DATABASE SYSTEMS-MINDTAPV2.0
DATABASE SYSTEMS-MINDTAPV2.0
13th Edition
ISBN: 9780357427873
Author: Coronel
Publisher: CENGAGE L
Question
100%
Book Icon
Chapter 3, Problem 26P
Program Plan Intro

Primary Key:

A Primary Key in a database table is a field in the table that uniquely identifies every row or record present in the database table.

Example:

Students in Universities are assigned a unique registration number.

Therefore, in a STUDENT database table, the attribute “reg_no” acts as primary key.

Foreign Key:

Foreign Key is a column in a relational database table which provides a relation between two tables. It provides a cross reference between tables by pointing to primary key of another table.

Example:

In STUDENT database table, the attribute “reg_no” acts as primary key and in COURSE database table in which the student selects his or her course, the same “reg_no” acts as foreign key for the STUDENT table.

Many to One Relationship:

When more than one record in a database table is associated with only one record in another table, the relationship between the two tables is referred as many to one relationship. It is also represented as M: 1 relationship.

One to Many Relationship:

When one record in a database table is associated with more than one record in another table, the relationship between the two tables is referred as one to many relationship. It is also represented as1: M relationship. This is the opposite of many to one relationship.

One to One Relationship:

When one record in a database table is associated with one record in another table, the relationship between the two tables is referred as one to one relationship. It is also represented as1: 1relationship.

RELATIONAL DIAGRAM:

Relational Diagram is also known as Entity Relational Diagram. It is used to define the conceptual view of the database as viewed by the end user. It is used to depict the database’s main components: entities, relationships and attributes. It describes how data is related to each other.

Expert Solution & Answer
Check Mark

Explanation of Solution

Given database tables:

Table Name: CHARTER

CHAR_TRIP CHAR_DATE CHAR_PILOT CHAR_COPILOT AC_NUMBER CHAR_DESTINATION CHAR_DISTANCE CHAR_HOURS_FLOWN CHA_HOURS_WAIT CHAR_FUEL_GALLONS CHAR_OIL_QTS CUS_CODE
10001 05-Feb-18 104   2289L ATL 936.0 5.1 2.2 354.1 1 10011
10002 05-Feb-18 101   2778V BNA 320.0 1.6 0.0 72.6 0 10016
10003 05-Feb-18 105 109 4278Y GNV 1574.0 7.8 0.0 339.8 2 10014
10004 06-Feb-18 106   1484P STL 472.0 2.9 4.9 97.2 1 10019
10005 06-Feb-18 101   2289L ATL 1023.0 5.7 3.5 397.7 2 10011
10006 06-Feb-18 109   4278Y STL 472.0 2.6 5.2 117.1 0 10017
10007 06-Feb-18 104 105 2778V GNV 1574.0 7.9 0.0 348.4 2 10012
10008 07-Feb-18 106   1484P TYS 644.0 4.1 0.0 140.6 1 10014
10009 07-Feb-18 105   2289L GNV 1574.0 6.6 23.4 459.9 0 10017
10010 07-Feb-18 109   4278Y ATL 998.0 6.2 3.2 279.7 0 10016
10011 07-Feb-18 101 104 1484P BNA 352.0 1.9 5.3 66.4 1 10012
10012 08-Feb-18 101   2289L MOB 884.0 4.8 4.2 215.1 0 10010
10013 08-Feb-18 105   4278Y TYS 644.0 3.9 4.5 174.3 1 10011
10014 09-Feb-18 106   4278V ATL 936.0 6.1 2.1 302.6 0 10017
10015 09-Feb-18 104 101 2289L GNV 1645.0 6.7 0.0 459.5 2 10016
10016 09-Feb-18 109 105 2778V MQY 312.0 1.5 0.0 67.2 0 10011
10017 10-Feb-18 101   1484P STL 508.0 3.1 0.0 105.5 0 10014
10018 10-Feb-18 105 104 4278Y TYS 644.0 3.8 4.5 167.4 0 10017

Table Name: AIRCRAFT

AC_NUMBER MODE-CODE AC_TTAF AC_TTEL AC_TTER
1484P PA23-250 1833.1 1833.1 101.8
2289L C-90A 4243.8 768.9 1123.4
2778V PA31-350 7992.9 1513.1 789.5
4278Y PA31-350 2147.3 622.1 243.2

Table Name: MODEL

MOD_CODE MOD_MANUFACTER MOD_NAME MOD_SEATS MOD_CHG_MILE
B200 Beechcraft Super KingAir 10 1.93
C-90A Beechcraft KingAir 8 2.67
PA23-250 Piper Aztec 6 1.93
PA31-350 Piper Navajao Chiettan 10 2.35

Table Name: PILOT

EMP_NUM PIL_LICENSE PIL_RATINGS PIL_MED_TYPE PIL_MED_DATE PIL_PTI35_DATE
101 ATP ATP/SEL/MEL/Instr/CFII 1 20-Jan-18 11-Jan-18
104 ATP ATP/SEL/MEL/Instr 1 18-Dec-17 17-Jan-18
105 COM COMM/SEL/MEL/Instr/CFI 2 05-Jan-18 02-Jan-18
106 COM COMM/SEL/MEL/Instr 2 10-Dec-17 02-Feb-18
109 COM ATP/SEL/MEL/SES/Instr/CFII 1 22-Jan-18 15-Jan-18

Table Name: EMPLOYEE

EMP_NUM EMP_TITLE EMP-LNAME EMP_FNAME EMP_INITIAL EMP_CODE EMP_HIRE_DATE
100 Mr. Kolrnycz George D 15-Jun-62 15-Mar-08
101 Ms. Lewis Rhonda G 19-Mar-85 25-Apr-06
102 Mr. Vandam Rhett   14-Nov-78 18-May-13
103 Ms. Jones Anne M 11-May-94 26-Jul-17
104 Mr. Lange John P 12-Jul-91 20-Aug-10
105 Mr. Williams Robert D 14-Mar-95 19-Jun-17
106 Mrs. Duzak Jeanine K 12-Feb-88 13-Mar-18
107 Mr. Deante George D 01-May-95 02-Jul-16
108 Mr. Wiesanbach Paul R 14-Feb-86 03-Jun-13
109 Ms. Travis Elizabeth K 18-Jun-81 14-Feb-16
110 Mrs. Genkazi Lieghla W 19-May-90 29-Jun-10

Table Name: EMPLOYEE

CUS_CODE CUS_LNAME CUS_FNAME CUS_INITIAL CUS_AREACODE CUS_PHONE CUS_BALANCE
10010 Ramas Alfred A 615 844-2573 0.00
10011 Dunne Leona K 713 894-1293 0.00
10012 Smith Kathy W 615 894-2285 896.54
10013 Owolski Paul F 615 894-2180 1285.19
10014 Orlando Myron 615 222-1672 673.21
10015 OBrian Amy B 713 442-3381 1014.86
10016 Brown James G 615 297-1228 0.00
10017 Williams George 615 290-2556 0.00
10018 Fariss Anne G 713 382-7185 0.00
10019 Smith Olette K 615 297-3809 453.98

PRIMARY KEY in the above tables:

For Table Name: CHARTER:

Primary Key: CHAR_TRIP

“CHAR_TRIP” acts as primary key of the table because the attribute “CHAR_TRIP” is a unique ID that is assigned to every individual trip by the charter plane. It also uniquely identifies every other row present in the database table.

For Table Name: AIRCRAFT:

Primary Key: AC_NUMBER

“AC_NUMBER” acts as primary key of the table because the attribute “AC_NUMBER” is a unique number that is assigned to every individual charter plane and is used to distinguish among them. It also uniquely identifies every other row present in the database table.

For Table Name: MODEL:

Primary Key: MOD_CODE

“MOD_CODE” acts as primary key of the table because the attribute “MOC_CODE” is a unique number that is assigned to every individual model of the charter plane and is used to distinguish models among them. It also uniquely identifies every other row present in the database table.

For Table Name: PILOT:

Primary Key: EMP_NUM

“EMP_NUM” acts as primary key of the table because the attribute “EMP_NUM” is a unique number that is assigned to every pilot that flies an aircraft. It also uniquely identifies every other row present in the database table.

For Table Name: EMPLOYEE:

Primary Key: EMP_NUM

“EMP_NUM” acts as primary key of the table because the attribute “EMP_NUM” is a unique number or ID that is assigned to every employee that works in the airline. It also uniquely identifies every other row present in the database table.

For Table Name: CUSTOMER:

Primary Key: CUS_CODE

“CUS_CODE” acts as primary key of the table because the attribute “CUS_CODE” is a unique code that is assigned to every customer that books a flight with the airline. It also uniquely identifies every other row present in the database table.

FOREIGN KEY in the above tables:

For Table Name: CHARTER:

Primary Key: CHAR_PILOT,CHAR_COPILOT,AC_NUMBER,CUS_CODE

“CHAR_PILOT” acts as foreign key of the table because the attribute “CHAR_PILOT” is also present in the table PILOT and it references PILOT and hence it forms a link between the two tables.

“CHAR_COPILOT” acts as foreign key of the table because the attribute “CHAR_COPILOT” is also present in the table PILOT and it references PILOT and hence it forms a link between the two tables.

“AC_NUMBER” acts as foreign key of the table because the attribute “AC_NUMBER” is also present in the table AIRCRAFT and it references AIRCRAFT and hence it forms a link between the two tables.

“CUS_CODE” acts as foreign key of the table because the attribute “CUS_CODE” is also present in the table CUSTOMER and it references CUSTOMER and hence it forms a link between the two tables.

For Table Name: AIRCRAFT:

Foreign Key: MOD_CODE

“MOD_CODE” acts as foreign key of the table because the attribute “MOD_CODE” is also present in the table MODEL and it references MODEL and hence it forms a link between the two tables.

“For Table Name: MODEL:

Foreign Key: None

There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.

For Table Name: PILOT:

Primary Key: EMP_NUM

“EMP_NUM” acts as foreign key of the table because the attribute “EMP_NUM” is also present in the table EMPLOYEE and it references EMPLOYEE and hence it forms a link between the two tables.

For Table Name: EMPLOYEE:

Foreign Key: None

There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.

For Table Name: CUSTOMER:

Foreign Key: None

There is no Foreign Key attribute present in the table because there is no attribute in the table except the primary key which is present in any other database table.

Relationship among the tables:

A CUSTOMER requests many CHARTER trips and more than one CHARTER trip can be requested by a single customer. Hence, the relationship between CUSTOMER and CHARTER is one to many or 1: M.

An AIRCRAFT can fly many CHARTER trips but that each CHARTER trip is flown by one AIRCRAFT. Hence, the relationship between AIRCRAFT and CHARTER is one to many or 1: M.

Each AIRCRAFT references a single MODEL but a MODEL references many AIRCRAFT. Hence, the relationship between AIRCRAFT and MODEL is many to one or M: 1.

Many CHARTER trips are flown by a single PILOT and with a single COPILOT but a PILOT can fly only one charter trip at a time. Hence, the relationship between CHARTER and PILOT is many to one or M: 1.

All PILOTS are EMPLOYEES, but not all EMPLOYEES are PILOTS – some are mechanics, accountants, and so on.

There is an optional (default) 1:1 relationship between EMPLOYEE and PILOT. It can be represented that EMPLOYEE is the “parent” of PILOT.

Relational Diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, PILOT, EMPLOYEE and CUSTOMER:

The Relational Diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, PILOT, EMPLOYEE and CUSTOMER is shown below:

DATABASE SYSTEMS-MINDTAPV2.0, Chapter 3, Problem 26P

The above relational diagram represents the one to many relationship between CUSTOMER represented as “1” and CHARTER represented as “∞”, one to many relationship between AIRCRAFT represented as “1”  and CHARTER represented as “∞” , many to one relationship between AIRCRAFT  represented as “∞” and MODEL represented as “1”  , many to one relation between CHARTER represented as “∞”  and PILOT represented as “1”  and an optional one to one relationship between PILOT represented as “1” and EMPLOYEE represented as “1”. A new entity PILOT_1 table is created to split the PILOT table and represent the many to one relationship between CHARTER represented as “∞”   and PILOT_1 represented as “1”.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
ut da Q4: Consider the LIBRARY relational database schema shown in Figure below a. Edit Author_name to new variable name where previous surname was 'Al - Wazny'. Update book_Authors set Author_name = 'alsaadi' where Author_name = 'Al - Wazny' b. Change data type of Phone to string instead of numbers. عرفنه شنو الحل BOOK Book id Title Publisher_name BOOK AUTHORS Book id Author_name PUBLISHER Name Address Phone e u al b are rage c. Add two Publishers Company existed in UK. insert into publisher(name, address, phone) value('ali','uk',78547889), ('karrar', 'uk', 78547889) d. Remove all books author when author name contains second character 'D' and ending by character 'i'. es inf rmar nce 1 tic عرفته شنو الحل e. Add one book as variables data? عرفته شنو الحل
Add a timer in the following code. public class GameGUI extends JPanel {    private final Labyrinth labyrinth;    private final Player player;    private final Dragon dragon;    private Timer timer;    private long elapsedTime;  public GameGUI(Labyrinth labyrinth, Player player, Dragon dragon) {        this.labyrinth = labyrinth;        this.player = player;        this.dragon = dragon;        String playerName = JOptionPane.showInputDialog("Enter your name:");        player.setName(playerName);        elapsedTime = 0;        timer = new Timer(1000, e -> {            elapsedTime++;            repaint();        });        timer.start(); }    @Override    protected void paintComponent(Graphics g) {        super.paintComponent(g);        int cellSize = Math.min(getWidth() / labyrinth.getSize(), getHeight() / labyrinth.getSize());}
Change the following code so that when player wins the game, the game continues by creating new GameGUI with the same player. However the player's starting position is same, everything else should be reseted. public static void main(String[] args) {        Labyrinth labyrinth = new Labyrinth(10);        Player player = new Player(9, 0);        Random rand = new Random();        Dragon dragon = new Dragon(rand.nextInt(10), 9);        JFrame frame = new JFrame("Labyrinth Game");        GameGUI gui = new GameGUI(labyrinth, player, dragon);         frame.setLayout(new BorderLayout());        frame.setSize(600, 600);        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);        frame.add(gui, BorderLayout.CENTER);        frame.pack();        frame.setResizable(false);        frame.setVisible(true);    } public class GameGUI extends JPanel {    private final Labyrinth labyrinth;    private final Player player;    private final Dragon dragon;    private Timer timer;    private long…

Chapter 3 Solutions

DATABASE SYSTEMS-MINDTAPV2.0

Ch. 3 - Prob. 11RQCh. 3 - Prob. 12RQCh. 3 - Use Figure Q3.13 to answer Questions 1317. FIGURE...Ch. 3 - Create the table that results from applying a...Ch. 3 - Write the relational algebra formula to apply an...Ch. 3 - Create the table that results from applying an...Ch. 3 - Using the tables in Figure Q3.13, create the table...Ch. 3 - Prob. 18RQCh. 3 - Prob. 19RQCh. 3 - Prob. 20RQCh. 3 - Identify and describe the components of the table...Ch. 3 - Identify the primary keys. FIGURE Q3.22 THE...Ch. 3 - Identify the foreign keys. FIGURE Q3.22 THE...Ch. 3 - Create the ERM. FIGURE Q3.22 THE CH03_THEATER...Ch. 3 - Create the relational diagram to show the...Ch. 3 - Prob. 26RQCh. 3 - What would be the conceptual view of the INDEX...Ch. 3 - Prob. 1PCh. 3 - Prob. 2PCh. 3 - Do the tables exhibit referential integrity?...Ch. 3 - Describe the type(s) of relationship(s) between...Ch. 3 - Prob. 5PCh. 3 - Prob. 6PCh. 3 - Prob. 7PCh. 3 - Prob. 8PCh. 3 - Create the relational diagram to show the...Ch. 3 - Prob. 10PCh. 3 - Prob. 11PCh. 3 - Create the relational diagram to show the...Ch. 3 - Prob. 13PCh. 3 - Do the tables exhibit referential integrity?...Ch. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - For each table, identify the primary key and the...Ch. 3 - Prob. 18PCh. 3 - Do the tables exhibit referential integrity?...Ch. 3 - Identify the TRUCK tables candidate key(s). FIGURE...Ch. 3 - For each table, identify a superkey and a...Ch. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Create the ERD. (Hint: Look at the table contents....Ch. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Create the table that would result from applying...Ch. 3 - Create the table that would result from applying...Ch. 3 - Create the table that would result from applying a...
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
  • Text book image
    Database Systems: Design, Implementation, & Manag...
    Computer Science
    ISBN:9781305627482
    Author:Carlos Coronel, Steven Morris
    Publisher:Cengage Learning
    Text book image
    Database Systems: Design, Implementation, & Manag...
    Computer Science
    ISBN:9781285196145
    Author:Steven, Steven Morris, Carlos Coronel, Carlos, Coronel, Carlos; Morris, Carlos Coronel and Steven Morris, Carlos Coronel; Steven Morris, Steven Morris; Carlos Coronel
    Publisher:Cengage Learning
    Text book image
    Oracle 12c: SQL
    Computer Science
    ISBN:9781305251038
    Author:Joan Casteel
    Publisher:Cengage Learning
  • Text book image
    A Guide to SQL
    Computer Science
    ISBN:9781111527273
    Author:Philip J. Pratt
    Publisher:Course Technology Ptr
    Text book image
    CMPTR
    Computer Science
    ISBN:9781337681872
    Author:PINARD
    Publisher:Cengage
Text book image
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781305627482
Author:Carlos Coronel, Steven Morris
Publisher:Cengage Learning
Text book image
Database Systems: Design, Implementation, & Manag...
Computer Science
ISBN:9781285196145
Author:Steven, Steven Morris, Carlos Coronel, Carlos, Coronel, Carlos; Morris, Carlos Coronel and Steven Morris, Carlos Coronel; Steven Morris, Steven Morris; Carlos Coronel
Publisher:Cengage Learning
Text book image
Oracle 12c: SQL
Computer Science
ISBN:9781305251038
Author:Joan Casteel
Publisher:Cengage Learning
Text book image
A Guide to SQL
Computer Science
ISBN:9781111527273
Author:Philip J. Pratt
Publisher:Course Technology Ptr
Text book image
CMPTR
Computer Science
ISBN:9781337681872
Author:PINARD
Publisher:Cengage