SUBJECT: DESIGN OF DATABSE NOTE: PLEASE PROVIDE SOLUTION TO THIS QUESTION THANKS
ID: 3856039 • Letter: S
Question
SUBJECT: DESIGN OF DATABSE
NOTE: PLEASE PROVIDE SOLUTION TO THIS QUESTION
THANKS
Description of Homework 1. Consider the following relations, and for each relation, a) Identify a primary key b) Identify the functional dependencies c) Is the relation in 2NF? If not, find a 2NF decomposition d) Is the relation in 3NF? If not, find a 3NF decomposition e) Is the relation in BCNF? If not, find a decomposition that is in BCNF. Identify any functional dependencies that are not preserved i. Workl(empld, empName, dateHired, jobTitle, jobLevel) Assumption: Each employee has unique empld. ii. Work2(empld, empName, jobTitle, ratingDate, raterName, rating) Assumption: Multiple ratings can be stored for each employee, but a maximum of one rating can be given to an employee on a given date ii. Work3(empld, empName, projectNo, projectName, projBudget, empManager, hoursAssigned) Assumption: Each employee can be assigned to more than one project. iv. Work4(empld, empName, schoolAttended, degreeAwarded, graduationDate) Assumption: More than one degree can be stored for an employee, however an employee never earns the same degree twice Work5(empld, empName, empSSN, dependentName, dependentAddress relationtoEmp) Assumption: An employee may have several dependents, and different dependents can have same name, however no two dependents with same relation to employee has same name v.Explanation / Answer
1)
i)
a)pk - empId
b)There are not fun. dependencies
c)2NF
d)3NF
e)BCNF
ii)
a)pk - (empId, ratingDate)
b)partial dependencies
empId -> empName
empId -> jobTitle
c) Not in 2NF
d) Not in 3NF
e) Not in BCNF
iii)
a)pk - (empId, projectNo)
b) partial dependencies
empId -> empName
projectNo -> projectName
empId -> empManager
c) Not in 2NF
d) Not in 3NF
e) Not in BCNF
iv) a)pk - (empId, degreeAwarded)
b) partial dependenciess
empId -> empName
c) Not in 2NF
d) Not in 3NF
e) Not in BCNF
v) a)pk - (empId, relationtoEmp)
b)partial dependencies
empId -> empName
c) Not in 2NF
d) Not in 3NF
e) Not in BCNF
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