DBMS Test - 5

280 Views
93 Downloads
PDF 5,322,578 Bytes

GATE
CSE

Database
ManagementSystem

Test-05Solutions

DATABASE MANAGEMENT SYSTEM
1. Simplest way of deadlock recovery is

a) Roil back
b) Preempt resource
c) Lock one of the processes
d) Kill one of the processes
Answer: (d)
Solution:
Recover from deadlock
Recovery though per-emption
Recovery though rollback
Recovery though killing process
Recovery though killing process is the crudest but simplest way
to break a deadlock
2. A schema describes:
a) data elements
b) records and files
c) record relationship
d) all of the above
Answer: (d)
3. An ER model of a database consists of entity types A and B.
These are connected by a relationship R which does not have its
own attribute. Under which one of the following conditions, can
the relational table for R be merged with that of A?

1

a) Relationship R is one-to-many and the participation of A in R
is total.

b) Relationship R is one-to-many and the participation of A in R
is Partial.

c) Relationship R is many-to-one and the participation of A in R
is total.

d) Relationship R is many-to-one and the participation of A in R
is Partial.

Answer: (c)
Solution:

Hence the relation R is merged with R.
4. A rule in a limited entry decision table is a

a) row.af the table consisting of condition entries
b) row of the table consisting of action entries
c) column of the table consisting of condition entries and

corresponding action entries
d) columns of the table consisting of conditions of the stub
Answer: (c)
Solution:
A rule in a limited entry decision table is a column of the table
consisting of the condition entries and the corresponding action
entries. A decision table is a tool to use in both testing and
requirements management. Essentially, it is a structured exercise

2

to formulate requirements when dealing with complex business
rules.
5. Relational database schema normalization is NOT for:
a) Reducing the number of joins required to satisfy a query.
b) Eliminating uncontrolled redundancy of data stored in the

database.
c) Eliminating number of anomalies that could otherwise occur

with inserts and deletes.
d) ensuring that functional dependencies are enforced
Answer: (a)
Solution:
Goal of normalization
(i) Zero present redundancy
(ii) Lossless join decomposition
(iii) Dependency preserving decomposition
6. The characteristic equation of T flip flop
a) +1 = * Qn+1 = TQ*n
b) +1 = + Qn+1 = T + Qn
c) +1 = Qn+1 = TQn
d) +1 = Qn+1 = TQ
The symbols used have the usual meaning
Answer: (a)

3

Solution:

Characteristic equation:
T Q(n+1)
0 Qn
1 Q�n

Excitation table:
T Q(n+1) T
00 0
01 1
10 1
11 0

The characteristic equation of a T-flip flop is
+1 = * Qn+1 = TQ*
7. Find the false statement:
a) The relationship construct known as the weak relationship

type was defined by Dey, Story& Barron (1999)
b) A weak relationship occurs when two relationship types are

linked by either Event-Precedent sequence or Condition
Precedent sequence.
c) Conceptual model is not accurate representation of "Universe
of interest"

4

d) Ternary, Quaternary and Quandary relationships are shown
through a series of application scenario's and vignettes.

Answer: (c)
Solution:
It is important that a conceptual model be an accurate
representation of the universe interest such an objective can only
be achieved though the careful evaluation of how wills the
developed conceptual model addressed the explicit and /or
implicit questions associated with the requirements
specifications
8. Consider the table R with attributes A, B and C. The f functional
dependencies that hold on B are: A  B, C  AB. Which of the
following statements is/are True?
i. L The decomposition of B into R1, (C, A) and

R2 (A, B) is lossless.
ii. The decomposition of F into B, (A, B) and

R2 (B, C) is lossy.
a) only I
b) only II
c) Both I and II
d) Neither I or II
Answer: (a)
Solution:

Given relation R(ABC)with FD’s
{A→B, C→AB}

5

A decomposition is lossless join if all attributes of relation are
covered in the decomposition and there should be some
common attributed in the decomposition relation which must be
the candidate key of either of the relation

R1(C, A) R2(A, B)
C→A A→B
Common attribute is A which is candidate key of R2 so, this
decomposition is lossless join
R1(A, B) R2(B, C)
A→B BC
Common attribute is B, which is not a candidate key of either
relation, so this decomposition is lossy
9. Consider two transactions T1 and T2 and four schedules S1, S2,
S3, S4, of T1 and T2 as given below:
T1: R1[x] W1[x] W1[Y]
T2: R2[x] R1[x] W2[Y]
S1: R1[x] R2[x] R2[Y]
W1[x] W1[x] W2[Y]
S2: R1[x] R2[x] R2[Y]
W1[x] W2[x] W1[Y]
S3: R1[x] W1[x] R2[Y]
W1[y] R2[y] W2[Y]
S4: R2[x] R2[y] R1[x]
W1[x] W1[y] W2[Y]
Which of the above schedules are conflicts serializable?

6

a) S1 and S2
b) S2 and S3
c) S3 Only
d) S4 Only
Answer: (b)
Solution:
In S1 R2(x) and W1(x) conflicts hence T2

Data Loading...

DBMS Test - 5 - PDF Flipbook