Special Edition for CSEDU
CSEDU Students

Students

TOUCH-N-PASS EXAM CRAM GUIDE SERIESDBMS – II

Prepared By

Sharafat Ibn Mollah Mosharraf
CSE, DU
12 Batch (20052005-2006)
2006)
th

Table of Contents

CHAPTER 9: OBJECT-BASED DATABASES ............................................................................................................................ 1
THEORIES ........................................................................................................................................................................................ 1
EXCERCISES .................................................................................................................................................................................... 6
CHAPTER 15: TRANSACTIONS ................................................................................................................................................ 11
CHAPTER 16: CONCURRENCY CONTROL ........................................................................................................................... 19
CHAPTER 18: DATA ANALYSIS AND MINING ..................................................................................................................... 29
DATA ANALYSIS AND OLAP ........................................................................................................................................................ 29
DATA WAREHOUSE....................................................................................................................................................................... 33
DATA MINING ............................................................................................................................................................................... 37
CHAPTER 20: DATABASE SYSTEM ARCHITECTURES ..................................................................................................... 42
CHAPTER 21: PARALLEL DATABASES ................................................................................................................................. 46
CHAPTER 22: DISTRIBUTED DATABASES .......................................................................................................................... 49

CHAPTER 9OBJECT-BASED DATABASES
Theories
9.1

What are the major drawbacks of relational data model?
OR, What is the motivation to use object based data model? [Marks: 3]
The major drawbacks of relational data model are:
1. Limited type system supported by the relational model:
Complex application domains require correspondingly complex data types, such as nested
record structures, multivalued attributes and inheritance, which are supported by traditional
programming languages. Such features had to be translated to simpler relational model datatypes.
2. Difficulty in accessing database data from programs written in object-oriented
programming languages:
Merely extending the type system supported by the database was not enough to solve thisproblem completely. Differences between the type system of the database and the type systemof the programming language make data storage and retrieval more complicated, and need tobe minimized. Having to express database access using a language (SQL) which is different
from the programming language again makes the job of the programmer harder. It is desirable,
for many applications, to have programming language constructs or extensions that permit
direct access to data in the database, without having to go through an intermediate language
such as SQL.

9.2

Compare object-oriented and object-relational databases. [2008. Marks: 3]
Object-Oriented Databases

Object-Relational Databases

1. Object-oriented databases are built 1. Object-relational databases are object-oriented
aroundpersistent
programming
databases built on top of the relational model.
languages.
2. Targeted towards applications that 2. Targeted towards making data modeling andhave high performance requirements.
querying easier by using complex data types.
3. Provide low-overhead and thus high 3. Impose significant performance penalty for certain
performance access to persistent data.
kinds of applications that run primarily in main
memory and that perform a large number ofaccesses to the database.
4. More susceptible to data corruption 4. Provide a good protection
by programming errors.
programming errors.

of

data

from

5. Typical applications include CAD 5. Typical applications include storage and querying
databases.
of complex data, including multimedia data.
9.3

How does the concept of an object in the object-oriented data model differ from the conceptof an entity in the entity-relationship model? [Marks: 4]
An entity is simply a collection of variables or data items. An object is an encapsulation of data as
well as the methods (code) to operate on the data. The data members of an object are directly visible
only to its methods. The outside world can gain access to the object’s data only by passing pre-defined
messages to it, and these messages are implemented by the methods.

9.4

How object-relational database features are implemented? [Marks: 5]
Storing complex data types:

1

The complex data types supported by object-relational systems can be translated to the simpler
system of relational databases. The techniques for converting E-R model features to tables can beused, with some extensions, to translate object-relational data to relational data at the storage level.
Storing subtables:
Subtables can be stored in an efficient manner, without replication of all inherited fields, in one of
two ways:•

Each table stores the primary key (which may be inherited from a parent table) and theattributes are defined locally. Inherited attributes (other than the primary key) do not need
to be stored, and can be derived by means of a join with the supertable, based on the
primary key.

•

Each table stores all inherited and locally defined attributes. When a tuple is inserted, it isstored only in the table in which it is inserted, and its presence is inferred in each of thesupertables. Access to all attributes of a tuple is faster, since a join is not required.

Storing array and multiset types:
Implementations may choose to represent array and multiset types directly, or may choose to use anormalized representation internally. Normalized representations tend to take up more space andrequire an extra join / grouping cost to collect data in an array or multiset. However, normalized
representations may be easier to implement.
9.5

Compare simple data types in relational model and complex data types in object-relationalmodel. [Marks: 3]
Simple Data Types in Relational Model

Complex Data Types in Object-Relational Model

1. The basic data items are records that are 1. Records are large and fields are not atomic.
fairly small and whose fields are atomic.

9.6

2. Database is in 1NF.

2. Database is not normalized.

3. There are only a few record types.

3. There are many user-defined records types.

Explain Complex Data Types with appropriate examples. [Marks: 3]
Complex data types are structured types which can represent E-R model concepts, such ascomposite attributes, multivalued attributes, generalization and specialization directly, without acomplex translation to the relational model.
For example, we can define the following structured type to represent a composite attribute namewith component attributes firstname and lastname:create type Name as (firstname varchar(20),lastname varchar(20))
Multivalued attributes can be represented as arrays and multisets:
create type Book as (title varchar(20),author_array varchar(20) array[10],pub_date date,
keyword_set varchar(20) multiset))
Inheritance of teacher from a type Person can be defined as follows:
create type Person (name varchar(20),address varchar(20)2)
create type Teacher under Person (salary integer,
department varchar(20))
9.7With example explain single and multiple inheritances of types. [Marks: 4]
A single type inheritance of the types Student and Teacher from Person can be defined as follows:
create type Person (name varchar(20),address varchar(20))
create type Student under Person (degree varchar(20),department varchar(20))
create type Teacher under Person (salary integer,
department varchar(20))
And a multiple inheritance of type TeachingAssistant from the types Student and Teacher can be
defined as follows:
create type TeachingAssistant under Student, Teacher
However, the attribute department is defined separately in Student and Teacher and thus conflictin TeachingAssistant. To avoid a conflict between the two occurrences of department, we can rename
them by using an as clause:
create type TeachingAssistantunder Student with (department as student_dept),
Teacher with (department as teacher_dept)

9.8

With example explain single and multiple inheritances of tables. [Marks: 4]
Single inheritance of the tables students and teachers from the table people can be created asfollows:
create table people of Person
create table students of Student under people
create table teachers of Teacher under people
When we declare students and teachers as subtables of people, every tuple present in students orteachers becomes also implicitly present in people. Thus, if a query uses the table people, it will find
not only tuples directly inserted into that table, but also tuples inserted into its subtables, namely
students and teachers. However, only those attributes that are present in people can be accessed.
Multiple inheritance of the table teaching_assistants from the tables students and teachers can becreated as follows:
create table teaching_assistants of TeachingAssistant under students, teachers
As a result of the declaration, every tuple present in the teaching_assistants table is also implicitly
present in the teachers and in the students table, and in turn in the people table.

9.9

How arrays and multiset types are defined in SQL? [Marks: 4]
An array of values can be created in SQL in this way:
Array[′Smith′,′Jones′]3

Similarly, a multiset of keywords can be constructed as follows:
Multiset[′parsing′,′analysis′]
A book type using array and multiset can be declared as follows:
create type Book as (title varchar(20),author_array varchar(20) array[10],pub_date date,publisher Publisher,
keyword_set varchar(20) multiset))
A tuple of the relation of type Book can be created as:(′Compilers′, array[′Smith′,′Jones′], new Publisher(′McGraw-Hill′,′New York′), multiset[′parsing′,′analysis′])9.10

With example define nesting and unnesting. [Marks: 5]
Unnesting
The transformation of a nested relation into a form with fewer (or no) relation-valued attributes iscalled unnesting.
Consider the following books relation:
title

author_arraypublisher (name, branch)

keyword_set

Compilers [Smith, Jones] (McGraw-Hill, New York) {parsing, analysis}
Networks[Jones, Frick](Oxford, London){Internet, Web}

Suppose that we want to convert the relation into a single flat relation, with no nested relations or
structured types as attributes. We can use the following query to carry out the task:
select title, A.author, publisher.name as pub_name, publisher.branch as pub_branch, K.keyword
from books as B, unnest (B.author_array) as A(author), unnest (B.keyword_set) as K(keyword)
The result of the preceding query is the following relation which is in 1NF:
title

author

pub_name

pub_branch keyword

Compilers

Smith

McGraw-Hill

New York

parsing

Compilers

Jones

McGraw-Hill

New York

parsing

Compilers

Smith

McGraw-Hill

New York

analysis

Compilers

Jones

McGraw-Hill

New York

analysis

Networks

Jones

OxfordLondonInternet

Networks

Frick

OxfordLondonInternet

Networks

Jones

OxfordLondonWeb

Networks

Frick

OxfordLondonWeb

Nesting
The reverse process of transforming a 1NF relation into a nested relation is called nesting.
The above 1NF relation can be converted back to the nested relation using the following query:select title, collect(author) as author_set, Publisher(pub_name, pub_branch) as publisher,collect(keyword) as keyword_set
from flat_books
group by title, publisher

4

9.11

Explain the distinction between a type x and a reference type ref(x). Under whatcircumstances would you choose to use a reference type?
If the type of an attribute is x, then in each tuple of the table, corresponding to that attribute, there
is an actual object of type x. If its type is ref(x), then in each tuple, corresponding to that attribute,there is a reference to some object of type x. We choose a reference type for an attribute, if that
attribute’s intended purpose is to refer to an independent object.

9.12

What do you mean by persistent programming language? How it differs from high level
languages with embedded SQL? [2008. Marks: 1 + 3]
A persistent programming language is a programming language extended with constructs tohandle persistent data.
Persistent programming languages differ from high level languages with embedded SQL in thefollowing ways:
1. With an embedded language, the type system of the host language usually differs from thetype system of the data-manipulation language. The programmer is responsible for any typeconversions between the host language and SQL.
In contrast, in a persistent programming language, the query language is fully integrated withthe host language, and both share the same type system. Objects can be created and stored in
the database without any explicit type or format changes; any format changes required are
carried out transparently.
2. The programmer using an embedded query language is responsible for writing explicit code tofetch data from the database into memory. If any updates are performed, the programmer must
write code explicitly to store the updated data back in the database.
In contrast, in a persistent programming language, the programmer can manipulate persistent
data without writing code explicitly to fetch it into memory or store it back to disk.

9.13

What are the advantages and drawbacks of persistent programming languages? [Marks: 4]
Advantages of persistent programming languages:
1. Objects can be created and stored in the database without any explicit type or format changes;
any format changes required are carried out transparently.
2. The programmer can manipulate persistent data without writing code explicitly to fetch it intomemory or store it back to disk.
Disadvantages of persistent programming languages:
1. Since the programming language is usually a powerful one, it is relatively easy to makeprogramming errors that damage the database.
2. The complexity of the language makes automatic high-level optimization, such as to reduce
disk I/O, harder.
3. Support for declarative querying is important for many applications, but persistent
programming languages currently do not support declarative querying well.

9.14

Suppose that you have been hired as a consultant to choose a database system for yourclient’s application. For each of the following applications, state what type of database system(relational, persistent programming language–based OODB, object relational; do not specify acommercial product) you would recommend. Justify your recommendation.
1. A computer-aided design system for a manufacturer of airplanes.
2. A system to track contributions made to candidates for public office.
3. An information system to support the making of movies.
1. An OODB system would be suitable for this. That is because CAD requires complex data
5

types, and being computation oriented, CAD tools are typically used in a programming
language environment needing to access the database.
2. A relational system would be apt for this, as data types are expected to be simple, and apowerful querying mechanism is essential.
3. Here there will be extensive use of multimedia and other complex data types. But queries
are probably simple, and thus an object relational system is suitable.

Exercises
9.1

A car-rental company maintains a vehicle database for all vehicles in its current fleet. For all
vehicles, it includes the vehicle identification number, license number, manufacturer, model,
date of purchase, and color. Special data are included for certain types of vehicles:••••

Trucks: cargo capacity
Sports cars: horsepower, renter age requirement
Vans: number of passengers
Off-road vehicles: ground clearance, drivetrain (four- or two-wheel drive)

Construct an SQL:1999 object-oriented database schema definition for this database. Use
inheritance where appropriate. [2008. Marks: 3]
create type Vehicles (vehicle-id int,
licence-number string,
manufacturer string,model string,purchase-date date,color string)
create type Trucks under Vehicles (cargo-capacity int)
create type SportCars under Vehicles (horsepower int,
renter-age-requirement int)
create type Vanss under Vehicles (num-passengers int)
create type OffRoadVehicles under Vehicles (ground-clearance real,
drivetrain int)
9.2

Consider a database schema with a relation Emp whose attributes are as shown below, withtypes specified for multivalued attributes.
Emp = (ename, ChildrenSet multiset(Children), Skillset multiset(Skills))
Children = (name, Birthday)
Birthday = (day, month, year)
Skills = (type, ExamSet multiset(Exams))
Exams = (year, city)6

a. Define the above schema with appropriate types for each attribute. [Marks: 4]
b. Using the above schema, write the following queries in SQL:i.
Find the names of all employees who have a child who has a birthday in March.
ii.
Find those employees who took an examination for the skill type “typing” in thecity “Dayton”.
iii. List all skill types in the relation Emp.
a. create type Exams (year char(4),city varchar(20))
create table exams of type Exams
create type Skills (type varchar(20),ExamSet Exams multiset)
create table skills of type Skills
create type Birthday (day int(2),month char(3),year int(4))
create type Children (name varchar(50),Birthday birthday)
create table children of type Children
create type Emp (ename varchar(50),ChildrenSet Children multiset,
SkillSet Skills multiset)
create table emp of type Emp
b.

9.3

i.

select enamefrom emp, emp.ChildrenSet as cwhere ′March′ in (select birthday.month from c)ii.

select enamefrom emp, emp.SkillSet as s, s.ExamSet as x
where s.type = ′typing′ and x.city = ′Dayton′iii.

select distinct s.type
from emp, emp.SkillSet as sConsider the following ERD.
a. Give an SQL:2003 schema definition corresponding to the ERD.
b. Give constructors for each of the structured types defined in (a).

7

a.

b.

8

9.4

Give an SQL:2003 schema definition of the E-R diagram below.

9

9.5

Consider the relational schema below. Give a schema definition in SQL:2003 corresponding
to the relational schema, but using references to express foreign-key relationships.
employee (person_name, street, city)
works (person_name, company_name, salary)company (company_name, city)manages (person_name, manager_name)10CHAPTER 15
TRANSACTIONS
15.1

What is transaction? [2008, In-course 2008-2009. Marks: 1]
A transaction is a unit of program execution that accesses and possibly updates various data
items.
For example, a transfer of funds from a checking account to a savings account consists of several
operations from the point of view of the database system. All these operations result into a single
transaction.

15.2

What are the ACID properties of transaction? [2008, In-course 2008-2009. Marks: 1]
OR, Explain the transaction properties that are required for the database system to ensure
integrity of the data? [Marks: 4]
OR, List the ACID properties. Explain the usefulness of each. [In-course. Marks: 6]
To ensure integrity of the data, we require that the database system maintain the following
properties of the transactions: Atomicity
Either all operations of the transaction are reflected properly in the database or none at all.
This is typically the responsibility of the application programmer who codes thetransactions.
 Consistency
Execution of a transaction in isolation (that is, with no other transaction executing
concurrently) preserves the consistency of the database.
Clearly, lack of atomicity will lead to inconsistency in the database.
 Isolation
Even though multiple transactions may execute concurrently, the system guarantees that,for every pair of transactions Ti and Tj, it appears to Ti that either Tj finished execution
before Ti started, or Tj started execution after Ti finished. Thus, each transaction is
unaware of other transactions executing concurrently in the system.
The user view of a transaction system requires the isolation property, and this is theproperty that concurrent schedules use to take the system from one consistent state toanother. These requirements are satisfied by ensuring that only serializable schedules ofindividually consistency preserving transactions are allowed.
 Durability
After a transaction completes successfully, the changes it has made to the databasepersist, even if there are system failures.

15.3

Database-system implementers have paid much more attention to the ACID properties thanhave file-system implementers. Why might this be the case?
Database systems usually perform crucial tasks whose effects need to be atomic and durable, andwhose outcome affects the real world in a permanent manner. Examples of such tasks are monetarytransactions, seat bookings etc. Hence the ACID properties have to be ensured. In contrast, mostusers of file systems would not be willing to pay the price (monetary, disk space, time) of supporting
ACID properties.

11

15.4

Draw the state diagram of a transaction and describe each state. [2008, In-course 20072008. Marks: 3]
A transaction must be in one of the following states: Active, the initial state; the transaction stays in thisstate while it is executing.
 Partially committed, after the final statement has
been executed.
 Failed, after the discovery that normal execution can
no longer proceed.
 Aborted, after the transaction has been rolled backand the database restored to its state prior to the startof the transaction.
 Committed, after successful completion.

15.5

Figure: State diagram of a transaction.

During its execution, a transaction passes through several states, until it finally commits oraborts. List all possible sequences of states through which a transaction may pass. Explain why
each state transition may occur.
The possible sequences of states are:
1. Active → partially committed → committed. This is the normal sequence a successfultransaction will follow. After executing all its statements it enters the partially committed
state. After enough recovery information has been written to disk, the transaction finallyenters the committed state.
2. Active → partially committed → aborted. After executing the last statement of thetransaction, it enters the partially committed state. But before enough recovery information is
written to disk, a hardware failure may occur destroying the memory contents. In this case thechanges which it made to the database are undone, and the transaction enters the aborted state.
3. Active → failed → aborted. After the transaction starts, if it is discovered at some point that
normal execution cannot continue (either due to internal program errors or external errors), it
enters the failed state. It is then rolled back, after which it enters the aborted state.

15.6

What are the advantages of concurrent execution of transactions? [2008, In-course 20072008. Marks: 3]
OR, Why do database systems support concurrent execution of transactions in spite of theextra programming effort needed to ensure that concurrent execution does not cause any
problems?
The advantages of concurrent execution of transactions are:
1. Improved throughput and resource utilization: I/O activity and CPU activity canoperate in parallel leading to better transaction throughput. One transaction can be using
the CPU while another is reading from or writing to the disk. The processor and disk
utilization also increase; the processor and disk spend less time idle.
2. Reduced waiting time and average response time: Short transactions need not wait
behind long ones. If the transactions are operating on different parts of the database, it isbetter to run them concurrently, sharing the CPU cycle and disk accesses among them. It
also reduces the average response time - the average time for a transaction to becompleted after it has been submitted.
For these reasons, database systems support concurrent execution of transactions in spite of theextra programming effort needed to ensure that concurrent execution does not cause any problems.

12

15.7

Justify the following statement: Concurrent execution of transactions is more importantwhen data must be fetched from (slow) disk or when transactions are long, and is lessimportant when data is in memory and transactions are very short. [In-course 2008-2009.
Marks: 3]
If a transaction is very long or when it fetches data from a slow disk, it takes a long time tocomplete. In absence of concurrency, other transactions will have to wait for longer period of timeand average response time will increase. Also when the transaction is reading data from disk, CPU isidle. So, resources are not properly utilized. Hence concurrent execution becomes important in thiscase.
However, when the transactions are short or the data is available in memory, these problems do
not occur. Hence concurrent execution is less important in these cases.

15.8

With example define schedule and serial schedule. [Marks: 4]
A schedule is a sequence of instructions that specify the chronological order in which instructionsof transactions are executed.
A serial schedule is a schedule which consists of a sequence of instructions from varioustransactions, where the instructions belonging to one single transaction appear together in thatschedule.
For example, the schedules S1 and S2 shown below are both schedules; but S2 is a serial schedule
whereas S1 is not.
Schedule S2

Schedule S1
T1
read(A)
write(A)T2

read(A)
write(A)read(B)
write(B)read(B)
write(B)
Schedule, but not serial schedule.

15.9

T1
read(A)
write(A)read(B)
write(B)T2

read(A)
write(A)read(B)
write(B)
Serial schedule.

Explain the distinction between the terms serial schedule and serializable schedule. [Incourse 2007-2008. Marks: 2]
A schedule in which all the instructions belonging to one single transaction appear together iscalled a serial schedule.
A schedule which is equivalent to a serial schedule is called a serializable schedule.
For example, the following schedule S1 is not a serial schedule, but it is a serializable schedule
which is equivalent to the serial schedule S2 below.
[See the schedules S1 and S2 in question 15.8]

15.10

With examples show the differences among schedule, serial schedule and serializable
schedule. [In-course 2008-2009. Marks: 5][Combine the answers of questions 15.8 and 15.9. Try yourself. ☺]15.11

How do you identify conflicting instructions to determine conflict serializability? [Marks: 3]
Let us consider a schedule S in which there are two consecutive instructions, li and lj (i ≠ j) of
transactions Ti and Tj respectively.
If li and lj, refer to different data item, then we can swap them without affecting the results of any
instruction in the schedule.
However, if li and lj refer to the same data item Q, then the order of the two steps may matter.
13

The following four cases need to be considered:
1. Ii = read(Q), Ij = read(Q).
The order of Ii and Ij does not matter, since the same value of Q is read by Ti and Tj,
regardless of the order.
2. Ii = read(Q), Ij = write(Q).
If Ii comes before Ij, then Ti does not read the value of Q written by Tj in instruction Ij.
If Ij comes before Ii, then Ti reads the value of Q written by Tj.
Thus, the order of Ii and Ij matters.
3. Ii = write(Q), Ij = read(Q).
The order of Ii and Ij matters for reasons similar to those of the previous case.
4. Ii = write(Q), Ij = write(Q).
Since both instructions are write operations, the order of these instructions does not affecteither of Ti or Tj.
However, the value obtained by the next read(Q) instruction of S is affected, since the resultof only the latter of the two write instructions is preserved in the database.
If there is no other write(Q) instruction after Ii and Ij in S, then the order of Ii and Ij directly
affects the final value of Q in the database state that results from schedule S.
Thus, we can say that two instructions in a schedule conflict if and only if there exists same item
Q accessed by both of them and at least one of these instructions is a write operation on Q.
15.12

What do you understand by conflict equivalent and view equivalent schedules? Explain
with example. [Marks: 5]
OR, Distinguish between conflict serializable schedule and view serializable schedule withproper example. [In-course. Marks: 6]
Conflict Equivalent Schedules:
If a schedule S can be transformed into a schedule S´ by a series of swaps of non-conflicting
instructions, we say that S and S´ are conflict equivalent.
Conflict Serializable Schedule:
We say that a schedule S is conflict serializable if it is conflict equivalent to a serial schedule.
For example, schedule 1 below is conflict equivalent to schedule 2. The sequence of swaps ofnon-conflicting instructions is also shown.
Schedule 1
T1
read(A)
write(A)T2

read(A)
write(A)read(B)
write(B)

Schedule 2
T1
read(A)
write(A)T2

read(A)T1
read(A)
write(A)read(B)read(B)read(A)
write(A)write(A)write(B)read(B)
write(B)T2

write(B)read(B)
write(B)read(B)
write(B)T1
read(A)
write(A)read(B)T2

read(A)
write(B)write(A)read(B)
write(B)T1
read(A)
write(A)read(B)
write(B)T2

read(A)
write(A)read(B)
write(B)Figure: Transforming Schedule 1 – which is conflict equivalent to serial schedule 2 – into Serial Schedule 2.

View Equivalent Schedules:
Let S and S´ be two schedules with the same set of transactions. The schedules S and S´ are said
to be view equivalent if the following three conditions are met:14

1. For each data item Q, if transaction Ti reads the initial value of Q in schedule S, thentransaction Ti must, in schedule S´, also read the initial value of Q.
2. For each data item Q, if transaction Ti executes read(Q) in schedule S, and that value wasproduced by a write(Q) operation executed by transaction Tj, then the read(Q) operationof transaction Ti must, in schedule S´, also read the value of Q that was produced by thesame write(Q) operation in transaction Tj.
3. For each data item Q, the transaction (if any) that performs the final write(Q) operation inschedule S must perform the final write(Q) operation in schedule S´.
View Serializable Schedule:
We say that a schedule S is view serializable if it is view equivalent to a serial schedule.
For example, Schedule 3 below is view serializable to serial schedule 4, but it is not conflict
serializable, since every pair of consecutive instructions conflict, and, thus, no swapping of
instructions is possible.
Schedule 3
T3
read(Q)

T4

Schedule 4
T5

write(Q)T3
read(Q)
write(Q)write(Q)T4

write(Q)write(Q)

15.13

T5

write(Q)

What are the conditions to determine view equivalence? [Marks: 3][See question 15.12]

15.14

Give an example of a schedule that is view equivalent but not conflict equivalent. Explain
with causes. [In-course 2008-2009. Marks: 4]
As an example, Schedule 1 below is view serializable to schedule 2, but it is not conflict
serializable, since every pair of consecutive instructions conflict, and, thus, no swapping of
instructions is possible.
Schedule 1
T3
read(Q)

T4

Schedule 2
T5

write(Q)T3
read(Q)
write(Q)write(Q)T4

T5

write(Q)write(Q)write(Q)A schedule is view equivalent but not conflict equivalent when it contains blind writes, i.e. theschedule contains transactions which perform write operations without having performed a read
operation.
15.15“Every conflict-serializable schedule is view serializable” – explain how? [2007. Marks: 3]
Every conflict-serializable schedule is also view serializable as all the conditions of viewequivalence also holds for conflict-serializable schedule. For example, Schedule 1 below is conflictequivalent to Schedule 2.
Schedule 1
T1
read(A)
write(A)T2

read(A)
write(A)read(B)
write(B)read(B)
write(B)

15

Schedule 2
T1
read(A)
write(A)read(B)
write(B)T2

read(A)
write(A)read(B)
write(B)Schedule 1 is also view equivalent to Schedule 2 because:
1. T1 reads the initial values of A and B in both schedules.
2. T2 reads the value of A and B which were written by T1 in both schedules.
3. T2 performs the final write(A) and write(B) operations in both schedules.
15.16

Since every conflict-serializable schedule is view serializable, why do we emphasize conflict
serializability rather than view serializability? [2008, In-course 2007-2008. Marks: 3]
Most of the concurrency control protocols (protocols for ensuring that only serializable schedules
are generated) used in practice are based on conflict serializability — they actually permit only asubset of conflict serializable schedules (to make the schedules recoverable and cascadeless). The
general form of view serializability is very expensive to test, and only a very restricted form of it isused for concurrency control. That is why we emphasize conflict serializability rather than view
serializability.

15.17

Consider a database with objects X and Y and assume that there are two transactions T1
and T2. Transaction T1 reads objects X and Y and then writes object X. Transaction T2 reads
objects X and Y and then writes objects X and Y.
i. Give an example schedule with actions of transactions T1 and T2 on objects X and Y
that results in a write-read conflict.
ii. Give an example schedule with actions of transactions T1 and T2 on objects X and Y
that results in a read-write conflict.
iii. Give an example schedule with actions of transactions T1 and T2 on objects X and Y
that results in a write-write conflict.
iv. For each of the above three schedules, show that Strict Two-Phase Lock (2PL)
protocol disallows the schedule. [2006. Marks: 3 × 2 + 4 = 10]
The transactions are as follows:
T1: read(X);read(Y);
write(X);T2: read(X);read(Y);
write(X);write(Y);i.

T1read(X)read(Y)T2

read(X)

ii.

T1

iii.

T1
read(X)read(Y)

write(X)write(Y)T2

read(X)read(Y)read(X)

write(X)read(Y)
write(X)write(Y)T2
read(X)read(Y)

write(X)read(Y)
write(X)write(X)write(Y)iv. If strict 2PL is used, then in (i) and (iii), T1 would acquire lock-X(X) and T2 would not be able
to acquire lock-X(X) and subsequently lock-X(Y) until T1 commits. As for the case in (ii), T2
would acquire lock-X(X) and hence T1 would not be able to acquire lock-X(X) until T2
commits. Thus strict 2PL disallows the above three schedules.
15.18

What is a recoverable schedule? [In-course 2007-2008. Marks: 2]
A recoverable schedule is one where, for each pair of transactions Ti and Tj such that transaction
Tj reads a data item previously written by a transaction Ti, the commit operation of Ti appears before
the commit operation of Tj.

15.19

Why is recoverability of schedules desirable?
Recoverable schedules are desirable because failure of a transaction might otherwise bring thesystem into an irreversibly inconsistent state.
16

15.20

Are there any circumstances under which it would be desirable to allow non-recoverable
schedules? Explain in detail. [Marks: 2]
Non-recoverable schedules may sometimes be needed when updates must be made visible early
due to time constraints, even if they have not yet been committed, which may be required for very
long duration transactions.

15.21

Suppose that there is a database system that never fails. Is a recovery manager required forthis system?
Even in this case the recovery manager is needed to perform roll-back of aborted transactions.

15.22

What is cascadeless schedule? [In-course 2007-2008, 2008-2009. Marks: 2]
A cascadeless schedule is one where, for each pair of transactions Ti and Tj such that Tj reads adata item previously written by Ti, the commit operation of Ti appears before the read operation of Tj.

15.23

What is the importance of cascadeless schedules? [Marks: 2]
OR, Why is cascadelessness of schedules desirable? [In-course 2008-2009. Marks: 1]
Cascadeless schedules are desirable because the failure of a transaction does not lead to theaborting of any other transaction. Of course this comes at the cost of less concurrency.

15.24

Are there any circumstances under which it would be desirable to allow noncascadelessschedules? Explain your answer. [In-course 2008-2009. Marks: 1]
If failures occur rarely so that we can pay the price of cascading aborts for the increased
concurrency, noncascadeless schedules might be desirable.

15.25

How conflict serializability is tested through precedence graph? [In-course 2008-2009.
Marks: 5]
Testing for Conflict Serializability:
1. Consider a schedule S of a set of transactions T1, T2, ..., Tn.
2. We construct a directed graph called precedence graph, from S. This graph consists of a pairG = (V, E), where V is the set of vertices and E is a set of edges.
The set of vertices consists of all the transactions participating in the schedule. The set ofedges consists of all the edges Ti → Tj for which one of the three conditions holds:
a. Ti executes write(Q) before Tj executes read(Q).
b. Ti executes read(Q) before Tj executes write(Q).
c. Ti executes write(Q) before Tj executes write(Q).
If an edge Ti → Tj exists in the precedence graph, then, in any serial schedule S´ equivalent toS, Ti must appear before Tj.
3. If the precedence graph has a cycle, then schedule S is not conflict serializable. If no cycle
exists, then it is conflict serializable.

15.26

Consider the following precedence graph. Is the corresponding schedule conflict
serializable? Explain your answer. [2007, In-course 2008-2009. Marks: 4]
T1

T2

T4

T3
T5

Yes, the corresponding schedule is conflict serializable, since the graph is acyclic. A possible
schedule is obtained by doing a topological sort, that is, <T1, T2, T3, T4, T5>.
17

15.27

Consider the following two transactions:
T1: read(A);read(B);
if A = 0 then B := B + 1;
write(B);T2: read(B);read(A);
if B = 0 then A := A + 1;
write(A);Let the consistency requirement be A = 0 ˅ B = 0, with A = B = 0 the initial values.
a. Show that every serial execution involving these two transactions preserves theconsistency of the database.
b. Show a concurrent execution of T1 and T2 that produces a nonserializable schedule.
c. Is there a concurrent execution of T1 and T2 that produces a serializable schedule?
a. There are two possible executions: T1T2 and T2T1.
Case 1:
A
Initially 0
After T1 0
After T2 0

B
0
1
1

Consistency met: (A = 0) ˅ (B = 0) ≡ (T ˅ F = T)
Case 2:A
Initially 0
After T2 1
After T1 1

B
0
0
0

Consistency met: (A = 0) ˅ (B = 0) ≡ (F ˅ T = T)
b. Any interleaving of T1 and T2 results in a non-serializable schedule.
T1

T2

read(A)read(B)read(A)read(B)
if A = 0 then B = B + 1
if B = 0 then A = A + 1
write(A)write(B)c. There is no parallel execution resulting in a serializable schedule. From part (a), we know thata serializable schedule results in A = 0 ˅ B = 0. Suppose we start with T1 read(A). Then when
the schedule ends, no matter when we run the steps of T2, B = 1. Now suppose we start
executing T2 prior to completion of T1. Then T2 read(B) will give B a value of 0. So when T2completes, A = 1. Thus (B = 1 ˄ A = 1) → ¬ (A = 0 ˅ B = 0).
Similar case happens for starting with T2 read(B).

18

CHAPTER 16
CONCURRENCY CONTROL
16.1

What is concurrency control scheme? [In-course 2008-2009. Marks: 2]
When several transactions execute concurrently in the database, to ensure the isolation property,
the system must control the interaction among the concurrent transactions and this control isachieved through one of the variety of mechanisms called concurrency-control schemes.

16.2

What is locking protocol?
OR, Define locking protocol. [In-course 2008-2009. Marks: 2]
A locking protocol is a set of rules followed by all transactions while requesting and releasing locks.

16.3

What are the modes of locking in lock-based protocols? [Marks: 2]
Data items can be locked in two modes:
1. Shared (S) Mode – data item can only be read. If a transaction Ti has obtained a shared-mode
lock on item Q, then Ti can read, but cannot write Q. S-lock is requested using lock-S
instruction.
2. Exclusive (X) Mode – data item can be both read and written. If a transaction Ti has obtained
an exclusive-mode lock on item Q, then Ti can both read and write Q. X-lock is requested
using lock-X instruction.

16.4
16.5

What are the functions of a lock manager? [In-course 2008-2009. Marks: 4]
How starvation arises in case of granting of locks? How starvation can be avoided? [2008.
Marks: 1 + 2]
Also, How concurrency control manager grants locks so that starvation does not arise?
[2008. Marks: 2]
A transaction may be waiting for an X-lock on an item, while a sequence of other transactions
request and are granted an S-lock on the same item. Thus the first transaction never makes progressand starvation arises.
To avoid starvation, when a transaction Ti requests a lock on a data item Q in a particular mode
M, the lock can be granted provided that:
1. There is no other transaction holding a lock on Q in a mode that conflicts with M.
2. There is no other transaction that is waiting for a lock on Q and that made its lock request
before Ti.

16.6

With example define two-phase locking protocol. [In-course 2008-2009. Marks: 3]
OR, How two-phase locking protocol works? Explain with example.
The two-phase locking protocol requires that each transaction issue lock and unlock requests in
two phases:
T1
1. Growing phase• Transaction may obtain locks• Transaction may not release any lock2. Shrinking phase• Transaction may release locks• Transaction may not obtain any new locks
Initially, a transaction is in the growing phase. The transaction acquires locks asneeded. Once the transaction releases a lock, it enters the shrinking phase, and it canissue no more lock requests.
For example, transaction T1 in the figure is two-phase.
19

lock-X(B)read(B)
B := B – 50
write(B)lock-X(A)read(A)A := A + 50
write(A)unlock(B)unlock(A)16.7

Consider the following two transactions:
T1: read(A);read(B);
if A = 0 then B := B + 1;
write(B).
T2: read(B);read(A);
if B = 0 then A := A + 1
write(A).
a. Add lock and unlock instructions to transactions T1 and T2, so that they observe thetwo-phase locking protocol. [2007. Marks: 4]
b. Can the execution of these transactions result in a deadlock?
a. Lock and unlock instructions:
T1: lock-S(A)read(A);lock-X(B);read(B);
if A = 0 then B := B + 1;
write(B);unlock(A);unlock(B).
T2: lock-S(B)read(B);
lock-X(A);read(A);
if B = 0 then A := A + 1;
write(A);unlock(B);unlock(A).
b.

Execution of these transactions can result in deadlock. For example, consider the following
partial schedule:
T1lock-S(A)T2lock-S(B)read(B)read(A)lock-X(B)lock-X(A)The transactions are now deadlocked.
16.8

Show that the two-phase locking protocol ensures conflict serializability and thattransactions can be serialized according to their lock points. [Marks: 3]
Suppose two-phase locking does not ensure serializability. Then there exists a set of transactions
T0, T1, ..., Tn-1 which obey 2PL and which produce a non-serializable schedule. A non-serializable
schedule implies a cycle in the precedence graph, and we shall show that 2PL cannot produce suchcycles. Without loss of generality, assume the following cycle exists in the precedence graph:
T0 → T1 → T2 → … → Tn-1 → T0
Let αi be the time at which Ti obtains its last lock (i.e. Ti’s lock point). Then for all transactions
such that Ti → Tj, αi < αj. Then, for the cycle, we have20

α0 < α1 < α2 < … < αn - 1 < α0
Since α0 < α0 is a contradiction, no such cycle can exist. Hence, 2PL cannot produce non-serializable
schedules.
Because of the property that for all transactions such that Ti → Tj, αi < αj, the lock point orderingof the transactions is also a topological sort ordering of the precedence graph. Thus transactions canbe serialized according to their lock points.
16.9

Define compatibility function/lock compatibility matrix. [Marks: 2]
Given a set of lock modes, we can define a compatibility function on them as follows.
Let A and B represent arbitrary lock modes. Suppose that a transaction Ti requests a lock of mode
A on item Q on which transaction Tj (Ti ≠ Tj) currently holds a lock of mode B. If transaction Ti canbe granted a lock on Q immediately, in spite of the presence of the mode B lock, then we say mode A
is compatible with mode B.
Such a function can be represented conveniently by a matrix, which is called the lockcompatibility matrix. For example, the compatibility relation between the shared and exclusivemodes of locking appears in the matrix below. An element (A, B) of the matrix has the value true if
and only if mode A is compatible with mode B.
S
true
false

S
X

16.10

X
false
false

What are the advantages and drawbacks of two-phase locking protocol? [In-course 20082009. Marks: 3]
Advantage: Ensures conflict serializability.
Drawbacks:
1. Doesn’t ensure freedom from deadlock.
2. Cascading rollback might occur.
3. Concurrency might become less.

16.11

What benefit is provided by strict two-phase locking? What disadvantages result? [2007.
Marks: 3]
Because strict 2PL produces only cascadeless schedules, recovery is very easy.
However, the set of schedules obtainable is a subset of those obtainable from plain two phaselocking, thus concurrency is reduced.

16.12

Most implementations of database systems use strict two-phase locking. Suggest threereasons for the popularity of this protocol.
OR, Why strict two-phase locking is preferred to other protocols in implementing databasesystems?
1. Relatively simple to implement.
2. Imposes low rollback overhead because of cascadeless schedules.
3. Usually allows an acceptable level of concurrency.

16.13

What benefit does rigorous two-phase locking provide? How does it compare with otherforms of two-phase locking?
Rigorous two-phase locking has the advantages of strict 2PL. In addition, it has the property thatfor two conflicting transactions, their commit order is their serializability order. In some systems
users might expect this behavior.

16.14

Can strict two-phase or rigorous two-phase locking protocols ensure freedom fromdeadlock? [In-course 2008-2009. Marks: 1]
No. No form of 2PL ensures freedom from deadlock.
21

16.15Compare two-phase, strict two-phase and rigorous two-phase protocol. [2008, In-course
2007-2008. Marks: 3]
In 2PL, each transaction issues lock and unlock requests in two phases:
1. Growing Phase – May obtain locks, but may not release locks
2. Shrinking Phase – May release locks, but may not obtain any new locks
In strict 2PL, in addition to above, all exclusive-mode locks taken by a transaction are held untilthat transaction commits.
Rigorous 2PL is like strict 2PL except that instead of exclusive-mode locks only, all locks taken
by a transaction are held until that transaction commits.

16.16

With example show how tree protocol works? [Marks: 4]
In the tree protocol, the only lock instruction allowed is lock-X. Each transaction Ti can lock adata item at most once, and must observe the following rules:
1. The first lock by Ti may be on any data item.
2. Subsequently, a data item Q can be locked by Ti only if the parent of Q is currently locked byTi.
3. Data items may be unlocked at any time.
4. A data item that has been locked and unlocked by Ti cannot subsequently be relocked by Ti.
Example:T1
lock-X(B)T2lock-X(D)lock-X(H)
unlock(D)lock-X(E)lock-X(D)unlock(B)unlock(E)unlock(H)unlock(D)16.17

What are the advantages and disadvantages of tree protocol? [2008, In-course 2007-2008.
Marks: 2]
Advantages:
1. Ensures conflict serializability.
2. Ensures freedom from deadlock.
Disadvantages:
1. Doesn’t ensure recoverability and cascadelessness.
2. In some cases, a transaction may have to lock data items that it does not access. Thisadditional locking results in increased locking overhead, the possibility of additional waiting
time, and a potential decrease in concurrency.
3. Without prior knowledge of what data items will need to be locked, transactions will have tolock the root of the tree, and that can reduce concurrency greatly.

16.18

Show by example that there are schedules possible under the tree protocol that are notpossible under the two-phase locking protocol and vice-versa. [In-course 2008-2009. Marks: 3]
Consider the tree-structured database graph given below.

22

Schedule possible under tree protocol but not under 2PL:

Schedule possible under 2PL but not under tree protocol:

16.19

Show by example that there are schedules possible under the timestamp protocol that are
not possible under the two-phase locking protocol and vice-versa.
A schedule which is allowed in the two-phase locking protocol but not in the timestamp protocol
is:This schedule is not allowed in the timestamp protocol because at step 7, the W-timestamp of B is 1.
A schedule which is allowed in the timestamp protocol but not in the two-phase locking protocol is:

This schedule cannot have lock instructions added to make it legal under two-phase locking
protocol because T1 must unlock(A) between steps 2 and 3, and must lock(B) between steps 4 and 5.
16.20

What are the advantages and disadvantages of timestamp-ordering protocol? [In-course
2008-2009. Marks: 4]
Advantages:
1. Ensures conflict serializability.
2. Ensures freedom from deadlock as no transaction ever waits.
Disadvantages:
1. Possibility of starvation of long transactions if a sequence of conflicting short transactions
causes repeated restarting of the long transaction.
2. Generates schedules that are not recoverable and may require cascading rollbacks.
2316.21

In timestamp ordering, W-timestamp(Q) denotes the largest timestamp of any transactionthat executed write(Q) successfully. Suppose that, instead, we defined it to be the timestamp of
the most recent transaction to execute write(Q) successfully. Would this change in wording
make any difference? Explain your answer.
It would make no difference. The write protocol is such that the most recent transaction to write
an item is also the one with the largest timestamp to have done so.

16.22

When a transaction is rolled back under timestamp ordering, is it assigned a newtimestamp? Why can it not simply keep its old timestamp? [Marks: 2 + 2]
A transaction is rolled back because a newer transaction has read or written the data which it was
supposed to write. If the rolled-back transaction is re-introduced with the same timestamp, the same
reason for rollback is still valid, and the transaction will have to be rolled back again. This willcontinue indefinitely.

16.23

Show, with an example, how Thomas’ write rule modifies the timestamp-ordering protocol?
[2007. Marks: 3]
The timestamp ordering protocol requires that Ti be rolled back if Ti issues write(Q) and TS(Ti) <W-timestamp(Q). However, in Thomas’ Write Rule, in those cases where TS(Ti) ≥ R-timestamp(Q),we ignore the obsolete write.
For example, in the following schedule, in case of timestamp-ordering
protocol, Ti would be rolled back. However, if Thomas’ write rule is applied, thewrite(Q) operation in Ti is ignored only.

16.24

Ti
read(Q)

Ti + 1
write(Q)write(Q)

Under a modified version of the timestamp protocol, we require that a commit bit be tested
to see whether a read request must wait. Explain how the commit bit can prevent cascading
abort. Why is this test not necessary for write requests?
Using the commit bit, a read request is made to wait if the transaction which wrote the data item
has not yet committed. Therefore, if the writing transaction fails before commit, we can abort thattransaction alone. The waiting read will then access the earlier version in case of a multiversion
system, or the restored value of the data item after abort in case of a single-version system. For
writes, this commit bit checking is unnecessary. That is because either the write is a “blind” write
and thus independent of the old value of the data item or there was a prior read, in which case the test
was already applied.

16.25

Consider the validation-based concurrency-control scheme. Show that by choosing
Validation(Ti), rather than Start(Ti), as the timestamp of transaction Ti, we can expect better
response time provided that conflict rates among transactions are indeed low.
In the validation-based concurrency control scheme, choosing Start(Ti) as the timestamp of Ti
gives a subset of the schedules allowed by choosing Validation(Ti) as the timestamp. Using Start(Ti)means that whoever started first must finish first. Clearly transactions could enter the validation
phase in the same order in which they began executing, but this is overly restrictive. Since choosing
Validation(Ti) causes fewer non-conflicting transactions to restart, it gives the better response times.

16.26

Why multiple granularities are used in case of locking? What is intention mode lock?
[Marks: 2 + 2]
There are circumstances where it would be advantageous to group several data items, and to treat
them as one individual synchronization unit. For example, if a transaction Ti needs to access theentire database, and a locking protocol is used, then Ti must lock each item in the database. Clearly,
executing these locks is time consuming. It would be better if Ti could issue a single lock request to
lock the entire database. What is needed is a mechanism to allow the system to define multiple levelsof granularity. Hence multiple granularities are used in case of locking.
Intention-mode lock is a class of locks using which the system using multiple granularities candetermine if a node can be locked. If a node is locked in an intention mode, it means explicit locking
is being done at a lower level of the tree.
2416.27In multiple-granularity locking, what is the difference between implicit and explicit
locking? [In-course 2007-2008. Marks: 2]
When a transaction explicitly locks a node in shared or exclusive mode, it implicitly locks all thedescendents of that node in the same mode. The transaction need not explicitly lock the descendent
nodes. There is no difference in the functionalities of these locks, the only difference is in the way
they are acquired, and their presence tested.

16.28

Although SIX mode is useful in multiple-granularity locking, an exclusive and intendshared (XIS) mode is of no use. Why is it useless?
An exclusive lock is incompatible with any other lock kind. Once a node is locked in exclusivemode, none of the descendents can be simultaneously accessed by any other transaction in any mode.
Therefore an exclusive and intend-shared declaration has no meaning.

16.29

What is deadlock? [In-course 2007-2008. Marks: 1]
A system is said to be in a deadlock state if there exists a set of transactions such that everytransaction in the set is waiting for another transaction in the set. More precisely, there exists a set of
waiting transactions {T0 , T1 , . . ., Tn} such that T0 is waiting for a data item that T1 holds, and T1 iswaiting for a data item that T2 holds, and …, and Tn-1 is waiting for a data item that Tn holds, and Tn
is waiting for a data item that T0 holds. None of the transactions can make progress in such asituation.

16.30

What are the methods to deal with deadlock problem? [Marks: 4]
There are two principal methods for dealing with deadlock problem
1. We can use a deadlock prevention protocol to ensure that the system will never enter adeadlock state.
There are two approaches to deadlock prevention:
1. Ensuring that no cycle waits can occur by ordering the requests for locks, or requiring all
locks to be acquired together.
2. Using preemption and transaction rollbacks.
The wait-die and wound-wait schemes use this approach.
2. Alternatively, we can allow the system to enter a deadlock state, and then try to recover byusing a deadlock detection and deadlock recovery scheme.
The wait-for graph is used to detect deadlocked transactions and they are rolled back using
the following actions:
1. Selection of a victim
2. Rollback
3. Handling starvation
There exists another method for dealing with deadlock problem – called the timeout-based
scheme. In this approach, a transaction that has requested a lock waits for at most a specified amountof time. If the lock has not been granted within that time, the transaction is said to time out, and it
rolls itself back and restarts.

16.31

Under what conditions is it less expensive to avoid deadlock than to allow deadlocks tooccur and then to detect them?
Deadlock avoidance is preferable if the consequences of abort are serious (as in interactive
transactions), and if there is high contention and a resulting high probability of deadlock.

16.32

Describe some of the prevention strategies/approach to restrict deadlock. [Marks: 5]
There are two approaches to deadlock prevention:
1. Ensuring that no cycle waits can occur by ordering the requests for locks, or requiring all25

locks to be acquired together.
2. Using preemption and transaction rollbacks.
The wait-die and wound-wait schemes use this approach.
[Now describe the wait-die and wound-wait schemes as in the following answer (16.32).]
16.33

Describe wait-die and wound-wait schemes for deadlock prevention.
OR, What are the deadlock prevention schemes using timestamps?
OR, How deadlock can be prevented using timestamps? [In-course 2008-2009. Marks: 4]
1. Wait-Die Scheme
When transaction Ti requests a data item currently held by Tj, Ti is allowed to wait only if
it has a timestamp smaller than that of Tj (i.e. Ti is older than Tj). Otherwise, Ti is rolledback (dies).
For example, suppose that transactions T22, T23 and T24 have timestamps 5, 10 and 15
respectively. If T22 requests a data item held by T23, then T22 will wait. If T24 requests adata item held by T23, then T24 will be rolled back.
2. Wound-Wait Scheme
When transaction Ti requests a data item currently held by Tj, Ti is allowed to wait only if
it has a timestamp larger than that of Tj (i.e. Ti is younger than Tj). Otherwise, Tj is rolledback (Tj is wounded by Ti).
Returning to our example, with transactions T22, T23 and T24, if T22 requests a data item
held by T23, then the data item will be preempted from T23, and T23 will be rolled back. If
T24 requests a data item held by T23, then T24 will wait.

16.34

If deadlock is avoided by deadlock avoidance schemes, is starvation still possible? Explain
your answer.
A transaction may become the victim of deadlock-prevention roll-back arbitrarily many times,
thus creating a potential starvation situation.

16.35

How wait-for graph is used for deadlock detection? [In-course 2007-2008. Marks: 2]
A deadlock exists in the system if and only if the wait-for graph contains a cycle. Eachtransaction involved in the cycle is said to be deadlocked. To detect deadlocks, the system needs tomaintain the wait-for graph, and periodically to invoke an algorithm that searches for a cycle in thegraph.

16.36

What are the steps used for deadlock recovery? [Marks: 5]
OR, What is the process of recovering from deadlock? [2008. Marks: 2]
When a detection algorithm determines that a deadlock exists, the system must recover from thedeadlock. The most common solution is to roll back one or more transactions to break the deadlock.
Three actions need to be taken:
1. Selection of a Victim
Given a set of deadlocked transactions, we must determine which transaction (or transactions) toroll back to break the deadlock. We should roll back those transactions that will incur the minimum
cost. Unfortunately, the term minimum cost is not a precise one. Many factors may determine thecost of a rollback, includinga. How long the transaction has computed, and how much longer the transaction willcompute before it completes its designated task.
b. How many data items the transaction has used.
c. How many more data items the transaction needs for it to complete.
d. How many transactions will be involved in the rollback.
26

2. Rollback
Once we have decided that a particular transaction must be rolled back, we must determine how
far this transaction should be rolled back.
1. Total Rollback: Abort transaction and then restart it.
2. Partial Rollback: Roll back the transaction only as far as necessary to break thedeadlock.
3. Starvation
In a system where the selection of victims is based primarily on cost factors, it may happen thatthe same transaction is always picked as a victim. As a result, this transaction never completes its
designated task, thus there is starvation. We must ensure that transaction can be picked as a victim
only a (small) finite number of times. The most common solution is to include the number ofrollbacks in the cost factor.
16.37

What are the factors for selecting a victim during deadlock recovery? [Marks: 5]a. How long the transaction has computed, and how much longer the transaction will compute
before it completes its designated task.
b. How many data items the transaction has used.
c. How many more data items the transaction needs for it to complete.
d. How many transactions will be involved in the rollback.

16.38

For each of the following protocols, describe aspects of practical applications that would
lead you to suggest using the protocol, and aspects that would suggest not using theprotocol:
•
•
•
•
•
•

•

Two-phase locking
Two-phase locking with multiple-granularity locking
The tree protocol
Timestamp ordering
Validation
Multiversion timestamp ordering
Multiversion two-phase locking•

Two-phase locking: Use for simple applications where a single granularity is acceptable. If
there are large read-only transactions, multiversion protocols would do better. Also, if deadlocks
must be avoided at all costs, the tree protocol would be preferable.

•

Two-phase locking with multiple granularity locking: Use for an application mix where some
applications access individual records and others access whole relations or substantial parts
thereof. The drawbacks of 2PL mentioned above also apply to this one.

•

The tree protocol: Use if all applications tend to access data items in an order consistent with aparticular partial order. This protocol is free of deadlocks, but transactions will often have tolock unwanted nodes in order to access the desired nodes.

•

Timestamp ordering: Use if the application demands a concurrent execution that is equivalent
to a particular serial ordering (say, the order of arrival), rather than any serial ordering. But
conflicts are handled by roll-back of transactions rather than waiting, and schedules are notrecoverable. To make them recoverable, additional overheads and increased response time haveto be tolerated. Not suitable if there are long read-only transactions, since they will starve.
Deadlocks are absent.

•

Validation: If the probability that two concurrently executing transactions conflict is low, thisprotocol can be used advantageously to get better concurrency and good response times with low
overheads. Not suitable under high contention, when a lot of wasted work will be done.

•

Multiversion timestamp ordering: Use if timestamp ordering is appropriate but it is desirable
27

for read requests to never wait. Shares the other disadvantages of the timestamp ordering
protocol.
•Multiversion two-phase locking: This protocol allows read-only transactions to always commit
without ever waiting. Update transactions follow 2PL, thus allowing recoverable schedules withconflicts solved by waiting rather than roll-back. But the problem of deadlocks comes back,though read-only transactions cannot get involved in them. Keeping multiple versions adds space
and time overheads though; therefore plain 2PL may be preferable in low conflict situations.

28

CHAPTER 18
DATA ANALYSIS AND MINING
Data Analysis and OLAP
18.1

What is the difference between transaction processing system and decision support system?
[Marks: 3]
TPS

DSS

1. Transaction-processing systems are 1. Decision-support systems aim to get high-levelsystems that record information about
information out of the detailed information storedtransactions.
in transaction-processing systems, and to use thehigh-level information to make a variety ofdecisions.
2. Contains day-to-day transaction data.

2. Contains historical data.

3. Maintains data required for business 3. Maintains data required for analysis to make
operations.
business decisions.
18.2

Compare ROLAP, MOLAP and HOLAP. [Marks: 3]
OR, Define the terms: OLAP (Online Analytical Processing), ROLAP (Relational OLAP)
and MOLAP (Multidimensional OLAP). [Marks: 6]
OLAP is an approach to swiftly answer multi-dimensional analytical queries. OLAP is part of thebroader category of business intelligence (a.k.a. decision-support system), which also encompasses
relational reporting and data mining.
MOLAP is the classic form of OLAP which stores data cubes in optimized multi-dimensional
array storages rather than in a relational database.
MOLAP generally delivers better performance due to specialized indexing and storage
optimizations. MOLAP also needs less storage space compared to ROLAP because the specialized
storage typically includes compression techniques.
ROLAP works directly with relational databases. The base data and the dimension tables are
stored as relational tables and new tables are created to hold the aggregated information. Thismethodology relies on manipulating the data stored in the relational database to give the appearance oftraditional OLAP's slicing and dicing functionality.
ROLAP is generally more scalable. However, large volume pre-processing is difficult toimplement efficiently so it is frequently skipped. ROLAP query performance can therefore suffer
tremendously. Again, since ROLAP relies more on the database to perform calculations, it has more
limitations in the specialized functions it can use.
Hybrid systems, which store some summaries in memory and store the base data and othersummaries in a relational database, are called HOLAP (Hybrid OLAP) systems.
HOLAP encompasses a range of solutions that attempt to mix the best of ROLAP and MOLAP. It
can generally pre-process swiftly, scale well, and offer good function support.

18.3

What is OLAP? What operations can be done using OLAP? [2008, In-course 2008-2009.
Marks: 1 + 3]
OLAP is an approach to swiftly answer multi-dimensional analytical queries. OLAP is part of thebroader category of business intelligence (a.k.a. decision-support system), which also encompasses
relational reporting and data mining.
The following operations can be done using OLAP:
1. Visualization of data as cross-tabs and data cubes.
2. Pivoting, i.e. changing the dimensions used in a cross-tab and getting responses online, within29

a few seconds.
3. Displaying data at any desired level of granularity.
4. Drill down, rollup, slicing and dicing operations.
18.4

What are the differences between Online Analytical Processing (OLAP) and Online
Transaction Processing (OLTP) systems? [Marks: 4]
OLAP

OLTP

Time Scale

This stores History data for analysis.

This stores current data.

Indexing

Optimizes ad hoc queries by including
lots of indices.

Optimizes update performance byminimizing the number of indices.

Normalization Possibly partially denormalized for
performance reasons as this is used forreporting.

18.5

This is fully normalized.

Organization

Data stored revolves around information
topics.

Data stored revolves around business
functions.

Stored Values

Stores descriptive data.

Stores typically coded data.

Homogeneity

Centralized in data warehouse.

Scattered among different databases
or DBMS.

Data Present
FormatData is present in multidimension format.

Data is present in 2D format.

What do you mean by measure attribute and dimension attribute of a given relation? [2006.
Marks: 1]
OR, For online analytical processing, define measure and dimension attributes using
example. [In-course 2008-2009. Marks: 3]
Given a relation used for data analysis, we can identify some of its attributes as measure attributes,
since they measure some value, and can be aggregated upon. For instance, consider the following
sales relation:
sales(item-name, color, size, number)
The attribute number of the sales relation is a measure attribute, since it measures the number ofunits sold.
Some (or all) of the other attributes of the relation are identified as dimension attributes, since they
define the dimensions on which measure attributes and summaries of measure attributes are viewed.
In the sales relation, item-name, color, and size are dimension attributes.

18.6

What is the multidimensional data model? [Marks: 1]
Multidimensional data model is a model where data can be modeled as dimension attributes andmeasure attributes.

18.7

With example define cross-tabulation/cross-tab/pivot table. [In-course 2008-2009. Marks: 3]
A cross-tab is a table where values for one attribute (say A) form the row headers, values foranother attribute (say B) form the column headers, and the values in an individual cell are derived asfollows:
1. Each cell can be identified by (ai, bj), where ai is a value for A and bj a value for B.
2. If there is at most one tuple with any (ai, bj) value, the value in the cell is derived from thatsingle tuple (if any); for instance, it could be the value of one or more other attributes of thetuple.
3. If there can be multiple tuples with an (ai, bj) value, the value in the cell must be derived byaggregation on the tuples with that value.
30

18.8

What is data cube? [In-course 2008-2009. Marks: 2]
The generalization of a cross-tab, which is 2-dimensional, to n dimensions can be visualized as ann-dimensional cube, called the data cube.

18.9

Design a data cube from a multidimensional data. [Marks: 4]
Consider the following relation:
sales(item-name, color, size, number)
Following are cross-tabs of sales by item-name and color with size = small and size = large
respectively:
color
color
white dark Total
white dark Total
item-name shirt
item-name shirt
20
15
35
10
25
35
pant
10
30
40
pant
5
10
15
Total
30
45
75
Total
15
35
50
The data cube is as follows:

18.10

Explain with suitable example the role of cross-tabulation for analyzing multidimensional
data. [2006. Marks: 2]
Cross-tabulation provides an easy, visual direct interpretation of multidimensional data.
For example, consider the following relation:
sales(item-name, color, size, number)
From the following cross-tab of sales by item-name and color with size = small, an analyst caneasily determine which items are selling quite well and which ones are not much popular.
color
white dark Total
item-name shirt
20
5
25
pant
2
30
32
Total
22
35
57

18.11

What do you mean by concept hierarchy? [Marks: 2]
A hierarchically organized collection of different levels of detail for an attribute (or concept) iscalled a concept hierarchy.

18.12

Given a relation schema sales(item-name, color, size, number), differentiate the cube androllup operations. [In-course 2008-2009. Marks: 4]
The cube operation computes the union of all possible different groupings of the relation, whereas
rollup generates aggregates at multiple levels of a hierarchy.
For example, a cube operation on the sales relation is as follows:
select item-name, color, size, sum(number)from sales
31

group by cube(item-name, color, size)
This query computes the union of eight different groupings of the sales relation:{(item-name, color, size), (item-name, color), (item-name, size), (color, size), (item-name), (color),(size), ()}A representative rollup construct isselect item-name, color, size, sum(number)from sales
group by rollup(item-name, color, size)
Here, only four groupings are generated:{(item-name, color, size), (item-name, color), (item-name), ()}18.13

Give an example of a pair ofgrouping that cannot be expressed byusing a single group by clause with cubeand rollup. [Marks: 6]
Consider an example of hierarchies on
dimensions in the figure beside.
We cannot express a query to seek
aggregation on groups (City, Hour of day)and (City, Date) using a single group byclause with cube and rollup.
Any single group by clause with cubeand rollup that computes these two groups
would also compute other groups also.

18.14

Show how to express group by cube(a, b, c, d) using rollup; your answer should have only
one group by clause. [2008, In-course 2008-2009. Marks: 2]
group by rollup(a), rollup(b), rollup(c), rollup(d)18.15

Given a relation S(Student, Subject, Marks), write a query to rank the top n students basedon the total marks by using ranking. [2008, In-course 2008-2009. Marks: 2]
We assume that multiple students do not have the same marks since otherwise the question is notdeterministic; the query below deterministically returns all students with the same marks as the nstudent, so it may return more than n students.
select student, sum(marks) as total, rank() over (order by (total) desc) as t_rank
from S
group by student
having t_rank <= n

18.16

Write a query to find cumulative balances, equivalent to that shown below, but without
using the extended SQL windowing constructs.
select account_number, date_time,sum(value) over(partition by account_number
order by date_time
Rows unbounded preceding)as balancefrom transaction
order by account_number, date_time[The query gives the cumulative balances on each account just before each transaction on
the account; the cumulative balance of the account is the sum of values of all earlier transactions
32

on the account.]
select t1.account-number, t1.date-time, sum(t2.value)
from transaction as t1, transaction as t2
where t1.account-number = t2.account-number and t2.date-time < t1.date-timegroup by t1.account-number, t1.date-timeorder by t1.account-number, t1.date-time
18.17

Consider the sales relation below.
sales(item-name, color, size, number)
Write an SQL query to compute the cube operation on the relation, giving the relation in the
figure beside. Do not use the with cube construct.
(select color, size, sum(number)from sales
group by color, size)
union(select color, ′all′, sum(number)from sales
group by color)
union(select ′all′, size, sum(number)from sales
group by size)
union(select ′all′, ′all′, sum(number)from sales)

Data Warehouse
18.18

What is a data warehouse? [2008, In-course 2008-2009. Marks: 1]
A data warehouse is a repository (or archive) of information gathered from multiple sources,
stored under a unified schema, at a single site.

18.19

Sketch the data warehouse architecture. [2008. Marks: 1]33

18.20What are the components of a data warehouse? [2006, Marks: 1. In-course 2008-2009,
Marks: 5]
Followings are the components of a data warehouse1:

18.21

•

Database Servers: Operational data accumulated during standard business must be extracted
and stored into a database. Most companies use a relational database stored on a mainframe
server.

•Queries/Reports: Querying is a broad term that encompasses all the activities of requestingdata from a data warehouse for analysis. Reports are then generated to display the results forthe specified query. Querying, obviously, is the whole point of using the data warehouse.

•

OLAP/Multi-dimensional analysis: Relational databases store data in a two dimensional
format; tables of data represented by rows and columns. Multi-dimensional analysis,
commonly referred to as On-Line Analytical Processing (OLAP), offer an extension to therelational model to provide a multi-dimensional view of the data. These tools allow users todrill down from summary data sets into the specific data underlying the summaries. Statisticalanalysis tools provide summary information and help determine the degree of relationshipbetween two factors.

•

Data Mining: Data mining is the process of analyzing business data in the data warehouse tofind unknown patterns or rules of information that one can use to tailor business operations.
Data mining predicts future trends and behaviors, allowing businesses to make proactive,knowledge driven decisions.

What are the features of a data warehouse? [Marks: 3]
A data warehouse has the following features2:•

Subject-oriented
In operational database, data is stored by individual applications. In the data sets for an orderprocessing application, the data for that particular application is kept. These data sets provide
the data for all the functions for entering orders, checking stock, verifying customer’s credit,
and assigning the order for shipment. But these data sets contain only the data that is needed
for those functions relating to this particular application.
In striking contrast, in the data warehouse, data is stored by business subjects, not by
applications. Business subjects differ from enterprise to enterprise. These are the subjects
critical for the enterprise. For a manufacturing company, sales, shipments, and inventory are
critical business subjects. For a retail store, sales at the check-out counter is a critical subject.

•

Integrated
For proper decision making, all the relevant data from the various applications need to bepulled together. The data in the data warehouse come from several operational systems.

•

Time-variant
For an operational system, the stored data contains the current values. A data warehouse,because of the very nature of its purpose – which is analysis and decision making, has tocontain historical data, not just current values.

•

Nonvolatile
In the data warehouse, the data are kept as snapshots over time. The data warehouse is not
updated every time a single order is processed. Data from the operational systems are moved
into the data warehouse at specific intervals.

1
2

Collected From: http://www.pharmacyinformatics.org/datawarehouse.htm#comp [Last accessed: 30-08-2010 07:15pm]
Collected From: http://www.datawarehousesolution.net/defining-features-of-data-warehouse/Data_Warehouse_for_Beginners[Last accessed: 30-08-2010 07:15pm]
34

•

Data Granularity
Data granularity in a data warehouse refers to the level of detail. In a data warehouse, data
are kept summarized at different levels. Depending on the query, a user can then go to theparticular level of detail and satisfy the query.

18.22

Describe benefits and drawbacks of a source-driven architecture for gathering of data at adata warehouse as compared to destination-driven architecture. [In-course 2008-2009. Marks:
3]
In a destination-driven architecture for gathering data, data transfers from the data sources to thedata-warehouse are based on demand from the warehouse, whereas in a source-driven architecture,
the transfers are initiated by each source.
The benefits of a source-driven architecture are:•Data can be propagated to the destination as soon as it becomes available. For a destinationdriven architecture to collect data as soon as it is available, the warehouse would have toprobe the sources frequently, leading to a high overhead.

•

The source does not have to keep historical information. As soon as data is updated, thesource can send an update message to the destination and forget the history of the updates. In
contrast, in a destination-driven architecture, each source has to maintain a history of data
which have not yet been collected by the data warehouse. Thus storage requirements at thesource are lower for a source-driven architecture.

On the other hand, a destination-driven architecture has the following advantages:

18.23

•

In a source-driven architecture, the source has to be active and must handle error conditions
such as not being able to contact the warehouse for some time. It is easier to implement
passive sources, and a single active warehouse. In a destination-driven architecture, eachsource is required to provide only a basic functionality of executing queries.

•

The warehouse has more control on when to carry out data gathering activities, and when toprocess user queries; it is not a good idea to perform both simultaneously, since they may
conflict on locks.

Describe the steps of building a data warehouse. [Marks: 5]
The major steps in developing a data warehouse are3:
1. Identify the data source:
The very first step before starting to develop a data warehouse, the data source should be
identified. The data that are required to be put into the data warehouse need to be figured out.
2. Build customized ETL4 tool
Each data warehouse has different requirements. Therefore, a customized ETL tool is thebetter solution in order to fulfill the requirements.
3. Extraction
This can be the most time consuming part where the data from various data source need to be
grabbed and store into the staging database. Much of the time and effort are needed in writinga custom program to transfer the data from sources into staging database. As a result, during
extraction, the database system that will be used for the staging area needs to be determined
and also the necessary data that are needed needs to be figured out before grabbing them.

3

Collected From: http://www.dotnetspider.com/attachments/Resources/38797-5129-term-paper-on-New-Trends-in-datawarehousing.pdf [Last accessed: 28-08-2010 10:45pm]
4
ETL Tool: The different steps involved in getting data into a data warehouse are called as extract, transform and load or ETLtasks; extraction refers to getting data from the sources, while load refers to loading the data into the data warehouse.
354. Transformation
After extracting the data from various data sources, transformation is needed to ensure thedata consistency. In order to transform the data into data warehouse properly, a way ofmapping the external data sources fields to the data warehouse fields needs to be figured out.
Transformation can be performed during data extraction or while loading the data into data
warehouse.
5. Loading
Once the extracting process, transforming and cleansing has been done, the data are loaded
into the data warehouse. The loading of data can be categorized into two types: the loading ofdata that currently contain in the operational database and the loading of the updates to thedata warehouse from the changes that have occurred in the operational database.
18.24

Describe the functions a user can perform on a data warehouse. [Marks: 3]

18.25

Distinguish between
i. Fact Table and Dimension Table
ii. Star Schema and Snow-Flake Schema [2006. Marks: 3]
i. Tables containing multidimensional data are called fact tables.
To minimize storage requirements, dimension attributes are usually short identifiers that areforeign keys into other tables called dimension tables.
For example, a fact table sales would have dimension attributes item-id, store-id, customer-id
and date, and measure attributes number and price. The attribute store-id is a foreign key intoa dimension table store, which has other attributes such as store location (city, state, country).

ii. A schema with a fact table, multiple dimension tables, and foreign keys from the fact table tothe dimension tables, is called a star schema.
More complex data warehouse designs may have multiple levels of dimension tables; forinstance, the item-info table may have an attribute manufacturer-id that is a foreign key intoanother table giving details of the manufacturer. Such schemas are called snowflake schemas.
18.26

What are the schemas used for data warehouse? Give example of any one of them. [Marks:
2 + 2]The schemas used for data warehouse are:
1. Star Schema
2. Snowflake Schema
Example of star schema:[See the figure in question 18.25]

36

Data Mining
18.27

What is data mining? [In-course 2008-2009. Marks: 1]
The term data mining refers loosely to the process of semi-automatically analyzing large
databases to find useful patterns.
Like knowledge discovery in artificial intelligence (also called machine learning), or statisticalanalysis, data mining attempts to discover rules and patterns from data.

18.28

What are the applications of data mining techniques? [Marks: 4]OR, What are the major areas of data mining? [Marks: 2]
The applications / major areas of data mining are as follows:
1. The most widely used applications are those that require some sort of prediction.
For instance, when a person applies for a credit card, the credit-card company wants topredict if the person is a good credit risk. The prediction is to be based on known attributes ofthe person, such as age, income, debts, and past debt repayment history. Rules for making theprediction are derived from the same attributes of past and current credit card holders,along with their observed behavior, such as whether they defaulted on their credit-card dues.
Other types of prediction include predicting which customers may switch over to acompetitor (these customers may be offered special discounts to tempt them not to switch),predicting which people are likely to respond to promotional mail (“junk mail”), or
predicting what types of phone calling card usage are likely to be fraudulent.
2. Another class of applications looks for associations.
For instance, books those tend to be bought together. If a customer buys a book, an online
bookstore may suggest other associated books. If a person buys a camera, the system maysuggest accessories that tend to be bought along with cameras.
Other types of associations may lead to discovery of causation. For instance, discovery of
unexpected associations between a newly introduced medicine and cardiac problems led to
the finding that the medicine may cause cardiac problems in some people. The medicine wasthen withdrawn from the market.

18.29

Explain a situation when data mining may be useful. [2007. Marks: 2]
For instance, when a person applies for a credit card, the credit-card company wants to predict if
the person is a good credit risk. The prediction is to be based on known attributes of the person, suchas age, income, debts, and past debt repayment history. Rules for making the prediction are derived
from the same attributes of past and current credit card holders, along with their observed
behavior, such as whether they defaulted on their credit-card dues. Data mining would be useful toautomate this rules derivation process.

18.30

Define support and confidence of association rules in data mining with suitable example.
[2006, In-course 2008-2009. Marks: 3]
Support is a measure of what fraction of the population satisfies both the antecedent and theconsequent of the rule. For instance, suppose only 0.001 percent of all purchases include milk andscrewdrivers. Thus, the support for the rule milk ⇒ screwdrivers is low.
Confidence is a measure of how often the consequent is true when the antecedent is true. For
instance, the rule bread ⇒ milk has a confidence of 80% if 80% of the purchases that include breadalso include milk.

18.31

Define minimum support (minsup) and minimum confidence (minconf). [Marks: 3]
An itemset A satisfies a minimum support if its frequency count is greater than or equal to a
specific support count, min_sup.
For example, suppose min_sup is 50%.
37

If sup(A, C) = 50% ≥ 50%, (A, C) is frequentIf sup(A, D) = 25% ≤ 50%, (A, D) is non-frequent
Similarly, an itemset A satisfies a minimum confidence if its frequency count is greater than orequal to a specific confidence count, min_conf.
18.32

What is the process of finding association rules using apriori algorithm? [Marks: 8]
There are two steps for finding association rules using apriori algorithm:
1. Find all frequent itemsets – the sets of item that have minimum support.
2. Generate strong association rules from the frequent itemsets found that have minimum
confidence.
Apriori performs a level-wise iterative search to find frequent itemset Lk from Lk – 1 where Li is afrequent itemset having i items. While finding a frequent itemset, the apriori property is maintained,that is, all non-empty subsets of a frequent itemset must also be frequent. Thus, the first step itself isdivided into two steps:
1. Join Step: Ck is generated by joining Lk-1with itself
2. Prune Step: Any (k – 1)-itemset that is not frequent cannot be a subset of a frequent kitemset
The pseudocode is as follows:
Ck: Candidate itemset of size k
Lk : frequent itemset of size k
L1 = {frequent 1-itemset};for (k = 1; Lk != ∅; k++) do begin
Ck+1 = candidates generated from Lk;
for each transaction t in database do
increment the count of all candidates in Ck+1 that are contained in t;
end
Lk+1 = candidates in Ck+1 with min_support;
end
return ∪k Lk;
The process of finding association rules is as follows:
1. For each frequent itemset I, generate all nonempty subsets of I.
2. For each non-empty subset S of I, output the rule S ⇒ (I – S) if
support_count(I) / support_count(S) ≥ minimum_confidence.

18.33

What are the major characteristics and limitations of Apriori algorithm in connection withassociation rule mining? [In-course 2008-2009. Marks: 5]
Major characteristics of Apriori algorithm:
1. Uses frequent (k – 1)-itemsets to generate candidate k-itemsets.
2. Uses database scan and pattern matching to collect counts for the candidate itemsets.
Major limitations of Apriori algorithm:
1. Generates huge candidate sets. For example, 104 frequent 1-itemset will generate 107
candidate 2-itemsets. Again, to discover a frequent pattern of size 100, e.g. {a1, a2, …, a100},
one needs to generate 2100 ≈ 1030 candidates.
2. Needs multiple scans of database. Specifically, needs (n + 1) scans, where n is the length ofthe longest pattern.

3818.34

Suppose half of all the transactions in a clothes shop purchase jeans, and one-third of alltransactions in the shop purchase T-shirts. Suppose also that half of the transactions thatpurchase jeans also purchase T-shirts. Write down all the (nontrivial) association rules you candeduce from the above information, giving support and confidence of each rule. [2007. Marks:
4]
The rules are as follows. The last rule can be deduced from the previous ones.
Rule
Support Confidence
50%
50%∀ transactions T, true ⇒ buys(T, jeans)33%
33%∀ transactions T, true ⇒ buys(T, t-shirts)
25%
50%∀ transactions T, buys(T, jeans) ⇒ buys(T, t-shirts)
25%
75%∀ transactions T, buys(T, t-shirts) ⇒ buys(T, jeans)

18.35

Consider the following transactional data set:
TID
T100
T200
T300
T400
T500
T600
T700
T800
T900

List of Items
I1, I2, I5
I2, I4I2, I3I1, I2, I4
I1, I3I2, I3I1, I3I1, I2, I3, I5
I1, I2, I3i) Use Apriori algorithm to find the frequent itemsets with their support count whenminimum support count is 2 (i.e. minsup = 2/9 = 22%).
ii) Generate the association rules with their corresponding confidences. [Marks: 12 + 4]i) Generating 1-itemset frequent pattern:
L1
Item Set Supp. Count
{I1}
6
{I2}
7
{I3}
6
{I4}
2
{I5}
2

C1
Item Set Supp. Count
{I1}
6
{I2}
7
{I3}
6
{I4}
2
{I5}
2
Generating 2-itemset frequent pattern:
Item Set{I1, I2}{I1, I3}{I1, I4}{I1, I5}{I2, I3}{I2, I4}{I2, I5}{I3, I4}{I3, I5}{I4, I5}

C2
Supp. Count
4
4
1
2
4
2
2
0
1
0

Item Set{I1, I2}{I1, I3}{I1, I5}{I2, I3}{I2, I4}{I2, I5}

39

L2
Supp. Count
4
4
2
4
2
2

Generating 3-itemset frequent pattern:
C3
Item Set Supp. Count{I1, I2, I3}
2{I1, I2, I5}2

L3
Item Set Supp. Count{I1, I2, I3}
2{I1, I2, I5}
2

Generating 4-itemset frequent pattern: C4 = φ∴ L1 ∪ L2 ∪ L3 is the frequent itemset with minimum support count of 2.
ii) Taking {I1, I2, I5}, its nonempty subsets are {I1, I2}, {I1, I5}, {I2, I5}, {I1}, {I2}, {I5}.
The resulting association rules with corresponding confidences are as follows:
Association RuleI1 ˄ I2 ⇒ I5
I1 ˄ I5 ⇒ I2
I2 ˄ I5 ⇒ I1
I1 ⇒ I2 ˄ I5
I2 ⇒ I1 ˄ I5
I5 ⇒ I1 ˄ I2

Confidencesc{I1, I2, I5} / sc{I1, I2} = 2 / 4 = 50%
2 / 2 = 100%
2 / 2 = 100%
2 / 6 = 33%
2 / 7 = 29%
2 / 2 = 100%

Taking {I1, I2, I3}, its nonempty subsets are {I1, I2}, {I1, I3}, {I2, I3}, {I1}, {I2}, {I3}.
The resulting association rules with corresponding confidences are as follows:
Association RuleI1 ˄ I2 ⇒ I3
I1 ˄ I3 ⇒ I2
I2 ˄ I3 ⇒ I1
I1 ⇒ I2 ˄ I3
I2 ⇒ I1 ˄ I3
I3 ⇒ I1 ˄ I2

Confidence
2 / 4 = 50%
2 / 4 = 50%
2 / 4 = 50%
2 / 6 = 33%
2 / 7 = 29%
2 / 6 = 33%

Similarly, for the rest of the itemsets,
Taking Association Rule
{I1, I2}
I1 ⇒ I2
I2 ⇒ I1{I1, I3}
I1 ⇒ I3
I3 ⇒ I1{I1, I5}
I1 ⇒ I5
I5 ⇒ I1{I2, I3}
I2 ⇒ I3
I3 ⇒ I2{I2, I4}
I2 ⇒ I4
I4 ⇒ I2{I2, I5}
I2 ⇒ I5
I5 ⇒ I2

Confidence
4 / 6 = 67%
4 / 7 = 57%
4 / 6 = 67%
4 / 6 = 67%
2 / 6 = 33%
2 / 2 = 100%
4 / 7 = 57%
4 / 6 = 67%
2 / 7 = 29%
2 / 2 = 100%
2 / 7 = 29%
2 / 2 = 100%40

18.36

Apply the Apriori algorithm to the following data set:
TID
101
102
103
104
105
106
107
108
109
110

Items Purchased
milk, bread, eggs
milk, juice
juice, butter
milk, bread, eggs
coffee, eggs
coffeecoffee, juice
milk, bread, cookies, eggs
cookies, butter
milk, bread

The set of items is { milk, bread, cookies, eggs, butter, coffee, juice }.
Use 0.2 for the minimum support value. [Marks: 6][Similar to 18.35. Try yourself… ☺]18.37

What is decision tree? [Marks: 2]
A decision tree is a decision support tool that uses a tree-like graph or model of decisions andtheir possible consequences, including chance event outcomes, resource costs, and utility.

41

CHAPTER 20DATABASE SYSTEM ARCHITECTURES
20.1

Describe the transaction server process structure.
A typical transaction server system today consists of multiple processes accessing data in sharedmemory, as in the figure. The processes that form part of the database system include•

Server processes: These are processes that receive user queries (transactions), execute them,and send the results back.

•Lock manager process: This process implements lock manager functionality, which includes
lock grant, lock release, and deadlock detection.

•

Database writer process: There are one or more processes that output modified buffer
blocks back to disk on a continuous basis.

•

Log writer process: This process outputs log
records from the log record buffer to stablestorage.

•

Checkpoint process: This process performs
periodic checkpoints.

•

Process monitor process: This process monitors
other processes, and if any of them fails, it takes
recovery actions for the process, such as aborting
any transaction being executed by the failed
process, and then restarting the process.

The shared memory contains all shared data, such as:

20.2

•

Buffer pool•

Lock table•

Log buffer, containing log records waiting to be
output to the log on stable storage•

Cached query plans, which can be reused if thesame query is submitted again

Define speedup and scaleup with respect to parallel database system. What are the factorsthat affect parallel operations as well as speedup and scaleup? [2008, In-course 2008-2009.
Marks: 3 + 2]
Running a given task in less time by increasing the degree of parallelism is called
speedup. For example, if the original system took 60 seconds to perform a task, and two parallelsystems took 30 seconds, then the value of speedup would be equal to 2.
Handling larger tasks by increasing the degree of parallelism is called scaleup. For example, if
the original system can process 100 transactions in a given amount of time, and the parallel system
can process 200 transactions in this amount of time, then the value of scaleup would be equal to 2.5
Factors that affect parallel operations as well as speedup and scaleup:
1. Startup costs
2. Interference
3. Skew

5

Improved response time is speedup, and enhanced throughput is scaleup.
42

20.3

20.4

Mention several architecture models for parallel machines.
•

Shared memory – All the processors share a common memory (Figure 18.8a).

•

Shared disk – All the processors share a common set of disks (Figure 18.8b).

•

Shared nothing – The processors share neither a common memory nor common disk
(Figure 18.8c).

•

Hierarchical – This model is a hybrid of the preceding three architectures (Figure 18.8d).

Mention the characteristics of shared-memory parallel database architecture.
In shared-memory architecture, the processors and disks have access to a common memory,
typically via a bus or through an interconnection network.
Benefit:
Extremely efficient communication between processors.
Drawbacks:
1. The architecture is not scalable beyond 32 or 64 processors because the bus or theinterconnection network becomes a bottleneck (since it is shared by all processors).
2. Maintaining cache-coherency becomes an increasing overhead with increasing number ofprocessors.

20.5

Mention the characteristics of shared-disk parallel database architecture. [2008. Marks: 2]
In the shared-disk model, all processors can access all disks directly via an interconnection
network, but the processors have private memories.
Advantages:
1. Since each processor has its own memory, the memory bus is not a bottleneck.
2. It offers a cheap way to provide a degree of fault tolerance: If a processor (or its memory)
fails, the other processors can take over its tasks, since the database is resident on disks thatare accessible from all processors.
Disadvantage:
Scalability – although the memory bus is no longer a bottleneck, the interconnection to the disk
subsystem is now a bottleneck.

43

20.6

Mention the characteristics of shared-nothing parallel database architectures.
In a shared-nothing system, each node of the machine consists of a processor, memory, and one
or more disks. The processors at one node may communicate with another processor at another node
by a high-speed interconnection network.
Advantages:
1. Since local disk references are serviced by local disks at each processor, the shared-nothingmodel overcomes the disadvantage of requiring all I/O to go through a single interconnection
network; only queries, accesses to nonlocal disks, and result relations pass through the
network.
2. The interconnection networks for shared-nothing systems are usually designed to be scalable,
so that their transmission capacity increases as more nodes are added. Consequently, sharednothing architectures are more scalable and can easily support a large number of processors.
Disadvantages:
1. High costs of communication.
2. High costs of nonlocal disk access, which are higher than in shared-memory or shared-diskarchitecture since sending data involves software interaction at both ends.

20.7

Compare different parallel database architectures. [In-course 2008-2009. Marks: 7]
1. Architecture:
In shared-memory architecture, the processors and disks have access to a common memory,
typically via a bus or through an interconnection network.
In the shared-disk model, all processors can access all disks directly via an interconnection
network, but the processors have private memories.
In a shared-nothing system, each node of the machine consists of a processor, memory, andone or more disks. The processors at one node may communicate with another processor atanother node by a high-speed interconnection network.
2. Scalability:
Shared-memory architecture is not scalable beyond 32 or 64 processors because the bus or
the interconnection network becomes a bottleneck (since it is shared by all processors).
In shared-disk model, although the memory bus is no longer a bottleneck, the interconnection
to the disk subsystem becomes a bottleneck.
The shared-nothing architecture is scalable. However, the communication between nodes andnonlocal disk access costs are very high.

20.8

What are the advantages and disadvantages of distributed database system? [2008. Marks:
3]
Advantages:
1. Sharing data – Users at one site may be able to access the data residing at other sites.
2. Autonomy – Each site is able to retain a degree of control over data that are stored locally.
3. Availability – If one site fails in a distributed system, the remaining sites may be able tocontinue operating if data items are replicated in several sites.
Disadvantages:
1. Added complexity (such as more costly software development, greater potential for bugs,
increasing processing overhead to achieve intersite coordination etc.) required to ensure
proper coordination among the sites.
2. Recovery from failure is more complex in distributed systems than in centralized systems.

44

20.9

Why is it relatively easy to port a database from a single processor machine to a
multiprocessor machine if individual queries need not be parallelized?
Porting is relatively easy to a shared memory multiprocessor machine. Databases designed forsingle-processor machines already provide multitasking, allowing multiple processes to run on thesame processor in a time-shared manner, giving a view to the user of multiple processes running inparallel. Thus, coarse-granularity parallel machines logically appear to be identical to singleprocessor machines, making the porting relatively easy.
Porting a database to a shared disk or shared nothing multiprocessor architecture is a little harder.

20.10

Transaction server architectures are popular for client-server relational databases, wheretransactions are short. On the other hand, data server architectures are popular for clientserver object-oriented database systems, where transactions are expected to be relatively long.
Give two reasons why data servers may be popular for object-oriented databases but not for
relational databases.
Data servers are good if data transfer is small with respect to computation, which is often the casein applications of OODBs such as computer aided design. In contrast, in typical relational databaseapplications such as transaction processing, a transaction performs little computation but may touch
several pages, which will result in a lot of data transfer with little benefit in a data server architecture.
Another reason is that structures such as indices are heavily used in relational databases, and will
become spots of contention in data server architecture, requiring frequent data transfer. There are nosuch points of frequent contention in typical current-day OODB applications such as computer aided
design.

20.11

Instead of storing shared structures in shared memory, an alternative architecture would
be to store them in the local memory of a special process, and access the shared data byinterprocess communication with the process. What would be the drawback of such anarchitecture?
The drawbacks would be:
1. Two interprocess messages would be required to acquire locks, one for the request and one
to confirm grant. Interprocess communication is much more expensive than memory access,so the cost of locking would increase.
2. The process storing the shared structures could also become a bottleneck.

20.12

What are the factors that can work against linear scaleup in a transaction processing
system? Which of the factors are likely to be the most important in each of the following
architectures: shared memory, shared disk, and shared nothing?
Increasing contention for shared resources prevents linear scale-up with increasing parallelism.
In a shared memory system, contention for memory (which implies bus contention) will result infalling scale-up with increasing parallelism.
In a shared disk system, it is contention for disk and bus access which affects scale-up.
In a shared-nothing system, inter-process communication overheads will be the main impeding
factor. Since there is no shared memory, acquiring locks, and other activities requiring message
passing between processes will take more time with increased parallelism.

45

CHAPTER 21
PARALLEL DATABASES
21.1

Discuss three basic data partitioning strategies. [2008. Marks: 4]
The three basic data partitioning strategies are as follows, assuming that there are n disks, D0, D1,…, Dn – 1, across which the data are to be partitioned.
1. Round-Robin:This strategy scans the relation in any order and sends the i-th tuple to disk number Di mod n.
The round-robin scheme ensures an even distribution of tuples across disks.
2. Hash partitioning:This declustering strategy designates one or more attributes from the given relation’s schemaas the partitioning attributes. A hash function is chosen whose range is {0, 1, . . . , n - 1}.
Each tuple of the original relation is hashed on the partitioning attributes. If the hash function
returns i, then the tuple is placed on disk Di.
3. Range partitioning:This strategy distributes contiguous attribute-value ranges to each disk. It chooses apartitioning attribute, A, as a partitioning vector. The relation is partitioned as follows. Let[v0, v1, …, vn - 2] denote the partitioning vector, such that, if i < j, then vi < vj. Consider a tuple
t such that t[A] = x. If x < v0, then t goes on disk D0. If x ≥ vn – 2, then t goes on disk Dn – 1. If vi≤ x < vi + 1, then t goes on disk Di + 1.

21.2

For each of the three partitioning techniques, namely round-robin, hash partitioning, andrange partitioning, give an example of a query for which that partitioning technique would
provide the fastest response.
Round robin partitioning: When relations are large and queries read entire relations, roundrobin gives good speed-up and fast response time.
Hash partitioning: For point queries, this gives the fastest response, as each disk can process aquery simultaneously. If the hash partitioning is uniform, even entire relation scans can be performed
efficiently.
Range partitioning: For range queries which access a few tuples, this gives fast response.

21.3

In a range selection on a range-partitioned attribute, it is possible that only one disk may
need to be accessed. Describe the benefits and drawbacks of this property.
If there are few tuples in the queried range, then each query can be processed quickly on a single
disk. This allows parallel execution of queries with reduced overhead of initiating queries on
multiple disks.
On the other hand, if there are many tuples in the queried range, each query takes a long time to
execute as there is no parallelism within its execution. Also, some of the disks can become hot-spots,
further increasing response time.

21.4

What factors could result in skew when a relation is partitioned on one of its attributes by:a. Hash partitioningb. Range partitioning
In each case, what can be done to reduce the skew? [2008, 2007. Marks: 3]
a. Hash-partitioning:
Too many records with the same value for the hashing attribute, or a poorly chosen hashfunction without the properties of randomness and uniformity, can result in a skewed
partition.
46

To improve the situation, we should experiment with better hashing functions for that
relation.
b. Range-partitioning:
Non-uniform distribution of values for the partitioning attribute (including duplicate values
for the partitioning attribute) which are not taken into account by a bad partitioning vector isthe main reason for skewed partitions.
Sorting the relation on the partitioning attribute and then dividing it into n ranges with equal
number of tuples per range will give a good partitioning vector with very low skew.
21.5

What form of parallelism (interquery, interoperation, or intraoperation) is likely to be themost important for each of the following tasks:a. Increasing the throughput of a system with many small queriesb. Increasing the throughput of a system with a few large queries, when the number ofdisks and processors is large [2007. Marks: 4]
a. When there are many small queries, inter-query parallelism gives good throughput.
Parallelizing each of these small queries would increase the initiation overhead, without any
significant reduction in response time.
b. With a few large queries, intra-query parallelism is essential to get fast response times. Given

that there are large number of processors and disks, only intra-operation parallelism can takeadvantage of the parallel hardware – for queries typically have few operations, but each one
needs to process a large number of tuples.
21.6

With pipelined parallelism, it is often a good idea to perform several operations in apipeline on a single processor, even when many processors are available.
a. Explain why.
b. Would the arguments you advanced in part a hold if the machine has a shared-memoryarchitecture? Explain why or why not.
c. Would the arguments in part a hold with independent parallelism? (That is, are there
cases where, even if the operations are not pipelined and there are many processors
available, it is still a good idea to perform several operations on the same processor?)a. The speed-up obtained by parallelizing the operations would be offset by the data transfer
overhead, as each tuple produced by an operator would have to be transferred to its
consumer, which is running on a different processor.
b. In shared-memory architecture, transferring the tuples is very efficient. So the above
argument does not hold to any significant degree.
c. Even if two operations are independent, it may be that they both supply their outputs to acommon third operator. In that case, running all three on the same processor may be better
than transferring tuples across processors.

21.7

Give two examples of a join that is not a simple equi-join for which partitioned parallelism
can be used. What attributes should be used for partitioning?
a. r ⋈(r.A = s.B) ˄ (r.A < s.C) s
Here we have extra conditions which can be checked after the join. Hence partitioned
parallelism is useful.
b. r ⋈ (r.A ≥ s.B * 20) ˄ (r.A < s.B + 10) s
This is a query in which an r tuple and an s tuple join with each other if they fall into thesame range of values. Hence partitioned parallelism applies naturally to this scenario.
For both the queries, r should be partitioned on attribute A and s on attribute B.
47

21.8

Suppose you have a histogram where values are between 1 and 100, and are partitioned
into 10 ranges, 1– 10, 11–20, ..., 91– 100, with frequencies 15, 5, 20, 10, 10, 5, 5, 20, 5, and 5,
respectively. Give a load-balanced range partitioning function to divide the values into 5partitions. [2007. Marks: 3]
A partitioning vector which gives 5 partitions with 20 tuples in each partition is: [21, 31, 51, 76].
The 5 partitions obtained are 1 - 20, 21 - 30, 31 - 50, 51 - 75 and 76 - 100.
The assumption made in arriving at this partitioning vector is that within a histogram range, eachvalue is equally likely.

21.9

Describe the benefits and drawbacks of pipelined parallelism.
Benefit:No need to write intermediate relations to disk only to read them back immediately.
Drawbacks:
1. Cannot take advantage of high degrees of parallelism, as typical queries do not have large
number of operations.
2. Not possible to pipeline operators which need to look at all the input before producing any
output.
3. Since each operation executes on a single processor, the most expensive ones take a longtime to finish. Thus speed-up will be low in spite of parallelism.
21.10

Some parallel database systems store an extra copy of each data item on disks attached to adifferent processor, to avoid loss of data if one of the processors fails. Why is it a good idea topartition the copies of the data items of a processor across multiple processors?
The copies of the data items at a processor should be partitioned across multiple other processors,rather than stored in a single processor, for the following reasons:
1. To better distribute the work that should have been done by the failed processor, among theremaining processors.
2. Even when there is no failure, this technique can to some extent deal with hot-spots createdby read only transactions.

48

CHAPTER 22
DISTRIBUTED DATABASES
22.1

Discuss the relative advantages of centralized and distributed databases.
A distributed database allows a user convenient and transparent access to data which is not stored
at the site, while allowing each site control over its own local data. A distributed database can bemade more reliable than a centralized system because if one site fails, the database can continuefunctioning, but if the centralized system fails, the database can no longer continue with its normal
operation. Also, a distributed database allows parallel execution of queries and possibly splitting one
query into many parts to increase throughput.
A centralized system is easier to design and implement. A centralized system is cheaper tooperate because messages do not have to be sent.

22.2

Explain how the followings differ: fragmentation transparency, replication transparency,
and location transparency.
With fragmentation transparency, the user of the system is unaware of any fragmentation thesystem has implemented. A user may formulate queries against global relations and the system will
perform the necessary transformation to generate correct output.
With replication transparency, the user is unaware of any replicated data. The system must
prevent inconsistent operations on the data. This requires more complex concurrency control
algorithms.
Location transparency means the user is unaware of where data are stored. The system must route
data requests to the appropriate sites.

22.3

How might a distributed database designed for a local-area network differ from one
designed for a wide-area network?
Data transfer on a local-area network (LAN) is much faster than on a wide-area network (WAN).
Thus replication and fragmentation will not increase throughput and speed-up on a LAN, as much as
in a WAN. But even in a LAN, replication has its uses in increasing reliability and availability.

22.4

When is it useful to have replication or fragmentation of data? Explain your answer.
Replication is useful when there are many read-only transactions at different sites wanting accessto the same data. They can all execute quickly in parallel, accessing local data. But updates become
difficult with replication.
Fragmentation is useful if transactions on different sites tend to access different parts of thedatabase.

22.5

Explain the notions of transparency and autonomy. Why are these notions desirable from ahuman-factors standpoint?
Autonomy is the amount of control a single site has over the local database. It is importantbecause users at that site want quick and correct access to local data items. This is especially true
when one considers that local data will be most frequently accessed in a database. Transparency
hides the distributed nature of the database. This is important because users should not be required toknow about location, replication, fragmentation or other implementation aspects of the database.

22.6

To build a highly available distributed system, you must know what kinds of failures canoccur.
a. List possible types of failure in a distributed system.
b. Which items in your list from part a are also applicable to a centralized system?
a. The types of failure that can occur in a distributed system include
i. Computer failure (site failure).
49

ii. Disk failure.
iii. Communication failure.
b. The first two failure types can also occur on centralized systems.
22.7

Consider a distributed system with two sites, A and B. Can site A distinguish among thefollowing?
•••

B goes down.
The link between A and B goes down.
B is extremely overloaded and response time is 100 times longer than normal.

What implications does your answer have for recovery in distributed systems? [2007.
Marks: 2]
Site A cannot distinguish between the three cases until communication has resumed with site B.
The action which it performs while B is inaccessible must be correct irrespective of which of thesesituations has actually occurred, and must be such that B can re-integrate consistently into the
distributed system once communication is restored.
22.8

Explain the difference between data replication in a distributed system and themaintenance of a remote backup site.
In remote backup systems all transactions are performed at the primary site and the data is
replicated at the remote backup site. The remote backup site is kept synchronized with the updates atthe primary site by sending all log records. Whenever the primary site fails, the remote backup sitetakes over processing.
The distributed systems offer greater availability by having multiple copies of the data at
different sites whereas the remote backup systems offer lesser availability at lower cost andexecution overhead.
In a distributed system, transaction code runs at all the sites whereas in a remote backup system itruns only at the primary site. The distributed system transactions follow two-phase commit to havethe data in consistent state whereas a remote backup system does not follow two-phase commit andavoids related overhead.

22.9

What are the advantages and disadvantages of data replication? [2007. Marks: 4]
Advantages of Data Replication:
1. Availability: failure of site containing relation r does not result in unavailability of r as
replicas exist.
2. Parallelism: queries on r may be processed by several nodes in parallel.
3. Reduced data transfer: The more replicas of r there are, the greater the chance that theneeded data will be found in the site where the transaction is executing. Hence, datareplocation minimizes movement of data between sites.
Disadvantages of Data Replication:
1. Increased overhead on update: each replica of relation r must be updated.
2. Increased complexity of concurrency control: concurrent updates to distinct replicas maylead to inconsistent data unless special concurrency control mechanisms are implemented.
22.10

Consider a relation that is fragmented horizontally by plant-number:
employee (name, address, salary, plant-number)
Assume that each fragment has two replicas: one stored at the New York site and one
stored locally at the plant site. Describe a good processing strategy for the following queriesentered at the San Jose site.
a. Find all employees at the Boca plant.
b. Find the average salary of all employees.
c. Find the highest-paid employee at each of the following sites: Toronto, Edmonton,
50

Vancouver, Montreal.
d. Find the lowest-paid employee in the company.
c. i. Send the query Πname(employee) to the Boca plant.
ii. Have the Boca location send back the answer.
d. i. Compute average at New York.
ii. Send answer to San Jose.
e. i. Send the query to find the highest salaried employee to Toronto, Edmonton, Vancouver,and Montreal.
ii. Compute the queries at those sites.
iii. Return answers to San Jose.
f. i. Send the query to find the lowest salaried employee to New York.
ii. Compute the query at New York.
iii. Send answer to San Jose.
22.11

Consider the relations
employee (name, address, salary, plant-number)machine (machine-number, type, plant-number)
Assume that the employee relation is fragmented horizontally by plant-number, and thateach fragment is stored locally at its corresponding plant site. Assume that the machine
relation is stored in its entirety at the Armonk site. Describe a good strategy for processing
each of the following queries. [2007. Marks: 4]
a. Find all employees at the plant that contains machine number 1130.
b. Find all employees at plants that contain machines whose type is “milling machine.”c. Find all machines at the Almaden plant.
d. Find employee ⋈ machine.
c. i. Perform Πplant-number(σmachine-number = 1130(machine)) at Armonk.
ii. Send the query Πname(employee) to all site(s) which are in the result of the previous query.
iii. Those sites compute the answers.
iv. Union the answers at the destination site.
b. i. This strategy is the same as (a), except the first step should be to performΠplant-number(σtype= “milling machine” (machine)) at Armonkc. i. Perform σplant-number = x (machine) at Armonk, where x is the plant-number for Almaden.
ii. Send the answers to the destination site.
d. Strategy 1:
i. Group machine at Armonk by plant number.
ii. Send the groups to the sites with the corresponding plant-number.
iii. Perform a local join between the local data and the received data.
iv. Union the results at the destination site.
Strategy 2:
Send the machine relation at Armonk, and all the fragments of the employee relation to
the destination site. Then perform the join at the destination site.
There is parallelism in the join computation according to the first strategy but not in thesecond. Nevertheless, in a WAN the amount of data to be shipped is the main cost factor. We
expect that each plant will have more than one machine, hence the result of the local join ateach site will be a cross-product of the employee tuples and machines at that plant. This crossproduct’s size is greater than the size of the employee fragment at that site. As a result thesecond strategy will result in less data shipping, and will be more efficient.

51

22.12For each of the strategies in the previous question-answer, state how your choice of astrategy depends on:a. The site at which the query was enteredb. The site at which the result is desireda. Assuming that the cost of shipping the query itself is minimal, the site at which the query was
submitted does not affect our strategy for query evaluation.
b. For the first query, we find out the plant numbers where the machine number 1130 is present,
at Armonk. Then the employee tuples at all those plants are shipped to the destination site.
We can see that this strategy is more or less independent of the destination site. The same canbe said of the second query. For the third query, the selection is performed at Armonk andresults shipped to the destination site. This strategy is obviously independent of thedestination site.
For the fourth query, we have two strategies. The first one performs local joins at all the plantsites and their results are unioned at the destination site. In the second strategy, the machine
relation at Armonk as well as all the fragments of the employee relation are first shipped tothe destination, where the join operation is performed. There is no obvious way to optimize
these two strategies based on the destination site. In the answer to the previous question, we
saw the reason why the second strategy is expected to result in less data shipping than thefirst. That reason is independent of destination site, and hence we can in general preferstrategy two to strategy one, regardless of the destination site.

22.13

Compute r ⋉ s for the relations of the following figure: [2007. Marks: 4]

Is ri ⋉ rj necessarily equal to rj ⋉ ri? Under what conditions does ri ⋉ rj = rj ⋉ ri hold?
r⋉s= A B C
1 2 3
5 3 2
In general, ri ⋉ rj ≠ rj ⋉ ri. This can be easily seen here where r ⋉ s ≠ s ⋉ r. r ⋉ s is as above whiles⋉r= C
3
3
2

D
4
6
3

E
5
8
2

By definition, ri ⋉ rj = ΠRi(ri ⋈ rj) and rj ⋉ ri = ΠRj(ri ⋈ rj), where Ri and Rj are the schemas of ri
and rj respectively. For ΠRi(ri ⋈ rj) to be always equal to ΠRj(ri ⋈ rj), the schemas Ri and Rj must be
the same.

52