Answers for test of Database Design and Development

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You have been sub-contracted to design the database system for Marlowe Interiors. Marlowe Interiors is a medium sized interior design company that does building work and decorating. An initial analysis of Marlowe Interiors has identified the following requirements:

ï‚· Marlowe Interiors keeps a record of the jobs it performs. Jobs are for particular customers and

classified by job type (Single Room, Part-house, Whole-house). The full cost of a job depends on

the labour used and the parts used. Records of parts and labour should be kept separately.

ï‚· Labour costs depend on the type of work done. A job may have different sorts of work and employ one or more types of worker (plumber, labourer, qualified builder, interior designer, electrician). A worker will work on more than one job. The hourly rate of pay for each of these types of worker will vary as shown below:

Type

Rate per Hour

Plumber

£50

Labourer

£20

Qualified Builder

£55

Interior Designer

£39

Electrician

£50

ï‚· A job may involve one or more parts. Examples of parts are: sink, bath, pipe, thermostat, wallpaper, decking, door, light-fitting. The costs of these parts will vary and suitable test values should be supplied as part of the initial implementation.

ï‚· Parts are supplied by various suppliers. The cost of a part could vary depending on the supplier.

ï‚· A customer record should be kept for the job with all the usual details such as name, address and phone-number. Study the above detail and then perform the following tasks:

Task 1

Task 1 - 30 Marks

Draw an Entity Relationship Data Model that describes the content and structure of the data held by Marlowe Interiors.

The technique of Entity-Relationship Diagram is to represent a conceptual model using constructs from the entity relationship data model. Entity Relationship Data Model is a data Model is a data model. It has three parts. They are entities, relationships and attributes.

Entity can be defined as a thing which an organization recognizes as being capable of an independent existence and which can be uniquely identified. To store data about many properties of something, then that thing is likely to be an entity. An entity may be a physical object, an event, a concept. Entities are by their very nature interesting things because entities are normally used to define logical data groupings.

There are four types about relationships. A relationship is some association between entities. The four type of the relationships are Optional (zero - one), Mandatory (one - one), Optional (zero - many) and Mandatory (one - many). Attributes means an characterized entity by a number of properties. Values assigned to attributes are used to distinguish one entity from another.

1.1 Entity Relationship Diagram

Figure ( 1 . 1 )Entity Relationship Diagram

1.2 Data Dictionary

1.2.1 Entity

Entity : Job

Description : Any job that is for particular customer

Identifying Attribute(s) : JobID, CusID, PartID, JobTypeID

Other Attributes(s) : -

Entity : Customer

Description : Any customer who make order for job

Identifying Attribute(s) : CusID

Other Attributes(s) : CusName, Adrs, PhNo, E.mail

Entity : Job Type

Description : Any job type that perform by labour

Identifying Attribute(s) : JobTypeID

Other Attributes : JobName, DurationofJob

Entity : Part

Description : Parts that is supplied by supplier

Identifying Attribute(s) : PartID

Other Attributes : PartName, PartMaintainCost, PartSuppliedCost

Entity : Supplier

Description : Supplier who supplies the part for job

Identifying Attribute(s) : SupID

Other Attributes : SupName, Adrs, PhNo, E.mail, PartID

Entity : Work

Description : Any work that perform by any labour

Identifying Attribute(s) : JobID, LabourID

Other Attributes: -

Entity : Labour

Description : Labour who works at the job

Identifying Attribute(s) : LabourID

Other Attributes : LabourName, Adrs, PhNo, E.mail

Entity : Labour Type

Description : Labour type that is classified by rate per hour

Identifying Attribute(s) : LabourTypeID

Other Attributes : LabourTypeName, RateperHour

1.2.2 Attribute

Attribute : JobID

Description : Unique identification number of job

Type : Integer

Attribute : CusID

Description : Unique identification number of customer

Type : Integer

Attribute : PartID

Description : Unique identification number of part

Type : Integer

Attribute : JobTypeID

Description : Unique identification number of job type

Type : Integer

Attribute : SupID

Description : Unique identification number of supplier

Type : Integer

Attribute : LabourID

Description : Unique identification number of labour

Type : Integer

Attribute : LabourTypeID

Description : Unique identification number of labour type

Type : Integer

1.2.3 Relationship

Relationship ID : RN1

Relationship Name(s) : Job, is Jobbed

Description : Associates customers with the job type they are currently jobbed upon

Participating Entity : customer

Cardinality : one

Optionality : Mandatory

Participating Entity : Job

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN2

Relationship Name(s) : Job, is jobbed

Description : Associates job type with the customer they are currently jobbed upon

Participating Entity : Job Type

Cardinality : One

Optionality : Mandatory

Participating Entity : Job

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN3

Relationship Name(s) : Part, is used

Description : Accessociates job with the supplier they are currently depended upon part

Participating Entity : Job

Cardinality : One

Optionality : Mandatory

Participating Entity : Part

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN4

Relationship Name(s) : Supplier, is supplied

Description : Accessociates part with the job they are currently supplied upon

Participating Entity : Part

Cardinality : One

Optionality : Mandatory

Participating Entity : Supplier

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN5

Relationship Name(s) : Work, is worked

Description : Accessociates job with the labour they are currently worked upon

Participating Entity : Job

Cardinality : One

Optionality : Mandatory

Participating Entity : Work

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN6

Relationship Name(s) : Labour, is performed

Description : Accessociates work with the labour type they are currently depended upon labour

Participating Entity : Work

Cardinality : One

Optionality : Mandatory

Participating Entity : Labour

Cardinality : Many

Optionality : Mandatory

Relationship ID : RN7

Relationship Name(s) : Labour Type, is performed

Description : Accessociates labour with the work they are currently depended upon labour type

Participating Entity : Labour

Cardinality : One

Optionality : Mandatory

Participating Entity : Labour Type

Cardinality : Many

Optionality : Mandatory

1.2.4 Integrity Constraints

The Integrity Constraints for this system are shown below:

A customer can make a number of job ordering at any time.

Every job belongs to at most one customer.

A job can have more than one part but must have at least one part.

A supplier has one more supply part and must have at least supply part.

A labour performs a different kind of job but must have at least one job.

Task 2

Task 2 - 20 Marks

Produce the resulting tables clearly indicating the primary and foreign keys.

2.1 Resulting Tables

The tables as shown below describe the name of the table, list of the attributes, the data type of each attributes, distinguishing primary key (PK) and foreign key (FK) and the descriptions of each attributes.

2.1.1 Customer

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

CusID

int(11)

PK

Unique identification number for every customer

2

CusName

Varchar(30)

Name of the customer

3

Adrs

Varchar(30)

Address of the customer

4

PhNo

Int(11)

Phone No of the customer

5

E.mail

Varchar(30)

E.mail address of the customer

2.1.2 Job

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

JobID

Int(11)

PK

Unique identification number of job

2

CusiD

Int(11)

FK

Unique identification number of customer

3

PartID

Int(11)

FK

Unique identification number of part

4

JobTypeID

Int(11)

FK

Unique identification number of job type

2.1.3 Job Type

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

JobTypeID

Int(11)

PK

Unique identification number of job type

2

JobName

Varchar(45)

Name of the job

3

DurationofJob

Varchar(10)

Duration of job

2.1.4 Labour

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

LabourID

int(11)

PK

Unique identification number for every labour

2

LabourName

Varchar(45)

Name of the labour

3

Adrs

Varchar(45)

Address of the labour

4

PhNo

Int(11)

Phone No of the labour

5

E.mail

Varchar(45)

E.mail address of the labour

2.1.5 Labour Type

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

LabourTypeID

Int(11)

PK

Unique identification number of labour type

2

LabourTypeName

Varchar(45)

Name of the labour type

3

RateperHour

Varchar(10)

Rate per hour for each labour type

2.1.6 Part

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

PartID

Int(11)

PK

Unique identification number of part

2

PartName

Varchar(45)

Name of the part

3

PartMaintainCost

Int(11)

Cost for part maintaining

4

PartSuppliedCost

Int(11)

Cost of supplied part

2.1.7 Supplier

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

SupID

int(11)

PK

Unique identification number for every supplier

2

SupName

Varchar(45)

Name of the supplier

3

Adrs

Varchar(45)

Address of the supplier

4

PhNo

Int(11)

Phone No of the supplier

5

E.mail

Varchar(30)

E.mail address of the supplier

6

PartID

int(11)

FK

Unique identification number for part

2.1.8 Work

No

Name

Data type

Primary key(PK)/

Foreign key(FK)

Description

1

JobID

Int(11)

PK

Unique identification number of job

2

LabourID

Int(11)

PK

Unique identification number of labour

Task 3

Task 3 - 20 Marks

Using a Database Management System (DBMS) of your choice, set-up all of the above normalised tables, and populate them with well-designed test data (minimum 5 records per table). Provide printouts of all tables.

Reasonable assumptions may be made with regard to data.

-First, DDDassignment SCHEMA was created.

CREATE SCHEMA IF NOT EXISTS `DDDassignment` ;

USE `DDDassignment` ;

Customer

CREATE TABLE IF NOT EXISTS `DDDassignment`.`Customer` (

`CusID` INT NOT NULL ,

`CusName` VARCHAR(30) NULL ,

`Adrs` VARCHAR(30) NULL ,

`PhNo` INT NULL ,

`E.mail` VARCHAR(30) NULL ,

PRIMARY KEY (`CusID`) )

ENGINE = InnoDB;

Figure(3.1)Customer table creation

Figure(3.2)Test data in customer table

Job

CREATE TABLE IF NOT EXISTS `DDDassignment` . `Job` (

`JobID` INT NOT NULL ,

`CusID` INT NULL ,

`PartID` INT NULL ,

`JobTypeID` INT NOT NULL ,

PRIMARY KEY (`JobID`, `PartID`, `CusID`, `JobTypeID`) ,

INDEX `fk_Job_Customer1` (`CusID` ASC) ,

INDEX `fk_Job_JobTypeID1` (`JobTypeID` ASC) ,

CONSTRAINT `fk_Job_Customer1`

FOREIGN KEY (`CusID` )

REFERENCES `DDDassignment` . `Customer` (`CusID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION,

CONSTRAINT `fk_Job_JobTypeID1`

FOREIGN KEY (`JobTypeID` )

REFERENCES `DDDassignment` . `JobTypeID` (`JobTypeID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION)

ENGINE = InnoDB;

Figure(3.3)Job table creation

Figure(3.4)Test data in job table

Job Type

CREATE TABLE IF NOT EXISTS `DDDassignment` . `JobTypeID` (

`JobTypeID` INT NOT NULL ,

`JobName` VARCHAR(45) NULL ,

`DurationofJob` VARCHAR(10) NULL ,

PRIMARY KEY (`JobTypeID`) )

ENGINE = InnoDB;

alter table `DDDassignment`.`JobType`

add constraint ck_JobType check(Position in('Single Room','Part-house','Whole-house')

Figure(3.5)Job Type table creation

Figure(3.6)Test data in job type table

Labour

CREATE TABLE IF NOT EXISTS `DDDassignment` . `Labour` (

`LabourID` INT NOT NULL ,

`LabourName` VARCHAR(45) NULL ,

`Adrs` VARCHAR(45) NULL ,

`PhNo` INT NULL ,

`E.mail` VARCHAR(45) NULL ,

PRIMARY KEY (`LabourID`) )

ENGINE = InnoDB;

Figure(3.7)Labour table creation

Figure(3.8)Test data in labour table

Labour Type

CREATE TABLE IF NOT EXISTS `DDDassignment` . `LabourType` (

`LabourTypeID` INT NULL ,

`RateperHour` INT NULL ,

`LabourID` INT NULL ,

PRIMARY KEY (`LabourTypeID`) ,

INDEX `LabourID` (`LabourID` ASC) ,

CONSTRAINT `LabourID`

FOREIGN KEY (`LabourID` )

REFERENCES `DDDassignment` . `Labour` (`LabourID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION)

ENGINE = InnoDB;

alter table `DDDassignment`.`LabourType`

add constraint ck_LabourType check(Position in('plumber','labourer','qualified builder','interior designer','electrician')

Figure(3.9)Labour Type table creation

Figure(3.10)Test data in labour type table

Part

CREATE TABLE IF NOT EXISTS `DDDassignment` . `Part` (

`PartID` INT NOT NULL ,

`PartName` VARCHAR(10) NULL ,

`PartMaintainCost` INT NULL ,

`PartSuppliedCost` INT NULL ,

PRIMARY KEY (`PartID`) ,

CONSTRAINT `fk_Part_Job1`

FOREIGN KEY (`PartID` )

REFERENCES `DDDassignment` . `Job` (`PartID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION)

ENGINE = InnoDB;

Figure(3.11)Part table creation

Figure(3.12)Test data in part table

Supplier

CREATE TABLE IF NOT EXISTS `DDDassignment` . `Supplier` (

`SupID` INT NULL ,

`SupName` VARCHAR(45) NULL ,

`Adrs` VARCHAR(45) NULL ,

`PhNo` INT NULL ,

`E.mail` VARCHAR(30) NULL ,

`PartID` INT NULL ,

PRIMARY KEY (`SupID`) ,

INDEX `PartID` (`PartID` ASC) ,

CONSTRAINT `PartID`

FOREIGN KEY (`PartID` )

REFERENCES `DDDassignment` . `Part` (`PartID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION)

ENGINE = InnoDB;

Figure(3.13)Supplier table creation

Figure(3.14)Test data in supplier table

Work

CREATE TABLE IF NOT EXISTS `DDDassignment`.`Work` (

`JobID` INT NULL ,

`LabourID` INT NULL ,

PRIMARY KEY (`JobID`, `LabourID`) ,

INDEX `fk_Work_Job1` (`JobID` ASC) ,

INDEX `fk_Work_Labour1` (`LabourID` ASC) ,

CONSTRAINT `fk_Work_Job1`

FOREIGN KEY (`JobID` )

REFERENCES `DDDassignment`.`Job` (`JobID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION,

CONSTRAINT `fk_Work_Labour1`

FOREIGN KEY (`LabourID` )

REFERENCES `DDDassignment` . `Labour` (`LabourID` )

ON DELETE NO ACTION

ON UPDATE NO ACTION)

ENGINE = InnoDB;

Figure(3.15)Work table creation

Figure(3.16)Test data in work table

Assumptions with regard to data

Task 4

Task 4 - 20 Marks

Set-up and test all of the following queries using Structured Query Language (SQL). Provide printouts of SQL code for each query and the output produced when you run the query in the database you have developed:

Display the names and addresses of the customers of Marlowe Interiors and the jobs and job types that have been carried out for them.

Display a full record of a job. This should include the name of the customer, the job type, the workers on the jobs and what type they are, the parts included in the job and the final overall cost for the job.

Queries using Structured Query Language (SQL)

(4.1)The names and addresses of the customers of Marlowe Interiors and the jobs and job types that have been carried out for them

create view Task4a as

Select c.CusName as CustomerName, c.Adrs as Address, j.JobID, t.JobTypeID, t.JobName From Customer c inner join Job j on c.CusID = j.CusID inner join JobType t on j.JobTypeID = t.JobTypeID

(4.2)A full record of job that include the name of the customer, the job type, the workers on the jobs and what type they are, the parts included in the job and the final overcall cost for the job

create view Task4b as

select job.jobid,`customer` . `CusName` AS `CustomerName` , `jobtype` . `JobTypeID` AS `JobTypeID` , `labour`.`LabourName` AS `LabourName` , `labourtype` . `LabourTypeID` AS `LabourTypeID` , `part`.`PartName` AS `PartName`,( ( (`jobtype`.`DurationofJob` * `labourtype` . `RateperHour`) + `part` . `PartMaintainCost`) + `part` . `PartSuppliedCost`) AS `Cost` from ( ( `labourtype` join ( ( ( ( `customer` join `job` on ( ( `customer` . `CusID` = `job` . `CusID` ) ) ) join `jobtype` on ( ( `job`.`JobTypeID` = `jobtype`.`JobTypeID` ) ) ) join `part` on ( ( `job`.`PartID` = `part`.`PartID` ) ) ) join `supplier` on ( ( `part`.`PartID` = `supplier`.`PartID` ) ) ) ) join ( `labour` join `work` on ( ( `labour` . `LabourID` = `work` . `LabourID`) ) ) ) order by jobid;

create view task4b2 as

select jobid,sum ( Cost ) as TotalCost from task4b group by jobid

Task 5

Task 5 - 10 Marks

Explain any assumptions you have made when analyzing, designing and implementing the above database, justify the approach you have taken and explain any alternative approaches you could have taken to any of the above tasks. Discuss any changes you would make to improve your work.

5.1 Assumptions from analyzing, designing and implementing

While I am analyzing, designing and implementing the database design and development, I establish based on the following concept:

Requirement Elicitation

Conceptual Modelling

Logical Modelling

Physical Modelling

By using the above concept, I want to present my assumption as shown below:

The main function of requirement elicitation is to support the technical requirement for database system. We need to elicit the formal or informal information from user and we would apply these information for database technical requirement. Most user may not clearly give their desire about the system they want. So, we need to get requirement specification by negotiating their requirements.

The assignment of database design and development subject is to establish the interior design company database development. The manual system keeps a record of the jobs it performs. So, job is an entity for entity relationship data model.

Jobs are for particular customers and they are classified by job type. By analyzing this, customer and job type are entities for entity relationship data model.

The full cost of job depends on the labour used and the parts used. Records of parts and labour should be kept separately. Labour and part entity will be need for entity relationship data model.

A job may have different sorts of work and employ one or more types of worker. A worker will work on more than one job. A job may involve one or more parts. The costs of these parts will vary and suitable test values should be supplied as part of the initial implementation. Parts are supplied by various suppliers. The cost of a part could vary depending on the supplier. This is why , supplier entity is needed for entity relationship data model.

By analyzing the information from user, I assume that there are eight entities for entity relationship data model. They are job, customer, job type, labour, part, supplier, job type and labour type. About them job and labour type are assumed that transaction file.

Entity Relationship Data Model :An entity is a thing about which an organization holds information. Entity modeling is a technique for showing relationships between entities. Entity analysis is the process by which the model is developed, and identifies the underlying structure of the data and relationships of those data. The diagram produced from entity modeling is called the Entity Relationship Diagram (ERD).

In task 1, I establish the entity relationship data model for database design and development. By analyzing the user requirement , I collect the data for entities. And I document the ERD that contain the entity name, the entity description, the key(s), attributes, including size and type and volumetrics. I want to present about the relations between the entities. First, the customer entity has the relation with job type entity. The customer can make order one or more type of job so customer have one to many relationship with job type. But the difficulities is job type has one to many relationship to customer too. So, I assumed that there need a transaction file between these two entities. I decided to assume job entity as transaction between customer entity and job type entity.

After that, job entity have one to many relationship with part entity and part entity have one to many relationship with supplier entity. There has no problem about these relationships about job entity, part entity and supplier entity. But the second difficulty is job entity has one to many relationship to labour entity and labour entity has one to many relationship to job entity too. So, there may be needed transaction file between these two entities. I decided that work entity is the transaction file between these two entities. After all, job entity have one to many relationship to work entity, work entity has many to one relationship with labour and labour have one to many relationship to labour type.

Normalization : Any collection of attributes can be said that un-normalized form . In order to obtain a set of normalized relations, the following steps should be applied in sequence: (1) gathering data ; (2) choosing a key ; (3) converting to first normal form ; (4) converting to second normal form ; (5) converting to third normal form ; (6) optimization.

Primary Key : A key or identifier is an attribute which uniquely identifies a single row within the relation. This key is often called the primary key.

Foreign Key : An attributes which is the primary key of one relation but appears as and attribute in another relation is called a foreign key.

In task 2, the resulting tables clearly indicating the primary key and foreign keys are needed to produce. This is why , I normalized by gathering data , choosing a key , converting to first normal form , converting to second normal form , converting to third normal form and optimization. At the part of gathering data, I ensure the designation of inputs, outputs and data structures for the correct representation of relationships between the data of new system . I care any relationships between the data from normalizing existing system which may no longer be required in new system and to modify the logical structure to represent the new system requirements.

The chosen key can ensure be uniquely identifies the particular grouping of data. First and second normal forms are not significant in themselves. They can be expected as stepping stone to third normal form. I collect the repeating group of data from un-normalized data and remove them to separate relation because conversion to first normal form involves the removal of repeating groups of data and the presence of repeating data in a file has many disadvantages. This is why, such data is removed to a separate relation.

The stage of second normal form is concerned with relations already in first normal form and relations with single attribute key are defined in second normal form. CustName can be determined uniquely from CusID, and will remain the same regardless of JobTypeID associated with it. The CusName is dependent on CusID but not on JobTypeID. As such it is removed from the original relation and new relation created and CusID as its key. In this case, DurationofJob is a more difficult to choose. If it remains the same regardless of JobTypeID, it is only dependent on CusID, and it too must be removed to the new relation with JobTypeID as its key. In this case , Durationof Job is found to be dependent on department code only. JobName can be determined uniquely from JobTypeID, and will remain the same regardless of PartID associated with it. The JobName is dependent on JobTypeID but not on PartID. As such it is removed from the original relation and new relation created and JobTypeID as its key. In this case, PartMaintainCost is a more difficult to choose. If it remains the same regardless of JobTypeID, it is only dependent on PartID, and it too must be removed to the new relation with PartID as its key. In this case , PartMaintainCost is found to be dependent on PartID only.

The stage of third normal form is to perform for relations already in second normal form and considers the dependencies between non-key items. In this stage, I recheck all relations to ensure that nothing has been missed and that the relations really are in third normal form. The stage of optimization is concerned with the consolidation of data from many sources, compiling a full set of attributes for each normalized relation.

In task 3 , I describe set-up all the normalized tables and populate them with well-designed test data by using a Database Management System (DBMS) of my choice,MySQL. And one table at least has 5 records and I will provide printouts of all tables.

In task 4, I set-up all queries that include the function to display the names and addresses of the customers of Marlowe Interiors and the jobs and job tylpes that have been carried out for them by using Structured Query Language (SQL). And the second function is to display a full record of a job and there should include the name of the customer, the job type, the workers on the jobs and what type they are, the parts included in the job and the final overall cost for job by using Structured Query Language (SQL).I provide the printouts of SQL code for each query and the output produced when I run the query in the database I have developed for these functions.

In addition, I made analyzing, designing and implementing with MySQL database system. First I made gathering data from given Scenario and make normalization step by step to get better performance for relation of each tables.

I think the relationships of the tables are most important for the system because the normalization is surely depend on the table relation and Entity Relationship Diagram.I use MySQL because it is the open source database software and it has the superior speed, reliability, and ease of use. It has become the preferred choice for Web, Web 2.0, SaaS, ISV, Telcom companies and forward-thinking corporate IT Managers because it eliminates the major problems associated with downtinme, maintenance and administration for modern, online applications. MySQL is the world's largest and fastest-growing organizations use MySQL to save time and money powering their high-volume web sites, critical business systems, and packed software. The flagship MySQL offering is MySQL Enterprise, a comprehensive set of production-tested software, proactive monitoring tools, and premium support services available in an affordable annual sunscription. MySQL is a key part of LAMP (Linux, Apache, MySQL,PHP / Perl / Python), the fast-growing open source enterprise software stack. More and more companies are using LAMP as an alternative to expensive proprietary software stacks because of its lower cost and freedom from platform lock-in.

I designed logical database as the process of constructing a business data model and it involves the reconciliation of the conceptual model to a relation schema. It is the implementation independent. The core of relational database theory is functional dependencies and normalization. Normalization is a process that consider the functional dependencies between attributes and decides which attributes belong together in relation.

And about physical database design, this provides insufficient information to enable effective physical database design. The performance parts of physical design are volume analysis, usage / transaction analysis, integrity analysis, control or security analysis and distribution analysis. The physical database design is the most important step when designing a schema since it affects the performance of the database system, in particular, the evaluation of queries. The one important aspect of physical database design is the controlled replication of data to provide more efficient data access.

5.2 changes that would make to improve work

After analyzing the whole designation, I desire to insert more details system in this task if possible and make more strong system than existing system. I will create the design of program more convenient and complete.

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