User Account

All Pages

query optimization, PL-SQL Programming

Assignment Help:
1.( /5 marks) Suppose that a B+-tree index with the search key (dept_name, building) is available on
relation department. What would be the best way to handle the following selection?
?(building = “Watson”) ? (budget < 55000) ? (dept_name = “Music”)(department)
a. 2 marks for clearly showing an equivalent relation algebra expression using the
appropriate, well identified rule supplied in the textbook (e.g., the derivation with rule #
applied).
b. 1.5 marks for clearly explaining how the B+ tree index would be used to evaluate the
conditions involving attributes that are specified as part of the index
c. 1.5 marks for clearly explaining how the remaining conditions that involve attributes that
are not specified as part of the index would be evaluated
2. ( /5 marks) Show how to derive the following relational algebra equivalences by a sequence of
transformations using the relational algebra equivalence rules in Section 13.2.1. Hint: make sure
you clearly define all rules you use; as well identify which rule is being applied and apply only one
rule for each line in your solution – show your work.
a. ??1??2??3 (E) = ??1 (??2 (??3 (E))) (2 marks)
b. ??1??2 (E1 ??3 E2) = ??1 (E1 ??3 (??2 (E2))), where ?2 involves only attributes from E2 (3
marks)
3. ( /5 marks) Consider the two expressions ?? (E1 E2) and ?? (E1 ? E2).
a. Show using a concrete example that the two expressions are not equivalent in general
(3 marks). Hint: first identify what the difference is between natural join and left outer join
(what is extra in the left outer join), and then make sure you give the concrete relation instances
for E1 and E2 that result in a different expression result.
b. Give a simple condition on the predicate ?, which if satisfied will ensure that the two
expressions are equivalent. Hint: Think about what extra tuples you would have in the left

1
Some of the problems are adapted from “Database System Concepts”, 6th Edition, 2011, Avis Silberschatz, Henry
F. Korth, S. Sudarshan.
outer join, how they joined tuple is constructed, and a condition that will exclude these extra
tuples.
4. ( /5 marks) Explain how to use a histogram to estimate the size of a selection of the form ?A=v(r).
Hint: Divide the distribution into n buckets {rj | j ? [1,n]} and consider the two cases: Part 1) the
selection value v is greater than or equal to the largest value of a given bucket ri (2 marks) and Part 2) the
selection value v is “part way” between the max and min values of bucket ri+1. (3 marks).
5. ( /5 marks) Give an example of a relational algebra expression defining a materialized view and
two situations (for the differentials) such that incremental view maintenance is better than
recomputation (recomputing the materialized view from scratch) in one situation (2.5 marks),
and recomputation is better in the other situation (2.5 marks). Hint: think about different values for
x between 0 and 100, where you insert x% tuples are deleted.

Related Discussions:- query optimization

Semidifference via except and join - sql, Semidifference via EXCEPT and JOI...

Semidifference via EXCEPT and JOIN - SQL SELECT * FROM (SELECT StudentId FROM IS_CALLED WHERE Name = 'Devinder' EXCEPT DISTINCT CORRESPONDING SELECT StudentId

Controlling cursor variables, Controlling Cursor Variables You use 3 s...

Controlling Cursor Variables You use 3 statements to control the cursor variable: OPEN-FOR, FETCH, & CLOSE. At First, you OPEN a cursor variable FOR a multi-row query. Then, y

In operator-comparison operators, IN Operator The operator IN tests the ...

IN Operator The operator IN tests the set membership. This means "equal to any member of." The set may have nulls, but they are ignored. For illustration, the statement below do

Pits, PITS Depressions in secondary cell wall is called pit. A pi...

PITS Depressions in secondary cell wall is called pit. A pit present on the free cell wall surface without its partner is called Blind pit. It consists of 2 parts -

Data types in sql - xml, Data Types in SQL - XML, Array, Row ...

Data Types in SQL - XML, Array, Row BINARY LARGE OBJECT for arbitrarily large bit strings. XML for XML documents and fragments. ARRAY types for arrays.

Sql queries-oracle , 1- You can check attribute names from each table in D...

1- You can check attribute names from each table in DBF11 by running for example:  desc dbf11.Member;  desc dbf11.Agent;  desc dbf11.Producer; Because some attribute names in

Using the collection methods, Using the Collection Methods The collecti...

Using the Collection Methods The collection methods below help to generalize the code and make collections easier to use and also make your applications easier to maintain:

PROCEDURE, Create a procedure named DDPROJ_SP that retrieves project inform...

Create a procedure named DDPROJ_SP that retrieves project information for a specific project based on a project ID. The procedure should have two parameters: one to accept a projec

Autonomous versus nested transactions, Autonomous versus Nested Transaction...

Autonomous versus Nested Transactions Though an autonomous transaction is started by the other transaction, it is not a nested transaction for the reasons shown below: (i)

Keyword and parameter description - forall statement, Keyword &Parameter De...

Keyword &Parameter Description: index_name: This is an undeclared identifier which can be referenced only within the FORALL statement and only as the collection subscript

Write Your Message!

Captcha
Order Assignment

Featured Services

Order Now

Popular Subjects

Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

Submit Assignment

Academics

Major Subjects

Majors

Get In Touch

whatsapp: +91-977-207-8620

Phone: +91-977-207-8620

Email: [email protected]

TERMS & POLICIES

HELP & SUPPORT

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd