TASK 01 (a) RELATIONAL SCHEMA Customer ( CustomerID , CustomerName, CustomerAddress, CustomerContactNum) Parcel ( ParcelID , ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status) Delivered (DeliverdDate) Returned (ReturnedDate) Lost (LostEntryDate) Invoice ( InvoiceID, InvoiceDate, InvoiceAmount) Claim ( ClaimID , ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate) Employee ( EmployeeID , EmployeeName, EmployeeStartDate) On the basis of relationship that exist between the entities, the table can be further broken down as; Since one customer can order for multiple parcels therefore, 1 to many relationship exists between them. · Parcel ( ParcelID , ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status, CustomerID )  Foreign Key: CustomerID Since there is 1 to1 relationship between Parcel and Delivered, Parcel and Returned, Parcel and Lost. · Delivered (ParcelID,DeliverdDate)   Primary Key: ParcelID · Returned (ParcelID,ReturnedDate)  Primary Key: ParcelID · Lost (ParcelID,LostEntryDate) Primary Key: ParcelID Since 1 to 1 relation exists between Parcel and Claim. · Claim ( ClaimID , ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate, ParcelID )  Foreign Key: parcelID Since 1 to 1 relation exists between LostParcel and Claim. · Lost ( LostId ,LostEntryDate, ParcelID )  Primary Key: LostID Foreign Key : ParcelID Since there is 1 to 1 relationship between Parcel and Invoice  1. Invoice ( InvoiceID, InvoiceDate, InvoiceAmount, ParcelID ) Foreign Key: parcelID Since there is 1 employee can be responsible for multiple Parcel. Therefore, 2. Parcel ( ParcelID , ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status, CustomerID, EmployeeID ) Foreign Key: CustomerID,EmployeeID Since 1 to many relationship exists between Employee and  · Delivered ( DeliveredID , ParcelID,DeliverdDate, EmployeeID )  Primary Key: DeliveredID · Returned ( ReturnedID , ParcelID,ReturnedDate, EmployeeID ) Primary Key: ReturnedID · Lost ( LostId ,ParcelID,LostEntryDate, EmployeeID )  Primary Key: LostId Foreign key: EmployeeID,ParcelID TASK 01 (b) NORMALIZATION 1. Parcel ( ParcelID , ParcelWeightInGrams, FromAddress, ToAddress, DateCollected, ParselValue, Charges, Status, CustomerID, EmployeeID ) In this one customer can order multiple parcels, therefore customerID is not enough to uniquely identify a tuple. There is no partial key dependency. The primary key is the only key that identifies the tuple, therefore, this table is 2NF. For 3NF, there must be a transitive dependency, but in this table here is no attribute that can further identify a tuple uniquely, therefor it is already in 3NF 2. Delivered ( DeliveredID , ParcelID,DeliverdDate, EmployeeID )  In this table, there is no partial key dependency, therefore it is already in 2NF. Every attribute is fully dependent on DeliveredID, there is no transitive dependen ...

1. TASK 01
(a) RELATIONAL SCHEMA
Customer (
CustomerID
, CustomerName, CustomerAddress, CustomerContactNum)
Parcel (
ParcelID
, ParcelWeightInGrams, FromAddress, ToAddress,
DateCollected, ParselValue, Charges, Status)
Delivered (DeliverdDate)
Returned (ReturnedDate)
Lost (LostEntryDate)
Invoice (
InvoiceID,
InvoiceDate, InvoiceAmount)
Claim (
ClaimID
, ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate)
Employee (
EmployeeID
, EmployeeName, EmployeeStartDate)
On the basis of relationship that exist between the entities, the
table can be further broken down as;

2. Since one customer can order for multiple parcels therefore, 1
to many relationship exists between them.
· Parcel (
ParcelID
, ParcelWeightInGrams, FromAddress, ToAddress,
DateCollected, ParselValue, Charges, Status,
CustomerID
) Foreign Key:
CustomerID
Since there is 1 to1 relationship between Parcel and Delivered,
Parcel and Returned, Parcel and Lost.
· Delivered (ParcelID,DeliverdDate) Primary Key: ParcelID
· Returned (ParcelID,ReturnedDate) Primary Key: ParcelID
· Lost (ParcelID,LostEntryDate) Primary Key: ParcelID
Since 1 to 1 relation exists between Parcel and Claim.
· Claim (
ClaimID
, ClaimLodgedDate, ClaimSettledAmount, ClaimPaidDate,
ParcelID
)
Foreign Key:
parcelID
Since 1 to 1 relation exists between LostParcel and Claim.
· Lost (
LostId
,LostEntryDate,

3. ParcelID
)
Primary Key: LostID
Foreign Key
: ParcelID
Since there is 1 to 1 relationship between Parcel and Invoice
1.
Invoice (
InvoiceID,
InvoiceDate, InvoiceAmount,
ParcelID
)
Foreign Key:
parcelID
Since there is 1 employee can be responsible for multiple
Parcel. Therefore,
2.
Parcel (
ParcelID
, ParcelWeightInGrams, FromAddress, ToAddress,
DateCollected, ParselValue, Charges, Status,
CustomerID, EmployeeID
)
Foreign Key:
CustomerID,EmployeeID
Since 1 to many relationship exists between Employee and

4. · Delivered (
DeliveredID
, ParcelID,DeliverdDate,
EmployeeID
) Primary Key: DeliveredID
· Returned (
ReturnedID
, ParcelID,ReturnedDate,
EmployeeID
) Primary Key: ReturnedID
· Lost (
LostId
,ParcelID,LostEntryDate,
EmployeeID
) Primary Key:
LostId
Foreign key:
EmployeeID,ParcelID
TASK 01
(b) NORMALIZATION
1.
Parcel (
ParcelID
, ParcelWeightInGrams, FromAddress, ToAddress,
DateCollected, ParselValue, Charges, Status,
CustomerID, EmployeeID
)
In this one customer can order multiple parcels, therefore
customerID is not enough to uniquely identify a tuple. There is

5. no partial key dependency. The primary key is the only key that
identifies the tuple, therefore, this table is 2NF.
For 3NF, there must be a transitive dependency, but in this table
here is no attribute that can further identify a tuple uniquely,
therefor it is already in 3NF
2.
Delivered (
DeliveredID
, ParcelID,DeliverdDate,
EmployeeID
)
In this table, there is no partial key dependency, therefore it is
already in 2NF.
Every attribute is fully dependent on DeliveredID, there is no
transitive dependency among the attributes because no one
attribute can uniquely identify any other attribute. Therefore,
we say it is already in 3NF
TASK 02 INTEGRITY CONSTRAINTS
Attribute
Data type
Integrity constraint implemented

6. Error message
DeliveredID
Autonumber
Primary Key
NULL
DeliverdDate
Date/Time
EmployeeID
Number
Foreign Key -Indexed

7. ParcelID
Number
Foreign Key -Indexed
Attribute
Data type
Integrity constraint implemented
Error message
ParcelID
Autonumber
Primary Key

8. ParcelWeightInGrams
Number
Decimal upto 2 places
FromAddress
LongText
ToAddress
LongText
DateCollected
Date/Time

9. Status
Shortext
In LOOKUP,
Display Control=List Box
Row source Type= Value Type
RowSource =
"Collected";"Transit";"Delivered";"Returned";"Lost"
EmployeeID,CustomerID
Number
Charges
Number

10. TASK 03IMPLEMENTATION REPORT
1.
Any two interesting things/procedures you have learnt by doing
this assignment.
How to add dropdown menu in status was one interesting thing
to learn about.
While creating relationship, how to link the primary and foreign
key with each other.
2.
Any one of the assignment tasks that was relatively difficult /
complex to solve but has been successfully completed by you.
Running queries were the most difficult work.
-- phpMyAdmin SQL Dump
-- version 3.3.9
-- http://www.phpmyadmin.net
--
-- Host: localhost
-- Generation Time: May 24, 2019 at 11:53 AM
-- Server version: 5.5.8

11. -- PHP Version: 5.3.5
SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
/*!40101 SET @
[email protected]
@CHARACTER_SET_CLIENT */;
/*!40101 SET @
[email protected]
@CHARACTER_SET_RESULTS */;
/*!40101 SET @
[email protected]
@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;
--
-- Database: `db2`
--
-- --------------------------------------------------------
--
-- Table structure for table `claim`

12. --
CREATE TABLE IF NOT EXISTS `claim` (
`ClaimID` int(11) NOT NULL AUTO_INCREMENT,
`ClaimLodgedDate` datetime NOT NULL,
`ClaimSettledAmount` int(11) NOT NULL,
`ClaimPaidDate` date NOT NULL,
`ParcelID` int(11) NOT NULL,
PRIMARY KEY (`ClaimID`),
KEY `ParcelID` (`ParcelID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=2 ;
--
-- Dumping data for table `claim`
--
INSERT INTO `claim` (`ClaimID`, `ClaimLodgedDate`,
`ClaimSettledAmount`, `ClaimPaidDate`, `ParcelID`) VALUES
(1, '2018-09-05 00:00:00', 1500, '2018-09-05', 5);

13. -- --------------------------------------------------------
--
-- Table structure for table `customer`
--
CREATE TABLE IF NOT EXISTS `customer` (
`CustomerID` int(11) NOT NULL AUTO_INCREMENT,
`CustomerName` varchar(255) NOT NULL,
`CustomerAddress` varchar(255) NOT NULL,
`CustomerContactNum` int(255) NOT NULL,
PRIMARY KEY (`CustomerID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=9 ;
--
-- Dumping data for table `customer`
--
INSERT INTO `customer` (`CustomerID`, `CustomerName`,
`CustomerAddress`, `CustomerContactNum`) VALUES

14. (1, 'Patrick A Evans', '3146 Hood Avenue, San Diego, 92123',
619),
(2, 'Russell T Foret', '1855 Colony Street,Stamford, 06901',
203),
(4, 'Byron J Christianson', '4412 Hide A Way Road, San Jose,
95118', 2147483647),
(5, 'Stanley D Alfonso', '1234 West Virginia Avenue, Colonie,
12205', 51833438),
(6, 'Ned S Clifford', '3321 Eagles Nest Drive,
Woodland,95695', 53090847),
(7, 'Nelson B Kelley', '2924 Stanton Hollow Road,
Randolph,02368', 78130881),
(8, 'Lawrence M Jaffe', '2426 John Calvin Drive,
CULEBRA,00775', 78745096);
-- --------------------------------------------------------
--
-- Table structure for table `delivered`
--
CREATE TABLE IF NOT EXISTS `delivered` (
`DeliveredID` int(11) NOT NULL AUTO_INCREMENT,

15. `ParcelID` int(11) NOT NULL,
`DeliveredDate` date NOT NULL,
`EmployeeID` int(11) NOT NULL,
PRIMARY KEY (`DeliveredID`),
KEY `ParcelID` (`ParcelID`),
KEY `EmployeeID` (`EmployeeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=2 ;
--
-- Dumping data for table `delivered`
--
INSERT INTO `delivered` (`DeliveredID`, `ParcelID`,
`DeliveredDate`, `EmployeeID`) VALUES
(1, 8, '2019-05-22', 4);
-- --------------------------------------------------------
--
-- Table structure for table `employee`

16. --
CREATE TABLE IF NOT EXISTS `employee` (
`EmployeeID` int(11) NOT NULL AUTO_INCREMENT,
`EmployeeName` varchar(255) NOT NULL,
`EmployeeStartDate` date NOT NULL,
PRIMARY KEY (`EmployeeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=7 ;
--
-- Dumping data for table `employee`
--
INSERT INTO `employee` (`EmployeeID`, `EmployeeName`,
`EmployeeStartDate`) VALUES
(1, 'Tommy M Horton', '2015-02-24'),
(2, 'Howard V Giddings', '2015-09-25'),
(3, 'Vincent L Figueroa', '2016-02-03'),
(4, 'Nathan M Ross', '2015-08-26'),
(5, 'Jeff C Mixson', '2015-03-14'),

17. (6, 'Robert G Cornejo', '2016-02-26');
-- --------------------------------------------------------
--
-- Table structure for table `invoice`
--
CREATE TABLE IF NOT EXISTS `invoice` (
`InvoiceID` int(11) NOT NULL AUTO_INCREMENT,
`ParcelID` int(11) NOT NULL,
`InvoiceDate` date NOT NULL,
`InvoiceAmount` int(11) NOT NULL,
PRIMARY KEY (`InvoiceID`),
KEY `ParcelID` (`ParcelID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=7 ;
--
-- Dumping data for table `invoice`

18. --
INSERT INTO `invoice` (`InvoiceID`, `ParcelID`,
`InvoiceDate`, `InvoiceAmount`) VALUES
(1, 4, '2017-08-11', 1500),
(2, 5, '2018-12-07', 2000),
(3, 6, '2019-05-30', 2550),
(4, 7, '2018-05-08', 2500),
(5, 3, '2018-02-02', 1000),
(6, 8, '2019-05-22', 1500);
-- --------------------------------------------------------
--
-- Table structure for table `lost`
--
CREATE TABLE IF NOT EXISTS `lost` (
`LostID` int(11) NOT NULL AUTO_INCREMENT,
`ParcelID` int(11) NOT NULL,
`LostEntryDate` date NOT NULL,

19. `EmployeeID` int(11) NOT NULL,
PRIMARY KEY (`LostID`),
UNIQUE KEY `ParcelID` (`ParcelID`),
KEY `EmployeeID` (`EmployeeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=2 ;
--
-- Dumping data for table `lost`
--
INSERT INTO `lost` (`LostID`, `ParcelID`, `LostEntryDate`,
`EmployeeID`) VALUES
(1, 5, '2018-09-05', 5);
-- --------------------------------------------------------
--
-- Table structure for table `parcel`
--

20. CREATE TABLE IF NOT EXISTS `parcel` (
`ParcelID` int(11) NOT NULL AUTO_INCREMENT,
`ParcelWeightInGrams` decimal(11,2) NOT NULL,
`FromAddress` varchar(2500) NOT NULL,
`ToAddress` varchar(2500) NOT NULL,
`DateCollected` date NOT NULL,
`ParselValue` int(11) NOT NULL,
`Charges` int(11) NOT NULL,
`Status`
enum('Collected','Transit','Delivered','Returned','Lost') NOT
NULL,
`CustomerID` int(11) NOT NULL,
`EmployeeID` int(11) NOT NULL,
PRIMARY KEY (`ParcelID`),
KEY `CustomerID` (`CustomerID`),
KEY `EmployeeID` (`EmployeeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=9 ;
--

21. -- Dumping data for table `parcel`
--
INSERT INTO `parcel` (`ParcelID`, `ParcelWeightInGrams`,
`FromAddress`, `ToAddress`, `DateCollected`, `ParselValue`,
`Charges`, `Status`, `CustomerID`, `EmployeeID`) VALUES
(3, '200.00', '2376 North Bend River Road, Somerset, 42501',
'4748 Bubby Drive,MORGANTOWN, 26501', '2018-02-02',
220, 1000, 'Collected', 4, 3),
(4, '352.00', '3469 Roosevelt Road,Dodge City,67801', '4269
Short Street,Austin,78723', '2017-08-11', 356, 1500, 'Returned',
6, 4),
(5, '198.00', '111 Lawman Avenue,Alexandria,22301', '2830
Hide A Way Road,Santa Clara,95050', '2018-12-07', 469, 2000,
'Lost', 2, 6),
(6, '146.00', '34 Sussex Court,Waco,76706', '1855 Hillcrest
Circle,Crystal,55429', '2019-05-30', 789, 2550, 'Transit', 7, 5),
(7, '436.00', '1049 Atha Drive,Bakersfield,93311', '1057 Froe
Street,TALMO,30575', '2018-05-08', 965, 2500, 'Collected', 1,
2),
(8, '716.00', '1057 Froe Street,TALMO,30575', '1049 Atha
Drive,Bakersfield,93311', '2019-05-22', 369, 1500, 'Delivered',
5, 5);
-- --------------------------------------------------------

22. --
-- Table structure for table `returned`
--
CREATE TABLE IF NOT EXISTS `returned` (
`ReturnedID` int(11) NOT NULL AUTO_INCREMENT,
`ParcelID` int(11) NOT NULL,
`ReturnedDate` date NOT NULL,
`EmployeeID` int(11) NOT NULL,
PRIMARY KEY (`ReturnedID`),
KEY `ParcelID` (`ParcelID`),
KEY `EmployeeID` (`EmployeeID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
AUTO_INCREMENT=2 ;
--
-- Dumping data for table `returned`
--
INSERT INTO `returned` (`ReturnedID`, `ParcelID`,
`ReturnedDate`, `EmployeeID`) VALUES

23. (1, 4, '2017-11-09', 4);
--
-- Constraints for dumped tables
--
--
-- Constraints for table `claim`
--
ALTER TABLE `claim`
ADD CONSTRAINT `claim_ibfk_1` FOREIGN KEY
(`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE
CASCADE ON UPDATE CASCADE;
--
-- Constraints for table `delivered`
--
ALTER TABLE `delivered`
ADD CONSTRAINT `delivered_ibfk_1` FOREIGN KEY
(`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE
CASCADE ON UPDATE CASCADE,

24. ADD CONSTRAINT `delivered_ibfk_2` FOREIGN KEY
(`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON
DELETE CASCADE ON UPDATE CASCADE;
--
-- Constraints for table `invoice`
--
ALTER TABLE `invoice`
ADD CONSTRAINT `invoice_ibfk_1` FOREIGN KEY
(`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE
CASCADE ON UPDATE CASCADE;
--
-- Constraints for table `lost`
--
ALTER TABLE `lost`
ADD CONSTRAINT `lost_ibfk_1` FOREIGN KEY
(`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE
CASCADE ON UPDATE CASCADE,
ADD CONSTRAINT `lost_ibfk_2` FOREIGN KEY
(`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON
DELETE CASCADE ON UPDATE CASCADE;
--

25. -- Constraints for table `parcel`
--
ALTER TABLE `parcel`
ADD CONSTRAINT `parcel_ibfk_1` FOREIGN KEY
(`CustomerID`) REFERENCES `customer` (`CustomerID`),
ADD CONSTRAINT `parcel_ibfk_2` FOREIGN KEY
(`EmployeeID`) REFERENCES `employee` (`EmployeeID`);
--
-- Constraints for table `returned`
--
ALTER TABLE `returned`
ADD CONSTRAINT `returned_ibfk_1` FOREIGN KEY
(`ParcelID`) REFERENCES `parcel` (`ParcelID`) ON DELETE
CASCADE ON UPDATE CASCADE,
ADD CONSTRAINT `returned_ibfk_2` FOREIGN KEY
(`EmployeeID`) REFERENCES `employee` (`EmployeeID`) ON
DELETE CASCADE ON UPDATE CASCADE;