MDA104 Introduction to Databases
5. Query Processing
Vlastislav Dohnal
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Query Processing
Overview
Evaluation of Expressions
Measures of Query Cost
Evaluation algorithms
Sorting
Join Operation
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Basic Steps in Query Processing
1. Parsing and translation
2. Optimization
3. Evaluation
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Basic Steps in Query Processing (Cont.)
Parsing and translation
Translate the SQL query into its internal form.
This is then translated into relational algebra.
Parser checks syntax, verifies relations
Optimization
Generate a query-evaluation plan and choose algorithms
for evaluating individual operations
Evaluation
The query-execution engine takes a query-evaluation plan,
executes that plan, and returns the answers to the query.
Basic Steps in Query Processing (Cont.)
Example of query:
List salary of all instructors that earn less than $75,000.
SQL query
SELECT salary FROM instructor WHERE salary < 75000
Conversion to rel. algebra
salary(salary75000(instructor))
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Basic Steps: Optimization
A relational-algebra expression may have many equivalent
expressions:
salary(salary75000(instructor))
salary75000(salary(instructor))
For a relational-algebra expression, an expression tree is
created
instructor
salary75000
salary
instructor
salary
salary75000
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Basic Steps: Optimization (Cont.)
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 execution-plan or evaluation-plan.
E.g., to find instructors with salary < 75000
use an index on salary, or
perform complete relation scan and discard instructors with
salary  75000
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Basic Steps: Optimization (Cont.)
Example of an evaluation-plan
instructor
salary75000
salary
Use index scan with index on salary
Eliminate duplicates by sorting
Use pipelining
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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.
There is a huge number of possible evaluation plans
Optimization uses some heuristics
1. Perform selection early
reduce the number of tuples (by using an index, e.g.)
2. Perform projection early
reduce the number of attributes
3. Perform most restrictive operations early
such as join and selection.
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Evaluation of Expressions
Alternatives for evaluating an entire expression tree
Materialization
Evaluate one operation at a time, starting at the lowest-level.
Use intermediate results materialized into temporary relations
to evaluate next-level operations.
Pipelining
pass on tuples to parent operations even as an operation is
being executed
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Evaluation of Expressions (Cont.)
Materialized evaluation
Compute σbuilding=`Watson’(department) and store it
Then read from stored intermediate result and compute its
join with instructor, store it
Finally read it and compute the projection on name and
output it.
This step can be conveniently evaluated using pipelining on
join result.
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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-read-cost
Number of blocks written * average-block-write-cost
Cost to write a block is greater than cost to read a block
Data is read back after being written to ensure that the write
was successful
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Measures of Query Cost (Cont.)
For simplicity we just use the number of block transfers from
disk and the number of seeks as the cost measures
tT – time to transfer one block
tS – time for one seek
Cost for b block transfers plus S seeks
b * tT + S * tS
We ignore CPU costs for simplicity
Real systems do take CPU cost into account
We do not include cost to writing output to disk in our cost
formulae
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Measures of Query Cost (Cont.)
Several algorithms can reduce disk I/O by using extra buffer
space
Amount of real memory available to buffer depends on
other concurrent queries and OS processes, known only
during execution
We often use worst case estimates, assuming only the
minimum amount of memory needed for the operation is
available
Required data may be buffer resident already, avoiding disk I/O
But hard to take into account for cost estimation
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Selection Operation
File scan (table / sequential scan) – no index structure is necessary
Scan each file block and test all records to see whether they satisfy
the selection condition.
Cost estimate = br block transfers + 1 seek
br denotes number of blocks containing records from relation r
If selection is on a key attribute, can stop on finding matching record
cost = (br /2) block transfers + 1 seek
Linear search can be applied regardless of
selection condition or
ordering of records in the file, or
availability of indices
Note: binary search generally does not make sense since data is not
stored consecutively
except when there is an index available,
and binary search requires more seeks than index search
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Selections Using Indices
Index scan – search algorithms that use an index
selection condition must be on search-key of index
Now, assume the sequential file is ordered by this key:
Algorithm for primary index & equality on primary key
Retrieve a single record that satisfies the corresponding
equality condition
Cost = (hi + 1) * (tS + tT)
hi – height of index i (for hashing hi =1)
+1 – for reading the actual record
Algorithm for primary index & equality on non-primary key
Retrieve multiple records.
Records will be on consecutive blocks
Let b = number of blocks containing all n matching records
Cost = hi * (tS + tT) + tS + tT * b
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Selections Using Indices
Algorithm for secondary index & equality on non-primary key
Sequential file is not ordered by this search key!
Retrieve a single record if the search-key is a candidate key
Cost = (hi + 1) * (tS + tT)
Retrieve multiple records if search-key is not a candidate key
Each of n matching records may be on a different block.
Cost = (hi + n) * (tS + tT)
Can be very expensive!
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Sorting Relations
We may build an index on the relation, and then use the index
to read the relation in the sorted order.
May lead to one disk block access for each tuple.
Use a sorting algorithm
For relations that fit in memory, techniques like quick-sort
can be used.
For relations that don’t fit in memory, external sort-merge
is a good choice.
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External Sort-Merge
Let M denote memory size (in pages/blocks):
1. Create sorted runs. Let i be 0 initially.
Repeatedly do the following till the end of the relation:
(a) Read M blocks of relation into memory
(b) Sort the in-memory blocks
(c) Write sorted data to run Ri ; increment i.
Let the final value of i be N
2. Merge the runs. (next slide)
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External Sort-Merge (Cont.)
2. Merge the runs (N-way merge).
We assume (for now) that N < M.
1. Use N blocks of memory to buffer input runs, and 1 block to
buffer output.
2. Read the first block of each run into its buffer page
3. repeat
1. Select the first record (in sort order) among all buffer pages
2. Write the record to the output buffer.
If the output buffer is full write it to disk.
3. Delete the record from its input buffer page.
If the buffer page becomes empty
then read the next block (if any) of the run into the buffer.
4. until all input buffer pages are empty.
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External Sort-Merge (Cont.)
If N  M, several merge passes are required.
In each pass, continuous groups of M - 1 runs are merged.
A pass reduces the number of runs by a factor of M -1, and
creates runs longer by the same factor.
E.g. If M=11, and there are 90 runs, one pass reduces the
number of runs to 9, each 10 times the size of the initial runs
Repeated passes are performed till all runs have been
merged into one.
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Example: External Sorting Using Sort-Merge
Available
memory
M=4
blocks.
One record
per block
g 24
a 19
d 31
c 33
b 14
e 16
r 16
d 21
m 3
p 2
d 7
a 14
initial
relation
a 19
c 33
d 31
g 24
b 14
d 21
e 16
r 16
a 14
d 7
m 3
p 2
runscreate
runs
(m-1)-way
merging
a 14
a 19
b 14
c 33
d 7
d 21
d 31
e 16
g 24
m 3
p 2
r 16
sorted
output
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External Sort-Merge (Cont.)
Cost analysis:
Total number of merge passes required: logM–1(br /M).
Block transfers for initial run creation as well as in each
pass is 2br
for final pass, we don’t count write cost
we ignore final write cost for all operations since the
output of an operation may be sent to the parent
operation without being written to disk
Thus total number of block transfers for external sorting:
br (2 logM–1(br / M) + 1)
Seeks: next slide
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Join Operation
Several different algorithms to implement joins
Nested-loop join
Block nested-loop join
Improved nested-loop join by reading records in blocks
Indexed nested-loop join
Improved by using an index to look up equal records
Merge-join
Hash-join
Choice based on cost estimate
For each of the variants a cost estimation can be stated.
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Nested-Loop Join
To compute the join r  s
for each tuple tr in r do begin
for each tuple ts in s do begin
test pair (tr ,ts) to see
if they satisfy the equality on shared attributes
if they do, add tr • ts to the result.
end
end
r is called the outer relation and s the inner relation of the join.
Requires no indices and can be used with any kind of join
condition.
Expensive since it examines every pair of tuples in the two
relations.
Cost = nr * (tS + tT) * (ns * (tS + tT) )
where nr = number of tuples in r
Nested-Loop Join (Cont.)
In the worst case, if there is enough memory only to hold one block of each
relation, the estimated cost is
nr  bs + br block transfers, plus
nr + br seeks
Example on student and takes
student (the smaller one) as the outer relation:
5000  400 + 100 = 2,000,100 block transfers,
5000 + 100 = 5,100 seeks
takes (the larger one) as the outer relation
10000  100 + 400 = 1,000,400 block transfers and 10,400 seeks
If the smaller relation fits entirely in memory, use that as the inner relation.
Reduces cost to br + bs block transfers and 2 seeks
Example: student fits entirely in memory
the cost estimate is 500 block transfers.
Block nested-loops algorithm (next slide) is preferable.
nstudent=5,000
bstudent=100
ntakes=10,000
btakes= 400
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Block Nested-Loop Join
Variant of nested-loop join in which every block of inner
relation is paired with every block of outer relation.
for each block Br of r do begin
for each block Bs of s do begin
for each tuple tr in Br do begin
for each tuple ts in Bs do begin
Check if (tr,ts) satisfy the join condition
if they do, add tr • ts to the result.
end
end
end
end
Cost: br * (1+bs) blocks; br * (1+1) seeks
For student (outer) and takes (inner):
100 + 100 * 400 = 40,100 block transfers
100 + 100 seeks
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nstudent=5,000
bstudent=100
ntakes=10,000
btakes= 400
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Merge-Join
1. Sort both relations on their join attributes
If not already sorted.
2. Merge the sorted relations to join them
Join step is similar to the merge stage of the sort-merge
algorithm.
Main difference is handling of
duplicate values in join attribute
Every pair with same value
on join attribute must be
matched
Merge-Join (Cont.)
Can be used only for equi-joins and natural joins
Each block needs to be read only once
assuming all tuples for any given value of the join attributes fit in
memory
Thus the cost of merge join is:
br + bs block transfers, and
max. 2*br / bb + 1 seeks
Assuming we read r in runs of bb blocks
+ the cost of sorting if relations are unsorted.
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Hash-Join
A hash function h is used to partition tuples of both relations
JoinAttrs are the common attributes of r and s used in r  s
h maps JoinAttrs values to {0, 1, ..., n}
r0 , r1 , . . . , rn denote buckets of r
Each tuple tr  r is put in bucket ri
where i = h(tr [JoinAttrs]).
s0 , s1 , . . . , sn denotes buckets of s
Each tuple ts  s is put in bucket si,
where i = h(ts [JoinAttrs]).
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Hash-Join (Cont.)
buckets ri of r buckets si of s
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Hash-Join (Cont.)
Tuples in ri need only to be compared with tuples in si
Need not be compared with s tuples in any other bucket, since:
a tuple of r and a tuple of s that satisfy the join condition will
have the same value for the join attributes.
If that value is hashed to some value i, the tuple of r has to be
in ri and the tuple of s in si.
Cost of hash join is 3(br + bs) block transfers
3*(100+400) for student  takes
nstudent=5,000
bstudent=100
ntakes=10,000
btakes= 400
Summary – Takeaways
Steps in query processing
Idea of query optimization
expression transformations (in parse tree)
selection of algorithm to evaluate an operator
index vs table scan
Algorithms
Sorting large relations (exceeding RAM allocation)
Joining relations
nested loops, merge join, hash join
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