Advanced Databases: Lecture 8 Query Optimization (III) 1 Query Optimization Advanced Databases By...

Advanced Databases: Lecture 8 Query Optimization (III)1

Query Optimization

Advanced DatabasesBy

Dr. Akhtar Ali


Overview of Query Optimization

• Logical Optimization: Using some transformation rules and algebraic equivalences– To choose different join orders

• R ⋈ S S ⋈ R (Commutative)• (R ⋈ S) ⋈ T R ⋈ (S ⋈ T) (Associative)

– To push selections and projections bellow of joins• (attr1 op value (R) ) ⋈ S attr1 op value (R ⋈ S ) attr1, attr2, attr3 ((attr1, attr2 (R) ) ⋈ S ) attr1, attr2,attr3 (R ⋈ S )

• Physical Optimization: For a given R.A expression– There could be several different plans possible using different

implementation of the R.A operators– Calculate the cost of these different plans and choose the best– Ideally, we want the best plan, but– Practically, we should avoid worst plans


Pushing Selection & Projection below Join

• A join is quite expensive operation.– The cost can be reduced by reducing the sizes of the input

relations.

• The sizes of the Input Relations can be reduced by applying:– Selection: restricting the input relation.

– Projection: reducing the number of columns.

• Usually Selection reduces the size of the input relation more than Projection.

• Projection before the Join should be done quite carefully as the cost of Projection could increase the overall cost if it does not reduce by a good factor the size of the input.


Estimating the Evaluation Cost of a Plan

• For each plan, we should be able to estimate the overall cost.

• For each node in the query tree, we estimate the cost of performing the corresponding operation;

• For each node in the query tree, we estimate the size of the result, which is used by the operation in the parent node;

• In order to correctly estimate the cost of each operation and the size of its result, the optimizer uses certain statistical information maintained by the DBMS.


Statistics Maintained by a DBMS

• Cardinality– The number of tuples for each relation.

• Size– The number of pages for each relation.

• Index Cardinality– The number of distinct key values for each index.

• Index Size– The number of pages for each index.

• Index Height– The number of non-leaf levels for each tree index.

• Index Range– The minimum present key value and the maximum present key

value for each index.


A motivating example

SELECT S.sname

FROM Reserves R, Sailors S

Where R.sid = S.sid

AND R.bid = 100

AND S.rating > 5

• Cost = 500 + 500 * 1000 I/Os (Using nested loops – page-oriented)– 500,500 I/Os

– This is not an efficient plan.

• We could have pushed selections down before join, no index is used.

• Goal of Optimization:– To find more efficient plans that compute

the same answer.

Plan

π sname

sid=sid

SailorsReserves

σ bid = 100 and rating > 5

(on-the-fly)

(on-the-fly)

(SNL join page-oriented)

RA Query Tree

π sname

sid=sid

SailorsReserves

σ bid = 100 and rating > 5


Optimization: Alternative 1 (no index)

• Push Selects before join

• Assuming that there are 1000 tuples in Reserves with bid=100 and 20000 tuples in Sailors with rating > 5. So Cost of Selections:

– Scan Reserves (1000 pages) and write the selected 1000 tuples to temp relation T1 (10 pages), so in total 1010 I/Os

– Scan Sailors (500 pages) and write the selected 20000 tuples to temp relation T2 (250 pages), so in total 750 I/Os

– Total cost = 1010 + 750 = 1760 I/Os so far

• Using SNL the cost = 10 + 10 * 250 = 2510 I/Os• Total cost = 1760 + 2510 = 4270 I/Os about 117 times less than the cost of

the initial plan i.e. 500,500 I/Os

Scan, write toTemp T2

Plan

π sname

sid=sid

SailorsReserves

(on-the-fly)


σ rating > 5σ bid = 100Scan, write to

Temp T1


Optimization: Alternative 2 (uses indexes)

• The same RA as alternative 1

• Using a clustered Hash index on Reserves:– A hash index will take 1 plus 10 to

retrieve the 1000 qualifying tuples

– Cost of Selection = 1 + 10 = 11 I/Os

– Cost of Writing to T1 = 10 I/Os

– Sub-Total = 11 +10 = 21

– The size of T1 = 10 pages

• Using a clustered B+ tree index on Sailors:– Cost of Selection = 2 (the constant cost) + 250 = 252 I/Os, Cost of Writing to T2 = 250

I/Os, Sub-Total = 252 + 250 = 502, and the size of T2 = 250 pages

• Using SNL the cost is:– Cost = 10 + 10 * 250 = 2510 I/Os

• Total cost = 21 + 502 + 2510 = 3033 I/Os, which is 165 times less than the cost of the initial plan i.e. 500,500 I/Os

Use B+ tree index,write to Temp T2

Plan

π sname

sid=sid

SailorsReserves

(on-the-fly)


σ rating > 5σ bid = 100Use Hash Index,write to Temp T1


Optimization: Alternative 3(using pipelining i.e. intermediate results are not written to disk)

• The same RA as alternative 1• Using a clustered Hash index on

Reserves:– A hash index will take 1 plus 10 to

retrieve the 1000 qualifying tuples– Cost of Selection = 1 + 10 = 11 I/Os– The size of T1 = 10 pages

• Using a clustered B+ tree index on Sailors:

– Cost of Selection = 2 (the constant cost) + 250 = 252 I/Os

– The size of T2 = 250 pages

• Using Block-Nested Loops join with 7 buffer pages:

– Cost =

• Total cost = 11 + 252 + 510 = 773 I/Os, which is 647 times faster than the initial plan and about 4 times faster than the previous one.

5102*2501027

10*25010

Use B+ tree index,do not write

result to Temp

Plan

π sname

sid=sid

SailorsReserves

(on-the-fly)

(Block-Nested Loops join)

σ rating > 5σ bid = 100

Use Hash Index,do not write

result to Temp


Summary

• Query optimization is very important component of a DBMS.

• We must understand the principals of optimization to know how it influences the performance of a database system.

• Two ways of optimization:– Logical: Push projection/selection below the join

– Physical: Using indexes and better implementation of relational operators.

Advanced Databases: Lecture 8 Query Optimization (III) 1 Query Optimization Advanced Databases By...

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