数据库ER图练习及答案

DB Modeling Exam Practical

Answer the following questions. 1.

Create an E-R schema for a database system used to manage account information at a community

bank. The bank has several branches with unique names. A customer may have one or more accounts in one or more branches. An account must belong to one and only one branch. Each account is operated on by transactions, which may be deposits to or withdrawals from some account. The database keeps track of all the transactions on each account, in addition to the balance of individual accounts and the assets of individual branches. For each entity, specify all its attributes, primary key, and alternate key(s). In your ER schema, be sure to capture the cardinality constraints and participation constraints of all relationships.

Make reasonable assumptions to complete the specification. Explicitly state all your assumptions. EVERY construct in your ER schema should be substantiated by either the specification above or your explicit assumptions. 2.

The following table stores information about which suppliers can supply which parts. The table captures

the fact that a part whose name is PartName and whose ID is PartID can be supplied by suppliers whose names are in SupplierName and whose IDs are in SupplierID. Note that a part can be supplied by many suppliers, and a supplier can supply many parts.

CAN_SUPPLY PartID PartName SupplierID SupplierName

1234 1234 2134 Nut Nut Bolt 223 224 223 ProMetal Biscayne ProMetal 3. 1. 2. 3. 4.

Perform the following tasks.

List the primary key. List all the FDs.

What normal form is the relation in? Explain.

Apply normalization to it incrementally, carrying the normalization process through each of the

higher normal forms possible up to 3NF. That is, if the relation were unnormalized, bring it to first normal form, then bring the first normal form you've just created to second normal form, and then bring the second normal form to third normal form.

For each transformation to the next higher normal form X,

? ? ?

other column(s).

Explain the steps you took to bring it to the normal form X.

Provide the normal form X's table structure, primary key(s), and the FDs.

Explain why you think it is in the normal form X. For example, if you think there is a

partial dependency, fully defend your conclusion by explaining how a column is partially dependent on some

That is, if the relation were in an unnormalized form, you would explain the transformation you performed to bring it to first, second, and third normal forms. You would also provide the table structure, the primary key and the FDs for the first, second, and third normal forms. You would also provide explanation for why you believe it is in first, second, and third normal forms. 4.

Convert the following E-R schema into a relational schema using the mapping algorithm specified in this

course. Specify key and referential integrity constraints, using directed arcs. Make sure you also identify alternate keys. Label each step of the mapping algorithm.

Answer:

1.

PhoneNum BankName BankPhone CustName CustID Bank 1 Customer 1 Open M 1 AofBranch N Has N AccountNo Branch Account 1 Balance BranchName Assets TofAccounOperationType BranchAddr BranchPhone N TID Transaction TDateTime

Entity:

1. Bank(BankName,BankPhone) (BankPhone is a multi-valued attribute.) PK: (BankName)

2. Cutomer(CustID, CustName, PhoneNum) PK: (CustID) AK: (PhoneNum)

3. Branch (BranchName, BranchAddr, BranchPhone, Assets) (BranchPhone is a multi-valued attribute.) PK: ( BranchName)

4. Account (AccountNo, Balance) PK: (AccountNo)

5. Transaction (TID, OperationType, TDateTime) PK: (TID) Relations:

1. Has: , 1:N, PARTIAL/ TOTAL;

2. Open: , 1:N, PARTIAL/ TOTAL; 3. AofBranch: , 1:N, PARTIAL/ TOTAL;

4. TofAccount: , 1:N, PARTIAL/ TOTAL; Assumptions:

1. A new bank can establish no branch.

2. One normal bank establishes one or more braches. 3. A bank has one or more telephones for customers. 4. A customer can open one or more Account.

5. An account must belong to one and only one branch.

6. One branch opens one or more accounts.

7. A branch has one or more telephones for customers. 8. An account belongs to just one branch;

2.

1.pk:(PartID, SupplierID)

2.FDs:

FD1: PartID->{PartName}

FD2: SupplierID->{SupplierName}

3. The relation is in the first normal form(1NF).

Each attribute of the relation allows a single atomic value, so it is in 1NF.

But some none-primary-key attributes, such as PartName and SupplierName, partially dependant on the primary key (as FD1 and FD2 show), so it is not in 2NF.

4. Normalization:

1) FD1: PartID->{PartName}

The relation can be decomposited into two relations:

PART(PartID, PartName),FDs={PartID->PartName}, PK:(PartID);

CAN_SUPLY(PartID, SupplierID, SuplierName), FDs={SupplierID->SupplierName), PK:(PartID,SupplierID). The relation PART is now in the third normal form because the only none-primary-key

attribute PartName, fully (not partially) and directly (not transively) dependants on the primary key PartId. The relation CAN_SUPPLY is still in the first normal because the only none-primary-key attribute SupplierName, partially dependants on the primary key (PartId,SupplierID).

2)CAN_SUPLY(PartID, SupplierID, SuplierName), FDs={SupplierID->SuplierName): For SupplierID->SuplierName, the relation can be decomposited into two relations: SUPPLIER(SupplierID,SupplierName),FDs={SupplierID->SuplierName}, PK:(SupplierID); CAN_SUPLY(PartID, SupplierID), FDs={}, PK:(PartID,SupplierID).

Both relations are in the third normal form, because for each one, no none-primary-key attribute patially or transively dependants on its primary key.

3) Three 3NF relations:

PART(PartID, PartName),FDs={PartID->PartName}, PK:(PartID);

SUPPLIER(SupplierID,SupplierName),FDs={SupplierID->SuplierName}, PK:(SupplierID); CAN_SUPLY(PartID, SupplierID), FDs={}, PK:(PartID,SupplierID).

3.

九步算法：

三种异常：修改异常、插入异常、删除异常。 S1：每一强实体用一个新表表示

S2：处理参与1：1标识联系的弱实体W S3：处理参与1：N标识联系的弱实体W

S4：处理每一二元1：1联系R,确定参与该联系的实体型对应的表S和T,将T的主码作为外码加入S,将R的所有简单属性和复合属性成分作为列加入S。

S5：处理每一二元1：N联系R,确定处于N端的实体表S和1端的实体表T,将T的主码作为外码加入S,将R的所有简单属性和复合属性成分作为列加入S.

S6:处理每一N元联系（包括二元M：N联系),对应新表T,将R的所有简单属性和复合属性成分作为列加入T,将参与联系的（强、弱）实体型的主码作为外码加入T,所有外码组合，共同构成T的主码.

S7：处理每一多值属性A,将A的所有简单属性和复合属性成分作为列加入T,将A所属的实体或联系型的主码作为外码加入T,将（上步得到的）外码和A对应的属性确定为T的主码. S8：处理每一非相交子类的特化. S9：处理每一相交子类的特化. S1:

T1: Coach(Name,Age) PK:(Name) T2: Team(Name) PK:(Name)

T3: Player(Name,Age) PK:(Name)

T4: Game(Number,Score,Time,Date) PK:(Number)

T5: Stadium(Name,Size,Location) PK:(Name)

S2:处理参与1:1标识符联系的弱实体

S3:处理参与1:N标识符联系的弱实体 S4:

T2: Team(Name,CoachName) PK:(Name)

FK:CoachName references Coach(Name) S5:

T3: Player(Name,Age,TeamName) PK:(Name)

FK:TeamName references Teach(Name) S6:

T6: Practice(TeamName,StadiumName,Date) PK:(TeamName,StadiumName)

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