
Concept explainers
a.
Normalization:
The process used to minimize data redundancy and dependency in a relational
Second normal form (2NF):
- The value of all non-primary key attributes should be dependent on the primary key attribute.
- If any attribute is depending on the partial primary key then it should determine the other attributes for an instance of the entity.
- The partial dependencies should be removed from the data model.
Third normal form (3NF):
- The value of any non-primary key attributes will not depend on any other non-primary key attributes.
- If any non-primary key attributes depend on any other non-primary key attribute then it should be moved or deleted.
- It is termed as transitive dependency.
Partial dependency:
A partial dependency exists at that time of an attributes depends only a part of primary key. This dependency is related with 1st normal form.
Transitive dependency:
A transitive dependency exists at that time of an attributes depends on another attribute which is not part of primary key.
Functional dependency:
An association between two attributes or two set of attributes in a same relational database table, which is having some constraints is known as functional dependency.
- In a table one attribute is functionally dependent on another attribute to take one value.
a.

Explanation of Solution
Construct the dependency diagram with all partial and transitive dependencies:
The relational schema for given STUDENT table is given below:
STUDENT(STU_NUM, STU_LNAME, STU_MAJOR, DEPT_CODE, DEPT_NAME,
DEPT_PHONE, ADVISOR_LNAME, ADVISOR_OFFICE, ADVISOR_BLDG, ADVISOR_PHONE, STU_GPA, STU_HOURS, STU_CLASS)
- Here, “STU_NUM” indicates the primary key.
The representation of dependency diagram with all transitive dependencies is shown below:
Explanation:
In the above dependency diagram,
- The transitive dependency is,
DEPT_CODE -> (DEPT_NAME, DEPT_PHONE, COLLEGE_NAME)
ADV_OFFICE -> (ADV_BUILDINGS)
STU_HOURS -> (STU_CLASS)
b.
Normalization:
The process used to minimize data redundancy and dependency in a relational database is known as normalization. The database table is divided into two or more tables and defines the relationship between those tables.
Second normal form (2NF):
- The value of all non-primary key attributes should be dependent on the primary key attribute.
- If any attribute is depending on the partial primary key then it should determine the other attributes for an instance of the entity.
- The partial dependencies should be removed from the data model.
Third normal form (3NF):
- The value of any non-primary key attributes will not depend on any other non-primary key attributes.
- If any non-primary key attributes depend on any other non-primary key attribute then it should be moved or deleted.
- It is termed as transitive dependency.
Partial dependency:
A partial dependency exists at that time of an attributes depends only a part of primary key. This dependency is related with 1st normal form.
Transitive dependency:
A transitive dependency exists at that time of an attributes depends on another attribute which is not part of primary key.
Functional dependency:
An association between two attributes or two set of attributes in a same relational database table, which is having some constraints is known as functional dependency.
- In a table one attribute is functionally dependent on another attribute to take one value.
b.

Explanation of Solution
Construct the dependency diagram:
The new dependency diagram is represented by removing all transitive dependencies in STUDENT table.
First table:
The relational schema for first table is given below:
STUDENT(STU_NUM, STU_LNAME, STU_MAJOR, DEPT_CODE, ADVISOR_NUM STU_GPA, STU_HOURS, STU_CLASS)
- Here, “STU_NUM” indicates the primary keys and “ADVISOR_NUM” indicates the foreign key.
- The relation is in third normal form (2NF), since there is transitive dependency in this table.
The representation of dependency diagram in first table is shown below:
Second table:
The relational schema for second table is given below:
MAJOR(MAJOR_CODE, DEPT_CODE, MAJOR_DESCRIPTION)
- Here, “MAJOR_CODE” indicates the primary key.
- The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.
The representation of dependency diagram in second table is shown below:
Third table:
The relational schema for third table is given below:
BUILDING(BLDG_CODE, BLDG_NAME, BLDG_MANAGER)
- Here, “BLDG_CODE” indicates the primary key.
- The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.
The representation of dependency diagram in third table is shown below:
Fourth table:
The relational schema for fourth table is given below:
DEPARTMENT(DEPT_NAME, DEPT_PHONE, COLL_CODE)
- Here, “DEPT_NAME” indicates the primary key.
- The relation is in third normal form (3NF) but it does not meet the BCNF, since there is no transitive dependency and no repeated attributes.
The representation of dependency diagram in fourth table is shown below:
Fifth table:
The relational schema for fifth table is given below:
COLLEGE(COLL_CODE, COLL_NAME)
- Here, “COLL_CODE” indicates the primary key.
- The relation is in third normal form (3NF), since there is no transitive dependency and no repeated attributes.
The representation of dependency diagram in fifth table is shown below:
Sixth table:
The relational schema for sixth table is given below:
ADVISOR(ADV_NUM, ADV_LASTNAME, ADV_OFFICE, ADV_BUILDING, ADV_PHONE)
- Here, “ADV_NUM” indicates the primary key.
- The dotted transitive dependency line specifies that dependency is interpretation.
- The relation is in third normal form (2NF), since there is transitive dependency in this table.
The representation of dependency diagram in sixth table is shown below:
c.
Explanation of Solution
The representation of Crow’s Foot Entity Relational Diagram (ERD) is shown below:
The following data model shows the solution for the given question.
Explanation:
- In the above data model, the college owns more than one department.
- The “COLLEGE” entity contains the “COLL_CODE” and “COLL_NAME” attributes.
- The primary key of “COLL_CODE” entity is “INV_NUM”.
- The “DEPARTMENT” entity contains the “DEPT_CODE”, “DEPT_NAME”, “DEPT_PHONE” and “COLL_CODE” attributes.
- The primary of this entity is “DEPT_CODE”.
- The foreign key of this entity is “COLL_CODE”.
- The “COLLEGE” entity contains the “COLL_CODE” and “COLL_NAME” attributes.
- The department offers more than one major.
- The “MAJOR” entity contains the “MAJOR_CODE”, “MAJOR_NAME”, and “DEPT_CODE” attributes.
- The primary key of “MAJOR_CODE” for this entity.
- The foreign key of this entity is “VEND_CODE”.
- The “MAJOR” entity contains the “MAJOR_CODE”, “MAJOR_NAME”, and “DEPT_CODE” attributes.
- The major attracts more than one student.
- The “STUDENT” entity contains the “STU_NUM”, “STU_LNAME”, “STU_CLASS”, “STU_HOURS”, “STU_GPA”, “MAJOR_CODE”, and “ADV_NUM” attributes.
- The primary key of “STU_NUM” entity.
- The foreign key of this entity is “MAJOR_CODE” and “ADV_NUM”.
- The “STUDENT” entity contains the “STU_NUM”, “STU_LNAME”, “STU_CLASS”, “STU_HOURS”, “STU_GPA”, “MAJOR_CODE”, and “ADV_NUM” attributes.
- The building houses more than one advisor.
- The “ADVISOR” entity contains the “ADV_NUM”, “DEPT_CODE”, “BLDG_CODE”, “ADV_LNAME”, “ADV_OFFICE”, and “ADV_PHONE” attributes.
- The primary key of “ADV_NUM” entity.
- The foreign key of this entity is “DEPT_CODE” and “BLDG_CODE”.
- The “BUILDING” entity contains the “BLDG_CODE”, “BLDG_NAME” and “BLDG_MANAGER” attributes.
- The primary key of “BLDG_CODE” entity.
- The “ADVISOR” entity contains the “ADV_NUM”, “DEPT_CODE”, “BLDG_CODE”, “ADV_LNAME”, “ADV_OFFICE”, and “ADV_PHONE” attributes.
- The advisor advises more than one student.
Want to see more full solutions like this?
Chapter 6 Solutions
Database Systems: Design, Implementation, & Management
- a database with multiple tables from attributes as shown above that are in 3NF, showing PK, non-key attributes, and FK for each table? Assume the tables are already in 1NF. [Hint: 3 tables will result after deducing 1NF -> 2NF -> 3NF]arrow_forwardIf a new entity Order_Details is introduced, will it be a strong entity or weak entity? If it is a weak entity, then mention its type (ID or Non-ID, also Justify why)?arrow_forwardWhich one of the 4 Entities mention in the diagram can have a recursive relationship? Order, Product, store, customer.arrow_forward
- Inheritance & Polymorphism (Ch11) There are 6 classes including Person, Student, Employee, Faculty, and Staff. 4. Problem Description: • • Design a class named Person and its two subclasses named student and Employee. • Make Faculty and Staff subclasses of Employee. • A person has a name, address, phone number, and e-mail address. • • • A person has a class status (freshman, sophomore, junior and senior). Define the status as a constant. An employee has an office, salary, and date hired. A faculty member has office hours and a rank. A staff member has a title. Override the toString() method in each class to display the class name and the person's name. 4-1. Explain on how you would code this program. (1 point) 4-2. Implement the program. (2 point) 4-3. Explain your code. (2 point)arrow_forwardSuppose you buy an electronic device that you operate continuously. The device costs you $300 and carries a one-year warranty. The warranty states that if the device fails during its first year of use, you get a new device for no cost, and this new device carries exactly the same warranty. However, if it fails after the first year of use, the warranty is of no value. You plan to use this device for the next six years. Therefore, any time the device fails outside its warranty period, you will pay $300 for another device of the same kind. (We assume the price does not increase during the six-year period.) The time until failure for a device is gamma distributed with parameters α = 2 and β = 0.5. (This implies a mean of one year.) Use @RISK to simulate the six-year period. Include as outputs (1) your total cost, (2) the number of failures during the warranty period, and (3) the number of devices you own during the six-year period. Your expected total cost to the nearest $100 is _________,…arrow_forwardWhich one of the 4 Entities mention in the diagram can have a recursive relationship? If a new entity Order_Details is introduced, will it be a strong entity or weak entity? If it is a weak entity, then mention its type (ID or Non-ID, also Justify why)?arrow_forward
- Please answer the JAVA OOP Programming Assignment scenario below: Patriot Ships is a new cruise line company which has a fleet of 10 cruise ships, each with a capacity of 300 passengers. To manage its operations efficiently, the company is looking for a program that can help track its fleet, manage bookings, and calculate revenue for each cruise. Each cruise is tracked by a Cruise Identifier (must be 5 characters long), cruise route (e.g. Miami to Nassau), and ticket price. The program should also track how many tickets have been sold for each cruise. Create an object-oriented solution with a menu that allows a user to select one of the following options: 1. Create Cruise – This option allows a user to create a new cruise by entering all necessary details (Cruise ID, route, ticket price). If the maximum number of cruises has already been created, display an error message. 2. Search Cruise – This option allows to search a cruise by the user provided cruise ID. 3. Remove Cruise – This op…arrow_forwardI need to know about the use and configuration of files and folders, and their attributes in Windows Server 2019.arrow_forwardSouthern Airline has 15 daily flights from Miami to New York. Each flight requires two pilots. Flights that do not have two pilots are canceled (passengers are transferred to other airlines). The average profit per flight is $6000. Because pilots get sick from time to time, the airline is considering a policy of keeping four *reserve pilots on standby to replace sick pilots. Such pilots would introduce an additional cost of $1800 per reserve pilot (whether they fly or not). The pilots on each flight are distinct and the likelihood of any pilot getting sick is independent of the likelihood of any other pilot getting sick. Southern believes that the probability of any given pilot getting sick is 0.15. A) Run a simulation of this situation with at least 1000 iterations and report the following for the present policy (no reserve pilots) and the proposed policy (four reserve pilots): The average daily utilization of the aircraft (percentage of total flights that fly) The…arrow_forward
- 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
- Database Systems: Design, Implementation, & Manag...Computer ScienceISBN:9781305627482Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningA Guide to SQLComputer ScienceISBN:9781111527273Author:Philip J. PrattPublisher:Course Technology PtrDatabase 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 Learning
- Principles of Information Systems (MindTap Course...Computer ScienceISBN:9781285867168Author:Ralph Stair, George ReynoldsPublisher:Cengage LearningFundamentals of Information SystemsComputer ScienceISBN:9781337097536Author:Ralph Stair, George ReynoldsPublisher:Cengage Learning





