SOLUTIONS TO THE FINAL EXAMINATION Introduction to Database Design Spring 2011

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1 SOLUTIONS TO THE FINAL EXAMINATION Introduction to Database Design Spring 2011 IT University of Copenhagen June 7, Database design (25 points) a) b) Figure 1 shows the ER diagram. create table rer ( varchar(30), date_of_birth date, team_ int not null, foreign key (team_) references team() create table team ( varchar(30) create table stage ( length int, start varchar(30), end varchar(30) create table mountain ( stage_ int, distance int, height int, int, primary key(stage_, distance), foreign key () references rer(), foreign key (stage_) references stage() 1

2 bet amount odds madeby customer tour team mountain represe nts sprint rer date of birth result time stage number length start end on mountain sprint distance height Figure 1: ER diagram 2

3 create table result ( stage_ int, rer_ int, result_time int, primary key(stage_, rer_), foreign key (stage_) references stage() foreign key (rer_) references rer() create table customer ( varchar(30), varchar(30) create table bet ( customer int not null, amount int not null, odds numeric(5,2) not null, foreign key (customer) references customer() create table tour bet ( int not null, foreign key () references bet(), foreign key () references rer() create table mountain bet ( int not null, stage_ int not null, distance int not null, foreign key () references bet(), foreign key () references team(), foreign key (stage_, distance) references mountain(stage_, distance) 3

4 2 SQL and relational algebra (30 points) a) select start, end, license_plate, description from car, category, reservation where car_ = car. and category_ = category.; b) select * from car left outer join reservation on car_ = car.; c) select customer., count(reservation.) from customer left outer join reservation on customer. = customer_ group by customer.; In this the left outer join (rather than regular join) is needed to ensure that also customers with zero reservations are counted. But this is a detail. The second query of this problem is: select customer_, count(reservation.) from customer, reservation group by customer_ having count(reservation.) >= 5; d) select car. from car, category where category. = category_ and description = sportscar and car. not in (select car_ from reservation where start <= 2011:06:01 and end >= 2011:06:01 e) customer.g count(reservation.) (customer reservation) f) σ count(reservation.) 5 ( customer. G count(reservation.) (reservation)) delete from reservation where start = 2011:08:01 and customer_ in (select from customer where phone_num =

5 3 Normal forms (20 points) There are two candate keys: AC and BC. BCNF decomposition: Since B is not a super key, B E is a BCNF violation. We split R into (B, E) and (A, B, C, D). In the second of these A + = ABD, and so A is not a superkey and A BD is a BCNF violation. We must therefore split (A, B, C, D) and we get the decomposition of R into (B, E), (A, C) and (A, B, D). The two first of these are BCNF because all 2-element tables are. The third is also BCNF because A BD is the only nontrivial functional dependency, and this says that A is a superkey. For the canonical cover, first note that E is extraneous in A BDE because A E follows from A B and B E. This leaves us with A BD B E BC A which is a canonical cover. The 3NF decomposition gives (A, B, D), (B, E), (A, B, C). We do not need to add any relations to this because (A, B, C) contains a superkey. 4 Transactions (15 points) Precedence graph: T 2 T 1 T 3 T 4 There are no cycles, so it is serializable. The schedule T 2 T 1 T 3 T 4 is equivalent. The schedule is not recoverable because the first read of T 3 is dirty, and so T 3 is dependent on T 2 and moreover, T 3 commits before T 2. 5 Indices (10 points) a) Queries (2) and (4) should become faster, but (1) and (3) should not. This is independent of whether the index is clustered or not. The join (4) might run a bit faster for a clustered index, but even for a non-clustered index the index will probably help. b) There is now a chance that (3) will run faster, but only if the index is clustered. 5