Please answer all question 1&2 then a,b,c,d Homework 6 Spring 2017-Q Microsoft W

Question

Please answer all question 1&2 then a,b,c,d Homework 6 Spring 2017-Q Microsoft Word non-commercial use AaBIcel Aabbod AaBbc, AaBboc AaB AaBbccu x. A- s A. a a E 1 Normal No spada. Heading 1 Heading 2 Title subtitle Style Paragraph objective: this assignment is designed to let you have a hands-on experience with normalization. A normalized schema provides a good starting point for further development of a database application and easier maintenance it later on. Normalization is a process of modifying an existing database schema to bring its component tables into compliance with a series of normal forms. The goal of the normalization is to ensure tha every non-key attribute in every table is directly dependent on the whole candidate key (nothing else but the key). 1. Consider a relation R with six attributes A, B, C, D, E, F. You are given the following dependences: C F, E BC- D, D B, and ED Isl (ABDE) a candidate key of this relation? If not, is (ADE)? our answer. Answer: 2. Consider the relation F with six attributes FR 1.DAY with the followino denen dences:

Explanation / Answer

Dependencies we have are:

C FE
BC D
D B
ED C

Since A is not present in RHS of any functional dependency(FD), A must be a part of candidate key

F cannot be a part of Candidate key, since F does not appear on LHS of any FD.

Therefore the candidate keys we have are:

ABC ABCDEF (Removal of B we cannot produce BD, and removal of C we cannot produce CDEF)

ACD ABCDEF (replace B with D, since we have D B)

ADE ABCDEF (Removal of D we cannot produce BCD, and removal of E we cannot produce CE)

Therefore ABDE is not a candidate key, since it can be reduced to ADE (because D B)

2.

For F,R,I,D,A,Y the FDs we have:

R I

RY F

FY A

FA R

a.

Before finding the all keys we need to determine the candidate keys:

observation : Since D,Y are not present in RHS, they should be present in the Candidate key
       And since I is not present in LHS of any FD, it cannot be present in the Candidate key

FDY FRIDAY
RDY FRIDAY

From the candidate keys we can generate other super keys:

FRDY, FIDY, FDAY, FRIDY, FRDAY, FIDAY,FRIDAY,
RIDY, RDAY, RIDAY

b.

Attribute set that belongs to any candidate key are called Prime Attributes.

Therefore {F,R,D,Y} are prime attribute
And {I,A} are non prime attribute

c.
R I      full functional dependency

RY F    full functional dependency

FY A   partial functional dependency // since FY is part of candidate key and A is non prime attribute

FA R   full functional dependency

d.
The table is in 1NF.
Reason: As per the 2NF there must not be any partial dependency of any column on primary key. It means that for a table that has compound primary key, each column in the table that is not part of the primary key must depend upon the entire concatenated key for its existence. If any column depends only on one part of the concatenated key, then the table fails Second normal form. (culprit: FY A)

I explained in detail as far as possible. If you face any problem with the logic and understanding of the problem, please let me know through the comment below. I shall be glad to guide you through the problem.

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