Solution of Erds
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This document discusses exercises related to entity relationship modeling. It includes ER diagrams for examples like a university registration system, car insurance claims, and sports team statistics. It also provides answers to various questions about identifying optimal ER diagrams, handling weak entities, and merging databases from two different banks. The key topics covered are ER modeling, identifying relationships, handling attributes and weak entities, and resolving conflicts when merging related databases.
Solution of Erds
- 1. C H A P T E R 2 Entity Relationship Model Exercises 2.2 Answer: See Figure 2.1 2.4 Answer: See Figure 2.2. In the answer given here, the main entity sets are student, course, course-offering, and instructor. The entity set course-offering is a weak entity set dependent on course. The assumptions made are : a. a class meets only at one particular place and time. This E-R diagram cannot model a class meeting at different places at different times. b. There is no guarantee that the database does not have two classes meeting at the same place and time. 2.5 Answer: model name license year person owns car participated accident address damage-amount report-number date location driver-id driver Figure 2.1 E-R diagram for a Car-insurance company. 5
- 2. 6 Chapter 2 Entity Relationship Model program time room course− offerings iid name dept title course courseno title credits syllabus prerequisite maincourse requires secno is offered student name grade teaches year semester enrols sid instructor Figure 2.2 E-R diagram for a university. secno time room course− offerings courseno exam name place time student program marks eid name year semester takes sid Figure 2.3 E-R diagram for marks database. a. See Figure 2.3 b. See Figure 2.4 2.7 Answer: See Figure 2.5
- 3. Exercises 7 secno time room course− offerings courseno program marks examof exam name place time student name year semester takes sid Figure 2.4 Another E-R diagram for marks database. date matchid stadium name age played match player season_score opponent own _score opp_score score Figure 2.5 E-R diagram for favourite team statistics. 2.13 Answer: By using one entity set many times we are missing relationships in the model. For example, in the E-R diagram in Figure 2.6: the students taking classes are the same students who are athletes, but this model will not show that. However, E-R diagrams are often split into parts so that each part will fit in a single page. In such a case, an entity set may appear on different pages, but to avoid repetition (and inconsistencies), its attributes should be specified at only one place. The absence of attributes at other places indicates that they are placeholders, and the full definition is present elsewhere. 2.14 Answer: a. See Figure 2.7 b. The additional entity sets are useful if we wish to store their attributes as part of the database. For the course entity set, we have chosen to include
- 4. 8 Chapter 2 Entity Relationship Model ss# name dept student ss# name takes class student plays sport courseno teamname Figure 2.6 E-R diagram with entity duplication. name section for time department c-number section of s-number enrollment course room in exam r-number capacity building exam-id Figure 2.7 E-R diagram for exam scheduling. three attributes. If only the primary key (c-number) were included, and if courses have only one section, then it would be appropriate to replace the course (and section) entity sets by an attribute (c-number) of exam. The reason it is undesirable to have multiple attributes of course as attributes of exam is that it would then be difficult to maintain data on the courses, particularly if a course has no exam or several exams. Similar remarks apply to the room entity set. 2.15 Answer: a. The criteria to use are intuitive design, accurate expression of the real-world concept and efficiency. A model which clearly outlines the objects and rela-tionships in an intuitive manner is better than one which does not, because it is easier to use and easier to change. Deciding between an attribute and an entity set to represent an object, and deciding between an entity set and relationship set, influence the accuracy with which the real-world concept is expressed. If the right design choice is not made, inconsistency and/or
- 5. Exercises 9 program time room course− offerings iid name dept title course courseno title credits syllabus prerequisite maincourse requires secno is offered student name grade teaches year semester enrols sid instructor Figure 2.8 E-R diagram for University(a) . loss of information will result. A model which can be implemented in an efficient manner is to be preferred for obvious reasons. b. Consider three different alternatives for the problem in Exercise 2.4. • See Figure 2.8 • See Figure 2.9 • See Figure 2.10 Each alternative has merits, depending on the intended use of the database. Scheme 2.8 has been seen earlier. Scheme 2.10 does not require a separate entity for prerequisites. However, it will be difficult to store all the prerequi-sites( being a multi-valued attribute). Scheme 2.9 treats prerequisites as well as classrooms as separate entities, making it useful for gathering data about prerequisites and room usage. Scheme 2.8 is in between the others, in that it treats prerequisites as separate entities but not classrooms. Since a regis-trar’s office probably has to answer general questions about the number of classes a student is taking or what are all the prerequisites of a course, or where a specific class meets, scheme 2.9 is probably the best choice. 2.16 Answer: a. If a pair of entity sets are connected by a path in an E-R diagram, the en-tity sets are related, though perhaps indirectly. A disconnected graph im-plies that there are pairs of entity sets that are unrelated to each other. If we split the graph into connected components, we have, in effect, a separate database corresponding to each connected component. b. As indicated in the answer to the previous part, a path in the graph between a pair of entity sets indicates a (possibly indirect) relationship between the two entity sets. If there is a cycle in the graph then every pair of entity sets
- 6. 10 Chapter 2 Entity Relationship Model ss# name program room_no building course− offerings name dept title course room courseno title credits syllabus prerequisite maincourse requires secno is offered meetsin iss# instructor student grade teaches year semester time enrols Figure 2.9 E-R diagram for University(b). ss# name program time room course− offerings name dept title course courseno title credits syllabus secno is offered prerequisite iss# instructor student grade teaches year semester enrols Figure 2.10 E-R diagram for University(c).
- 7. Exercises 11 A RA R B E C R C B Figure 2.11 E-R diagram to Exercise 2.17b. A RA R B E C R C B Figure 2.12 E-R diagram to Exercise 2.17d. on the cycle are related to each other in at least two distinct ways. If the E-R diagram is acyclic then there is a unique path between every pair of entity sets and, thus, a unique relationship between every pair of entity sets. 2.18 Answer: a. Let E = {e1, e2}, A = {a1, a2}, B = {b1}, C = {c1}, RA = {(e1, a1), (e2, a2)}, RB = {(e1, b1)}, and RC = {(e1, c1)}. We see that because of the tuple (e2, a2), no instance of R exists which corresponds to E, RA, RB and RC. b. See Figure 2.11. The idea is to introduce total participation constraints be-tween E and the relationships RA, RB, RC so that every tuple in E has a relationship with A, B and C. c. Suppose A totally participates in the relationhip R, then introduce a total participation constraint between A and RA. d. Consider E as a weak entity set and RA, RB and RC as its identifying rela-tionship sets. See Figure 2.12. 2.19 Answer: The primary key of a weak entity set can be inferred from its relation-ship with the strong entity set. If we add primary key attributes to the weak entity set, they will be present in both the entity set and the relationship set and they have to be the same. Hence there will be redundancy. 2.24 Answer: A inherits all the attributes of X plus it may define its own attributes. Similarly C inherits all the attributes of Y plus its own attributes. B inherits the
- 8. 12 Chapter 2 Entity Relationship Model attributes of both X and Y. If there is some attribute name which belongs to both X and Y, it may be referred to in B by the qualified name X.name or Y.name. 2.26 Answer: In this example, we assume that both banks have the shared identifiers for customers, such as the social security number.We see the general solution in the next exercise. Each of the problems mentioned does have potential for difficulties. a. branch-name is the primary-key of the branch entity set. Therefore while merg-ing the two banks’ entity sets, if both banks have a branch with the same name, one of them will be lost. b. customers participate in the relationship sets cust-banker, borrower and de-positor. While merging the two banks’ customer entity sets, duplicate tuples of the same customer will be deleted. Therefore those relations in the three mentioned relationship sets which involved these deleted tuples will have to be updated. Note that if the tabular representation of a relationship set is obtained by taking a union of the primary keys of the participating entity sets, no modification to these relationship sets is required. c. The problem caused by loans or accounts with the same number in both the banks is similar to the problem caused by branches in both the banks with the same branch-name. To solve the problems caused by the merger, no schema changes are required. Merge the customer entity sets removing duplicate tuples with the same social-security field. Before merging the branch entity sets, prepend the old bank name to the branch-name attribute in each tuple. The employee entity sets can be merged directly, and so can the payment entity sets. No duplicate removal should be performed. Before merging the loan and account entity sets, whenever there is a number common in both the banks, the old number is replaced by a new unique number, in one of the banks. Next the relationship sets can be merged. Any relation in any relationship set which involves a tuple which has been modified earlier due to the merger, is itself modified to retain the same meaning. For example let 1611 be a loan number common in both the banks prior to the merger, and let it be replaced by a new unique number 2611 in one of the banks, say bank 2. Now all the relations in borrower, loan-branch and loan-payment of bank 2 which refer to loan number 1611 will have to be modified to refer to 2611. Then the merger with bank 1’s corresponding relationship sets can take place. 2.27 Answer: This is a case in which the schemas of the two banks differ, so the merger becomes more difficult. The identifying attribute for persons in the US is social-security, and in Canada it is social-insurance. Therefore the merged schema cannot use either of these. Instead we introduce a new attribute person-id, and use this uniformly for everybody in the merged schema. No other change to the schema is required. The values for the person-id attribute may be obtained by several ways. One way would be to prepend a country code to the old social-security or social-insurance values (“U” and “C” respectively, for instance), to get the corresponding person-id values. Another way would be to assign fresh
- 9. Exercises 13 numbers starting from 1 upwards, one number to each social-security and social-insurance value in the old databases. Once this has been done, the actual merger can proceed as according to the answer to the previous question. If a particular relationship set, say borrower, in-volves only US customers, this can be expressed in the merged database by spe-cializing the entity-set customer into us-customer and canada-customer, and mak-ing only us-customer participate in the merged borrower. Similarly employee can be specialized if needed.