Database Concepts Teaching - Learning and Assessment

Question: Discuss about the Database Concepts: Teaching, Learning and Assessment. Answer: Person (Id, TFN, Name, Phone) Policy Owner (PersonId, PolicyNumber, TFN, Start-Date) Insurance Policy (PolicyNumber, AgentNumber, PremiumAmount, CoverageAmount) Agent (AgentNumber, Name, Phone, OfficeNumber) The above relational schema explains that if a person insurance fund through an agent then the information of the insurance owner and the insurance will be saved in the database. The information of the person who will borrow the fund will be stored in the person table. PolicyOwner table will hold the data of the policy regarding that person. The PersonID will be composed of the Id and TFN attributes of the Person entity and it will be a foreign key. In the agent table, the information of the agent will be stored. The InsurancePolicy entity will be holding the PolicyOwner information along with the agent number. In addition, it will comprise of Premium and Coverage amount against the insurance. Now each database must have some primary key for identifying individual rows. Foreign keys are for connecting two entities. In the above relation schema, Id and TFN will be a unique identifier in the Person entity. The ID and TFN together will be uniquely identifying each row or information of the individual person. The PersonId and PolicyNumber will be the primary key of the PolicyOwner table. The TFN and the PolicyOwner together are the candidate key of that table. The Start-Date indicates the initiation date of the insurance. The candidate key will also be able to identify the rows uniquely as it has the potential to be a primary key. The AgentNumber will be the primary key of the Agent entity. The InsurancePolicy will have the primary key, PolicyNumber and it will indicate all the rows uniquely. The AgentNumber is the foreign key, reference of the Agent. AgentNumber. It is crucial to remember that the foreign key must be a primary or candidate key of another table. Insurance Policy: The non-trivial functional dependency indicates that if a non-key attribute depends on a key attribute, then the non-key attribute has to be a subset of the key attribute. Taken as an example, id Y is dependent on X, X-Y then Y has to be a subset of X. So if the concept is applied to the entity InsurancePolicy, then all the trivial and non-trivial dependencies can be found. Thanking the primary key first, the PolicyNumber is a superset of PolicyNumber so this dependency, PolicyNumber - PolicyNumber is trivial. The given condition is that various PolicyNumber can have different AgentNumber, so the PolicyNumber- AgentNumber is a non-trivial FD. The PolicyNumber - PremiumAmount and PolicyNumber - CoverageAmount is also non-trivial. If any update or modification have to be made in the Agent or InsurancePolicy entity, then the updates will be only in the individual entity. The Agent number is will not require being updated which is installed in both the entities. It specifies that the relation does not comprise of any update anomalies. Person: In the Person Table, the primary key Id is not a composite key. So the attributes of the entity such as TFN, Name, and Phone will be a non-trivial functional dependency. It is because ID is not a superset of TFN, ID is not a superset of Name and ID is not a superset of Phone. Another candidate key is TFN. In the case of Id, TFN- Id it is a trivial and Id, TFN- TFN is trivial. But for Id, TFN- Name and Id, TFN- Phone is non-trivial. Now the question may come that why the Id, TFN with name and phone is even FD. It is because that the Id and TFN every time will create a new attribute value. The below diagram will explain the concept clearly. So from the above diagram, it is clear that the Id, TFN- Name and Id, TFN- Phone is an FD, but it is non-trivial. As the composite values of Id and TFN is never repeating. The entity Person is in 1st Normal Form as all the attributes are atomic, not a composite attribute. The relation is in 2nd Normal Form as the non-key attributes are completely dependent on the key attributes. The entity is also in 3rd normal form as the attribute in the entity is not more dependent on the on the other attribute than the key attribute. The entity is not in BCNF or Boyce-Codd Normal Form as the rows can be defined through the TFN, a candidate key or composite of Id and TFN. So decompose it into BCNF it is required to create two separate table. The table is as following. Person (Id, Name, Phone) Person-TFN (Id, TFN) PolicyOwner: The entity PolicyOwner is in 1st and 2nd Normal form. But it is not in 3rd normal form as the start-date is more depended on the PolicyNumber than TFN and PersonId. For that reason, the decomposed table will be PolicyOwner (PersonId, PolicyNumber, TFN) and PolicyOwner-Date (PolicyNumber, start-date). In the case of BCNF the PersonId, TFN - PolicyNumber so the decomposed table PolicyOwner is in BCNF and PolicyOwner-Dates case PolicyNumber- start-date so the both determinant is correct and the entities are in BCNF. 1: 2: 3: 4: 5: Account Attribute Data Type Length Description Key Type Account_No. Number Unique Identifier of the Account Entity Primary Balance Number 20 The Balance of Customer Account Customer Attribute Data Type Length Description Key Type CustId Number Unique Identifier of the Customer Entity Primary C-FName VARCHAR2 20 Customers First Name C-LName VARCHAR2 20 Customers Last Name Customer-Account Attribute Data Type Length Description Key Type CustId Number Unique Identifier of the Customer Entity Foreign Key (Customer.CustId) Account_No. Number 20 Unique Identifier of the Account Entity Foreign Key (Account. Account_No.) Employee Attribute Data Type Length Description Key Type EmpId Number Unique Identifier of the Employee Entity Primary BranchCode NUMBER Unique Identifier of the Branch Entity Foreign Key (Branch. BranchCode) E_ FName VARCHAR2 20 Employees First Name E_ LName VARCHAR2 20 Employees Last Name Branch Attribute Data Type Length Description Key Type BranchCode NUMBER Unique Identifier of the Branch Entity Primary Address VARCHAR2 100 Location of individual branches Product Attribute Data Type Length Description Key Type Pro_code Number Unique Identifier of the Product Entity Primary Name VARCHAR2 20 Name of the individual product Dependent Attribute Data Type Length Description Key Type SeqNo Number Unique Identifier of the Dependent Entity Primary Key CustId Number Unique Identifier of the Customer Entity Foreign Key (Customer.CustId) D_ FName VARCHAR2 20 First Name of Dependent D_ LName VARCHAR2 20 Last Name of Dependent School VARCHAR2 20 School Name of the dependent Promotion-Outcome Attribute Data Type Length Description Key Type Pro_code Number Unique Identifier of the Promotion-Outcome Entity Primary EmpId Number Unique Identifier of the Employee Entity Foreign Key (Employee. EmpId) CustId Number Unique Identifier of the Customer Entity Foreign Key (Customer. CustId) Outcome Number The Outcome of promotion regarding each product Figure 1: The Relational Database Schema (Source: Created by Author) Distinct Return Date for Each Book: As per the given database while a librarian inputs and person name against the books he or she intends to borrow. As the borrow_copy entity connects to book_copy and in return it connects the borrow entity with book entity, it not possible to enter different dates for books that have been borrowed at the same. For giving the option of returning the books that have been borrowed at the same transaction the boorowdate, duedate and returndate will be stored in another entity, borrwobookdate. The entity will have a unique identifier called dateId. The dateId will be a unique identifier in the book_copy. This way the returndate will be inserted as per each borrowed book. The return will not be dependent on the transactionId anymore. The return date will be inserted as per each book. Borrow (transactionID, personID*) book_copy (bookID, bookdescID*, bookId*) book (bookdescID, title, subtitle, edition, voltitle, volnumber, language, place, year, isbn, dewey, subjectID*) borrow_copy (transactionID*, bookID*) borrwobookdate (bookId, borrowdate, duedate, returndate) Loan Extension: As the borrwobookdate will individually handle the duedate the process of updating the duedate will not create any update anomalies. But for restricting the dudate update, suppose for, at maximum for three times, the borrwobookdate will consist of an attribute called renewal_limit. At every time of due date modification, an update_trigger will be called which will automatically store the value of the renewal_limit will be updated. At the initial value of the attribute will be zero and it will incremented by one every time update query will be executed. It will only allow the borrower to update the due date up to three times. 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