Network final exam 2010 model answer

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Oct 23, 2013 (3 years and 11 months ago)

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Network final exam 2010 model answer

Question 1 (Answer
s
):

A)

Ƭ

: end to end propagation delay ,
Ƭ

=


(

)







(



)











T:
time of packet transmission

, T =


(

)






(


)














a :

normalized end to end propagation delay , a=











write the vulnerable time of ALOHA, CSMA/CD


Pure ALOHA = 2T

ALOHA



Slotted ALOHA = T

CSMA/CD vulnerable time =

2 T


B)

Pseudo code for the one bit sliding window protocol

S
1

: s
end seq =0


Rec seq = 0

S
2

: From host (Buffer)

S
3

: generate Frame to be sent having


S.Info= Buffer


S.seq=sendseq


S.ack=1
-

recseq

S
4

: F(S)

S
5
: start timer (S.seq)

S
6

: wait ( event ={ Frame arrival| ChkSumErr|Timeout})

S
7

: event <>

frame Arrival S
12 |
S
4


S
8

: get F( r )

S
9

: if r.seq= rec.seq then ( toHost (
r.Info
)
; rec.seq=rec.seq
)

S
10

: if r.ack=send.seq then ( fromHost ( Buffer); sendseq=sendseq )

S
11

: S.info = buffer


S.seq=send.seq

S
13
: startTimer ( s.seq )

S
14

:
goto S
6


Explain the term piggybacking: when data frame arrive at an IMP ,instead of immediately sending a separate
control ( ack ) frame .the IMP at the reciver wait until the host has data message

to be send to and the acknowledge is
attac
hed to the outgoing data frame ( using ack field ) in the frame header.

C)

1
-

A gets( 0,1,B0)


B sends(0,1,B0)


A sends (0,0,A0)


B gets (0,0,A0)


to Host

2
-

A gets (0,1,B0)


B sends( 1,0,B1)


A sends (1,1,A1)


B gets ( 1,1,A1)


to host

3
-

A gets (0,1,B2)


B sends( 0,1,B2)


A sends (0,0,A2)


B gets ( 0,0,A2)

to host

4
-

A gets (1,0,B3)


B sends( 1,0,B3)


A sends (1,1,A3)


B gets ( 1,1,A3)

to host

5
-

A gets (0,1,B4)


B sends( 0,1,B4)




Question 2
(Answers)
:

a)

Channel allocation techniques

Free of collision







collision

Static









Dynamic

-

Frequency division multiple access (FDMA)



ALOHA

(Pure


Slotted )


-

Time division multiple access (TDMA)



CSMA


CSMA/CD

-

Polling

(Hub
polling


roll call polling)



TDMA : user assigned to a fixed number of predetermined packet time slot

D
TDMA
= 1 + N [ S/2(1
-
S)+1/2]

-

TDMA is better in average delay by (N/2
-
1) compared to FDMA

-

Disadvantage:

in low traffic there is low traffic utilization



FDMA : user assigned to a fraction w/n channel bandwidth and by assumption of infinite delay it can seen
that system is modeled
for N sub channel
each of them viewed M1D1 queue.

M: input rate in passion distribution.

D: deterministic service

1: one server

D
FDMA

= N[ 1+ S/2(1
-
S)]


-

Disadvantage: waste bandwidth due

to gap inter

between both TDMA and FDMA is inefficient channel
utilization
because bandwidth and time slot for user is independent of his needs.



Polling:


-

hub polling: control station just send po
lling message
to 1
st

terminal only and 1
st

terminal after its turn send
it to 2
nd

terminal and so on.



Dynamic( collision ): pure ALOHA
-
> S
max

=0.36 Slotted ALOHA
-
> S
max

=0.76


b)

The throughput for the slotted ALOHA :
-


P
k
(t) =
(

)












S=Throughput = InputRate * prob. Of success = p
0
(t) T=1

P
0
(t)=e
-
G


S=Ge
-
G

To get max through
put

Ds/dG = G (
-
1) e
-
G
+e
-
G
(1)=0

G
max
=1

S
max
= e
-
1
=

1/e = 0.36

C)channel capacity = maximum Throughput

Channel capacity

0.96

CSMA/CD

0.76

CSMA

0.36

Slotted ALOHA

And the best technique is hub polling which has the maximum throughput

D ) The protocol CSMA/CD : Carier

Sense multiple Access with collision detection if you have something to send
sense the channel firstly and if it's idle send your message ,still sensing the channel and if it's not idle wait
random time and try sending again. And the vulnerable time = 2
Ƭ


Question 3 (Answers)
:

A )

-

presentation layer

-

Network layer

-

Data Link layer

B)

in distributed system:
-

the existence of multiple autonomous computer is transparent (not visible) to the user
.

The operating system of distributed system selects the
best processor to do the job,finds and transport all the input
files to that processor and puts the results in an appropriate place . it looks like a virtual uni processor

-

In computer network: the user must explicitly log on to one machine and submit jobs

remotely, move files
around and generally handle all of the network management personally.


C )


Reliable means data never lost ,its done through acknowledge sent from the receiver to the sender.

-

Acknowledge process introduces overheads and delayes

-

Some
times acknowledge not accepted such as digitized voice traffic

Unreliable : means
( no acknowledge) connectionless services are often called datagram service.

Question 4 (A
nswers):

a)

Wireless topology
:

advantage: misuse

of cables and easy management of host
s, simplicity of trouble

shooting
.

Disadvantage: signal interference, interception

and blockage.

b)

10Base2 : it means the network architecture runs over 10MbPS using the Baseband communication
channel and
over a max distance = to 200 meter.

1000BaseC
X :

it means the network architecture runs over 1000MbPS using the Baseband communication
channel and over Coaxial Cables.

C )

PC to Networking Device






PC to PC

1________1








1________3

2________2








2________6

3________3








3____
____1

4________4








4________4

5________5








5________5

6________6








6________2

7________7








7________7

8________8








8________8


D )
Hubs: is central station part that connect all devices to it and we have t
w
o types of hub:



-
passive Hub
-
>broadcasted



-
Active Hub
-
>Regenerate data before sending it again to all destinations on the network
.

Switches: responsible for sending and receiving data from one device to multiple devices on the network,

through a smart mechanism which
called the MAC table inside the switch.


Gateways: responsible for translating information from one format to another. Runs over OSI layer according
to the information that to be translate

Questions 5 (Answers
):

a)

Bandwidth F=32 KHz ,n=16

BitRate (bps)

T
ime of Transmission T[msec]
=n/b * b
3

Frequency of 1
st

harmonic f=1/t ( Hz)

# of harmonics passed =f*t

300

53.33 16/300 10
3

18.75

1706.67

600

26.66

37.5

853.33

1200

13.33

75

426.67

2400

6.667

150

213.33

38000

0.421

2375

13.47


Conclusion: max bit
Rate bandwidth for 32 KHz is 38000 bps.



b)

ISO
-
OSI Reference model: the OSI reference model is based on a proposal of develop by ISO (international
standard organization ) of the first stop toward. The OSI (open system interconnection ) because it deal with

open system that system is open for communication with other systems.

-
ISO
-
OSI model has 7 layers which implement the following:

1
-

physical layer: transmit stream of bits over communication channel.

2
-

Data link layer: create and recognize frame
boundaries.

3
-

network layer : determine routing stratigies between IMPs.

4
-
Transport layer: which may be one of two main types:

-

Virtual circuit: deliver packets inorder.

-

Datagram: deliver packets randomly

5
-
session layer: user interface with the network

6
-

presentation layer: (text compression
-
encryption
-
incompatible formats or files of different hosts)
.

7
-
Application layer: manages the distributed DB or distributed processing.


c)

Different switching techniques used in computer networks:
-

1
-

Circuit switching.

2
-

Message switching (packet switching).

Packet switching

Circuit switching

-
doesn’t setup a physical connection between
sender and receiver in advance.

-

Se瑵p a physical connec瑩on in advance.

-
no was瑥 in bandwid瑨.

-

Bandwid瑨

is was瑥d.

-

䍩rcuit



never
defeca瑩ng


-
circui琠is

defeca瑩ng
.

-
doesn’t depend on cost between two ends.

-
doesn’t depend on traffic.

-
packe瑳 are received ou琠of order.

-
浥ssage received in

order
.

-
IMP providing buffer in 浡in 浥浯ry (in
-
s瑯re
-
and
-
forward
-
rou瑩ng).

-
buffering in secondary s瑯rage for receiving and
sending only.