEL conditions, even better. If index is
  Btree and GROUP-BY attrs form a prefix, can
  avoid sorting!
• Does it matter if this index is clustered or not?
  
• (what is a dense index? and what is a sparse
  index?)
• Note we can keep attr-values as data entries
  even if they are not indexed! E.g., index on
  genre, but data entry -- (genre, year). Why
  /When would this help?

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Plans w/ indexes Example.

• Revisit query data entries (key, rid) pairs.
  Btree index on year, hash index on genre, Btree
  index on (year, genre, sname).
• 1. Fetch data entries using hash index on genre
  then records then apply remaining ops.
• Does clustering make a difference?
• 2. Use both Btree index on year and hash index
  on genre intersect rids and fetch data.
• 3. Use only Btree index on year (in sorted
  order). Helps greatly for GROUP-BY.
• 4. Scan only the Btree index on (year, genre,
  sname) fetch data entries with yeargt2000 and in
  sorted order. Then apply remaining ops. (index
  could be unclustered.) index must be dense for
  this to be work!

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Queries Over Multiple Relations

• Fundamental decision in System R only left-deep
  join trees are considered.
• As the number of joins increases, the number of
  alternative plans grows rapidly we need to
  restrict the search space.
• Left-deep trees allow us to generate all fully
  pipelined plans.
• Intermediate results not written to temporary
  files.
• Not all left-deep plans are fully pipelined
  (e.g., Sort-Merge join).

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Enumeration of Left-Deep Plans

• Left-deep plans differ only in the order of
  relations, the access method for each relation,
  and the join method for each join.
• Enumerated using N passes (if N relations
  joined)
• Pass 1 Find best 1-relation plan for each
  relation.
• Pass 2 Find best way to join result of each
  1-relation plan (as outer) to another relation.
  (All 2-relation plans.)
• Pass N Find best way to join result of a
  (N-1)-relation plan (as outer) to the N-th
  relation. (All N-relation plans.)
• For each subset of relations, retain only
• Cheapest plan overall, plus
• Cheapest plan for each interesting order of the
  tuples.
• Where does this help?

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Enumeration of Plans (Contd.)

• ORDER BY, GROUP BY, aggregates etc. handled as a
  final step, using either an interestingly
  ordered plan or an additional sorting operator.
  Of course, hashing is an option too.
• An N-1 way plan is not combined with an
  additional relation unless there is a join
  condition between them, unless all predicates in
  WHERE have been used up.
• i.e., avoid Cartesian products if possible.
• In spite of pruning plan space, this approach is
  still exponential in the of tables.

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Example
Songs B tree on year. Hash on
sid Ratings B tree on uid

• Pass1
• Songs B tree matches yeargt2000, and is
• probably cheapest. However, if this selection
• is expected to retrieve a lot of tuples, and
• index is unclustered, file scan may be cheaper.
• Still, B tree plan kept (because tuples are in
• year order).
• Ratings B tree on uid matches uid100 perhaps
  cheapest.

• Pass 2
• We consider each plan retained from Pass 1 as
  the outer, and consider how to join it with the
  (only) other relation.
• e.g., Ratings as outer Hash index can be used
  to get Songs tuples
• that satisfy sid outer tuples sid value.
  What other plans might be considered?

32
Example (contd.)

• If we have gt 2 relations to join, for pass 2 ?
  pass 3 (indeed, for pass k ? pass k1), the same
  reasoning applies.
• Use same criteria as before for deciding which
  plans to keep for the join of each subset of k
  relations. (k1)th rel inner for join plan.
• For each k relations Join chosen (k1)th
  rel, consider all possible strategies and choose
  cheapest plans and/or plans w/ interesting tuple
  orders.

33
Nested Queries
SELECT S.sname FROM Songs S WHERE S.sid IN
(SELECT R.sid FROM Ratings R
WHERE R.ratinggt7 AND R.time gt S.year)

• Nested block is optimized independently, with the
  outer tuple considered as providing a selection
  condition.
• Outer block is optimized with the cost of
  calling a nested block computation taken into
  account.
• Implicit ordering of these blocks means that some
  good strategies are not considered. The
  non-nested version of the query is typically
  optimized better.
• Not all queries can be flattened this way!

Nested block to optimize SELECT R.sid FROM
Ratings R WHERE R.ratinggt7 AND R.timegt
outer value
Equivalent non-nested query SELECT S.sname FROM
Songs S, Ratings R WHERE S.sidR.sid AND
R.ratinggt7 AND R.time gt S.year
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Summary

• Query optimization is an important task in a
  relational DBMS.
• Must understand optimization in order to
  understand the performance impact of a given
  database design (relations, indexes) on a
  workload (set of queries).
• Two parts to optimizing a query
• Consider a set of alternative plans.
• Must prune search space typically, left-deep
  plans only.
• Must estimate cost of each plan that is
  considered.
• Must estimate size of result and cost for each
  plan node.
• Key issues Statistics, indexes, operator
  implementations.

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Summary (Contd.)

• Single-relation queries
• All access paths considered, cheapest is chosen.
• Issues Selections that match index, whether
  index key has all needed fields and/or provides
  tuples in a desired order. Is the index
  dense/sparse, clustered/unclustered. (Not all of
  this matters all the time.)
• Multiple-relation queries
• All single-relation plans are first enumerated.
• Selections/projections considered as early as
  possible.
• Next, for each 1-relation plan, all ways of
  joining another relation (as inner) are
  considered.
• Next, for each 2-relation plan that is
  retained, all ways of joining another relation
  (as inner) are considered, etc.
• At each level, for each subset of relations, only
  best plan for each interesting order of tuples is
  retained, in addition to the global cheapest
  plan (which may or may not come with an
  interesting order).