Given the following relation R and its functional dependencies: R(workerNumber, repairNumber, workerName, machineNumber, spentTime, repairDate, machineName, workshopNumber, workshopName ) workerNumber à workerName repairNumber à repairDate, machineNumber machineNumber à machineName, workshopNumber workshopNumber à workshopName Explain the types of all the dependencies in the relation R. Create a database whose tables are at least in 3NF, showing the dependency diagrams for each table.

Given the following relation R and its functional dependencies: R(workerNumber, repairNumber, workerName, machineNumber, spentTime, repairDate, machineName, workshopNumber, workshopName ) workerNumber à workerName repairNumber à repairDate, machineNumber machineNumber à machineName, workshopNumber workshopNumber à workshopName Explain the types of all the dependencies in the relation R. Create a database whose tables are at least in 3NF, showing the dependency diagrams for each table.

A Guide to SQL

9th Edition

ISBN:9781111527273

Author:Philip J. Pratt

Publisher:Philip J. Pratt

Chapter2: Database Design Fundamentals

Section: Chapter Questions

Problem 11RQ

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DATABASE SYSTEM A relation R is given which is R(A,B,C,D,E). It has the following functional dependencies: BC -> ADE , D -> B. Answer the following based on this information. Which normal forms does this relation satisfy? Which type of dependencies does this relation violate? Normalize the relationship so that it decomposes to 3NF.
Database Management A relation R is given which is R(A,B,C,D,E,F). It has the following functional dependencies: AB -> CDE, B -> F. Answer the following based on this information. Is it in 3NF? If not, decompose it until 3NF.
Given the following relation R and its functional dependencies: R(workerNumber, repairNumber, workerName, machineNumber, spentTime, repairDate, machineName, workshopNumber, workshopName ) workerNumber à workerName repairNumber à repairDate, machineNumber machineNumber à machineName, workshopNumber workshopNumber à workshopName 1)Create a database whose tables are at least in 3NF, showing the dependency diagrams for each table
The following relations kind a part of a on-line database (the relations area unit self-explanatory): CUSTOMER/CLIENT/SHOPPERS/BUYERS(Cust#, CAddress, Ctown, Balance,Cname) ORDER(Order.#, Date, Cust#.) ORDERLINE(Order#, Part#., Amount-Ordered) PART(Part#., Desc, Stock, Item-Class, Unit -Price) Design a physical ER for the above scenario. With respect to this relative schema, write SQL statements for the subsequent queries: List the name of the client United Nations agency has placed Order 123 (i.e. Order: = 123). List the names of shoppers United Nations agency in Cardiff and have ordered half 456 (i.e. Part. 456). List the order numbers (Order) for orders that contain a minimum of three elements and also the quantity ordered are minimum of ten. List the outline (Desc), of these elements that the typical quantity ordered is bigger than the number ordered for half .6 (Part. = 456) so as 123.(Order. .123).
Let the database has two relations P(a, b) and Q(m, a, n) in which a is the primary key of the relation P and m is the primary key of the relation Q. Instance of P has 70 Tuples and Instance of Q has 42. What is the maximum number of rows in the new instance produced by P Q? ( is natural join)
TASK 1 1. The InstantRide Management team founded a new team for car maintenance. The new team is responsible for the small maintenance operations for the cars in the InstantRide system. The main idea is to take actions faster and minimize the time spent for the maintenance. Therefore, the Car Maintenance team wants to store MAINTENANCE_TYPE_ID (char(5)) and a MAINTENANCE_TYPE_DESCRIPTION (varchar(30)) in the database. Using MAINTENANCE_TYPE_ID as the PRIMARY KEY, cook up a new table, MAINTENANCE_TYPES, and send the table description with the column names and types to the Car Maintenance team. Task: Make a new table to store car maintenance types. Task 2: The Car Maintenance team also wants to store the actual maintenance operations in the database. The team wants to start with a table to store CAR_ID (CHAR(5)), MAINTENANCE_TYPE_ID (CHAR(5)) and MAINTENANCE_DUE (DATE) date for the operation. Make a new table named MAINTENANCES. The PRIMARY_KEY should be the combination of the…
Part 3 5. Database design often involves decisions about the storage of attributes. For example a Telephone Number can be stored as a one attribute or split into three attributes (one for each of the three hyphen-delineated groups of numbers in a Telephone Number (XXX)-XXX-XXXX, I.e (123)-456-7890). However, Telephone Number is usually stored in one attribute. The decision is usually based on how the database will be used. This exercise asks you to think about specific situations where dividing the Telephone Number is useful. List and describe the different situations, give examples.
Based on the following database schemas:Homestay (homestayNo (PK), homestayName, address, telNo, city)Room (roomNo, homestayNo (PK), type, price)Booking (homestayNo, guestNo (PK), dateFrom, dateTo, roomNo)Guest (guestNo (PK), guestName, guestAddress, guestTelNo)Assume the following indexes exist:a) a hash index with no overflow on the roomNo and homestayNo as a composite primary key in Room;b) a clustering index on the foreign key attributes homestayNo in Room;c) a B+-tree index on the price attribute in Room;d) a secondary index on the attribute type in Room.nTuples(Room) = 20000 bFactor(Room) = 200nTuples(Homestay) = 100 bFactor(Homestay) = 40nTuples(Booking) = 100000 bFactor(Booking) = 60nDistincthomestayNo(Room) = 50 nDistinctroomNo(Booking) = 150nDistincttype(Room) = 10 nDistincthomestayName(Homestay) = 50nDistinctprice(Room) = 500 maxprice(Room) = 500minprice(Room) = 200 nLevelstype(I) = 2nLevelshomestayNo(I) = 2 nLfBlocksprice(I) = 50nLevelsprice(I) = 2 Log2100 = 6.64Log250 =…
DATABASE SYSTEM A relation R is given which is R(A,B,C,D,E). It has the following functional dependencies: BC -> D, AC -> BE, B->E . Answer the following based on this information. What is the candidate key of this relation? Is it in 2NF? Explain why or why not.
Based on the following database schemas: Homestay (homestayNo (PK), homestayName, address, telNo, city)Room (roomNo, homestayNo (PK), type, price)Booking (homestayNo, guestNo (PK), dateFrom, dateTo, roomNo) Guest (guestNo (PK), guestName, guestAddress, guestTelNo) Assume the following indexes exist:a) a hash index with no overflow on the roomNo and homestayNo as a composite primary key in Room;b) a clustering index on the foreign key attributes homestayNo in Room;c) a B⁺-tree index on the price attribute in Room;d) a secondary index on the attribute type in Room. nTuples(Room) = 20000 bFactor(Room) = 200nTuples(Homestay) = 100 bFactor(Homestay) = 40nTuples(Booking) = 100000 bFactor(Booking) = 60nDistincthomestayNo(Room) = 50 nDistinctroomNo(Booking) = 150nDistincttype(Room) = 10…
Based on the following database schemas: Homestay (homestayNo (PK), homestayName, address, telNo, city)Room (roomNo, homestayNo (PK), type, price)Booking (homestayNo, guestNo (PK), dateFrom, dateTo, roomNo) Guest (guestNo (PK), guestName, guestAddress, guestTelNo) Assume the following indexes exist:a) a hash index with no overflow on the roomNo and homestayNo as a composite primary key in Room;b) a clustering index on the foreign key attributes homestayNo in Room;c) a B⁺-tree index on the price attribute in Room;d) a secondary index on the attribute type in Room. nTuples(Room) = 20000 bFactor(Room) = 200nTuples(Homestay) = 100 bFactor(Homestay) = 40nTuples(Booking) = 100000 bFactor(Booking) = 60nDistincthomestayNo(Room) = 50 nDistinctroomNo(Booking) = 150nDistincttype(Room) = 10 nDistincthomestayName(Homestay) = 50nDistinctprice(Room)…
Based on the following database schemas: Homestay (homestayNo (PK), homestayName, address, telNo, city)Room (roomNo, homestayNo (PK), type, price)Booking (homestayNo, guestNo (PK), dateFrom, dateTo, roomNo) Guest (guestNo (PK), guestName, guestAddress, guestTelNo) Assume the following indexes exist:a) a hash index with no overflow on the roomNo and homestayNo as a composite primary key in Room;b) a clustering index on the foreign key attributes homestayNo in Room;c) a B⁺-tree index on the price attribute in Room;d) a secondary index on the attribute type in Room. nTuples(Room) = 20000 bFactor(Room) = 200nTuples(Homestay) = 100 bFactor(Homestay) = 40nTuples(Booking) = 100000 bFactor(Booking) = 60nDistincthomestayNo(Room) = 50 nDistinctroomNo(Booking) = 150nDistincttype(Room) = 10…

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