Fall 2004 CS 186 Discussion Section Exercises - Week 2 ending 9/10

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1 Fall 00 CS 86 Discussion Section Exercises - Week ending 9/0 RELATIONAL MODEL ) Design relational tables to represent the following information. a. A company has employees working in departments. b. Name, age, salary and manager are employee information. c. Name, budget and chief manager are department information. Use simple data types (INTEGER, CHAR). Employee(EmpId: INT, EmpName: CHAR(8), Age: INT, Salary: INT, Manager: INT); Department(DepId: INT, DepName: CHAR(8), Budget: INT, Chief: INT); Works(EmpId: INT, DepId: INT); Trivia: a) There can be another simpler answer as well if we assume that an employee can work in only one department. Employee(EmpId: INT, EmpName: CHAR(8), Age: INT, Salary: INT, Works: INT, Manager: INT); Department(DepId: INT, DepName: CHAR(8), Budget: INT, Chief: INT); In this case the Works column in the Employee table store any of the values of DepId column of the Deparment table. The former choice however gives the power of allowing an employee to work in multiple departments. It is also a better normalized model, something we will learn later in the course. The rest of the answers assume the first design. b) Is there anything wrong in the choice of Age field? Use of Age field is poor design. As time goes by the Age field does not update itself in the database to reflect the correct age. A good design choice is to use DOB. ) Write CREATE TABLE statements for creating the schema.

2 CREATE TABLE Employee(EmpId INTEGER, EmpName CHAR(8), Age INTEGER, Manager INTEGER); CREATE TABLE Department(DepId INTEGER, DepName CHAR(8), Budget INTEGER, Chief INTEGER); CREATE TABLE Works(EmpId INTEGER, DepId INTEGER); ) Rewrite the CREATE TABLE statements and establish primary key and foreign key constraints. CREATE TABLE Employee(EmpId INTEGER PRIMARY KEY, EmpName CHAR(8), Age INTEGER, Manager INTEGER); CREATE TABLE Department(DepId INTEGER PRIMARY KEY, DepName CHAR(8), Budget INTEGER, Chief INTEGER); CREATE TABLE Works(EmpId INTEGER, DepId INTEGER) PRIMARY KEY (EmpId, DepId), FOREIGN KEY (EmpId) REFERENCES Employee, FOREIGN KEY (DepId) REFERENCES DepId; Trivia: a) Should the Works table have a primary key or not? - The PRIMARY KEY as specified should be present if the employee is allowed to work in only one department. Otherwise the Works table need not have a primary key. This is totally up to the requirements of the application domain. It is something application domain experts have to decide. ) How to ensure that EmpName field will not be null? Add a NOT NULL constraint to the create table statement, as in: CREATE TABLE Employee(EmpId INTEGER PRIMARY KEY, EmpName CHAR(8) NOT NULL, Age INTEGER, Manager INTEGER); Same can be applied to DepName in Department table. ) How to ensure that EmpId field will not be a negative integer?

3 Add a CHECK constraint to the create table statement, as in: CREATE TABLE Employee(EmpId INTEGER PRIMARY KEY, EmpName CHAR(8) NOT NULL, Age INTEGER, Manager INTEGER) CHECK (EmpId > 0); Trivia: Same can be applied to DepId in Department table. Similar check constraints can be applied for Age, etc. 6) How to ensure that Manager field in Employee is also an Employee. It is possible to have a foreign key referring to itself: CREATE TABLE Employee(EmpId INTEGER PRIMARY KEY, EmpName CHAR(8) NOT NULL, Age INTEGER, Manager INTEGER REFERENCES Employee); Further Exercises: 7) How to ensure that nobody can be his or her own boss? Add a check constraint EmpId NOT =Manager to the Employee table. 8) How to ensure that EmpId and DepId always have 6 digits? Add a check constraint EmpId > AND EmpId < to the Employee table. Add a similar check constraint for the DepId in the Department table. II. STORING DATA - DISKS & FILES ) Why is the access method used by disks is partly random and partly sequential? It is partly random because any particular track can be reached by moving the disk head across. It is partly sequential because after a particular track is reached, a particular sector has to be read by sequentially rotating the disk head over the track.

4 ) Which among the following are best suited for databases and why? Answer in one or two sentences. a. RAM Even though provides random access, it is very expensive, is volatile has cannot recover from a system crash. Not well suited. b. Hard disk Persistent, provides random access, less expensive. Best suited. c. Magnetic tape Persistent, but provides only sequential access, though least expensive. Not well suited. d. CD-RW Persistent, provides random access, limited space and very slow compared to hard disk. Not well suited. ) Operating Systems provide support for random access files. Isn t that sufficient for a DBMS to store its data on a disk? Justify your answer with a couple of reasons. Operating systems are lazy in disk writes. They do not guarantee when a disk write will actually make it to the disk. Also, a DBMS can lay the pages on disk more efficiently as it knows the page access patterns, which an Operating System does not have. ) Why does a DBMS need buffer management when an Operating System can provide virtual memory? Operating systems cannot predict which page will be needed next. A DBMS can very well predict the page access pattern. ) What is a frame and what is a page? A frame is a partition of available main memory (buffer pool) used by the DBMS for loading and modifying data of a database. A page is a partition in the data stored in the disk. Disk space manager manages pages, while buffer manager manages frames. 6) Number of frames in a DBMS buffer pool will be: a. Always greater than number of pages in disk. b. Must be equal to number of pages in disk. c. Typically less than the number of pages in disk. c) 7) What does pinning and unpinning a frame mean? When a transaction results in a request for a page from the buffer manager, the buffer manager will ensure it is loaded into the buffer pool from the disk and return it to the calling process. While doing so it will keep track that the page is in use by incrementing a variable pin_count by. This is called pinning a frame. When the

5 calling process releases the page after its use, the buffer manager will decrement the pin_count variable by. This is called the unpinning of a frame. 8) When two transactions modify distinct portions of a frame, should the dirty flag be set to? Justify your answer in one sentence. The dirty flag is used only to reflect that the page has been modified, not how many times or how many processes modified it. It is enough if the dirty flag of a frame is treated as a boolean value and set to. When the dirty flag is set, the buffer manager has to write the frame to the disk before it can replace it with another page or an upper layer process asks it to write the frame forcibly (for ex for crash recovery purposes). 9) What will happen when all the frames in a buffer pool are pinned by one or more transactions and a transaction makes a new request for a new page to be loaded from the disk? Typically, the transaction making the new request will fail. However, database implementors may choose other techniques like time outs, etc. 0) Assume that a database has pages and frames. Assume that LRU is the page replacement policy. Which of the following sequences of page references will cause a sequential flooding? a.,,,,,,,,,,,,,,, b.,,,,,,,,,,,,,,, c.,,,,,,,,,,,,,,, d.,,,,,,,,,,,,,,, a, b, & c. Not only the repetition of sequence,,,, will cause sequential flooding, any sequence that repeats itself will cause sequential flooding! ) Assume that a database has pages and frames. Assume that you start with an empty buffer pool. In each of the following tables, a. The page reference vector is specified in the top row. b. The frame numbers are specified in the left most column. c. The value in (i, j) empty cells should represent the page id in the i th frame when the j th page reference is made. d. The bottom most row should represent whether a page fault occurred or not (y/n). Fill in the empty cells in following tables for each of the page replacement policies mentioned. Fill in the empty cells in following tables for each of the page replacement policies mentioned. NOTE: Bold numbers are in slots whose page was replaced. Underscored numbers are pages that were accessed (pinned and unpinned).

6 Frames FIFO A B C Fault? Y Y Y N N Y N N Y N N Y N Y N LRU Frames Total number of page faults = 7 A B C Fault? Y Y Y N N Y N N N Y N N Y Y Y Frames Total number of page faults = 8 LIFO A B C Fault? Y Y Y N N Y N N N Y N N Y Y N Frames Total number of page faults = 7 MRU A B C Fault? Y Y Y N N Y N N N N N N Y N N Total number of page faults = 6

7 Frames CLOCK NOTE: The subscripts show the value of reference bit. The < bracket shows the current position of clock hand. Clock hand cycles in the column from top to bottom and cycling back to the top. A. <. <. <. <. <....<.0.0..<.<. B <. 0<.<.0...<.0..0< C <.< Fault? Y Y Y N N Y N N Y Y N Y Y N Y Total number of page faults = 9

8 Frames Frames A ) Do the following assuming a buffer size of frames and disk space size of pages. SQ (Simplified Q) For Simplified Q assume that the A and Am queues have a threshold of 0% of buffer size each. NOTE: FIFO queue A threshold is and LRU queue Am threshold is. The subscript, A_, represent the A FIFO queue. The first in item has lower index () and is the tail, the last in item has highest index and is the head. The subscript Am_, Am_, represent the Am LRU queue. The least recently used item has lower index (Am_), while the most recently used item has highest index. B A_ C A_ A_ D A_ A_ A_ Am_ A_ A_ Am_ A_ Am_ Fault? Y Y Y Y N N Y Y N N Y N Y Total number of page faults = 8 FQ (Full Q) For Full Q assume Kin is % of the page slots and Kout should hold as many identifiers for as many pages as would fit on 0% of the buffer. NOTE: Kin = and Kout= The subscript, A_, represent the Ain FIFO queue. The first in item has lower index (), the last in item has highest index. The values in Aout queue are separately listed in a new row. A A_.A_.A_.A_.A_.A_.A_ B.A_.A_.A_.A_.A_.....Am_.Am_.Am_ C.A_.A_.A_.A_.A_.A_.A_.A_..Am_.Am_ D.A_.A_.A_.A_.A_.A_.A_.A_.. Fault? Y Y Y Y N N Y N N N Y Y N A out,,, Total number of page faults = 7 A_ A_ Am_ A_ A_ Am_ A_ Am_ Am_ Am_ Am_ Am_ Am_ Am_ A_ Am_ Am_ Am_ Am_ A_ Am_