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CS712 Midterm Current Paper 16 Dec 2017

CS712 Midterm Current Paper 16 Dec 2017

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,A4,A2)=2(135+11865+225)=2(11775)=23550

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