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,A2) = 0
Aff(A1,A3) = 15 + 20 + 10
= 45
Aff(A1,A4) = 0
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,A4,A2)=2(135+11865+225)=2(11775)=23550
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,A4,A2)=2(135+11865+225)=2(11775)=23550
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