
Concept explainers
Primary Key:
A Primary Key in a
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.

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.
Want to see more full solutions like this?
Chapter 3 Solutions
DATABASE SYSTEMS-MINDTAPV2.0
- Why is JAVA OOP is really difficult to study?arrow_forwardMy daughter is a Girl Scout and it is time for our cookie sales. There are 15 neighbors nearby and she plans to visit every neighbor this evening. There is a 40% likelihood that someone will be home. If someone is home, there is an 85% likelihood that person will make a purchase. If a purchase is made, the revenue generated from the sale follows the Normal distribution with mean $18 and standard deviation $5. Using @RISK, simulate our door-to-door sales using at least 1000 iterations and report the expected revenue, the maximum revenue, and the average number of purchasers. What is the probability that the revenue will be greater than $120?arrow_forwardQ4 For the network of Fig. 1.41: a- Determine re b- Find Aymid =VolVi =Vo/Vi c- Calculate Zi. d- Find Ay smid e-Determine fL, JLC, and fLE f-Determine the low cutoff frequency. g- Sketch the asymptotes of the Bode plot defined by the cutoff frequencies of part (e). h-Sketch the low-frequency response for the amplifier using the results of part (f). Ans: 28.48 2, -72.91, 2.455 KS2, -54.68, 103.4 Hz. 38.05 Hz. 235.79 Hz. 235.79 Hz. 14V 15.6ΚΩ 68kQ 0.47µF Vo 0.82 ΚΩ V₁ B-120 3.3kQ 0.47µF 10kQ 1.2k0 =20µF Z₁ Fig. 1.41 Circuit forarrow_forward
- a. [10 pts] Write a Boolean equation in sum-of-products canonical form for the truth table shown below: A B C Y 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 0 1 1 1 1 0 1 1 0 a. [10 pts] Minimize the Boolean equation you obtained in (a). b. [10 pts] Implement, using Logisim, the simplified logic circuit. Include an image of the circuit in your report.arrow_forwardUsing XML, design a simple user interface for a fictional app. Your UI should include at least three different UI components (e.g., TextView, Button, EditText). Explain the purpose of each component in your design-you need to add screenshots of your work with your name as part of the code to appear on the interface-. Screenshot is needed.arrow_forwardQ4) A thin ring of radius 5 cm is placed on plane z = 1 cm so that its center is at (0,0,1 cm). If the ring carries 50 mA along a^, find H at (0,0,a).arrow_forward
- 4. [15 pts] A logic function F of four variables a; b; c; d is described by the following K-map. Derive the fully minimized SOP logic expression form of F. cd ab 00 01 11 10 00 0 0 0 1 01 1 0 0 1 11 1 0 1 1 10 0 0 1 1arrow_forward2. [20 pts] Student A B will enjoy his picnic on sunny days that have no ants. He will also enjoy his picnic any day he sees a hummingbird, as well as on days where there are ants and ladybugs. a. Write a Boolean equation for his enjoyment (E) in terms of sun (S), ants (A), hummingbirds (H), and ladybugs (L). b. Implement in Logisim, the logic circuit of E function. Use the Circuit Analysis tool in Logisim to view the expression, include an image of the expression generated by Logisimarrow_forwardHow would I go about creating this computer database in MariaDB with sql? Create a database name "dbXXXXXX" Select the database using the "use [database name]" command. Now you are in the database. Based on the above schema from Enrolment System database, create all the tables with the last 6 digits of "123456", then the table name for table Lecturer should be "123456_Lecturer". Refer to basic SQL lecture note to create table that has primary keys and Foreign Keys. Provide the datatype of each attributes. Add a column called "Department" with datatype "VARCHAR(12)" to the table "Lecturer". Shows the metadata of the updated "Lecturer" table. (Use Describe command) Drop the "Department" column from the table "Lecturer", and show the metadata of the updated "Lecturer" table. Insert three (3) data to each of the table in the tables created. Note: If you have foreign key issues, please disable foreign key constraints before inserting the data, see below SET FOREIGN_KEY_CHECKS=0;…arrow_forward
- CSE330 Discrete Mathematics 1. In the classes, we discussed three forms of floating number representations as given below, (1) Standard/General Form, (2) Normalized Form, (3) Denormalized Form. 3. Consider the real number x = (3.395) 10 (a) (b) Convert the decimal number x into binary format up to 7 binary places (7 binary digits after decimal) Convert the calculated value into denormalized form and calculate fl(x) for m=4 Don't use any Al tool show answer in pen a nd paper then take pi ctures and sendarrow_forwardSimplify the following expressions by means of a four-variable K-Map. AD+BD+ BC + ABDarrow_forwardCSE330 Discrete Mathematics 1. In the classes, we discussed three forms of floating number representations as given below, (1) Standard/General Form, (2) Normalized Form, (3) Denormalized Form. 2. Let ẞ 2, m = 6, emin = -3 and emax = 3. Answer the following questions: Compute the minimum of |x| for General and Normalized form (a) Compute the Machine Epsilon value for the General and Denormalized form. If we change the value of emax to 6 then how will it affect the value of maximum scale invariant error for the case of Normalized form? Explain your answer. show answer in pen a Don't use any Al tool nd paper then take pi ctures and sendarrow_forward
- Programming with Microsoft Visual Basic 2017Computer ScienceISBN:9781337102124Author:Diane ZakPublisher:Cengage Learning
- Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781305627482Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningDatabase Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781285196145Author:Steven, Steven Morris, Carlos Coronel, Carlos, Coronel, Carlos; Morris, Carlos Coronel and Steven Morris, Carlos Coronel; Steven Morris, Steven Morris; Carlos CoronelPublisher:Cengage LearningNp Ms Office 365/Excel 2016 I NtermedComputer ScienceISBN:9781337508841Author:CareyPublisher:Cengage



