MECHANICAL ENGINEERING DEPARTMENT

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1

MECHANICAL ENGINEERING DEPARTMENT

LAB MANUAL


SUJECT



COMPUTER

APPLICATION



II


VI SEMESTER


1.

Study of SQL (Structured Query Language).

2.

Use of SQL for
creation and modification of table.

3.

Use of SQL for creation of table with constraints.

4.

Use o
f SQL for
insertion of data in the table.

5.

Use of
ER (Entity Relationship Model).

6.

Use of SQL
or single table retrieval.

7.

Use of SQL
set functions, concatenation.

8.

Use of SQL
having & group by clause.

9.

Use of SQL
joins and correlations.

10.

Use o
f SQL nested queries.

11.

Use of OSI table updation.

12.

Use of OSI (Open System Interconnection) reference model.

13.

Study of indexing and hashing.

14.

Study

of various network topologies.













2

Experiment No. 1


Aim:

-

Introduction to SQL

Theory:
-

(1)

History of SOL



SQL was deve
loped by IBM to
provide

on interface to the relational
database
they were developing in their
SAN Jose research Laboratory in the middle to
lot 703 .since then
SOL has become the standard Lan
guage to access most relationa
l
database. Because
of this, scripts of SQL statements ate more or less
portable
between operating system & different relation databases.



(2)

What is
SQL?



SQL is more than a query Language not with standing its name.
Within the SQL Language there
are

t
wo major components

DDL (Data definition language)

DML (Data manipulation language)



SQL

is programming language but it is not a procedural language
(white loops, if statements etc)
Variables generally SQL works on sets of date

tables
restricted by the co
nstraint included in the statement. A single SQL Statement can
retrieve a single item or thousands of items from the database depending on what the
uses requests in the SQL statement.


(3)

Categories of Languages.



The database system is an intermediate
between physical data base,
operator, operating system & user. In order too provided various facilities to different
types of uses the DBMS provides
a specialized programming language called
database language.



1)

Data definition language.

2)

Data Manipulation

language

3)

Data control language.


3

1)

Data definition language:


Data definition language is used to describe the details of data. The DBMS
will have a complier

whose function is to process DDL statement in order to identify
the description of
scheme (summar
y). The storage structure and access methods used
by the database system are
specified by a set of defamations in a social

by of DDL
called a data storage and data definition language. The result of compilation
of
this
definition

is a set of instructions t
o specify the implementation details of the data base
schemas details are usually hidden from the user.


2)

Data Manipulation Language:


Once the database summary is complied user most have some means for
manipulating
the database, by manipulation means,


1)

Retrieval of information stored in database.

2)


Insertion

of

new

information

3)

Deletion of information

4)

Modification of information

This
language enables users to access or manipulate data is called
data
manipulation language. It contains basic select state
ment to retrieve the data
from the tables. Commands like select, update, delete & inserts are used in
DML.


3)

Data control Language.


It
has a command to deal with the authorization of user. It has command

to
specify access right for relation & views.


4
)

Data types:


Following are the domain types in SQL.

i)

Char (n):
-

It is a fixed length character string with uses specified
length. The full
from character can be used instead.


4

ii)

Varchar 2 (
n
)
:
-

It

is

a variable length character string, with
user
specified m
aximum length. The full from character varying is
equivalent.

iii)

Int:
-
I
t is
an

integer
(a

finite subject of the triggers that is machine
depended) the
full form integer is equivalent.

iv)

Number
:
-

(p,d)

It
is a fixed point number with user specified
precision
. The
number consists of P digit (plus a sign) and d of P digit
is to the right of the decimal
point. Thus, number (3,1) allows 44.5 to
be store exactly, but neither 444.5 or 0.45 can be stored exactly in a
field of this type.

v)

Date
:
-

It is a
calendar date

containing a (four digit) year month and
day of month.


Common syntax used

1)

Insert:
-


Sql>


Insert into table name (attr 1, attr 2)




Values
(&

attr & att

r
2)


2)

Select:
-


Sql>

Select salter 1, attr 2 from table name.


3)

Update:
-


Sql>

Update tabl
e name


Set alter 1


=’




Where attr 1 = ‘ ’

4)

Delete:
-


Sql> Delete table name


Where attr 1


=’ ‘





5

Experiment No 2





Aim:
-

Creation

and Modification of table.


1)

Table
name:
-




Client master


Description:
-



Use
to store informatio
n about clients.


Creation of table.

Sql>

CREATE TABLE CLIENT MASTER


(Client _no
varchar 2 (6) PRIMARY KEY.


Name varchar (20) NOT NULL
,


Address varchar
2

(30),


Address carchar 2 (30),


City varchar 2 (15),


Pincode number (6)


Bol due number (10
,
2
)


Sql>

TABLE CREATED (Out put).


Sql>

ALTER TABLE Client _master


ADD CONSTERAINT DK Client_no PRIMARY KEY


(CLIENT No);


Sql>

ALTER TABLE client _master


ADD CONSTRAINT Ch_clientno CHECK


(Client no like ‘cy’)



6

2)

Table name:
-




Product master,


De
scription:
-



Use to store information about produce.



SqL>


C
REATE TABLE PRODUCT MASTER



(Product no


varcher 2 (6)


PRIMARY KEY,


d
escription


Varcher 2 (20)

NOT N
ULL


P
rpfit_precent

n
umber (5
, 2
)


NOT NULL


Unit_meadure
Varchar (10)


NOT NULL


Q
ty_on_hand


number (8)


NOT NULL


Reorder Iv
l


number (8)


NOT NULL
,


Sell_price


number (8
, 2
)


NOT NULL
,


Cost_price


number (8
, 2
)


NOT NULL
,


Table created



Sql>

ALTER


TABLE


product_master


ADD

CONSTRAINT DK
_production PRIMARY KEY


(Product n
o):

Sql>

ALTER TABLE


product_master



ADD CONSTRANT Ch_product no


CHECK (product_no like ‘Py”)



7

3)

Table Name
:
-



salesman _master


Description:
-



Use to store information about salesman

Working

in the company.


SqL >

CREATE

TABLE

S
alesman _mas
ter



(Salsman_no



varchar 2 (6)



PRIMARY KEY,


Address 1



varchar

2(
30
)



NOT NULL,


Address 2



varchar2 (
30)





City




varchar2 (
20
),



Pincode



varchar2 (
6
),


State




varchar2 (20),


S
al

amt



number

(6
, 2
)



NOT NULL,


Tgt_to_get



number (
6, 2)



NOT NULL
,


Ytd
-
sales



number (6, 2)



NOT NULL


Remarks



Vachar 2 (60)





);


SqL >

ALTER TABLE:
-



Salesman_master


ADD CONSTRANT



Ch_salesman no.


CHECK (Salesman

no Like ‘SY’
)













8

Experiment No 3


Aim:
-

C
reation of table with co
nstraint


1)

Table name:
-



Sales
-
order

Sql> CREATE TABLE



Sales_order




( S_order_n
o




Varchar 2 (6)


PRIMARY KEY,


S_order_date




Date,


Client_no




varchar 2 (6)


REFERENCE
S

Client_master




(client
-
no)




Dely_addr




varchar 2 (25)


Salesman_
no




varchar 2 (6)


REFERENCES

Salesman_master



(Salesman_no)


Dely_type




char (1)


DEFAULT ‘F'


Billed_yn




char (1)




Dely_date




Date,

Order_Status




varchar (10) CHECK

(
Oarder
-
status in (in process ‘FULFILLED’

CONSTRAINT Chk s_order_no CHECK

S
_order_noLike ‘0%’)

CONSTRAINT Chk_delay_
type CHECK

(Dely_typeIn(‘P’, ‘F’)

CONSTRAINT Chk_delay
-
type CHECK

(Delay_date > s_oeder_date)

);

Table created



9

2)

Table name:
-



Sales
-
order_detals

Sql>


CREATE TABLE



Sales_order
-
detals



(S_order
-
no



varchar

2(6)


REFEREN
C
ES



Sales_order



(S_order_no)



Product_no



Varchar 2(6)


REFEREN
C
ES



Product_master


(Product no)



Qty_ordered



number (8),


Qty_disp



number (8)

Prodyct rate



number (10,2)

CONSTRAINT PK_S_order_no_product no

PRIMARY KEY (S_order_n
o, Product_no));

Table created.


3)

Table name
:
-




Challan_Header

Sql>

CREATE TABLE



Challan_Header


(
Challan_no



varchar 2(6)


PRIMARY KEY


S_
order_no



varchar2 (6)


references


Sales_order



(S.order
-
no)




Challan
-
date



Date



NOT NULL,



Billed_y
n



char (10


DEFAULT ‘N’


CONSTRAINT Chk_Challan_no CHECK


(Challan_no like ‘CHK’)


CONSTRAINT Chk
-
billed
-
yn CHECK


(Billed_yn I n (‘Y’ ‘N’)

);

Table created



10

4)

Table Name
:
-


Challan_Details


Description:
-



Use to store information about challan deta
ils.

Sql >

CREATIVE TABLE


Challan_Details


(Challan_no


varchar 2 (6)


REFERIENCES


Challan_header

(Challan
-
no)



Product_no


vachar 2 (6)


REFERENCES


Product master (Product no)





Qty_disp


number (4,2)


NOT NULL<


CONSTRAINT PK. Challan
_no_product
_no


PRIMARY KEY (Challab_no,product_no)


);



Table created






















11

Experiment No.
4


Aim:
-

Insertion of values is a table


Table name
:
-



Client master



To insert the values in the table

We ca
n insert the values in the table in three wa
ys & by using insert into table
name command.



1)

Sql > INSERT INTO


Client master



(
Client_no, name, city, pincode, state,
bal
-
due)


V
alues (‘c 00001’
, ’Ivan

Bayross’,



Bombay ‘
, 400054
,
Maharashtra’
, 15000
);

This above method is only used for inserti
ng single value at a time.



2)

Sql>

INSERT INTO

Client master




(Client_no, name, city, pincode, state, bal_due)


VALUES (&
client_no, & name, & city, & pincode, & state, &
vol
-
due);


Enter value for
client no



: ‘c00001’

Enter value for name



: ‘Ivan

Bayross’

Enter value for city




: ‘Bomay ‘

Enter value for

pincode



: 400054.

Enter value for state




: ‘Maharashtr
a


Enter value fo
r
bal
-
due



: 15000

1 row created

T
he above method is used for inserting multiple values by just typing and enter and
a
gain the same process is repeated.





12

3)

INSERT INTO


Client_master



Client no, name, city, pin code, state, bal
-
due)

VALUES (‘& client no’, & name
’ & city’, & pin code’, & state’, &’
bal
-
due’

Enter value for clent_no

: c00001

Enter value for name

: Iv
an

Bayross



Enter value for city


: ‘Bomay



Enter value for pincode

: 400054.



Enter vale for state


: ‘Maharashtr
a




Enter value for bal
-
due

: 15000



1 row created


I
f we give the single
code in the values row then there is no
need

of giving
single
code to the varchar data while entering the value. The above method is used for
inserting multiple values by simply typing the ‘/’ immediately next to above
operation.



13

Experiment No. 5


Aim
:
-

To study
Entity Relationship Diagram,


Theory
:
-



Entity

An

entity is an object which is distinguishable from other object. An
entity is thing in a real work with an independent existence.


Entity set

it

is a set of entity of some type that shares some properties of same
type
.


Ex. All student of 6
th

SEMESTER MEC
HANICAL (A)
is an entity set.
Because it consist of number of student entity.


Attributes

are descriptive property process by each member of an entity set.
Ex Roll no semester name are attribute o student entity set.


Domain

For

every attribute there is

a set of permitted values called Domain or
vale set of that attribute.

Ex
-
Domain of attribute customer name might set of are string of certain length.




Type
of

Attributes


1)


Simple and composite attribute.

2)


Single

valued & multiple valued.

3)


Hul
l

value attribute.

4)


Derived

attribute.



Mapping Constraint
-
Mapping constraint express the number of entity to which
another entity can be associated via a
relationship set.



14

Type of Constraint
-


1)

One to One


An
entity

A is associated with at the most

one entity in B, &
entity B is associated with at most one entity in A.


2)

One to Many


An entity in A
is associate
d

with any number of entitles ij
B
.
and

entity
B

however can
be associated with
at most one
entity in A.



3)

Many to one
-

An entity in B is as
sociated with any no of entities in B. An
entity in a however can be associated with entity in B.




4)

Many
to

Many

A
entity in B is associated with any no of entities in B. An
entity in a however can be associated with any no of entities in B.


Keys
-

Concept of keys allows making destination

1)

Primary Key

2)

Foreign Key

3)

Super Key

4)

Candidate Key.


1)

Super Key
:
-

Super key is a set of one or more attribute which
taken collectively allow to identity uniquely an entity in entity set.


2)

Candidate Key
:

-

A super
key for where there is no subject is
called as candidate key.



3)

Primary Key
:
-

Primary key is a candidate key chosen by database
designer to identify entities identity set.


15



4)

Foreign Key:
-

A Foreign key is a set of attributes of one relation
whose vale at
e required to much
value

of

some key attribute of
another relation.



Entity relationship

diagram.


Overall logical structure of database can be expressed graphically by E
-
R
diagram relative simplicity and pictorial drearily of these diagramming teen
rues

may well account in large part of wide spread use of entity

relationship
diagram.


1)




Represent entity set


2)




Attribute


3)




Relationship set







4)




Links

attributes to entity set



5)


M
ultivalue attributes.



Specialization



The
process of specialization is a process of identifying subset of an entity set
that share some distinguishing character eristic.


16

Generalization
-


The process is inverse of specialization wher
e is with identifying
some
common characters of an entity set and creating a new entity set that contains entities
processing
common

characteristics.


Example


Consider ER diagram for database

of an department store. There ate various
departments in the s
tore. One department sales many items some items may be sold
by more than one department. A department has many employees. An employee can
belong to at most one department. A manager is an employee who may look after
more than one department but a departme
nt may be looking after by only one
manager
.

A

unique number called item

nooks assigned to
very

item.

Every item is
supplied by only one supply at a time. Make ER diagram for database intentioned,
indicting primary key.




















17

Experiment. No.
6


Aim
:
-

Use of SOL for single table gets retrieval.


1)


F
ind out name of all clients.

Sql>

SELECT NAME


FROM CLIENT_MASTER;

Output


NAME




IVAN BAYROSS




VANDANA SAITWAL




PRAMADA JAGUSTE




BASIS NAVINDGI




RAVI SREEDHARAN




RUKMINI


2) P
rint the

entire client master table

SQL>SELECT

FROM CLIENT
-
MASTER

OUT PUT
-

CLIENT CITY

NAME & STATE

ADDRES

1

ADDRESS
-
2







PINCODE


BAL.
-
DUE

00001



IVAN BAYROSS

BOMBAY


MAHARASHTRA

400054


15000

C00002


VANDANA




SAI
TWAL

MADRAS


TAMILNADU

780001


0

C00003


PRAMADA




JAGUSTE

BOMBAY


MAHARASHTRA

4000057


5000

C00003


BASU NAVINDGI



18

BOMBAY


MAHARASHTRA

400056


0

C00005


RAVI




SREEDHAPAN

DELHI





1
00001


2000


C00006


RUKMINI


BOMBAY


MAHARASHTRA

400060


0


3)

Re
trieve the list of names & cities of all th
e clients.


Sql > SELECT NAME, CITY FROM CLIENT

MASTER;


OUTPUT
-

NAME




CITY

IVAN BAYROSS


BOMBAY

VANDANA SAITWAL

MADRAS

PRAMADA JAGUSTE

BOMBAY

BASU NAVINDGI


BOMBAY

RAVI SREEDHARAN

DELHI

RUKMINI



BOMBAY



4)

List
the various produce available from th
e produce master table.

Sql > SELECT DESCRIPTION


FROM PRODUCT
-
MASTER;


OUTPUT
-

DESCRIPTION

1.44 FLOPPIES

MONITORS

MOUSE


19

1.22 FLOPPIES

KEYBOARDS

CD DRIVE

540 HDD

1.44 DRIVE

1.22 DRIVE


5)


Find the name of all client having ‘o’
as second letter in
their name.


SQL> SELECT NAME


FROM CLIENT MASTER


WHERE NAME LIKE
-
AY;


OUTPUT
-

NAME


VANDANA SAITWAL

BASU NAVINDGI



RAVI SREEDHARAN


6)

Find out the client is who
stay in

city whose second letter is ‘o’

SQL > SELECT NAME

FROM CLIENT
-
MASTER

WHERE CIT
Y LIKE ‘
-
AY’’;

OUTPUT

NAME

VANDANA SAITWAL


7)

Find out the list of all client who stay in
a city ‘Bombay’ or city ‘Delhi’ or
City ‘Madras’

SQL > SELECT NAME

FROM CLIENT
-
MASTER

WHERE CITY ‘BOMBAY’ OR CITY
-

‘DELHI’ OR CITY
-

‘MADRAS’;


20

OUTPUT
-

NAME

IVAN BAY
ROSS

VANDANA SAITWAL

PRAMADA JAGUSTE

BASU NAVINDGI

RAVI SREEDHARAN

RUKMINI.


8)

List all
client who are located in
A Bombay
.

SQL

> S
E
LECT NAME

FROM CLIENT

MASTER

WHERE CITY


‘BOMBAY’;

OUT
PUT
-

NAME

IVAN BAYROSS

PRAMADA JAGUSTE

BASU NAVINGI

RUKMINI


9
)

P
rint the list of client whose bal.
-
due greater than value 10,000.

SQL
-
SELECT NAME

FROM CLIENT
-
MASTER

WHERE BAL
-
DUE > 10, 000.

OUTPUT
-

NAME


AVAN BAYROSS.




10)

Print the information from sales
-
order table of order placed in month

JANUARY

SQL > SELECT

F
ROM SALES_ORDER


21

WHERE TO
-
CHAR (S
-
ORDER
-
DATE,’MON’)0JAN.


OUTPUT
-

S
-
ORDER

S
-
ORDER
-
D.

CLIENT

DELY.ADDR.


SALES NO.

D B

DELEY.DATE

19001


12
-
JAN
-
96

C00002


S00001

F

N

20
-
JAN
-
96

19002


25
-
JAN
-
96

C00002


S00002

P

N

27
-
JAN
-
96


11)

Display the order information for

client
-
no.’Coooo1’& ‘C00002
’.

SQL > SELECT FROM SALES
-
ORDER

WHERE CLIENT
-
NO.’ C00001’ AND CLIENT NO.
-
C00002’


OUTPUT

S
-
ORDER

S
-
ORDERP
-
D

CLIENT

DELEY
ADDR.

SALES
NO.

D B

DELY.DA.
TE

ORDER
STA.T

019001

12
-
JAN
-
96

C00001


S00001

F N

20
-
JAN
-
96

I P.

019002

25
-
JAN
-
96

C00002


S00002

P N

27JAN
-
96

C

019003

03
-
APR
-
96

C00001


S00001

F N

07
-
APR
-
96

F



22


12)

F
inds the products with description as ‘1.44
drives

& ‘1.22 drive’.

SQL > SELECT FROM PRODUCT
-
MASTER

WHERE DESCRIPTION = ‘1.44 DRIVE’ OR DESCRIPTION = ‘1.22

DRIVE’
;


OUTPUT

PRODUC

DESCRIPTION

PROFIT
-
PERCENT

UNIT
-
MEASU.

QTY.ON
HAND

REORDER

SELL.PRICE

P07975

1.44

DRIVE

5

PIECE

10

3

1050

P08865

1.22 DRIVE

5

PIECE

2

3

1050


13)

Find the product whose selling price is grater than 2000 & less than or
equal to 5000.


SQL > SELECT DECRIPTION FROM PRODUCT
-
MASTER WHERE

SELL
-
PRICE > 2000 AND SELL
-
PRICE=5000;


OUTPUT
-

NAME



MONITORS



KEY BOARDS



CD DRIVE



540HDD

14)

Find the product whose sell price is more than 1500 and also find new
selling price is

SQL > SELECT DE
SCRIPTION, SELL
-
PRICE IS FROM PRODUCT
-
MASTER WHERE SELL
-
PRICE > 1500;




OUTPUT
-

DESCRIPTION



SELL
-
PRICE IS



MONITORS


18000


23



KEY BOARDS



CD DRIVE


78750



540HDD


126000


1
6
)

FIND THE PRODUCT WHOSE COST PRICE IS LESS THAN 1500

SQL > SELECT PRODUCT NO
.

FROM PRODUCT
-
MASTER

WHERE COST
-
PRICE < 1500;


OUTPUT
-


PRODUCT NO.



P00001



P06734



P07805



P07975



P08865


17)

List the product in shorted order of their description.

SQL > SELECT DESCRIPTION

FROM
PRODUCT
-
MASTER

ORDER BY DESCRIPTION

OUTPUT_

DESCRI
PTION

1.22 DRIVE

1.22 FLOPPIES

1.44 DRIVE

1.44 FLOPPIES

540 HDD

CD DRIVE

KEYBOARD

MONITORS

MOUSE


24

18)

Calculate the square root of price of each product.

SQL > SELECT SQRT (SELL0PRICE0

FROM PRODUCT
-
MASTER;

OUTPUT
-

SQRT (SELL
-
PRICE)

22.912878

109.54451

32
.406703

22.912878

56.12861

72.456884

91.651514

32.403703

32.403703



25

Experiment

No
.7


Aim
-

Use
of SQL for set function & concatenation.


23) Count the total number of orders.

SQL > SELECT COUNT (n)

FROM SALES_ORDER;

OUTPUT
-

COUNT (
*)




6


24)

Calculate

average price of all product

SQL > SELECT AVG (SELLA_PRICE)

FROM PRODUCT _MASTER;

OUTPUT
-
AVG (SELL_PRICE)



3666.6667


25)

Calculate minimum price of products.

SQL > SELECT MIN (SELL
-
PRICE)

FROM PRODUCT
-
MASTER;

OUTPUT


MIN (SELL
-
PRICE)



525


26)

Determi
ne the maximum & minimum product price Rename little as

max
-
price and min
-
price

respectively.

SQL > SELECT MAX (SELL_PRICE) MAX_PRICE,

MIN (SELL_PRICE) MIN_PRICE

FROM PRODUCT _MASTER;


OUTPUT



MAX_PRICE


MIN.PRICE



12000



525



26

27)

Count the number of pr
oduct having price greater than o
r

equal to 1500.

SQL > SELECT COUNT (*)

FROM PRODUCT_MASTER

WHERE SELL_PRICE >=1500


OUT
PUT
:

-


COUNT (
*)




4


28)

Find products whose Qty_on_hand is less than recorder level.

SQL > SELECT DECRIPTION


FROM PRODUCT_MASTER


WHERE QTY_ON_HAND < REORDER_LVL

OUTPUT:
-



DESCRIPTION



1.22 DRIVE


29)

Print the information of client_master, product_master, sales_order table in
following format of all records.


{cust_name} has placed order {order_no.} on {s_order_no.}



SELECT C. | |N
AME HAS PLACED ORDER | |

S.S ORDER.NO. | |*ON’ | | S.S ORDER
-
DATE

FROM CLIENT MASTER C, SALES ORDER S

WHERE C CLIENT NO. = S
CLIENT.NO.

OUTPUT
-

C.NAME | |’ HAS PLACED PRDER’ 11 S.S ORDER
-
NO | |’ ON’ | | S.S

ORDER DATE

IVAN BAYROSS HAS PLACED ORDER 019001

ON 12
-
JAN
-
96

VANDANA SAITWAL HAS PLACED ORDER 019002 ON 25
-
JAN
-
96


27

PRAMAD JAGUSTE HAS PLACED ORDER 046865 ON 18
-
FEB
-
96

IVAN OAYROSS HAS PLACED ORDER
019003 ON 03
-
APR
-
96

BASU NAVJHDGI HAS PLACED ORDER 04866 ON 20 MAY
-
96

RAVI SEEDHARAN HAS PLACED ORDER 01000
8 ON 24
-
MAY
-
96





























28

Experiment

No

.8


Aim
:
-
U
se the SQL for having & croupy.


30)

Print the description & total Qty sold for each product

SELECT P. DESCTRIPTION, SUM (S.QTY.DISP) TOTAL

FROM PRODUCT
-
MASTER P, SALES
-
ORDER
-
DETAILS SD

WH
ERE P. PRODUCT NO. SD
-
PRODUCT. NO.

GROUP BY SD. PRODUCT NO. P. DESCRIPTION ;




OUTPU
T

DESCRIPTION


TOTAL

1.44 FLOPPIES


19

MONITOR



06

MOUSE



01

KEY BO
ARD


03

CD DRIVE



02

540HDD



1

1.44 DRIVE



3


31)

Find
value of each product sold.

SQL> SELECT P.
DESCRIPTION,
SUM (
S.QTY.DISP) TOTAL

FROM PRODUCT_MASTER p, SALES
-
ORDER_DETAILS SD

WHERE P. PRODUCT_NO.SD
.
PRODUCT_NO.

GROUP BY SD. PRODUCT NO. P. DESCRIPTION.


OUTPUT
-


DESCRIPTION




TOTAL



1.44 FLOPPIES




19



MONITORS



6



MOUSE



1


29

KEY BOARD


3

CD DR
IVE



2

540HDD



1

1.44 DRIVE



3


32)

Cal. The avg. qty. Sold for each client that has a
maximum

order value of
1
5000.00.

SQL > SELECT C. NAME AVG (SD.QUTY_DISP)

FROM

CLIENT_MASTER

&

SALES_ORDER

S,

LASES

WHERE S.S_
ORDER NO. = SD. S_ORDER NO.

ABD C. CLI
ENT NO.=S. CLIENT NO.

GROUP BY C. NAME, S. CLIENT NO.

HAVING SUM (SD.PRODUCT RATE * SD. QTY
-
ORDER) <=15000,


OUT
P
UT
-

NAME







AVG

(SD.QTY
-
DISP)



BASU NAVINGI





0



RAVI SREEDHARAN




4



VANDANA SAITWAL




0


33
)
Find out total sales amount receivabl
e for the month of JAN. It
wills

the sum
total sum total of all
the billed order for month.

SQL >
SELECT SUM (QTY
-
DISP*PRODUCT
-
RATE)

FROM SALES
-
ORDERS, SALES
-
ORDER
-
DETAILS SD

WHERE S.S. ORDER NO. =SD.S
-
ORDER
-
NO.

AND TO
-
CHAR (S.DELY
-
DATE,’MON’) =’JAN.’

AND
S.BILLED
-
YN=’N’;

OUTPUT
-
SUM (QTY
-
DISP

*PRODUCT
-
RATE)




15750




30

34)

Print the information of product
-
master, order
-
details table in following
format for all records; description worth Rs
$
total sales for product was order in
month.

SQL > SELECT P. DESCRIP
TION 11’WIRTG RS, | |

SUM (QTY
-
DISP*PRODUCT
-
RATE) | |’WASORDER IN MONTH’ | |

TO
-
CHAR (S.ORDER
-
DATE, ‘MONTH | |’ WAS SOLD ‘||

FROM PRODUCT
-
MASTER P, SALES
-
ORDER S, SALES
-
OR

DER
-

DETAILS SD

WHERE P. PRODUCT NO. = SD. PRODUCT NO.

AND SD. S
-
ORDER NO. = S.S
-
ORD
ER NO.

GROUP BY SD. PRODUCT NO. P. DESCRIPTION, TO
-
CHAR

(S
-
ORDER
-
DATE, ‘MON’);



OUTPUT

DESCRIPTION |

| WORTH RS.’ | |
SUM (QTY. DISP*
PRODU. RATE)

11, WAS PRDER IN MONTH”






1.44 FLOPPIES

WORTH RS.




5250


HAS

ORDERED IN

MONTH FEB

1.44 FLOPPIES

WORTH R
S.




2100


HAS

ORDERED IN MONTH JAN






1.44 FLOPPIES
WORTH RS
.




2625


Has

ORDERED IN MONTH MAY

MONITORS




WORTH RS.


2100


HAS

ORDERED IN MONTH APR





MONITORS




WORTH RS.


4200

HAS

ORDERED IN MONTH

FEB

MOUSE




12000


HAS ORDERED IN MONTH



FEB

KE
YBOARD




WORTH RS.


9450


HAS

ORDERED IN MONTH

FEB


31

CD DRIVE


WORTH RS.




5250


HAS

ORDERED IN MONTH

FEB

CD DRIVE


WORTH RS.




5250


HAS

ORDERED IN MONTHFEB

540 HDD


WORTH RS.




8400


HAS

ORDERED IN MONTH JAN

540 HDD


WORTH RS.




0


HAS

ORDERED IN MONT
H MAY

1.44 DRIVE


WORTH RS




3150


HAS

ORDERED IN MONTH MAY


35)

Print

information of product
-
master, order detail table in following format
for all record.

D
escription worth Rs. Total sales for product was sold.

SQL > SELECT P. DESCRIPTION 11 ‘WORTH RS
.’ | |

SUM (QTY
-
DISP*PRODUCT
-
RATE) 11 ‘WAS SLOD’ | |

FROM ORIDYCT
-
MASTER P, SALES ORDER S, SALES ORDER

DETAILS SD.

WHERE P
. PRODUCT NO. =SD.PRODUCT NO.

AND
SD.S
-
ORDER NO.=S.S
-
ORDER NO.

G
ROUP BY SD. PRODUCT NO. P.DESCRIPTION;



OUTPUT
-

1.44 FLOPPIES WORTH
RS. 9975 WAS SLOD

MONITOR WORTH RS
.
6300 WAS SOLD

MOUSE WORTH RS.12000 WAS SOLD

KEYBOARD WORTH RS.
9450 WAS SOLD

CD DRIVE WORTH RS.
10500 WAS SOLD

540 HDD WORTH RS.
8400

WAS SOLD



1.44 DRIVE WORTH RS.
3150 WAS SOLD
.


32

Experiment No
. 9


Aim
-

Use of SQL for jo
ins & correlation.


36)

find product which has been sold to IV AN BAYROSS’.

SQL > SELECT DESCRIPTION

FROM PRODUCT _MASTER P, SALES_ORDER S,

WHERE_ORDER_DETAILS SD, CLIENT_MASTER C,

AND S. CLIENT NO. = C. CLIENT NO.

AND C. NAME = ‘IVAN BAYROSS’.

AND S.S
-
ORDER NO. = SD.
-
ORDER NO
;


OUTPUT
-

DESCRIPTION



1.44

FLOPPIES



5.40HDD



C
D DRIVE



MONITORS



MOUSE


37)

F
ind out product & there description i.e. Quantity that will have to
delivered.

SQL > SELECT DESCRIPTION (QTY
-
ORDER PER
-
QTY
-
DISP)

FROM PRODUCT
-
MASTE
R P, SALES ORDER S, SALES
-
ORDER
-

DETAILS SD.

WHERE P.
PRODUCT NO
. = SD. PRODUCT NO. ANDS.S
-
ORDER

NO. = SD. S
-
ORDER NO. AND S.DELY
-
TYPE = ‘P’;


OUTPUT
-

DESCRIPTION

(QTY. ORDER
-
QTY
-
DISP)



1.44 FLOPPIES

10



540HDD


1



1.44 DRIVE


1


33

40)

L
ist the product no
& s
-
order no. of customers having Qty
-
ordered less than
s from the order details table for product 1.44 floppies.

SELECT P. PRODUCT NO, S.S
-
ORDER NO.

FROM PRODUCT
-
MASTER P, SALES
-
ORDER S,

SALES
-
ORDER
-
DETAILS SD

WHERE

P.

DESCRIPTION
-
1.44

FLOPPIES’

AND

SD.

QTY
-

ORDERED <S AND

S.S_ORDER NO. = SD.S
-
SORDER NO
. AND P. PRODUCT NO. = SD.

PRODUCT NO;

OUTPUT
:

-

PRODUCT_
NO





S_ORDER



P00001






O19001


41)

Find product and their quantities for order placed by Vandana Saitwal &
Ivan Bayross.

SQL > SELECT P. DESCR
IPTION, SD. QTY
-
ORDERED

FROM CLIENT
-
MASTER C, PRODUCTION
-
MASTER P,

SALES
-
ORDER S, SALES
-
ORDER
-
DETAILS
SD

WHERE (C.NAME = ‘IVAN BAYROSS OR C. NAME = ‘VANDANA
SAITWAL’)

AND P.PRODUCT NO. = SD. PRODUCT NO.

AND
S. S
-
ORDER NO. = SD. S
-
ORDER NO.

AND C. CLIENT NO
. = S. CLIENT NO.

OUTPUT
-


DESCRIPTION


QUTY ORDERED



1.44 FLOPPIES


4



540HDD



2

CD DRIVE



2



1.44 FLOPPIES


10



MONITORS



2



MOUSE



1



34

42)

Find
product and there quantities for order placed by client no. ‘C00001’ &
‘C00002’.

SQL > SELECT P. DESC
RIPTION, SD. QTY
-
ORDERED

FROM CLIENT
-
MASTER C, PRODUCT
-
MASTER P,

SALES
-
PRDER S, SALWS
-
ORDER
-
DETAILS SD

WHERE (C.CLIENT NO.=C00001’ OR C. CLIENT NO. = ‘C00002’)

AND P. PRODUCT NO = SD. PRODUCT NO.

AND S.S
-
. ORDER NO = SD. S
-
ORDER NO.

AND C. CLIENT NO. = S
. CLIENT NO.

OUTPUT
-

DESCRIPTION


QTY. ORDERED

1.44 FLOPPIES


4

540HD



2

CD DRIVE



2

1.44 FLOPPIES


10

MONITORS



2

MOUSE



1














35

Experiment

No
.10


Aim
:
-

Use
Of SQL For Nested Queries.


43)

F
ind the product no and description of non moving prod
uct.

SQL > SELECT PRODUCT NO, DESCRIPTION

FROM PRODUCT
-
MASTER

WHERE

NO. PRODUCT


NO


IN

(SELECT

PRODUCT

NO FROM SALES
-
ORDER
-
DETAILS);




OUTPUT
-

PRODUCT NO

DESCRIPTION

P 07865


1.22 FLOPPIES

P 08865


1.22 DRIVE


44)

Find customer named address
1, address 2, city & pincode for client no. has
placed order no ‘019001’.

SQL

>

SELECT NAME, ADDRESS 1, ADDRESS2, CITY PINCODE

FROM CLIENT
-
MASTER

WHERE CLIENT
NO. = (SELECT CLIENT NO FROM SALES
-
ORDER
WHERES
-
ORDER NO. = ‘019001’);



OUTPUT
-

NAME

ADD
RESS
-
1


ADDRESS
-
2

CITY


PINCODE


IVAN BAYROSS

BOMBAY

400054


36

45)

F
ind client name who have placed order before month of May
-
96.

SQL > SELECT NAME FROM CLIENT
-
MASTER

WHERE CLIENT NO. IN (SELECT CLIENT NO FROM SALES
-

ORDER
WHERE S
-
ORDER
-
DATE < ’01
-
MAY
-
96
)

OUTPUT
-


NAME




IVAN BAYROSS




VANDANA SAITWAL




PRAMADA JAGUSTE





46)

Find out if product (1.44 drive) is ordered by an client and paint client no,
name to whom it is was sold.

SQL>
SELECT CLIENT NO, NAME FROM CLIENT MASTER


WHERE CLIENT NO. IN (C
LIENT NO FROM SALES
-

ORDER WHERE S
-
ORDER NO.IN (SELECT S
-
ORDER NO FROM

SALES

ORDER
-
DETAILS WHERE PRODUCT NO (SELECT


PRODUCT NO FROM PRODUCT MASTER WHERE DESCRIPTION

= 1.44 FLOPPIES);


OUTPUT
-

CLIENT NO.



NAME

C00001



IVAN BAYROSS

C00002



V
ANDANA SAITWAL

C00003



PRAMADA JAGUSTE

C00005



RAVI SREEDHARAN





37

47)

Find names of client who name placed order with rs. 10000 or more

SQL> SELECT NAME FROM CLIENT
-
MASTER WHERE CLIENT

NO IN (SELECT CLIENT NO FROM SALES
-
ORDER WHERE S
-

ORDER NO
. IN (SEL
ECT S
-
ORDER NO FROM SALES
-
ORDER
-

DETAILS WHERE PRODUCT
DATE #

QTY
-
ORDERED > = 10000);

OUTPUT
-

NAME

IVAN BAYROSS

PRAMADA JAGUSTE






















38

Experiment No
. 11 (a)


AIM
:
-

Use of
SQL for queries using dates.


48)

Display the order number day on wh
ich client placed there order

SQL> SELECT S
-
ORDER NO, TO
-
CHAR (S
-
ORDER
-
DATE,’ DAY’)

FROM SALES
-
ORDER;

OUTPUT
-


S
-
ORDER NO


TO
-
CHAR


(S
-
ORDER

DATE, DAY




019001


Thursday




019002


Wednesday




046865


Saturday




019003


Tuesday




046866


Sunday




01
00008


Thursday


49)

Display the month (in alphabet)
and done when order must be delivered.

SQL> SELECT S_ORDER NO.

TO
-
CHAR (DELY_DATE,’ MONTH’DO)


FROM SALES
-
ORDER;

OUTPUT
-


S_ORDER NO.

TO


CHAR (
DELY
-
DATE

‘MONTH’DD





019001


JANUARY


20




019002


JAN
UARY


27




046865


FEBRUARY

20




019003


APRIL


07




046866


MAY


22




0100008


MAY


26



39

Experiment
No
. 11 (b)


Aim
:
-

USE

OF SQL for Table updating.


53)

Change the S
-
order
-
date of client no ‘
c00001’ to 24/07196

SQL> UPDATE SALES
-
ORDER

SET S
-
ORDER
-
DAT
E = 24 JUL.96

WHERE CLIENT NO. = ‘C00001’;



54)

Change the sell price of 1.44 floppy drive to rs. 1150.00

SQL> UPDATE PRODUCT
-
MASTER

SET SELL
-
PRICE = 1150

WHERE DESCRIPTION = ‘1.44 FLOPPY DRIVE’;


55)

delete

the record with order number ‘019001’ from orde
r table.

SQL> DELETE FROM SALES
-
ORDER

WHERE S
-
ORDER NO = ‘019001’;


56)

Delete all record having delivery date before 10
th

July 96

SQL> DELETE FROM SALES
-
ORDER

WHERE DELY
-
DATE <’ 10
-
JUL
-
96’;


57)

Change

city of client no ‘C00005’ to ‘Bombay’;

SQL> UPDATE CLIEN
T
-
MASTER

SET CITY = ‘BOMBAY’

WHERE CLIENT NO

= ‘c00005’;


58)

Change delivery date of order number ‘010008’ to 16
-
08
-
96

SQL> UPDATE SALES
-
ORDER

SET DELY
-
DATE = ’

16
-
AUG
-
96’


40

WHERE CLIENT NO. = ‘010008’



59)

Change the due of
client no ‘C00001’ to 1000

SQL
> UPDATE CLIENT MASTER

SET BAL DUE= 1000

WHERE CLIENT NO ‘C00001’









41

Experiment No. 12


Aim:
-

To
study the

OSI reference model.


Theory



The
OSI model is as shown in Fig (Minus PHYSICAL MEDIUM) this model
is based on a personal developed by intern
ational standards organization of the
protocols used in various layers. The model is called ISO OSI [
Open System Inter
Connedction
]
R
eference model because it deals with connecting open system that
are
open for communication wit other system. It is usually

called OSI model. The OSI
model has seven layers, he principle that user applied to arrive at seven layers. Ate as
follows
-

1)

A layer should be created where a difference law of abstraction is needed

2)

Each layer should perform a well defined in with an eye t
owards defining
internationally standard protocol.

3)

The function of each layer should chose with an eye towards defining
internationally standardized protocols.

4)

The layers boundaries should be larger enough that distinct in need
not be
thorn

together

in Sa
m layer out of necessity & small enough that are
tincture does not become unwieldy.

Each layer of model in turn is discussed starting

a bottom layer, OSI model itself is
not a network architecture because it does not specify the exact services & protocol
t
aken used in earn layer. It just tells hat e
ach layer should do however ISO has also
produced standards for all the layers. Although these ate not part of reference mode
itself. Each one has been published as a separate
international standard.



Physical
L
ayer:


T
he physical layer is concerned with transmitting row bits
outer a
communication channel. The design issues have to do with making sure that when
one side sends a bit it is retched by other sides as a 1 bit and not as a o bit. Typical
Question here
are how many volts should be used to represent a1 & how many ao,

42

how many microseconds a bit lasts whether transmission may proceed simultaneously
in both direction now the initial connection is establishing and how it is turn down
when both sales ate fin
wished and now many pins the network connector has and
what earn pin the network co
nnector has and what earn with mechanical electrical &
procedural interfaces & physical transmission medium which lies below physical
layer.


Data Link
Layer:


The main

tas
k of data link layer is to take a new transmission facility &
transform it into a line that appears free of unaccredited transmission error

to network
layers.
It accomplsties

this taste by having stoned break. The input data up to process
the acknowledgeme
nt frames sent back by receives. Since the physical layer merely
accepts and
transmits,
steams of bit with our any regards
to meaning or structure, it is
up to data link layer to create and
recognize frame

foundries. This can be
accomplished by attached sp
ecifically bit
patterns to beginning& of frame. It these bit
patterns can accidentally course in date, special care must be taken to make sure these
patters are not incorrect inter prefer as proem delimiters.


A noise burst on the line can destroy a frame

completely. In this case, the data
link layer. Software on source M/C can retransmit the frame introduction the
possibility of duplicate frames. A duplicate frame could be sent it acknowledgement

frame from
receives back to sender where lost. It is up to
this layer to solve the
problem

caused by damaged, cost and duplicate frames. The date link layer may offer
secular different survives classes to the network layer, each of different quality &
with different price.



Another issue that arises in the date
link layer has to keep a Fast transmitter
from drawing a slow receiver in data. Some traffic regulation me
chanism must be
employed
to let the transmitter know how much buffer space the
receiver

has at the
moment. Frequently, this flow regulation and
our

ha
ndling are integrated.


If the line can be used to transmit data in both direction. This introduces a new
complication that the data link. Layer software must ideal with the

problem

is that in
acknowledgement frames from A to B traffic. Complete for the u
se of line with data,

43

transfer B to A traff
ic R driver both has be devised. Broadcast networks have
additional issue in data link layer, how to control access to shared channel A special
sub layer of the date link layer, medium access sub layer deals with
this pro
b.


Network Layer:


T
he network layer is concerned with controlling the operation

of subnet
. A
key design issue is determine how packets are routed from source to
destination
router

c
an be base on Static tables that are “wired into” the network
and

rarely
changed
.

They can also be determined at the start of each conversion. Ex
-
terminal
session finally, they can highly dynamic being determined a new for each pocket to
connect network pad.


I
f too many pockets are present in subnet at same time, they

will get in each
other

way, forming bottle necks. The control of such congestion also belonged to
network layer.


Since operation of subnet may expert remuneration for their efforts, there is
often
some accounting function built intonetwork layer. At lea
st software must count
how many packets or characters or bits are sent by
e
ach customer to produce
billing
information when a pocket crosses
border, with different rates on each side, the
accounting can become complicated.



When a pocket has to travel fr
om one network to another target its destination,
may problems can arise. The addressing used by second network may be different
form first one. The second one may not accept packet at all because it is too large.
The protocols may differ as so on. It is u
p to the network layer to overcome all thus
problem to allow net erogenous network
to be inter connected.


In
broadcast

network the receiving problems simple, so network layer is offer
thin as even non
-
existent.


Transport
La
y
er
:


T
he basic function of tra
nsport layer is to accept data from session layer split
it up smaller units if need be pass these to network layer & ensure that plies all arrive
correctly at other end further more, all this must b done efficiently and in a way that


44

isolates upper layer
from the inevitable changes in hardware technology.


Under the normal condition the transport layer relates a distant network
connection for transport connection required a high through put however
the transport
layer might create multiple network connect
ion dividing the data among the network
connection to improve

thro

o
u
t on
other hand, if outing or maintaining a network
connection is expensive the transport layer might
multiple several transport
connection on to the some network connection to reduce the

cost. In all case the
transport layer i
s required to make multiplexing transport to session layer.


The transport layer also determine what type of secure to provide session layer
& ultimately u
ses of network the most popular type of transport connection

in an
error free point to point channel that delivers message or
bytes in order in which they
user sent however other possible kinds or transport service are transport of isolated
messages with
of

grantee about order
of

delivery of broadcasting of message
s to
multiple destination the type
of

service is determined when connection is established.


Session
Layer:


The session

layer allows user or deferent m
/
e to established session between
then a sessions allows ordinary data transport as does the transport
layer but it also
produced enhances services useful in some application. Accession might be used to
allow a user to log into a remote time snaring but ot to transport a file between two
m/c.


One of the services of session layer is to manage dialogue contr
ol session
layer is to allow traffic to go in both direction at sometime only one direction at a
time id traffic can
only

go one way at a time session layer can help keep
track

of
whose turn i
t is.


A related session services is taken management for some p
rotocols it is
essential that both sates do not attempt some operation at same time. To manage

these
activities the session layer produces taken that can be exchanged only the side
holding taken many perform certifies operation


45

Presentation
Layer:


The pr
esentation layer
reforms certain function that are requested sufficiently
often

to
warrant

find a general solution for then,
rather than setting each user to solve
the problem. In particularly unlike all lower layer which ate just interested in moving
bits

reliably from here to there, the presentation layer is concerned with syntax &
scantiest of information transmitted.


A typical example of protestation services is encoding data in
standard argues
upon way most use programs don’t
exchange random binary b
it strings. They
exchange thing such as people names, dates

amount of money & invoices these items
are represented ants character

string imagers floating point number & data structure
composed
of several

simple items. Different computers

have foddering cod
es for
representing
character string

integers and soon. In order to make it possible for
computer will defined representation to communication the data structure to be
exchange can be defined in an abstract way. Along

with standard encoding data
structure
& converts
from

the deprecation used inside computer to network STD
representation and back.















46

Experiment No.13


Aim
:
-

Computer of ordered indexing &
hashing.


Theory
:
-

Servable ordered indexing schemes and servable hashing
scheme have seen
fi
les of records can be organized as ordered files using index sequent ional
organization
or using B tree organization alternately files can be organized
using
hashing. Finally they can be organized a sheep files can be organized as heap files
where recorded

ate not ordered in any particular way.



Each s
cheme has an advantage in certain
situation a database
sy
stem

implements cold produce many schemes leaving the final decision of where schemes
to use to the database designer However such an approach require
d the
implementer
to write more code adding both to cost of
sy
stem

and to space that the system
occupies. Thus most database
sy
stem

use only a few or just one form of ordered
indexing to
hashing
is
acceptable
?

What is relative’s freq
of insertion & deletio
n
:

Is it desirable to optimize our access time at the expense of increasing the worst case
access time?

What queries
are likely to places by user?

When it comes to make choice that 1 % compare ordered indexing & hashing the
faith i.e. expected type of quer
y is most critical.

If most quires ate of form

Select a 1 A2 An

From r

Where Ai =c



Then to prices this query the syt will perform a look upon ordered index or a
hash structure
for

attribute A
for values. For queries of these form a hashing scheme
is pr
eferable Am ordered index look up required time proportional to the leg of no of
values in r for A In a hash structure however the average

look up tine is a constant

47

independent of the size o database.

The
only advantage
to the logoff number of values
in ‘
r’ for Ai by constrict of hashing is used worst case look up time is preoperational
to no of values in r for ai However the work case look up time is unlikely to
occur

with hashing

and is preferable in thus case ordered index technique

ate preferable to
ha
shing in case where a range of values is specified in quarry.

Select Ail A2 an

From r

Where A1 < =C2 and A1 >=C1

There proceeding query finds all records with Ai values between C1 and C2 Consider
how this quarry is processed using on ordered index First
a look up is performed on
value C1 once the buckets for value C1

once the buckets for values C1 has been
found the pointer chain in index is followed to read next bucket in order and this is
continued until C2 is reached.


F instead of an ordered index a
hash structure is used then a lilt upon on C12
can be performed & the corresponding bucket is locked
but it easy in gentle to
determine

the next bucket that must be examined. The difficulty arises because

agreed hash function assign values randomly to buck
et. Thus there is no simple
notion of “next buckets in stored order”

The reason buckets cannot
be

chained
together in sorted area. On a is that each bucket is assigned many search key
values
.
Since values ate scattered randomly by hash in values
in

specifi
ed range are likely to
be scatter a across many or all of bucket therefore all buckets
must

be read to find
required search keys.



Result
:
-

An
overall comparison between the ordered indexing & hashing
showed that a wise choice between the two ultimately

depends upon what type


are
user likely t cash




48

Experiment No. 14


Aim
:
-

T
o
study various network topologies


Theory:


T
he ten
networks

are

used for a system which allows two or more computer
microprocessor to be linked for the
interchange

of data.

The logical form of the links
is known as Network topology. The term
NODE is used for a point a point

in a
network where one or more communication lined terminates as a unit is connected to
communication lines commonly used form are

1)

Bus
Topology:


Th
is has a linear but into
which all station ate plugged. This system is often
used for multipoint terminal cluster. It is generally preferred mentioned for
distant

between nodes of more than loom.

2)

Star Topology
:


This is has dedicated channel between eac
h station % a central switching hub
Norwich all communication must pass. This is

a type of network used in Telephone
services (private branch exchanges in many companies. All lines passes through
control exchange this system is also
often used to connect r
emote & local terminals to
control main fare computer one major disadvantage is that if control nub

fails then
entire system fails.

3)

Hierarchy of
tree:


This consists of a screw of branches converging indirectly to a point at head of
tree. With this sy
stem there is only one transmission path between any two stations.
This arrangement may be formed from number between nodes of more than loom.

4)

Ring
Topology:


This is very popular method for local area network involving each station
being connected to a

ring. The distance between nodes is generally less than loom.
Data put intoning system continue to circulate round to ting until some system
ramous it. The data is available to the station.

5)

Mesh Topology
:


49


This method has not formal pattern to connecti
on between stations and there
will be multiple data point between them. The term local area and there will be
multiple data
point

between them. The term local area network over a local
geographical area such as a building or a group of building on one site
. For which the
bus, stat ring mesh etc topology ate done a wide area networks (WAN)

is one that
connects computers terminals & local area network outer a national or
international

level terminals & local area
network

outer a national or international leve
l.



Conclusion:


The various networks Topology have been studies with implied meaning fully
comp bended.