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1. Gain insight for running pre-defined decision trees and explore results

using MS OLAP Analytics.

Solution:

The purpose of this experiment is to generate a decision tree for a given data set. We can

either write our own data set or use a predefined data set provided to us as in this case is the

bank example.

We start off by opening the weka explorer window.

The above screen would appear, click on the Explorer button to begin.

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Once the explorer is opened, we click on open file and select the appropriate data set.

REMEMBER the data sets if written by you it should be saved as .arff file. For now we

import an already existing file by using the open option. As shown below,

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Once we import the data set, the software itself generates the following,

Weka displays all the attributes of the imported dataset and shows some statistics based

graph.

Click on the classify tab and choose the algorithm J48 as shown below,

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After selecting J48, we can see the defaults are selected that is Cross-Validation and in the

drop down box, Nominal attribute pep is selected as default. Click on the start button and

an output screen as below will be seen,

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In the result list Right Click trees.J48 & choose the option visualize tree, as seen below,

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The output as below will be generated,

We also save the result buffer as follows,

The buffer result gives information about the TP RATE, FP RATE, PRECISION,

RECALL, F-MEASURE and the CONFUSION MATRIX.

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Buffer result.arff (Includes Time taken, Decision tree details, confusion matrix and all other

details)

Upon opening the buffer result.arff file we get,

=== Run information ===

Scheme: weka.classifiers.trees.J48 -C 0.25 -M 2Relation: bankInstances: 300Attributes: 9

agesexregion

incomemarriedchildrencarmortgagepep

Test mode: 10-fold cross-validation

=== Classifier model (full training set) ===

J48 pruned tree------------------

children = YES| income


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| | car = YES: NO (50.0/15.0)| | car = NO| | | married = YES| | | | income 13106.6| | | | | mortgage = YES: YES (12.0/3.0)

| | | | | mortgage = NO| | | | | | income 18923: NO (10.0/3.0)| | | married = NO: NO (22.0/6.0)| income > 30099.3: YES (59.0/7.0)children = NO| married = YES| | mortgage = YES| | | region = INNER_CITY| | | | income 39547.8: NO (4.0)| | | region = RURAL: NO (3.0/1.0)| | | region = TOWN: NO (9.0/2.0)

| | | region = SUBURBAN: NO (4.0/1.0)| | mortgage = NO: NO (57.0/9.0)| married = NO| | mortgage = YES| | | age 39: NO (11.0)| | mortgage = NO: YES (20.0/1.0)

Number of Leaves : 17

Size of the tree : 31

Time taken to build model: 0.09 seconds

=== Stratified cross-validation ====== Summary ===

Correctly Classified Instances 206 68.6667 %Incorrectly Classified Instances 94 31.3333 %Kappa statistic 0.3576Mean absolute error 0.379Root mean squared error 0.4816Relative absolute error 76.2791 %

Root relative squared error 96.6145 %Total Number of Instances 300

=== Detailed Accuracy By Class ===

TP Rate FP Rate Precision Recall F-Measure ROC AreaClass

0.536 0.185 0.712 0.536 0.612 0.683YES

0.815 0.464 0.673 0.815 0.737 0.683NOWeighted Avg. 0.687 0.336 0.691 0.687 0.68 0.683

=== Confusion Matrix ===

a b


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74 64 | a = YES30 132 | b = NO

This completes the first experiment.

2. Design a data mart from scratch to store the credit history of customers of

a bank. Use this credit profiling to process future loan applications.

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Solution:

The purpose of this experiment is the prediction of attribute values for the future unknown

attribute values using the known class labeled values of the records.

Start the software SPRINT (WELLY), as shown below,

Click on the Explorer button to open the Welly Explorer,

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Use the open file button to import the bank data set,

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Select the classify tab and choose the J48 algorithm and as before click the start button.

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After this, in the test options section select the supplied test set option and click the set

button and provide the test set file bank-new.arff.

Once we have provided the test set i.e. bank-new.arff we right click on the result list andchoose the option as visualize classifier errors, as shown below,

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The output screen as below will be seen,

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On this page click on the save option and save the file as shown below,

The saved file contains the predicted values. We show this by comparing all the 3 datasets

being used here, Bank.arff (main dataset), Bank-new.arff (supplied test set) and bank-

predicted.arff (generated output).

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Bank.arff:

Bank-new.arff:

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Bank-predicted.arff:

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Thus the values have been predicted. As we can see that the question marks have been

replaced with a yes or no value.

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3. For a given dataset generate the Association rules using weka and based

on these association rules describe which rules are Strong and which

rules are Weak.

Solution:

The purpose of this experiment is to generate the associate rules for a given dataset in this

case we use the contact lenses dataset. Point to remember here is that association rules

can be only generated for nominal attributes (that is only attributes which have a choice asin [yes,no] or [male,female] etc). Based on the generated rules we have to calculate the

support and confidence values and then describe which rules are Strong or Weak.



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Now provide the contact-lenses data set using the open file option,

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Once we have done this, click on the associate tab and choose the Apriori algorithm which

should be chosen by default,

And then click the start button, the associate rules will be generated as below,

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Scheme: weka.associations.Apriori -N 10 -T 0 -C 0.9 -D 0.05 -U 1.0 -M0.1 -S -1.0 -c -1Relation: contact-lensesInstances: 24Attributes: 5

agespectacle-prescrip

astigmatismtear-prod-ratecontact-lenses

=== Associator model (full training set) ===

Apriori=======

Minimum support: 0.2 (5 instances)Minimum metric : 0.9Number of cycles performed: 16

Generated sets of large itemsets:

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Size of set of large itemsets L(1): 11



Best rules found:

1. tear-prod-rate=reduced 12 ==> contact-lenses=none 12 conf:(1)2. spectacle-prescrip=myope tear-prod-rate=reduced 6 ==> contact-lenses=none 6 conf:(1)3. spectacle-prescrip=hypermetrope tear-prod-rate=reduced 6 ==> contact-lenses=none 6 conf:(1)4. astigmatism=no tear-prod-rate=reduced 6 ==> contact-lenses=none 6conf:(1)5. astigmatism=yes tear-prod-rate=reduced 6 ==> contact-lenses=none 6conf:(1)6. contact-lenses=soft 5 ==> astigmatism=no 5 conf:(1)7. contact-lenses=soft 5 ==> tear-prod-rate=normal 5 conf:(1)

8. tear-prod-rate=normal contact-lenses=soft 5 ==> astigmatism=no 5conf:(1)9. astigmatism=no contact-lenses=soft 5 ==> tear-prod-rate=normal 5conf:(1)10. contact-lenses=soft 5 ==> astigmatism=no tear-prod-rate=normal 5conf:(1)

Confidence values are already given. Support values must be calculated by dividng the value in

each rule by the total no. of instances i.e.

For example,

1st rule says that there are 12 instances where, for tear-prod-rate attribute the value is reduced

and for all that contact-lenses attribute value is none so support is calculated as 12 divided by

total no. of instances which is 24.

Hence , 12/24 = 0.5. Thus the support for rule 1 is 0.5 and confidence is 1.

Likewise, support and confidence values should be calculated for all the rules. Based on the

question given we can decide as to whether the rule is strong or weak.( In the question it would

be mentioned as rules with support value 0.5 or above and confidence value 1 are all strong

rules, so based on such a question we must calculate the values and demonstrate which rules

are strong and which rules are weak.)

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4. To understand ETL (Extract Transform Load) processes.

Solution:

The purpose of this experiment is to create 2 tables and show inner join operation on these

tables using MySQL 5.0 based on query.

We begin by starting the MySQL command prompt. The command prompt asks for a

password. As shown below,

Enter password: ******* (i.e. root123)

Welcome to the MySQL monitor. Commands end with ; or \g.

Your MySQL connection id is 1

Server version: 5.1.42-community MySQL Community Server (GPL)

Type 'help;' or '\h' for help. Type '\c' to clear the current input statement.

mysql> show databases;

+--------------------+

| Database |

+--------------------+

| information_schema |

| mysql |

| test |

| useless |

+--------------------+

4 rows in set (0.03 sec)

mysql> use information_schema;

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| time_zone_leap_second |

| time_zone_name |

| time_zone_transition |

| time_zone_transition_type |

| user |

+---------------------------+


mysql> create database jkl;

Query OK, 1 row affected (0.00 sec)

mysql> use jkl;

Database changed

mysql> create table orders(orderid varchar(5),productid varchar(5),quantity int(

5),unitsaleprice int(5),discountprice int(5),numoffreeservice int(3));

Query OK, 0 rows affected (0.05 sec)

mysql> insert into orders values('O101','P121',10,5000,10,2);








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mysql> SELECT * FROM ORDERS;

+---------+-----------+----------+---------------+---------------+------------------+

| orderid | productid | quantity | unitsaleprice | discountprice | numoffreeservice |

+---------+-----------+----------+---------------+---------------+------------------+

| O101 | P121 | 10 | 5000 | 10 | 2 |

| O101 | P01 | 1 | 234 | 5 | 3 |

| O101 | P180 | 2 | 4000 | 12 | 3 |

| O102 | P02 | 5 | 2500 | 3 | 2 |

| O102 | P122 | 2 | 2800 | 3 | 2 |

+---------+-----------+----------+---------------+---------------+------------------+


mysql> create table products(productid varchar(5) primary key,companyid varchar(

5),productname varchar(10),producttype varchar(10),productprice int(5),productdo

m date,productinstock int(5));


mysql> insert into products values('P01','30','CABLE',121,234,'1999-01-09',25);


mysql> insert into products values('P02','28','OPCABLE',122,500,'1998-08-04',35)

;


mysql> insert into products values('P121','12','MONITOR',147,5000,'2001-09-25',1

9);

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mysql> insert into products values('P122','11','BATTERY',124,1400,'2003-08-15',1

5);


mysql> insert into products values('P180','2','FAX',168,2100,'2003-03-12',12);


mysql> SELECT * FROM PRODUCTS;

+-----------+-----------+-------------+-------------+--------------+------------+----------------+

| productid | companyid | productname | producttype | productprice | productdom |

productinstock |

+-----------+-----------+-------------+-------------+--------------+------------+----------------+

| P01 | 30 | CABLE | 121 | 234 | 1999-01-09 | 25 |

| P02 | 28 | OPCABLE | 122 | 500 | 1998-08-04 | 35 |

| P121 | 12 | MONITOR | 147 | 5000 | 2001-09-25 | 19 |

| P122 | 11 | BATTERY | 124 | 1400 | 2003-08-15 | 15 |

| P180 | 2 | FAX | 168 | 2100 | 2003-03-12 | 12 |

+-----------+-----------+-------------+-------------+--------------+------------+----------------+


mysql> select p1.productid,p1.productname,p2.quantity,p2.orderid from products as

p1 INNER JOIN orders as p2 on p1.productid=p2.productid;

+-----------+-------------+----------+---------+

| productid | productname | quantity | orderid |

+-----------+-------------+----------+---------+

| P01 | CABLE | 1 | O101 |

| P02 | OPCABLE | 5 | O102 |

| P121 | MONITOR | 10 | O101 |

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| P122 | BATTERY | 2 | O102 |

| P180 | FAX | 2 | O101 |

+-----------+-------------+----------+---------+


Various options such as show databases;, show tables;, use database-name; can be

used as well.

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5. Generate a report using the Report Studio of cognos 8 using attributes

from provided tables.

Solution:

The purpose of this experiment is to show the generation of a report using the report studio

of cognos using the tables provided.

We start of opening the internet explorer and typing in the address,

192.100.100.150/cognos 8

A page as below will appear,

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Click on Quick Tour and in here select the MJCET public folder among the options and

click on the report studio option present at the top right of the screen. As shown below,

Once u click on the report studio, the report studio opens as below,

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Choose the create new report option and begin,

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You can choose any of the mentioned options, in this example I show using a list, so click

on the list option and press OK.

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Using the inset table options present on the left of the report studio, from the mypck folder

we can use any of the tables. Here we use the customer table and import some of its

attributes onto the right side as shown above. We simply drag and drop the attribute onto

the right side to generate the view as above.

We then use the play/run button on the tools horizontal menu and click on it to generate the

report.

The run button is the 12th button from the left. After clicking it we have to enter a user idand password ( sa and lab4 respectively). And hit ok to generate the report. As shown

below,

The generated report is as shown below,

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6.Generate Query based reports using Query Studio which performs thefollowing operations.

i. Pivot

ii. Group

iii. Ungroup

iv. Filter

Solution:

The purpose of this experiment is to show the generation of a query based reports using the

query studio of cognos using the tables provided.

We start of opening the internet explorer and typing in the address,

192.100.100.150/cognos 8

A page as below will appear,

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Click on Quick Tour and in here select the MJCET public folder among the options and

click on the query studio option present at the top right of the screen. As shown below,

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The query studio begins and we enter the username and password as below,

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Select the attributes by merely just drag and drop from left to right side or use the Insert

button provided on the left side of the query studio as shown in the above image. Use the

buttons in the tools to generate the reports for the corresponding operations.

In query studio as it is in report studio we have a list of tools as shown below,

These tools include the operations such as run, group, ungroup, pivot and filter buttons as

shown above.

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i. Pivot:

Here in query studio we choose order and product tables attributes to show the operations.

In pivot we show the relation between attributes as shown below,

As we can see we have selected orderId, productId and Quantity from the 2 tables

mentioned above and we have selected orderId as the attribute upon which we apply the

pivot operation this is highlighted by the yellow colour. We then use the pivot button

provided in the tools options and we generate the below,

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ii. Group:In group operation we show the forming of groups for a particular inserted value, here we

use orderId, productId and productName attributes and group on orderId attribute. As we

can see below,

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Selection of orderId as the attribute on which group operation is applied is signified by

yellow colour.

iii. Ungroup:

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In ungroup operation we show the opposite of grouping operation we remove the crosstab

formed using the group operation. As we can see below,

Selection of orderId as the attribute on which group operation is applied is signified by

yellow colour.

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iv. Filter:

In Filter operation we can choose which values of a particular selection of attributes we

want to view. As we can see below,

We have selected orderId, productId and Quantity as the attributes and select orderId as our

attribute on which filter operation is applied signified by the yellow color.

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Here we select the orderId values for which the report should be generated as we can see in

the previous figure. We then hit the ok button and we generate the report below,

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7. Design and build a DataWareHouse using bottom-up approach titled

Citizen Information System. This should be able to serve the analytical needs

of the various Government Departments and also provide a global integrated

view.

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Objective: Created three tables( birthcertf, college, addrproof) and also onemaster table ( citizeninfosystem ) joining these three tables. The objective is to

provide complete information related to citizen using MySQL 5.0. based on query.

mysql> create database prog8;


mysql> use prog8;

Database changed

mysql> create table birthcertf(name varchar(30), fathersname varchar(30),birth DATE,location varchar(30), SSN varchar(3),

PRIMARY KEY(SSN));


mysql> insert into birthcertf values('KLM','STR','1988-03-27','DELHI','665');


mysql> insert into birthcertf values('ABC','XYZ','1989-05-03','HYD','667');


mysql> insert into birthcertf values('GHI','MNO','1990-11-17','BNGLR','645');


mysql> insert into birthcertf values('DEF','PQR','1987-06-07','HYD','785');


mysql> insert into birthcertf values('GTH','THR','1989-06-01','CHNNI','765');


mysql> insert into birthcertf values('RTU','YHN','1994-07-24','DELHI','378');

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mysql> select * from birthcertf;

+------+-------------+------------+----------+-----+

| name | fathersname | birth | location | SSN |

+------+-------------+------------+----------+-----+

| RTU | YHN | 1994-07-24 | DELHI | 378 |

| GHI | MNO | 1990-11-17 | BNGLR | 645 |

| KLM | STR | 1988-03-27 | DELHI | 665 |

| ABC | XYZ | 1989-05-03 | HYD | 667 |

| GTH | THR | 1989-06-01 | CHNNI | 765 |

| DEF | PQR | 1987-06-07 | HYD | 785 |

+------+-------------+------------+----------+-----+


mysql> create table college (name varchar(30),fathersname varchar(30),birth DATE

,SSN varchar(3) REFERENCES birthcertf(SSN),rollno varchar(3));


mysql> insert into collge values('ABC','XYZ','1989-05-03','667','10');

ERROR 1146 (42S02): Table 'prog8.collge' doesn't exist

mysql> insert into college values('ABC','XYZ','1989-05-03','667','10');


mysql> insert into college values('KLM','STR','1988-03-27','665','11');


mysql> insert into college values('GHI','MNO','1990-11-17','645','15');


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mysql> insert into college values('DEF','PQR','1987-06-07','785','14');


mysql> insert into college values('GTH','THR','1989-06-01','765','12');


mysql> insert into college values('RTU','YHN','1994-07-24','378','19');


mysql> select * from college;

+------+-------------+------------+------+--------+

| name | fathersname | birth | SSN | rollno |

+------+-------------+------------+------+--------+

| ABC | XYZ | 1989-05-03 | 667 | 10 |

| KLM | STR | 1988-03-27 | 665 | 11 |

| GHI | MNO | 1990-11-17 | 645 | 15 |

| DEF | PQR | 1987-06-07 | 785 | 14 |

| GTH | THR | 1989-06-01 | 765 | 12 |

| RTU | YHN | 1994-07-24 | 378 | 19 |

+------+-------------+------------+------+--------+


mysql> create table addrproof ( name1 varchar(30),name2 varchar(30),

address varchar(30), SSN varchar(30)

REFERENCES birthcertf(SSN));


mysql> insert into addrproof values('ABC','XYZ','HYD','667');


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mysql> insert into addrproof values('RTU','YHN','HYD','378');


mysql> insert into addrproof values('GHI','MNO','DELHI','645');


mysql> insert into addrproof values('KLM','STR','BNGLR','665');


mysql> insert into addrproof values('GTH','THR','BNGLR','765');


mysql> insert into addrproof values('DEF','PQR','CHNNI','785');


mysql> select * from addrproof;

+-------+-------+---------+------+

| name1 | name2 | address | SSN |

+-------+-------+---------+------+

| ABC | XYZ | HYD | 667 |

| RTU | YHN | HYD | 378 |

| GHI | MNO | DELHI | 645 |

| KLM | STR | BNGLR | 665 |

| GTH | THR | BNGLR | 765 |

| DEF | PQR | CHNNI | 785 |

+-------+-------+---------+------+


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mysql> create table CitizenInfoSystem as( select birthcertf.name,

-> birthcertf.fathersname,birthcertf.birth,

-> birthcertf.SSN,

-> (birthcertf.location) as birthloc,

-> addrproof.address,

-> college.rollno

-> from birthcertf INNER JOIN addrproof

-> on birthcertf.name=addrproof.name1

-> INNER JOIN college

-> on birthcertf.name=college.name)

-> ;


Records: 6 Duplicates: 0 Warnings: 0

mysql> select * from CitizenInfoSystem;

+------+-------------+------------+-----+----------+---------+--------+

| name | fathersname | birth | SSN | birthloc | address | rollno |

+------+-------------+------------+-----+----------+---------+--------+

| ABC | XYZ | 1989-05-03 | 667 | HYD | HYD | 10 |

| KLM | STR | 1988-03-27 | 665 | DELHI | BNGLR | 11 |

| GHI | MNO | 1990-11-17 | 645 | BNGLR | DELHI | 15 |

| DEF | PQR | 1987-06-07 | 785 | HYD | CHNNI | 14 |

| GTH | THR | 1989-06-01 | 765 | CHNNI | BNGLR | 12 |

| RTU | YHN | 1994-07-24 | 378 | DELHI | HYD | 19 |

+------+-------------+------------+-----+----------+---------+--------+


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8.Case study for drawing datawarehouse schema using star or snowflake

schema or fact constellation schema.

Solution:

Star Schema:

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Snowflake Schema:

Fact Constellation Schema:

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9. Create an example of a data set and form its decision tree.

SOLUTION:

car.arff

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Supply the car.arff data set as the input.

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Click on the classify tab, and choose the J48 algorithm. Then click on start.

The following J48 pruned tree information is created with the confusion matrix.


Scheme: weka.classifiers.trees.J48 -C 0.25 -M 2Relation: carInstances: 10Attributes: 5

colortyresnumberplateACquality

Test mode: 10-fold cross-validation

=== Classifier model (full training set) ===

J48 pruned tree------------------

color = red: average (4.0/2.0)color = black: average (3.0/1.0)color = white: poor (3.0/1.0)

Number of Leaves : 3

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Size of the tree : 4

Time taken to build model: 0 seconds

=== Stratified cross-validation ====== Summary ===

Correctly Classified Instances 0 0 %Incorrectly Classified Instances 10 100 %Kappa statistic -0.5385Mean absolute error 0.563Root mean squared error 0.6536Relative absolute error 117.8295 %Root relative squared error 128.5972 %Total Number of Instances 10

=== Detailed Accuracy By Class ===

TP Rate FP Rate Precision Recall F-MeasureROC Area Class

0 0.571 0 0 00 good

0 0.833 0 0 00.229 average

0 0.143 0 0 00.286 poorWeighted Avg. 0 0.548 0 0 0

0.177

=== Confusion Matrix ===

a b c


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Documents

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