Midterm 2 - University of California, Irvine · 2020. 11. 17. · 34/34 Questions Answered Saved on Jun 4 at 4:49 PM TIME REMAINING: Midterm 2 Q1 Preliminaries 1 Point Instructions

34/34 Questions AnsweredSaved on Jun 4 at 4:49 PM

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Midterm 2

Q1 Preliminaries1 Point

Instructions

The allowed time for the exam is 60 minutes (50 minutes plus a 10 minute

technology buffer). Be sure to pay attention to time and budget your time

accordingly!

The exam is open pre-prepared cheat sheet, open book, open notes, open

web browser, and even open MySQL. You are just not allowed to communicate

with or otherwise interact with other students (or friends) during the course of

the exam, and this includes your HW brainstorming buddy. This exam is to be a

solo effort!

Read each question carefully, in its entirety, and then answer each part of the

question.

STUDENT NAME

Search students by name or email…

If you don't understand something, make your best guess; if you find

ambiguities in a question, note the interpretation that you are taking.

Acknowledgement: I certify that I am taking this exam myself, on my own, with honesty and integrity, without interaction with others during the exam, and

without having obtained any information about the exam's content from others

prior to taking it.

Reference Data

The exam questions that involve query-writing and data will all be based on the

following HR schema about employees, departments, work assignments, and

managers. (You will need to scroll back here periodically to answer some of the

questions.)

CREATE SCHEMA HR; USE HR; CREATE TABLE Emp ( eid integer PRIMARY KEY, ename varchar(50), age integer, salary integer, CHECK ((eid MOD 2) 0)); CREATE TABLE Dept ( did integer PRIMARY KEY, dname varchar(50), budget integer, eid integer, -- dept manager's id ename varchar(50), -- dept manager's name FOREIGN KEY (eid) REFERENCES Emp(eid)); CREATE TABLE Works ( eid integer, did integer, pcttime integer, PRIMARY KEY (eid, did), FOREIGN KEY (eid) REFERENCES Emp(eid)); INSERT INTO Emp (eid, ename, age, salary) VALUES (1, 'Dustin', 50, 60000), (3, 'Emily', 30, 30000),

True

False

(5, 'John', 20, 20000), (7, 'Win', NULL, 45000), (9, 'Arthur Jr', 22, 40000); INSERT INTO Dept (did, dname, budget, eid, ename) VALUES (101, 'Accounting', 50000, 1, 'Dustin'), (102, 'Sales', 50000, 3, 'Em'), (103, 'Legal', 100000, 7, 'Win'); INSERT INTO Works (eid, did, pcttime) VALUES (1, 101, 80), (1, 102, 20), (3, 102, 100), (5, 102, 50), (7, 101, 33), (7, 102, 33), (7, 103, 33), (9, 104, 100);

Correct

Save Answer Last saved on Jun 04 at 8:42 AM

Q2 Short and Two the Point(s)30 Points

For each of the following questions, you should

SELECT best(answer) FROM question LIMIT 1.

Q2.12 Points

A query language is said to be relationally complete if it has at least the expressive power of:

Correct


Q2.22 Points

Join is a core (necessary) operation in the relational algebra.

Correct


Q2.32 Points

Union is a core (necessary) operation in the relational algebra.

Correct


SQL

the relational algebra

the relational calculus

Python

True

False

True

False

Q2.42 Points

Division is a core (necessary) operation in the relational algebra.

Correct


Q2.52 Points

Based on the current HR reference tables, what is the cardinality of the result of

the relational algebra query: Emp x Dept

Correct


Q2.62 Points

Based on the HR reference data, what is the cardinality of the result of the

relational calculus query: { e | (e ∉ Emp) ∧ e.ename = 'Win' }

True

False

0

3

5

15

∞

Correct


Q2.72 Points

What would be the most appropriate MySQL mechanism to employ to ensure

that the budget of a Dept is always at least 10000?

Correct


Q2.82 Points


that the budget of a Dept is always sufficient to cover its manager's salary?

0

1

5

11

∞

FOREIGN KEY CONSTRAINT

CHECK CONSTRAINT

UNIQUE CONSTRAINT

TRIGGER

VIEW

Correct


Q2.92 Points


that no two employees who work in a given department have the same name?

Correct


Q2.102 Points

What would be the most appropriate MySQL mechanism to employ to enable

the secretarial staff to see the ids and names of employees, but not their ages

or salaries, along with the ids and names of the departments that they work in?


CHECK CONSTRAINT

UNIQUE CONSTRAINT

TRIGGER

VIEW


CHECK CONSTRAINT

UNIQUE CONSTRAINT

TRIGGER

VIEW

Correct


Q2.112 Points


that no two employees in the company make exactly the same amount of

money?

Correct


Q2.122 Points

What would the following MySQL query return given the current HR reference

tables?


CHECK CONSTRAINT

UNIQUE CONSTRAINT

TRIGGER

VIEW


CHECK CONSTRAINT

UNIQUE CONSTRAINT

TRIGGER

VIEW

SELECT (e1.age + e2.age) / 2FROM Emp e1, Emp e2 WHERE e1.eid = 7 AND e2.eid = 9;

Correct


Q2.132 Points

What would the following MySQL query return given the current HR reference

tables?

SELECT AVG(e.age) FROM Emp e WHERE e.eid >= 7;

Correct


Q2.142 Points

0

11

22

NULL

0

11

22

NULL

Consider the following MySQL aggregate query:

SELECT e.age, e.name, COUNT(*) FROM Emp e GROUP BY e.age HAVING COUNT(*) > 1;

Is this a legal SQL query?

Correct


Q2.152 Points

Consider the following MySQL UPDATE query:

UPDATE Emp e SET e.eid = e.eid + 1WHERE e.age > 21;

Look at the Emp table's CREATE TABLE statement again. Will this update query

succeed?

Correct


Q3 Taking an Equivalence Class

Yes

No

Yes

No

30 Points

Consider once again the tables of our HR reference database:

CREATE SCHEMA HR; USE HR; CREATE TABLE Emp ( eid integer PRIMARY KEY, ename varchar(50), age integer, salary integer, CHECK ((eid MOD 2) 0)); CREATE TABLE Dept ( did integer PRIMARY KEY, dname varchar(50), budget integer, eid integer, -- dept manager's id ename varchar(50), -- dept manager's name FOREIGN KEY (eid) REFERENCES Emp(eid)); CREATE TABLE Works ( eid integer, did integer, pcttime integer, PRIMARY KEY (eid, did), FOREIGN KEY (eid) REFERENCES Emp(eid));

For each of the following problems, you will be given a relational query along

with a set of additional queries. For each problem, indicate which query or

queries in the additional query set are equivalent to the given query. (Note that

equivalent means that it will produce the same result as the original query for

all possible database states permitted by the HR schema's CREATE TABLE statements.) Note: Be sure to consider each query carefully!

Q3.16 Points

Here is the given query:

SELECT DISTINCT d.dname, e.ename FROM Emp e JOIN Dept d ON e.eid = d.eid;

Select the equivalent query or queries (if any):

Correct


Q3.26 Points


(π eid, did (Works)) ÷ (π did ( σ budget = 50000 (Dept)))


Correct


π dname, ename (Emp ⨝ Dept)

SELECT DISTINCT d.dname, d.ename FROM Dept d;

{ t(dname, ename) | ∃e ∈ Emp (t.ename = e.ename ∧ ∃d ∈ Dept (e.eid =

d.eid ∧ t.dname = d.dname) }

{ t(eid) | ∃e ∈ Emp (t.eid = e.eid ∧ ∀d ∈ Dept (d.budget = 50000 (∃w ∈

Works (w.eid = e.sid ∧ d.bid = w.bid))))}

{ t(eid) | ∃e ∈ Emp (t.eid = e.eid ∧ ∀d ∈ Dept (d.budget = 50000 ∧ (∃w ∈

Works (w.eid = e.sid ∧ d.bid = w.bid))))}

{ t(eid) | ∃e ∈ Emp (t.eid = e.eid ∧ ¬(∀d ∈ Dept (d.budget ≠ 50000 ∨ (∃w ∈

Works (w.eid = e.sid ∧ d.bid = w.bid)))))}

Q3.36 Points


SELECT DISTINCT ename FROM Emp e WHERE e.salary = (SELECT MAX(salary) FROM Emp);


Correct


Q3.46 Points


π eid, ename (σ pcttime < 100 (Works ⨝ Emp))


{ t(ename) | ∃e ∈ Emp (t.ename = e.ename ∧ ∄m ∈ Emp (e.salary <

m.salary) }

π ename (Emp ⨝ salary ≥ salary2 (ρ eid2←eid, ename2←ename,

age2←age, salary2←salary (Emp)))

SELECT DISTINCT ename FROM Emp e WHERE 0 = (SELECT COUNT(*)

FROM Emp p WHERE p.salary > e.salary);

Correct


Q3.56 Points


{t(ename) | ∃e ∈ Emp (t.ename = e.ename ∧ e.salary < 25000 ∧ e.age < 25)}


Correct


SELECT e.eid, e.ename FROM Works w, Emp e WHERE w.pcttime < 100

AND w.eid = e.eid;

SELECT e.eid, e.ename FROM Emp e WHERE e.eid in (SELECT eid

FROM Works WHERE pcttime < 100);

SELECT DISTINCT e.eid, e.ename FROM Works w LEFT OUTER JOIN

Emp e ON w.eid = e.eid WHERE w.pcttime < 100;

SELECT DISTINCT ename FROM Emp WHERE salary < 25000 OR age <

25;

(π ename (σ age < 25 (Emp))) ∩ (π ename (σ salary < 25000 (Emp)))

SELECT DISTINCT ename FROM Emp e WHERE age < 25 AND eid IN

(SELECT eid FROM Emp WHERE salary < 25000);

Q4 Query Me This, Batman18 Points

It's time to check the box that says "I'm tired of checking boxes!" In this

problem you will actually write or "hand execute" a couple of queries of your

own! Exciting, right?

Note: When asked to show queries' answers below, feel free to show them in a

CSV-like format. For example, suppose the given query is:

SELECT * FROM Dept WHERE budget < 100000;

In this case you could denote the query's answer as follows (since Gradescope

won't let you hand-draw pictures of tables with rows and columns):

did, dname, budget, eid, ename 101, 'Accounting', 50000, 1, 'Dustin' 102, 'Sales', 50000, 3, 'Em'

In preparation for what follows, here's a repeat of the HR database's tables and

their current contents:

CREATE SCHEMA HR; USE HR; CREATE TABLE Emp ( eid integer PRIMARY KEY, ename varchar(50), age integer, salary integer, CHECK ((eid MOD 2) 0)); CREATE TABLE Dept ( did integer PRIMARY KEY, dname varchar(50), budget integer, eid integer, -- dept manager's id ename varchar(50), -- dept manager's name FOREIGN KEY (eid) REFERENCES Emp(eid)); CREATE TABLE Works ( eid integer, did integer, pcttime integer,

PRIMARY KEY (eid, did), FOREIGN KEY (eid) REFERENCES Emp(eid)); INSERT INTO Emp (eid, ename, age, salary) VALUES (1, 'Dustin', 50, 60000), (3, 'Emily', 30, 30000), (5, 'John', 20, 20000), (7, 'Win', NULL, 45000), (9, 'Arthur Jr', 22, 40000); INSERT INTO Dept (did, dname, budget, eid, ename) VALUES (101, 'Accounting', 50000, 1, 'Dustin'), (102, 'Sales', 50000, 3, 'Em'), (103, 'Legal', 100000, 7, 'Win'); INSERT INTO Works (eid, did, pcttime) VALUES (1, 101, 80), (1, 102, 20), (3, 102, 100), (5, 102, 50), (7, 101, 33), (7, 102, 33), (7, 103, 33), (9, 104, 100);

Q4.14 Points

Consider the following SQL query Q:

WITH -- define some useful building blocks for Q WorksInDept (eid, pct, did, dname, budget, mgrid, mgrname) AS (SELECT e.eid, w.pcttime, d.did, d.dname, d.budget, d.eid, d.ename FROM Emp e LEFT OUTER JOIN Works w ON e.eid = w.eid LEFT OUTER JOIN Dept d ON w.did = d.did), ManagesDept (eid, ename, mgrname, did, dname, budget) AS (SELECT e.eid, e.ename, d.ename, d.did, d.dname, d.budget FROM Emp e LEFT OUTER JOIN Dept d ON e.eid = d.eid) SELECT ... -- rest of query Q goes here

Show the answer that would result from executing this query if the rest of query

Q is:

SELECT * FROM WorksInDept;

eid,pct,did,dname,budget,mgrid,mgrname

1,80,101,Accounting,50000,1,Dustin

7,33,101,Accounting,50000,1,Dustin

1,20,102,Sales,50000,3,Em

3,100,102,Sales,50000,3,Em

5,50,102,Sales,50000,3,Em

7,33,102,Sales,50000,3,Em

7,33,103,Legal,100000,7,Win

9,100,NULL,NULL,NULL,NULL,NULL

Save Answer Last saved on Jun 04 at 4:46 PM

Q4.22 Points

What problem or issue does this query reveal with the schema (the CREATE

TABLE statements) or the current base tables' contents (the data in Emp, Dept,

or Works)?

It's currently possible for an employee to be assigned to work in a non-

existent department.


Q4.34 Points

Now show the answer that would result from executing the above query if the

rest of query Q is:

SELECT * FROM ManagesDept;

eid,ename,mgrname,did,dname,budget

1,Dustin,Dustin,101,Accounting,50000

3,Emily,Em,102,Sales,50000

5,John,NULL,NULL,NULL,NULL

7,Win,Win,103,Legal,100000

9,"Arthur Jr",NULL,NULL,NULL,NULL

Save Answer Last saved on Jun 04 at 4:49 PM

Q4.42 Points

And what problem or issue does this query reveal with the current schema or base table contents?

It's currently possible for an employee's name to be different than what's

recorded in the department table.


Q4.56 Points

Show what query Q's SELECT ending should be if your task is, for each

employee, to print their eids along with the dids and budgets of the

departments that they work in but do not manage. I.e., starting with SELECT ... , write the remainder of the query. The FROM clause(s) in your

answer should only reference the temporary tables introduced in the WITH

clause (and should not need or use any of the base tables):

SELECT w.eid, w.did, w.budget

FROM WorksInDept w

WHERE NOT EXISTS

(SELECT * FROM ManagesDept m

WHERE m.eid = w.eid AND m.did = w.did);


Q5 Overcoming SELECTion Bias20 Points

It's time to move past SELECT statements and put some of SQL's more

advanced language features to use on the HR database.

Q5.16 Points

Complete the following CREATE VIEW statement to aid the HR analysts in

keeping an eye on age-related salary equity for employees. Anticipating a

future with more employees, they want to be prepared to keep track of how

many employees there are of each age and what the age groups' salary

statistics are.

CREATE VIEW EmpStats (agegroup, numemps, lowsal, avgsal, highsal) AS

SELECT age AS agegroup, COUNT(*) AS numemps, MIN(salary) AS lowsal,

AVG (salary) AS avgsal, MAX(salary) AS highsal

FROM Emp

GROUP BY age;


Q5.25 Points

One of the senior IT staff in the IT department spotted one of the problems that

you identified in problem 4 and has proposed a PROCEDURE-based cure. From

now on she wants applications to use this procedure to assign managers to

departments:

DELIMITER // CREATE PROCEDURE NewManager (IN newdid integer, IN neweid integer) BEGIN UPDATE Dept SET eid = neweid, ename = (SELECT ename FROM Emp WHERE eid = neweid) WHERE did = newdid; END;

Show what the effect would be on the Dept table of running the following SQL

statement given the existing data:

CALL NewManager(102, 3);

Dept (did, dname, budget, eid, ename)

(101, 'Accounting', 50000, 1, 'Dustin'),

(102, 'Sales', 50000, 3, 'Emily'),

Show what each of the following three SQL statements would produce after

this trigger has been defined (starting with the existing data) if they are

executed one after another in sequence:

Here's the first statement in the sequence:

SELECT COUNT(*) AS prior FROM Works;

8


Q5.42 Points

Here's the next statement in the sequence:

INSERT INTO Works (eid, did, pcttime) VALUES (7, 104, 1);

Warning: Bad assignment!


Q5.51 Point

And here's the last statement in the sequence:

SELECT COUNT(*) AS after FROM Works;

8


Q5.63 Points

Look at the HR schema's CREATE TABLE statements. Is there a better way to

tackle the problem that the new hire's trigger is attempting to solve? If so,

explain briefly what you would recommend doing instead.

Yes.

Add a FOREIGN KEY constraint FROM Works(did) to Dept(did)!


Q5.72 Points

A SQL database can be made more secure by combining its access control capabilities (i.e., GRANT and REVOKE) with which of the following other SQL

DDL features? (Check all that apply.)

Correct

Primary keys

Views

Foreign keys

Stored procedures

Check constraints

Unique constraints


Q6 Last and also Least1 Point

Take one last careful look at the tables and columns in the HR schema and

data. What strikes you as quite literally being the oddest column in the entire HR database? :-) (Note: Your answer should be of the form Tblname.colname .)

Emp.eid

Correct


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Midterm 2 - University of California, Irvine · 2020. 11. 17. · 34/34 Questions Answered Saved on Jun 4 at 4:49 PM TIME REMAINING: Midterm 2 Q1 Preliminaries 1 Point Instructions