MIE253S Midterm Exam

Transcription

1 Mechanical and Industrial Engineering University of Toronto MIE253S Midterm Exam February 25 th 2016 Instructor: Mariano Consens DO NOT OPEN THIS BOOKLET UNTIL INSTRUCTED TO DO SO This is a closed book, notes, neighbors and no aids exam. This handout and the test booklets are the only papers allowed during the exam, and both have to be handed back at the end of the exam. This midterm represents 25% of the course grade. You have 90 minutes to complete the exam. Write legibly. Print your First Name Print your Last Name Print your Student Number I acknowledge and SIGN that I am submitting booklets. Problem Points Maximum Points Total 100 Problem 1 [30 points; 12 for 1.1, 18 for 1.2] An agricultural business maintains a relational database to keep track of crop yields for different seeds tested in plots set up at several farm locations, with the following relational schema: Plots(pid, farmid, area, soilquality) Neighbour(pid, farmid, direction, npid, nfarmid) Crop(pid, farmid, seed, sowdate, harvestdate, cropyield) A tuple (P,F,A,Q) in the relation Plots gives the numeric identifier P of a plot in a farm with numeric identifier F, the area A of the plot in (fractional) acres, and the soil quality Q (described only as high, average, or low ). All the data for a given plot in the database is always known. The plots are identified by consecutive numbers within a farm, always starting with 10 (and there are never more than 50 plots in a given farm), and farms are identified by 3 digit numbers. A tuple (P,F,D,NP,NF) in the relation Neighbour provides, for a plot P in farm F, the neighbour plot NP in (the same or in an adjacent) farm NF located in direction D (one of N, E, S, W, NE, NW, SE, SW ). There is only one neighbour in a given direction

2 The relation crop has tuples (P,F,S,SD,HD,Y) that initially record that the seed S (described by 12 letters) has been used to sow a given plot P in farm F (one seed always covers an entire plot) on the date SD. The same tuple is updated months later to record that the crop was harvested on the date HD producing a crop yield Y (a ratio between 0 and 5.5). To diversify the conditions under which seeds are tested, a given seed is never sowed twice on the same date Give the SQL domain creation statements for the 7 domains used in the relational schema above (the database uses the SQL data type DATE for dates). CREATE DOMAIN Dpid INTEGER -- npid has same domain CHECK (VALUE >= 10 AND VALUE <= 60) CREATE DOMAIN Dfarmid INTEGER -- nfarmid has same domain CHECK (VALUE >= 100 AND VALUE <= 999) CREATE DOMAIN Darea DOUBLE -- or REAL, FLOAT, DECIMAL CREATE DOMAIN Dquality VARCHAR(7) or CHAR(7), TEX CHECK (VALUE IN ( high, average, low )) CREATE DOMAIN Ddirection CHAR(2) CHECK (VALUE IN ( N, E, S, W, NE, NW, SE, SW )) CREATE DOMAIN Dseed CHAR(12) CREATE DOMAIN Dyield DOUBLE -- or REAL, FLOAT, DECIMAL CHECK (VALUE >= 0 AND VALUE <= 5.5) 1.2. Give the SQL table creation statements for the three relations above (including keys, foreign keys, as well as null and uniqueness constraints), using the 7 domains you defined above and the DATE data type. CREATE TABLE Plots ( area Darea NOT NULL, -- incorrect if NULL soilquality Dquality NOT NULL, -- incorrect if NULL PRIMARY KEY (pid, farmid)) CREATE TABLE Neighbour ( direction Ddirection, npid Dpid NOT NULL, -- acceptable if NULL nfarmid Dfarmid NOT NULL, -- acceptable if NULL PRIMARY KEY (pid, farmid, direction), FOREIGN KEY (pid, farmid) REFERENCES Plots, FOREIGN KEY (npid, nfarmid) REFERENCES Plots (pid, farmid)) CREATE TABLE Crop ( seed Dseed NOT NULL, -- incorrect if NULL sowdate DATE, - 2 -

3 harvestdate DATE, -- incorrect if NOT NULL cropyield Dyield, -- incorrect if NOT NULL PRIMARY KEY (pid, farmid, sowdate), UNIQUE (seed, sowdate), FOREIGN KEY (pid, farmid) REFERENCES Plots) Problem 2 [26 points; 10 for 2.1, 10 for 2.2, 5 for 2.3, (1 + up to 5 bonus) for 2.4] Using the agricultural business relational database of Problem 1, answer the following questions Write in relational algebra the query that lists the seeds (and the corresponding crop yield) that have been harvested before 30/4/2014 from plots with low soil quality. HBC= HD< 30/4/2014 Crop[P,F,S,SD,HD,Y] -- crop harvested before given date Answer21 = S,Y ( Q= low (P HBC) ) 2.2. Write in relational algebra the query that lists pairs of neighbouring plots (both the plot and farm identifier) that have the same soil quality and have been sowed with different seeds. The first plot in the pair should have been harvested during 2012, while the second should have been harvested during The area of the first plot should be smaller than the area of the second plot in the pair. HC12= HD>= 1/1/2012 AND HD<= 31/12/2012 Crop[P,F,S,SD,HD,Y] -- crop harvested during 2012 N=Neighbour[P,F,D,P2,F2] P2=Plots[P2,F2,A2,Q] -- Notice same soil quality Q HC13= HD>= 1/1/2013 AND HD<= 31/12/2013 Crop[P2,F2,S2,SD2,HD2,Y2] -- second crop harvested during 2013 Answer22 = P,F,P2,F2 ( S<>S2 AND A<A2 (P HC12 N P2 HC13) ) 2.3. Clearly describe in English the query answered by the relation Answer23 defined by the following relational algebra expressions NS = DS= S Neighbour [P,F,DS,PS,FS] CropS = Crop [PS,FS,S,SDS,HDS,YS] NSE = DSE= SE Neighbour [P,F,DSE,PSE,FSE] CropSE = Crop [PSE,FSE,S,SDSE,HDSE,YSE] Answer23 = F (P) - F (NS CropS NSE CropSE) - 3 -

4 List those farms where there is no plot in the farm that such that both the S and SE neighbour plots sowed with the same seed as the plot itself. 2.4 Write in SQL the query that lists the farms that have been seeded with all of the seeds used to seed the farm with farmid 157. Two possible solutions below SELECT DISTINCT C.farmid FROM Crop C WHERE NOT EXISTS -- if no rows, then C.farmid is not missing any seeds used to seed 157 (SELECT DISTINCT seed -- set of seeds used in farmid 157 FROM Crop C157 WHERE C157.farmid = 157) (SELECT DISTINCT seed -- set of seeds used in C.farmid FROM Crop CS WHERE CS.farmid = C.farmid) - notice correlated variable (SELECT DISTINCT C.farmid FROM Crop C) (SELECT DISTINCT farmid FROM ((SELECT DISTINCT farmid, seed FROM (SELECT farmid FROM Crop), (SELECT seed FROM Crop WHERE farmid = 157) ) (SELECT DISTINCT farmid, seed FROM Crop C) ) ) The second variation can be derived from the algebra solution using division FS = F,S (Crop[P,F,S,SD,HD,Y] ) S157 = S ( F=157 Crop[P,F,S,SD,HD,Y] ) Answer24 = FS / S157 = F ( FS ) - F ( ( F ( FS ) S157 ) - FS ) Problem 3 [14 points, 9 SQL DDL and DML, 5 explanation, plus 2 or 3 as bonus] The SQL database of a company has a relation Bonus(EMP_ID,MONTH,AMOUNT) with primary key (EMP_ID,MONTH) that keeps a record of the bonus AMOUNT paid to an employee (identified by EMP_ID) for a given MONTH. The DBA issues the following SQL DDL REVOKE SELECT, INSERT, DELETE, UPDATE ON Bonus FROM Analyst_Role; hoping to prevent users authenticated as Analyst_ Role to gain access to the information about the bonus paid to the employees. However, the users authenticated as - 4 -

5 Analyst_ Role have been granted all the remaining privileges and are able to create new tables in the database. Explain how users authenticated as Analyst_ Role are still able to find information such as whether an employee with EMP_ID will receive a bonus for the month of February, writing all the SQL DDL and DML needed to achieve this. Add comments to clearly indicate why the DDL and DML that you give will not violate SQL authorizations and to explain exactly how information is leaked by executing your SQL statements. CREATE TABLE TestBonus ( EMP_ID INTEGER, MONTH STRING, FOREIGN KEY (EMP_ID, MONTH) REFERENCES Bonus) Users authenticated as Analyst_ Role can create tables and the REFERENCES access mode privilege on table Bonus has not been revoked, so the new table can have the foreign key referencing Bonus. INSERT INTO TestBonus VALUES (12345, February ) If the insert succeeds then the employee and month must exists as primary key values in the Bonus table and hence a bonus was awarded, if it fails due to a foreign key violation then there was no bonus awarded. Problem 4 [30 points] Consider the following Java/JDBC program (continued in the next page). 4.1 (2) points for each explanation (8) 4.2 (2) points for R2 instance 4.3 (2) points for each output line (10) 4.4 (2) points for each explanation (6) 4.5 (2) points for R2 instance 4.6 (2) points for writing the missing line of code - 5 -