Synonyms:
When more than one name is assigned to same attribute, the attribute names are referred as “synonyms’. It exists when the same attribute has more than one name.
Example:
Suppose in table STUDENT, one of the attribute names is “STU_NUM” which displays the student registration number. Also, another attribute is “STU_ID” which also displays the student registration number. Then the attribute names are named as “Synonyms”.
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.
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.
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
There is an optional (default) 1:1 relationship between EMPLOYEE and PILOT. It can be represented that EMPLOYEE is the “parent” of PILOT.
Elimination of Homonyms:
In the above tables, there are two attributes that are homonyms. The attributes are CHAR_PILOTS and CHAR_COPILOTS.
The two homonyms attributes are eliminated by modifying the CHARTER table and deleting the CHAR_PILOTS and CHAR_COPILOTS attributes.
A new table CREW is added, which is a composite table and it acts as a link between the CHARTER and EMPLOYEE tables. One CHARTER requires many CREW members and hence there is a one to many relations between them. Many CREWS are employees but one EMPLOYEE can be a part of one crew and hence it represents a many to one relationship between them.
Relational diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, CREW, EMPLOYEE, PILOT and CUSTOMER:
The Relational diagram to represent relationship between CHARTER, MODEL, AIRCRAFT, CREW, EMPLOYEE, PILOT and CUSTOMER is shown below:
Explanation:
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 CREW table is created which is a composite table and it acts as a link between the CHARTER and EMPLOYEE tables. One CHARTER requires many CREW members and hence there is a one to many relation between them. Many CREWS are employees but one EMPLOYEE can be a part of one crew and hence it represents a many to one relationship between them. ”. A new entity RATING is created”. Which creates an M: N relationship between PILOT and RATING as a PILOT can earn many RATINGs but a RATING can be earned by many PILOTs. The M: N relationship is broken down into two 1: M relationships using EARNEDRATING entity.
Want to see more full solutions like this?
Chapter 3 Solutions
Database Systems: Design, Implementation, & Management
- The following is code for a disc golf program written in C++: // player.h #ifndef PLAYER_H #define PLAYER_H #include <string> #include <iostream> class Player { private: std::string courses[20]; // Array of course names int scores[20]; // Array of scores int gameCount; // Number of games played public: Player(); // Constructor void CheckGame(int playerId, const std::string& courseName, int gameScore); void ReportPlayer(int playerId) const; }; #endif // PLAYER_H // player.cpp #include "player.h" #include <iomanip> Player::Player() : gameCount(0) {} void Player::CheckGame(int playerId, const std::string& courseName, int gameScore) { for (int i = 0; i < gameCount; ++i) { if (courses[i] == courseName) { // If course has been played, then check for minimum score if (gameScore < scores[i]) { scores[i] = gameScore; // Update to new minimum…arrow_forwardIn this assignment, you will implement a multi-threaded program (using C/C++) that will check for Prime Numbers and Palindrome Numbers in a range of numbers. Palindrome numbers are numbers that their decimal representation can be read from left to right and from right to left (e.g. 12321, 5995, 1234321). The program will create T worker threads to check for prime and palindrome numbers in the given range (T will be passed to the program with the Linux command line). Each of the threads works on a part of the numbers within the range. Your program should have some global shared variables: • numOfPrimes: which will track the total number of prime numbers found by all threads. numOfPalindroms: which will track the total number of palindrome numbers found by all threads. numOfPalindromic Primes: which will count the numbers that are BOTH prime and palindrome found by all threads. TotalNums: which will count all the processed numbers in the range. In addition, you need to have arrays…arrow_forwardHow do you distinguish between hardware and a software problem? Discuss theprocedure for troubleshooting any hardware or software problem. give one reference with your answer.arrow_forward
- You are asked to explain what a computer virus is and if it can affect computer’shardware or software. How do you protect your computer against virus? give one reference with your answer.arrow_forwardDistributed Systems: Consistency Models fer to page 45 for problems on data consistency. structions: Compare different consistency models (e.g., strong, eventual, causal) for distributed databases. Evaluate the trade-offs between availability and consistency in a given use case. Propose the most appropriate model for the scenario and explain your reasoning. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qoHazb9tC440AZF/view?usp=sharing]arrow_forwardOperating Systems: Deadlock Detection fer to page 25 for problems on deadlock concepts. structions: • Given a system resource allocation graph, determine if a deadlock exists. If a deadlock exists, identify the processes and resources involved. Suggest strategies to prevent or resolve the deadlock and explain their trade-offs. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qoHazb9tC440 AZF/view?usp=sharing]arrow_forward
- Artificial Intelligence: Heuristic Evaluation fer to page 55 for problems on Al search algorithms. tructions: Given a search problem, propose and evaluate a heuristic function. Compare its performance to other heuristics based on search cost and solution quality. Justify why the chosen heuristic is admissible and/or consistent. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qoHazb9tC440 AZF/view?usp=sharing]arrow_forwardRefer to page 75 for graph-related problems. Instructions: • Implement a greedy graph coloring algorithm for the given graph. • Demonstrate the steps to assign colors while minimizing the chromatic number. • Analyze the time complexity and limitations of the approach. Link [https://drive.google.com/file/d/1wKSrun-GlxirS3IZ9qoHazb9tC440 AZF/view?usp=sharing]arrow_forwardRefer to page 150 for problems on socket programming. Instructions: • Develop a client-server application using sockets to exchange messages. • Implement both TCP and UDP communication and highlight their differences. • Test the program under different network conditions and analyze results. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qo Hazb9tC440AZF/view?usp=sharing]arrow_forward
- Refer to page 80 for problems on white-box testing. Instructions: • Perform control flow testing for the given program, drawing the control flow graph (CFG). • Design test cases to achieve statement, branch, and path coverage. • Justify the adequacy of your test cases using the CFG. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS3IZ9qo Hazb9tC440 AZF/view?usp=sharing]arrow_forwardRefer to page 10 for problems on parsing. Instructions: • Design a top-down parser for the given grammar (e.g., recursive descent or LL(1)). • Compute the FIRST and FOLLOW sets and construct the parsing table if applicable. • Parse a sample input string and explain the derivation step-by-step. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qoHazb9tC440 AZF/view?usp=sharing]arrow_forwardRefer to page 20 for problems related to finite automata. Instructions: • Design a deterministic finite automaton (DFA) or nondeterministic finite automaton (NFA) for the given language. • Minimize the DFA and show all steps, including state merging. • Verify that the automaton accepts the correct language by testing with sample strings. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qo Hazb9tC440AZF/view?usp=sharing]arrow_forward
- Database System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSON
- C How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education