COP4540 MidtermReview (2)

8/8/2019 COP4540 MidtermReview (2)

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COP4540 Database

Management SystemMidterm Review

Reviewed byRamakrishna

Slides created by Fernando & edited by Ramakrishna


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AGENDA Ch1. Overview of DBMSs

Ch2. Database Design Ch3. Relational Model

Ch4. Relational Algebra

Ch19. Normal Forms Ch5. SQL


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CH1 EXERCISES 1.4. Explain the difference between external, internal, andconceptual schemas. How are these different schema layers

related to the concepts oflogical and physical dataindependence?

External schemas:

Allow data access to be customized at the level of individual usersor groups of users using different VIEWSof the same conceptualschema.

Views are not stored in DBMS but they generated on-demand.

In the data model of DBMS.

Conceptual (logical) schemas:

Describes all the data in terms of the data model. In a relational

DBMS, it describes all relations stored. While there are several views for a given database, there is exactly

one conceptual schema to allusers.

Internal (physical) schemas:

Describes how the relations described in the conceptual schemaare actually stored on disk (or other physical media).


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CH1 EXERCISES Logical Data Independence

Protection from changes in Logical Structure

of Data (The Conceptual Schema) Provided by External Schema (Views)

Physical Data Independence

Protection from changes in Physical Structureof Data

Provided by Conceptual Schema


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CH2 EXERCISES 2.2. A university database contains information

aboutprofessors (id. by SSN) and courses (id. by courseid). Professors

teach courses; each of the following situations concerns the Teaches

relationship set. For each situation, draw an ER diagram thatdescribes it (assuming no further constraints hold).

1. Professors can teach the same course in several

semesters, and each offering must be recorded.


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CH2 EXERCISES 2.2. CONT

2. Professors can teach the same course in several

semesters, and only the most recent such offering needs to

be recorded. (Assume this condition applies in allsubsequent questions.)


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3. Every professor must teach some course.


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4. Every professor teaches exactly one course (no more, noless).


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5. Every professor teaches exactly one course (no more, no

less), and every course must be taught by someprofessor.


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CH2 EXERCISES

2.2. CONT

6. Certain courses can be

taught by a team of

professors jointly, but it is

possible that no one

professor in a team can

teach the course. Model

this situation, introducing

additional entity sets and

relationship sets if

necessary.


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CH2 EXERCISES 2.4

A company database needs to store information about

employees (identified byssn

, withsalaryandpho

neas attributes),departments (identified by dno, with dnameand budgetas attributes),

and children of employees (with nameand ageas attributes).

Employees workin departments; each department is managed byan

employee; a child must be identified uniquely by namewhen the parent

(who is an employee; assume that only one parent works for the

company) is known. We are not interested in information about a childonce the parent leaves the company.


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CH2 EXERCISES


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Ch2. Database Design

Ch3. Relational Model & SQL




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CH3 EXERCISES 3.2. How many distinct tuples are in a

relation instance with cardinality 22?

Since a relation is formally defined as a setof

tuples, if the cardinality is 22 (i.e., there are 22

tuples), there must be 22 distinct tuples.


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CH3 EXERCISES 3.4. What is the difference between a

candidate key and the primary key for a

given relation? What is a superkey?

Theprimary keyis the key selected by the

DBA from among the group ofcandidate

keys, all of which uniquely identify a tuple. A

superkeyis a set of attributes that contains a

key.


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CH3 EXERCISES 3.8. Answer each of the following questions briefly. Thequestions are based on the following relational schema:

Emp(eid: integer,ename: string,age: integer,salary: real)Works(eid: integer,did: integer,pcttime: integer)Dept(did: integer,dname: string,budget: real,managerid:

integer)

1. Give an example of a foreign key constraint that involves theDept relation. What are the options for enforcing this constraintwhen a user attempts to delete a Dept tuple?

Consider the following example. It is natural to require that the didfield

of Works should be a foreign key, and refer to Dept.CREATE TABLE Works ( eid INTEGER NOT NULL ,

did INTEGER NOT NULL ,pcttime INTEGER,PRIMARY KEY (eid, did),FOREIGN KEY (did) REFERENCES Dept )


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CH3 EXERCISES OPTIONS for maintaining referential

integrity

ON DELETE {CASCADE, SET DEFAULT,

SET NULL, NO ACTION}

ON UPDATE {CASCADE, SET DEFAULT,SET NULL, NO ACTION}


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CH3 EXERCISES 3.8.

2. Write the SQL statements required to create the preceding

relations, including appropriate versions of allprimary and

foreign key integrity constraints.CREATE TABLE Emp (

eid INTEGER,

ename CHAR(10),

age INTEGER,

salary REAL,

PRIMARY KEY (eid)

)

CREATE TABLE Works (

eid INTEGER,

did INTEGER,

pcttime INTEGER,

PRIMARY KEY (eid, did),

FOREIGN KEY (did)

REFERENCES Dept,

FOREIGN KEY (eid)

REFERENCES Emp,

ON DELETE CASCADE

)

CREATE TABLE Dept (

did INTEGER,

budget REAL,

managerid INTEGER ,

PRIMARY KEY (did),

FOREIGN KEY (managerid)

REFERENCES Emp,

ON DELETE SET NULL

)


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CH3 EXERCISES 3.8.

3. Define the Dept relation in SQL so thatevery department is guaranteed to have a

manager. CREATE TABLE Dept (

did INTEGER,budget REAL,managerid INTEGER NOT NULL ,

PRIMARY KEY (did),FOREIGN KEY (managerid) REFERENCES Emp)


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CH3 EXERCISES 3.8.

4. Write an SQL statement to add John

Doe as an employee with eid= 101,age = 32 and salary= 15, 000.

INSERT INTO Emp (eid, ename, age, salary)

VALUES (101, John Doe, 32, 15000)


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CH3 EXERCISES 3.8.

5. Write an SQL statement to give every

employee a 10 percent raise.

UPDATE Emp E

SET E.salary = E.salary * 1.10


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CH3 EXERCISES 3.8.

6. Write an SQL statement to delete the Toy

department. Given the referential integrityconstraints you chose for this schema,

explain what happens when this statement is

executed.

DELETEFROM Dept D

WHERE D.dname = Toy


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CH5 EXERCISES 5.2. Consider the following schema:Suppliers(sid: integer,sname: string,address: string)Parts(pid: integer,pname: string,color: string)Catalog(sid: integer,pid: integer,cost: real)

The Catalog relation lists the prices charged forparts bySuppliers. Write the following queries in SQL:

1. Find the pnames ofparts for which there is some supplier. SELECT DISTINCT P.pname

FROM Parts P, Catalog CWHERE P.pid = C.pid

5. Find the sids of suppliers who charge more for some part than

the average cost of that part (averaged over all the supplierswho supply that part). SELECT DISTINCT C.sid

FROM Catalog CWHERE C.cost >( SELECT AVG (C1.cost)

FROM Catalog C1WHERE C1.pid = C.pid )


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CH5 EXERCISES8. Find the sids of Suppliers who supply a red parts and a greenpart

SELECT DISTINCT C.sid

FROM Catalog C, Parts P

WHERE C.pid = P.pid AND P.color = Red

INTERSECT

SELECT DISTINCT C1.sid

FROM Catalog C1, Parts P1

WHERE C1.pid = P1.pid AND P1.color = Green

9. Find the sids of Suppliers who supply a red parts or a green

partSELECT DISTINCT C.sid FROM Catalog C, Parts P

WHERE C.pid = P.pid AND P.color = Red

UNION

SELECT DISTINCT C1.sid FROM Catalog C1, Parts P1

WHERE C1.pid = P1.pid AND P1.color = Green


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Ch3. Relational Model

Ch4. RelationalAlgebra



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CH4 EXERCISES 4.2. Given two relations R1 and R2, whereR1 contains N1

tuples,R2 contains N2 tuples, and N2 >N1 >0, give the minand maxpossible sizes for the resulting relational algebraexpressions:(1)R1UR2, (2)R1R2, (3)R1R2, (4)R1R2, (5)a=5(R1), and (6)

a(R1)


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CH4 EXERCISES 4.4. Consider the Supplier-Parts-Catalog schema from the

previous question. State what the following queries compute:

Find the Supplier names of the suppliers who supply a red part that costs less

than 100 dollars.

This Relational Algebra statement does not return anything


than 100 dollars and a green part that costs less than 100 dollars.


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CH4 EXERCISES

4.4. Consider the Supplier-Parts-Catalog schema from the

previous question. State what the following queries compute:

Find the Supplier ids of the suppliers who supply a red part that costs less than

100 dollars and a green part that costs less than 100 dollars.


than 100 dollars and a green part that costs less than 100 dollars.


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CH4 EXERCISES

4.6. What is relational completeness? Ifa query language is relationally

complete, can you write any desiredquery in that language?

Relational completenessmeans that a querylanguage can express all the queries that can

be expressed in relational algebra. It does notmean that the language can express anydesired query.


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AGENDA

Ch1. Overview of DBMSs


Ch3. Relational Model


Ch19. Normal Forms

Ch5. SQL


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CH19EXERCISESA relation R is in third normal form if for

every functional dependency of the form X

A one of the following statements is true:

A X that is, A is a trivial functionaldependency , or (1)

X is a superkey, or (2)

A is part of some key forR (3)

A relation R is in BCNF if (1) or (2)


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CH19EXERCISES

19.2. Consider a relation Rwith fiveattributesABCDE. You are given thefollowing dependencies:A B,BC E,

and ED A.1. List all keys forR.

CDE, ACD, BCD

2. Is Rin 3NF?

R is in 3NF because B, E and A are all parts of keys.3. Is Rin BCNF?

R is not in BCNF because none of A, BC and EDcontain a key.


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GOOD LUCK!!

RAMAKRISHNA

[email protected]

ECS 234

COP4540 MidtermReview (2)

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