CS712 Midterm Current Paper 16 Dec 2017

CS712 Midterm Current Paper 16 Dec 2017

Cs712 current paper

16 dec ,2017  11:am

Mcqs:10   (10marks)             total marks:30 marks     duration:1 hour

Q1:Is DDMS is benificiant in term of cost and access?    (5 marks)

Acess: Types of Access of DDBS:

Local Access: the access by the users connected to a site and accessing the data from the same site.

Remote Access: a user connected to a site lets say site 1 and accessing the data from site 2.

Global Access: no matter where ever the access is made, data will be displayed after being collected from all locations.

Q2:Distributed Database Design Strategies: (5marks)

·         Top-down approach

It is used when a database is being designed from scratch.

Issues: Fragmentation & allocation

·         Bottom Up approach

Integration of existing database.

Issues: Design of the export and global schemas     

Q3:mixture of below 2 example.(5marks)

 Example:1

Consider the relation Proj(jNo, jName, budget, loc). Assume that the following applications are defined to run on this relation. In each case we also give the SQL specification.

q1:       SELECT BUDGET FROM PROJ WHERE JNO=Value

q2:       SELECT JNAME, BUDGET FROM PROJ

q3:       SELECT JNAME FROM PROJ WHERE LOC=Value

q4:       SELECT SUM(BUDGET) FROM PROJ WHERE LOC=Value

Solution:

Let A₁=jNo, A₂=jName, A₃=budget, A₄=loc

So, the Attribute Usage Matrix will be

Example:2

Let us assume that ref1(qk) = 1 for all qk and S1. If the application frequencies are

Acc1(q1) = 15             Acc2(q1) = 20             Acc3(q1) = 10

Acc1(q2) = 5               Acc2(q2) = 0               Acc3(q2) = 0

Acc1(q3) = 25             Acc2(q3) = 25             Acc3(q3) = 25

Acc1(q4) = 3               Acc2(q4) = 0               Acc3(q4) = 0

Solution:

The Frequency Matrix will be

And the Attribute Usage Matrix is also given, such that

Now we find out the Attribute Affinity Matrix

Aff(A₁,A₁) = 15 + 20 + 10 = 45
Aff(A₁,A₂) = 0

Aff(A₁,A₃) = 15 + 20 + 10 = 45
Aff(A₁,A₄) = 0

Aff(A₂,A₂) = 5 + 25 + 25 + 25 = 80

Aff(A₂,A₃) = 5

Aff(A₂,A₄) = 25 + 25 + 25 = 75

Aff(A₃,A₃) = 15 + 20 + 10 + 5 + 3 = 53

Aff(A₃,A₄) = 3

Aff(A₄,A₄) = 25 + 25 + 25 + 3 = 78

So, the Attribute Affinity Matrix will be

Q4: Let us consider the AA matrix given as under and study the contribution of moving attribute A4 between A1 and A2.

Solution:

Given Attribute Affinity is
Bond(A₁,A₄)=45*0 + 0*75 + 45*3 + 0* 78=135
Bond(A₄,A₂)=0*0 + 75*80 + 3*5 + 78* 75=6000+15+5850=11865
Bond(A₁,A₂)=45*0 + 0*80 + 45*5 + 0* 75=225
So,
Cont(A₁,A_4,A₂)=2(135+11865+225)=2(11775)=23550