CSCE 4523 Introduction to Database Management Systems Final Exam Spring I have neither given, nor received,unauthorized assistance on this exam.

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1 CSCE 4523 Introduction to Database Management Systems Final Exam Spring 2017 I have neither given, nor received,unauthorized assistance on this exam. Signature Printed Name: Attempt all of the following questions. The test consists of 27 multiple choice/fill in the blank/short answer questions. For multiple choice questions, indicate which you consider the single most appropriate answer. Indicate your selection circling your selection. For fill in the blank questions, provide the missing word or fill in the letter for the most appropriate option, when options are given. The exam questions total 71 points. It will be graded out of 70 (1 bonus point built in). If you have any questions about the questions, write your assumptions in the margin and answer the question as you understand it. You may bring in any database textbook (printed or electronic) and 1 page of paper with anything written on it, front and back. If you are using an electronic version of the textbook, please sit at the front of the classroom and turn Internet and all other applications (except the PDF viewer) off. If you do not bring in a textbook, you may instead bring in 3 pages of paper with anything written on them, front and back. 1

2 1. In relational algebra, σ p q r(r) can be rewritten as: a. σ p (σ q (σ r (R))) b. σ q (σ p (σ r (R))) c. σ p (σ r (σ q (R))) d. σ r (σ p q(r)) e. All of the above 2. (4 points) Consider R some_kind_of_join S. Fill in the blanks using each letter exactly once: SEMI JOIN THETA JOIN NATURAL JOIN a. Join R and S based on any predicate. b. The results include data from all tuples in both tables. c. Only produces non-zero results if R and S have attributes with the same name FULL JOIN d. The results are a subset of tuples from R. 3. (4 points) Consider a database that includes two relations F (film) and T (type) F T F_ID Name T_ID Sales 111 Rogue One ,056M 222 Finding Dory ,029M 333 Capt America ,150M 444 The Martian M ID Type 0001 Animated 0002 SciFi 0003 Action Fill in the SQL expression that would return the following result: Type Total SciFi 1,686M Action 1,150M Animated 1,029M 2

3 SELECT Type, (Sales) AS Total FROM F, T WHERE F_ID = Type ORDER BY ; 4. (3 points) Given a database: Customer(Cust_no, Name, Address) Order(Order_no, Cust_no, C_Date, Completed) Make(Order_no, Maker_no, Dress_style, Colour) Fill in the blanks for the SQL command that would return the name and address of all customers whose orders have been already completed for Dress_style 123. SELECT, Address FROM Customer JOIN Order NATURAL JOIN Make WHERE Dress_style = AND = yes ; 5. Assume you have a sorted file that we want to be able to do binary search on. Each record has the following fields: Name (max 20 chars), SSN (9 chars), Address (max 30 chars). Assume that the file is stored on Linux, we use 1 byte per character, and we separate the fields with blanks and the records with newlines. How big is each record? a) 60 bytes b) 61bytes c) 62 bytes d) 63 bytes e) The record sizes will vary based on the string lengths. 6. In JDBC and ODBC, there is a class named Statement. Which the following methods in Statement should be used to run SQL INSERT and DELETE commands and return nothing? a) execute b) executeupdate c) update d) executequery e) query 3

4 7. (6 points) Fill in the blanks below based on the diagram above. Use each letter exactly once: Registers a. Entity type. dateadvert LeasedBy Branch Supervises staffno in Staff b. Ternary relationship type c. Binary relationship type. d. Relationship attribute. e. Primary key. f. Recursive relationship type. 4

5 8. What is the cardinality of the Has relationship diagrammed in the semantic net above? a) 1:1 b) 1 : N c) M : N d) N : 1 9. (5 points) Consider a relation Enrollment(CourseID, StudentID, SemesterID, CourseName, StudentName, Room, Capacity). Fill in the blanks for the following functional identities: -> CourseName StudentID -> CourseID, -> Room, Capacity What should be the primary key for the Enrollment relation? a) CourseID b) StudentID c) CourseID, StudentID d) CourseID, SemesterID e) CourseID, StudentID, SemesterID Enrollment is in a) UNF b) 1NF c) 2NF d) 3NF e) BCNF or above. 5

6 10. (2 points) If we transform the database into: Enrollment (CourseID, StudentID, SemesterID) Student (StudentID, StudentName) Section (CourseID, SemesterID, CourseName, Room, Capacity) With the same functional dependencies as in the previous question, Section is in: a) UNF b) 1NF c) 2NF d) 3NF e) BCNF or above. If we further transfrom the database into: Enrollment (CourseID, StudentID, SemesterID) Student (StudentID, StudentName) Course (CourseID, CourseName) Section (CourseID, SemesterID, Room, Capacity) With the same functional dependencies as in the previous question, the relation Section is in: a) UNF b) 1NF c) 2NF d) 3NF e) BCNF or above. 11. (6 points) Fill in the blanks based on what schema transformation you need to do to move from one normal form to the next. Use each letter exactly once: UNF -> 1NF 1NF -> 2NF 2NF -> 3NF 3NF -> BCNF BCNF -> 4NF 4NF -> 5NF a. Remove dependencies on non-candidate keys. b. Remove partial dependencies c. Remove transitive dependencies. d. Decompose into as many relations as possible. e. Remove multiple valued attributes. f. Remove multi-valued dependencies. 12. The first network was called? a) ARPANET b) USENET c) INTERNET d) NSFNET 6

7 13. Which an advantage of the CGI-programming? a) Efficiency b) Tracks user sessions well c) Security d) Programming language independence 14. Which a disadvantage of the CGI-programming? a) Complexity b) Many processes created c) Only works on linux servers d) Can only use scripting languages 15. (4 points) Fill in the blanks to match the property name to the property definition for the ACID properties for transactions: Atomicity Consistency Isolation Durability a. Once transaction completes, its effects are never lost. b. Entire transaction is done, or nothing is done. c. Effects of incomplete transactions not visible to other transactions. d. Database starts and ends in a valid state. 16. When a successfully completed update is overridden by another transaction: a) Lost Update b) Dirty Read c) Inconsistent Analysis d) Non-serializable e) None of the above 17. Consider following transactions T1 T2 T3 Read (N) Read (M) Write(N) Read(N) Write(M) Write(N) Write(N) 7

8 Which of the following schedules below is the correct serialization (i.e., an equivalent non-serial schedule) of the above? a) T1 - >> T2 ->> T3 b) T2 ->> T1 ->> T3 c) T1 ->> T3 ->> T2 d) T3 ->> T2 ->> T1 e) This is a non-serializable schedule. 18. Rather than avoiding conflicts using locking or timestamping, some systems use optimistic scheduling techniques because: a) They are just naturally optimistic database designers b) Invalid data in the database is unimportant c) Rollbacks are inexpensive d) Conflicts are rare e) None of the above 19. Which is NOT true of Deferred Updates: a) Updates are not written to the database until after a transaction has reached its commit point. b) If transaction fails before commit, it will not have modified database and so no undoing of changes required. c) May be necessary to redo updates of committed transactions as their effect may not have reached database. d) They are all true. 20. (2 points) Draw the following B+ tree after the insertion of 5: 8

21. (2 points) Draw the following B+ tree after the deletion of 16: 22. Which is one property of generic B trees that B+ trees do NOT share: a) The tree is balanced.

9 21. (2 points) Draw the following B+ tree after the deletion of 16: 22. Which is one property of generic B trees that B+ trees do NOT share: a) The tree is balanced. b) Internal nodes can store at most M-1 keys and M pointers. c) Internal nodes are at least half full. d) The internal nodes store all the keys and all the associated records. e) Inserting into the tree is O(log M ). 23. Which is not true of using a B+ tree for DBMS records: a) They can grow and shrink adaptively as records are added/deleted. b) The records can be output in sorted order by the primary key efficiently. c) Records can be inserted efficiently, O(log M ).. d) Records can be located very efficiently, O(c). e) All of the above are true. 24. Generally, the most efficient ordering of relational algebra operations in a relational algebra tree is to do selects first because: a) Selects are the least efficient operation. b) Selects decrease the result set size by decreasing rows. c) Selects decrease the result set size by decreasing columns. d) Selects are the most efficient operation. e) Order doesn t matter. 25. When implementing JOIN on relations R and S with r and s tuples respectively, if the only attribute with the same name in both relations is the primary key, and a B+ tree is built for the primary keys, the join can be implemented in time: a) O(r * s) b) O(r + s) c) O( r * log(s)). d) O (r). 9

10 26. When implementing JOIN on relations R and S with r and s tuples respectively, if there is no index structure on the attributes in R and S involved in the join, the join can be implemented in time: a) O(r * s) b) O(r + s) c) O( r * log(s)). d) O (r). 27. (17 points) For each of the questions below, circle either or False False In 5NF, we decompose all relations into as many relations as possible without introducing errors. False In Two Phase Locking, there is a growing phase and a shrinking phase. False Web forms always send values to the server using the POST protocol. False HTTP is a stateless protocol. False CGI programs run on the server. False Nonserial schedules run one transaction at a time. False All serializable schedules produce the same results. False If the Precedence Graph for a schedule contains a cycle, then the schedule is non-serializable. False In Strict Two-Phase Locking, all locks are removed as soon as possible. False Strict Two-Phase Locking guarantees serializability. False The finer grained the locked items are, the more concurrency possible. False Deadlock prevention increases system overhead. False Query optimization evaluates all possible execution plans and picks the most efficient one. False Only one relational algebra trees (RATs) can be constructed for a single SQL query. False Query optimizers estimate the cost for all possible execution trees and then pick the least expensive. False During semantic analysis, relation connection graphs check for incorrectly formatted queries. False During semantic analysis, if the normalized attribute connection graph contains a negative value cycle, the query is not correctly formulated. 10

CSCE 4523 Introduction to Database Management Systems Final Exam Fall I have neither given, nor received,unauthorized assistance on this exam.

CSCE 4523 Introduction to Database Management Systems Final Exam Fall 2016 I have neither given, nor received,unauthorized assistance on this exam. Signature Printed Name: Attempt all of the following