Solutions to Midterm Examination

Prof. Li-Yan Yuan
CMPUT 391: Database Management Systems
Solutions to Midterm Examination
February 28, 2007
It is a close-book examination and the time for the test is 50 minutes. There are six (6) questions. The value of
each question is indicated in [ ] and the total is 90. Good luck to all of you.
1. One of the user-defined data-types, as specified by SQL’99, is a distinct type that is based on a
built-in data type, such as INTEGER. Give one simple example to demonstrate the benefit of
using such a user-defined data type. [10]
solution: A simple example is to create two distinct types: canadian dollar and us dollar, based
on numeric.
2. Indicate whether the following table satisfies
(a) the functional dependency A → C; and
(b) the multivalue dependency B →→ C. [10]
A B C
1 2 3
3 3 2
2 2 3
3 3 4
solution: (a) No. (b) Yes.
3. One of the inference rules for both FD and MVD is that X → A |= X →→ A.
Prove or disprove that X →→ A |= X → A. [15]
Solution: The table in (2) provides a counter example to the claim for it satisfies B →→ C but
violates B → C.
4. Consider a database consisting of the following tables.
employee( e_id, e_name, position, salary )
works( e_id, d_id, pct_time )
dept( d_id, budget, manager_id )
The first table indicates the job title and salary for each employee, the second shows the percentage
of time that a given employee works for a given department, and the last one indicates the budget
and boss of each department. Note that a manager must be an employee and his/her manager id
is the corresponding e id.
Write SQL triggers (in the Oracle style) to ensure that no employee will work over time (i.e., the
total of pct time is over 100), except managers. [20]
Solution: First, we can see that the following events may lead to the violation of the constraint:
INSERT and UPDATE on works
INSERT, UPDATE and DELETE on dept
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Therefore, we need to create triggers for all the above events.
One trigger for INSERT and one for UPDATE on work are given below.
CREATE TRIGGER no_hard_working_1
BEFORE INSERT ON works
FOR EACH ROW
WHEN new.e_id NOT IN ( SELECT manager_id FROM dept )
DECLARE total INT;
BEGIN
SELECT sum(pct_time) INTO total
FROM works w
WHERE w.e_id = new.e_id;
IF (total > 100 - new.pct_time )
THEN raise_application_error(-2000, ’working too hard’);
ELSE IF
END
CREATE TRIGGER no_hard_working_2
BEFORE UPDATE ON works
FOR EACH ROW
WHEN new.e_id NOT IN ( SELECT manager_id FROM dept )
DECLARE total INT;
BEGIN
SELECT sum(pct_time) INTO total
FROM works w
WHERE w.e_id = new.e_id;
END
IF (total > 100 - new.pct_time + old.pct_time )
THEN raise_application_error(-2000, ’working too hard’);
ELSE IF
5. Consider a relation schema with attributes R = ABCGW XY Z and the set of dependencies
F = {XZ → ZY B, Y A → CG, C → W, B → G, XZ → G}.
(a) Find a minimal cover for F.
After right-reducing, we have
XZ → B
XZ → G
XZ → Y
Y A → C
Y A → G
C → W
B → G
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It is easy to see that the left-had sides XZ and YA (as well as C and B) cannot be reduced
because there are no FSs of the form X → ..., Y → ..., Z → ..., and A → ....
Further, XZ → G is redundant and none of the others. So, the minimal cover consists of
the above set minus XZ → G.
(b) Decompose R into a join-lossless 3NF database schema. [20]
D = {XZBY, Y ACG, CW, BG, XZA} is a join-lossless 3NF database schema.
Note that XZA is added because no other table in D that contains a key of R.
6. Consider R = ABCDEF GH and F = {BE → GH, G → F A, D → C, F → B}. [15]
(a) Can there be a key that does not contain D Explain.
Solution: No, there cannot be a key that does not contain D because D does not appear in
the right-hand side of any FD in F.
(b) Is R in BCNF Explain.
Solution: No, R is not in BCNF with respect to F. This is because F → B and F is not a
key of R.
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