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Question:

Published on: 23 January, 2022

Consider the following relations and write down expressions for the following queries:

EMP (eid, ename, age, address, salary)

Works (eid, did, hours)

Dept (did, dname, managerid)

List the names of employees who work for the ‘Research’ department for 8 hours using Relational Algebra.
List the name and address of all employees with department number 5 using Relational Calculas.
Find the managerid of managers who manage only departments with budgets greater than 1 lac using SQL.
Find the employee who has highest salary using SQL.

Answer:

Insufficient information. We are just assuming the Relation Works as Works (eid, did, hours, budget).

Hence the answer will be as follows:

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Software Engineering C Programming and Practices Data Structures & Algorithm Mobile computing Computer Graphics and Multimedia Compiler Design

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Let R (ABCDE) be a relation schema and consider the following functional dependencies F = { AB→ E, AD→B, B→ C, C →D}. Find out the candidate key.

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Find out the closure of attribute set (AG), i.e., (AG)+ in R. Set of FD’s are as given: R= {A,B,C,G,H,I}, F= {A → B,A → C,CG → H,B → H}. Is (AG) a super key of R?

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What is the need for normalization? Explain 2nd and 3rd normal form with example.

Consider the following relation:

EMP_PROJ { SSn, Pnumber, Hours, Ename, Pname, Plocation}

Assume { SSn, Pnumber} as primary key.

The dependencies are {SSn, Pnumber}→ Hours; SSn →Ename; Pnumber → {Pname, Plocation}.

Normalize the above relation into 2NF.

23 January, 2022

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