Database Systems: Design, Implementation, & Management
Database Systems: Design, Implementation, & Management
12th Edition
ISBN: 9781305627482
Author: Carlos Coronel, Steven Morris
Publisher: Cengage Learning
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Chapter 3, Problem 24P
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.

Candidate Key:

A set of attributes (minimal) which can uniquely identify a record is known as candidate keys. A attribute which could’ve been a primary key but for was not chosen as primary key for some reason is a candidate key.

The value of candidate is not null for every record in database and in unique.

The candidate key can be of composite attributes and there can be more than one candidate key in a relation.

SUPER KEY:

A set of one or more columns or attributes to uniquely identify rows in a table is called super key.

A Super Key is a candidate key containing redundant attributes.

All candidate keys are super keys as candidate keys are derived from super keys.

Any primary key plus any attribute is a super key.

SECONDARY KEY:

Secondary keys are the set of attributes which are not selected as primary key but are considered to be candidate keys for the primary key of the table.

Numbers of attributes that constitute secondary key are arbitrary.

Secondary keys are also known as alternate keys.

After selecting the attributes from candidate key to form a primary key, the remaining attributes of candidate key are called secondary keys.

Expert Solution & Answer
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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.

Super key and Secondary Key in the above tables:

For Table Name: CHARTER:

Super Key:

CHAR_TRIP+CHAR_DATE: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

CHAR_DATE+AC_NUMBER+CHAR_DESTINATION: This combination is an alternate key which will identify every other record present in the table is unique and it is less likely that one aircraft on same date will travel to the same destination twice.

For Table Name: AIRCRAFT:

Super Key:

AC_NUM+MOD_CODE: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

MOD_CODE: This is an alternate key which will identify every other record present in the table and is unique.

For Table Name-MODEL:

Super Key:

MOD_CODE+MOD_NAME: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

MOD_MANUFACTURER+MOD_NAME: This combination is an alternate key which will identify every other record present in the table and it is less likely that two models with same name and same manufacturer exist.

For Table Name-PILOT:

Super Key:

EMP_NUM+PIL_LICENSE: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

PIL_LICENSE+PIL_MED_DATE: This combination is an alternate key which will identify every other record present in the table and it is less likely that two pilots with same license and same medical certificate exist.

For Table Name-EMPLOYEE:

Super Key:

EMP_NUM+EMP_DOB: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

EMP_LNAME+EMP_FNAME+EMP_DOB: This combination is an alternate key which will identify every other record present in the table and it is less likely that two employees with same first name, same last name and same date of birth exist.

For Table Name-CUSTOMER:

Super Key:

CUS_CODE+CUS_LNAME: This combination of attributes can uniquely identify every other record present in the table.

Secondary Key:

CUS_LNAME+CUS_FNAME+CUS_PHONE: This combination is an alternate key which will identify every other record present in the table and it is less likely that two customers with same first name, same last name and same phone number exist.

Candidate keys present in the above table:

For table CHARTER:

Candidate Key: None

No practical candidate keys are possible. For example:

CHAR_DATE + CHAR_DESTINATION + AC_NUMBER + CHAR_PILOT + CHAR_COPILOT will not necessarily yield unique matches, because it is possible to fly an aircraft to the same destination twice on one date with the same pilot and copilot.

For table AIRCRAFT:

Candidate Key: None

No practical candidate keys are possible as no combination of attributes will yield unique matches, because it is possible to fly a same aircraft to the same destination twice on one date with the same pilot and copilot.

For table MODEL:

Candidate Key: None

No practical candidate keys are possible as no combination of attributes will yield unique matches, because it is possible to fly a same model of aircraft to the same destination twice on one date with the same pilot and copilot.

For table PILOT:

Candidate Key: None

No practical candidate keys are possible as no combination of attributes will yield unique matches, because it is possible to fly a same model of aircraft to the same destination twice on one date with the same pilot and copilot.

For table EMPLOYEE:

Candidate Key: EMP_LNAME + EMP_FNAME + EMP_INITIAL + EMP_DOB

The combinations of the above attributes will yield a unique outcome and hence is acceptable as candidate key.

For table CUSTOMER:

Candidate Key: CUS_LNAME + CUS_FNAME + CUS_INITIAL + CUS_PHONE

The combinations of the above attributes will yield a unique outcome and hence is acceptable as candidate key.

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Chapter 3 Solutions

Database Systems: Design, Implementation, & Management

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...
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