1. 1. Explain the following concepts of the ER data model:
Entity type and Entity type classification.
Attribute and attribute classification.
Constraint and constraint classification.
Entity type: the structure description of some kind of entities.
Entity type classification: strong entities and weak entities.
Attribute: a property of some kind of entities with the same
structure.
Attribute type classification: simple attribute, multi-valued
attribute, composite attribute, complex attribute, derived
attribute, and stored attribute.
Sept. 2004
Yangjun Chen

2. Constraint: constraints on relationships between or among
entity types.
Constraint classification:
cardinality constraints: 1:1, 1:M, N:M.
Participation constraints: total constraint
partial constraint
Sept. 2004
Yangjun Chen

3. 2. Draw an ER-diagram to describe the following real world problem.
At a university, instructors offer courses during particular semesters. Each instructor can teach some courses. Some courses are taught during a specific semester and some instructors teach during a specific semester. Each instructor has a name and a number. Each course has a name and a number. Each semester can be identified by the year and the season (spring, summer, fall and winter) when the semester is held.
Sept. 2004
Yangjun Chen

4. Taught_During M Iname year N N M Offers Instructor Semester M P M
season
Can_Teach
Offered_During N
Course N
Cnumber
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Yangjun Chen

5. - collision resolution strategy: chaining
3. Linear Hashing
- collision resolution strategy: chaining
- split rule: load factor > 0.7
- initially M = 4 (M: size of the primary area)
- hash functions: hi(key) = key mod 2i  M (i = 0, 1, 2, …)
- bucket capacity = 2
Trace the insertion process of the following keys into a linear
hashing file:
3, 2, 4, 1, 8, 14, 5, 10, 7, 24, 17, 13, 15.
Sept. 2004
Yangjun Chen

6. when inserting the sixth record we would have
The first phase – phase0
when inserting the sixth record we would have
but the load factor 6/8= 0.75 > 0.70 and so bucket 0 must be split (using h1 = Key mod 2M): 4 8 1 2 14 3
n=0 before the split
(n is the point to the bucket to be split.) 8 1 2 14 3 4
n=1 after the split
load factor: 6/10=0.6
no split
Sept. 2004
Yangjun Chen

7. 8 1 2 14 3 4 0 1 2 3 4 n=1 load factor: 7/10=0.7 no split 8 1 5 2 14 3
insert(5) 8 1 2 14 3 4
n=1
load factor: 7/10=0.7
no split 8 1 5 2 14 3 4
Sept. 2004
Yangjun Chen

8. 8 1 5 2 14 3 4 0 1 2 3 4 n=1 load factor: 8/10=0.8 split using h1. 8 1
insert(10) 8 1 5 2 14 3 4
n=1
load factor: 8/10=0.8
split using h1. 8 1 5 2 14 3 4
overflow 10
Sept. 2004
Yangjun Chen

9. 8 1 2 14 3 4 5 0 1 2 3 4 5 10 n=2 load factor: 8/12=0.66 no split
overflow 10
n=2
load factor: 8/12=0.66
no split
Sept. 2004
Yangjun Chen

10. 8 1 2 14 3 4 5 n=2 load factor: 9/12=0.75 split using h1. 10 8 1 2 14
insert(7) 8 1 2 14 3 4 5
n=2
load factor: 9/12=0.75
split using h1.
overflow 10 8 1 2 14 3 7 4 5
overflow 10
Sept. 2004
Yangjun Chen

11. 8 1 2 10 3 7 4 5 14 n=3 load factor: 9/14=0.642 no split. 8 1 2 10 3 7
insert(24) 8 1 2 10 3 7 4 5 14
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Yangjun Chen

12. 8 24 1 2 10 3 7 4 5 14 n=3 load factor: 10/14=0.71 split using h1. 8
Sept. 2004
Yangjun Chen

13. The second phase – phase1 n = 0; using h1 = Key mod 2M to insert and 8 24 1 2 10 3 4 5 14 7
n=4
The second phase – phase1
n = 0; using h1 = Key mod 2M to insert and
h2 = Key mod 4M to split.
insert(17) 8 24 1 2 10 3 4 5 14 7
Sept. 2004
Yangjun Chen

14. 8 24 1 17 2 10 3 4 5 14 7 n=0 load factor: 11/16=0.687 no split. 8 24
insert(13) 8 24 1 17 2 10 3 4 5 14 7
Sept. 2004
Yangjun Chen

15. 8 24 1 17 2 10 3 4 5 13 14 7
n=0
load factor: 12/16=0.75
split bucket 0, using h2. 1 17 2 10 3 4 5 13 14 7 8 24
Sept. 2004
Yangjun Chen

16. insert(15) 1 17 2 10 3 4 5 13 14 7 8 24 1 17 2 10 3 4 5 13 14 7 15 24
n=1
load factor: 13/18=0.722
split bucket 1, using h2. 1 17 2 10 3 4 5 13 14 7 15 8 24
Sept. 2004
Yangjun Chen

17. Here, we assume that each internal node can contain at most two
4. Given the following B+-tree, trace the deletion sequence: 8, 3, 5, 6.
Here, we assume that each internal node can contain at most two
keys and each leaf node can contain at most two value/point pairs. 3 2 5 1 2 3 5 6 8
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Yangjun Chen

18. 6 5 1 2 3 6 5 1 3 2
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Yangjun Chen

19. 3 1 5 1 2 6 1 5 1 2 6 1 2 1
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Yangjun Chen

20. fname, minit, lname, ssn, bdate, address, sex, salary, superssn, dno
5. Given the relation schemas shown in Fig. 2, construct expressions
(using relational algebraic operations) to evaluate the following query:
Find the names of employees who work on all the projects controlled by a
department that controls a project ’Web-DB’ and some other projects. Each
project is controlled by one department. But each department can control more
than one projects.
EMPLOYEE
fname, minit, lname, ssn, bdate, address, sex, salary, superssn, dno
DEPARTMENT
Dname, dnumber, mgrssn, mgrstartdate
PROJECT
WORKS_ON
Pname, pnumber, plocation, dnum
Essn pno, hours
Sept. 2004
Yangjun Chen

21. : DN Dnum(Pname = ‘Web-DB’(PROJECT))
DEPT_P PNUMBER(DNDN.Dnum = PROJECT.Dnum(PROJECT))
EMP_PNOS  ESSN,PNO(WORK_ON)
SSNS EMP_PNOS :
DEPT_P
RESULT  FNAME, LNAME(SSNS * EMPLOYEE)
Sept. 2004
Yangjun Chen

22. 6. Given the relation schemas as shown in Fig. 2, construct a SQL
clause to evaluate the following query:
Find department names, and the number of employees in each
department, who take part in at least one project.
SELECT Dname, count(distinct SSN)
FROM DEPARTMENT, EMPLOYEE, WORKS_ON
WHERE DNUMBER = DNO AND
SSN = ESSN
Group By Dname
Sept. 2004
Yangjun Chen