Query optimaization

1. QUERY
OPTIMIZATION
P R E S E N T E D BY A B D U L W A H A B

2. INTRODUCTION
• What is Query Optimization?
• Suppose you were given a chance to visit 15 pre-selected
different cities in Pakistan. The only constraint would be ‘Time’
-> Would you have a plan to visit the cities in any order?

3. • Plan:
-> Place the 15 cities in different groups based
on their proximity to each other.
-> Start with one group and move on to the
next group.
Important point made over here is that you
would have visited the cities in a more
organized manner, and the ‘Time’ constraint
mentioned earlier would have been dealt with
efficiently.

4. • Query Optimization works in a similar way:
There can be many different ways to get an answer
from a given query. The result would be same in all
scenarios.
DBMS strive to process the query in the most
efficient way (in terms of ‘Time’) to produce the
answer.
Cost = Time needed to get all answers

5. STEPS IN A QUERY
OPTIMIZATION
1. Parsing
2. Transformation
3. Implementation

6. QUERY FLOW
Parser
Optimizer
Code
Generator/I
nterpreter
Processor
SQL

7. • Query Parser – Verify validity of the SQL statement.
Translate query into an internal structure using
relational calculus.
• Query Optimizer – Find the best expression from
various different algebraic expressions. Criteria used
is ‘Cheapness’
• Code Generator/Interpreter – Make calls for the
Query processor as a result of the work done by the
optimizer.
• Query Processor – Execute the calls obtained from
the code generator.

8. PROJECTION EXAMPLE:
• Projections produce a result tuple for every argument
tuple.
• What is the change?
• Change in the output size is the change in the length
of tuples
Let’s take a relation ‘R’
Relation (20,000 tuples): R(a, b, c)
Each Tuple (190 bytes): header = 24 bytes, a = 8 bytes,
b = 8 bytes, c = 150 bytes
Each Block (1024): header = 24 bytes

9. We can fit 5 tuples into 1 block
- 5 tuples * 190 bytes/tuple = 950 bytes can fit
into 1 block
- For 20,000 tuples, we would require 4,000
blocks (20,000 / 5 tuples per block = 4,000
With a projection resulting in elimination of
column c (150 bytes), we could estimate that
each tuple would decrease to 40 bytes (190 –
150 bytes)

10. Now, the new estimate will be 25 tuples in 1
block.
- 25 tuples * 40 bytes/tuple = 1000 bytes will
be able to fit into 1 block
- With 20,000 tuples, the new estimate is 800
blocks (20,000 tuples / 25 tuples per block =
800 blocks)
Result is reduction by a factor of 5

11. QUERY OPTIMIZATION: ALGEBRAIC
EXPRESSIONS
If we had the following query-
SELECT count(id) from emp;
Instead of
SELECT count(*) from emp;

12. Suppose, In DLD, We have that expression.
X = A + ABC +AB + AC + AA’ ------------------(1)
X = A(1+BC+B+C+A’)
X = A(1)
X = A ---------------------------------------------------(2)
So, Tell me Which one is better 1 and 2. Both have
same result.

13. THANK
YOU