2016 question bank Computer Networks.doc

8/17/2019 2016 question bank Computer Networks.doc

1/17

SHANMUGHA ARTS, SCIENCE, TECHNOLOGY & RESEARCH ACADEMY

(SASTRA UNIVERSITY)

1. What are the five im!rta"t #!m!"e"t$ !f %ata #!mm"i#ati!"'

Transmitter, Receiver, Medium, data and protocol.

(. What are the three f"%ame"ta) #hara#teri$ti#$ %etermi"e the effe#tive"e$$ !f the %ata#!mm"i#ati!" $*$tem'

Te e!!ectiveness o! te data communication s"stem depends on tree !undamental

caracteristics#

$eliver"# Te s"stem must deliver data to te correct destination.

Accurac"# Te s"stem must deliver data accuratel".Timeliness# Te s"stem must deliver data in a timel" manner.

+. Defi"e r!t!#!).

A net%or& protocol de!ines rules and conventions !or communication 'et%een net%or&

devices. rotocols !or computer net%or&in all enerall" use pac&et s%itcin tecni*ues to send

and receive messaes in te !orm o! pac&ets.

. What i$ %ata-ram "et!r/'

A dataram is a 'asic trans!er unit associated %it a pac&et+s%itced net%or& in %ic

te deliver", arrival time and order are not uaranteed. A dataram consists o! eader and dataareas, %ere te eader contains in!ormation su!!icient !or routin !rom te oriinatin

e*uipment to te destination %itout rel"in on prior ecanes 'et%een te e*uipment and te

net%or&.

0. Defi"e $it#hi"-.

A net%or& s%itc is a small ard%are device tat -oins multiple computers toeter

%itin one local area net%or& (AN). Tecnicall", net%or& s%itces operate at la"er t%o ($atain& a"er) o! te /SI model. Tere are t%o !undamental approaces to%ards 'uildin a

net%or& core# circuit s%itcin and pac&et s%itcin.

. Defi"e #!mter "et!r/$.

A computer net%or&, o!ten simpl" re!erred to as a net%or&, is a collection o! computers

and devices connected '" communications cannels tat !acilitates communications amon osts(users) and allo%s osts to sare resources %it oter osts.

2. What i$ a "!%e'

In data communication, a p"sical net%or& node ma" eiter 'e a data circuit+terminatin

e*uipment ($0E) suc as a modem, u', 'ride or s%itc1 or a data terminal e*uipment ($TE)

suc as a diital telepone andset, a printer or a ost computer, !or eample a router, a%or&station or a server. I! te net%or& is te Internet, man" p"sical net%or& nodes are ost

computers, also &no%n as Internet nodes, identi!ied '" an I address, and all osts are p"sical

net%or& nodes. 2o%ever, data lin& la"er devices suc as s%itces, 'rides and 3AN access

points do not ave an I ost address (ecept sometimes !or administrative purposes), and arenot considered as Internet nodes, 'ut as p"sical net%or& nodes or AN nodes.


2/17

3. Defi"e m)ti)e4i"-.

In computer net%or&s, multiplein diital data streams are com'ined into one sinal

over a sared medium. Te aim is to sare an epensive resource. Multiplein at send ost isaterin data !rom multiple soc&ets, envelopin data %it eader later used !or demultiplein

%ic is deliverin received sements to te correct soc&ets at receive ost.

5. State a"* f!r me#ha"i$m$ t! a#hieve re)ia6)e %ata tra"$fer.

i) A04 5 NA4 ii) Timeout event iii) 0ec&sum iv) Retransmission v) Se*uence num'er

17. What are the /e* %e$i-" i$$e$ !f a #!mter Net!r/'

a. 0onnectivit" '. 0ost+e!!ective Resource Sarin

c. Support !or common Services d. er!ormance

11. Name the fa#t!r$ that affe#t the erf!rma"#e !f the "et!r/.

Num'er o! Users, T"pe o! transmission medium, 2ard%are, So!t%are

1(. What i$ the r!)e !f #!"#e"trat!r$ i" "et!r/i"-'

It com'ines multiple cannels onto a sinle transmission medium in suc a %a" tat all

te individual cannels can 'e simultaneousl" active. 6or eample, ISs use concentrators tocom'ine teir dial+up modem connections onto !aster T+7 lines tat connect to te Internet.

0oncentrators are also used in local+area net%or&s (ANs) to com'ine transmissions !rom a

cluster o! nodes.

1+. 8!r " %evi#e$ i" a "et!r/, hat i$ the "m6er !f #a6)e )i"/$ re9ire% f!r a me$h a"% ri"-

t!!)!-*'

Mes topolo" 8 n (n+7)9:. Rin topolo" 8 n

1. What are the f"#ti!"$ !f a DTE' What are the f"#ti!"$ !f a DCE'$ata terminal e*uipment is a device tat is an in!ormation source or an in!ormation sin&. It is

connected to a net%or& trou a $0E. $ata circuit+terminatin e*uipment is a device used asan inter!ace 'et%een a $TE and a net%or&.

10. Li$t !t the %i$a%va"ta-e$ !f ara))e) %ata tra"$mi$$i!".

o Re*uires separate lines !or eac 'it o! a %ord

o 0ostl" to run lon distances due to multiple %ires

o Su!!ers !rom electromanetic inter!erence

o 0a'le lents more limited tan a serial ca'le

1. Name the fa#t!r$ that affe#t the erf!rma"#e !f the "et!r/.

a. Num'er o! Users

'. T"pe o! transmission mediumc. 2ard%are

d. So!t%are

12. Differe"tiate %ata-ram a"% virta) #ir#it $it#hi"-.


3/17

S.No $ataram ac&et S%itcin Virtual 0ircuit ac&et S%itcin

7 Route esta'lised !or eac pac&et Route esta'lised !or entire conversation

: ac&et transmission dela" 0all setup dela"1 pac&et transmission dela"

; Net%or& ma" 'e responsi'le !or individual pac&ets

Net%or& ma" 'e responsi'le !or pac&etsse*uences

UNIT : II ; C!mter Net!r/$

1.What is the need for frame relay?It is a connection-oriented network with no error control and no fowcontrol. Because it was connection-oriented, packets were delivered inorder (i they were delivered at all). The properties o in-order deliverymake rame relay akin to a wide area L!. Its most importantapplication is interconnectin" L!s at multiple company o#ces.

2.Under light load which LAN has a smaller delay Ethernet or tokenring ? Explain

Answer:

(a) Ethernet has smaller delay under a light load. In Ethernet under a light

load, there is little or no contention for the channel, the delay incurred is close

to the frame transmission time. In token ring, however, there is always the

additional delay incurred from circulating the token around the ring.

(b) Token ring has smaller delay under a high load. In Ethernet there is more

contention for the channel, much of the time is spent in collision and backoff,

so on average the frames experience longer delay and higher delay variability.

In comparison, token ring provides each station with an orderly and round-

robin access to the channel by passing the token around. When the number of

frames transmitted per token is limited, and frames are kept at a fixed length,

token ring can guarantee a maximum delay for each station.

3.A scanner has a resolution of 600 x 600 pixels/suare inch! "owmany #its are produced #y an $%inch x &0inch image ifscanning uses $ #its/pixel ? '( #its/pixel ?

4. Why does a conventional telephone still work when the electrical power is

out?


4/17

Answer: The telephone company supplies each of its telephone lines with power

at the central office. This power is stored in the form of wet batteries that can

alternately be charged by a backup battery in the event of power failure at the

central office. These huge batteries occupy entire floors in telephone offices.

5. Suppose the signal has twice the power as a noise signal that is added to it.Find the SNR in decibels. Repeat if the signal has 10 times the noise

power? 2n times the noise power? 10k times the noise power?

Answer:

SNR dB = 10log10 x2/n

2 = 10log102 = 3.01 dB

SNR dB = 10log1010 = 10 dB

SNR dB = 10log102n = 10n log102 = 3.01n dB

SNR dB = 10log1010k = 10k dB

6. A 10 kHz baseband channel is used by a digital transmission system. Ideal

pulses are sent at the Nyquist rate and the pulses can take 16 levels. What is

the bit rate of the system?

Answer: Nyquist pulses can be sent over this channel at a rate of 20000 pulses

per second. Each pulse carries log216 = 4 bits of information, so the bit rate is

80000 bits per second.

7.$ompare and contrast %.&' and rame elay networks.

%.&' rame elay*rror $orrection !ode to !ode !one+ropa"ationelay

i"h Low

*ase oImplementation

i#cult *asy

ood or

Interactivepplications

Too /low 0es

ood or 1oice !o 0esood or L! ile

Transer/low 0es

8.)tate the characteristics of A*+ networks.


5/17

9.,hat are the di-erent types of ser.ices o-ered #y A*+ networksa. $B ($onstant Bit ate)2. 1B (1aria2le Bit ate)c. B (vaila2le Bit ate) low $ontrol, ate-2ased, $redit-2asedd. 3B (3nspeci4c Bit ate) !o low control.

10. ,hat were the maor applications of computer networks?5. Business pplications&. ome pplications6. 7o2ile 3sers

11. Give any two networks which will work under lower layers of OSI.


6/17

• 188.69.44.3

• PDF document

• 5-5-5-5-5-5

• TCP 80

• 10.6.9.7

• Bit pattern

Answer: Layer 4, Layer 3, Layer 7, Layer 2, Layer 4, Layer 3, Layer 1

' Whi#h !f the f!))!i"- term$ i$ $e% $e#ifi#a))* t! i%e"tif* the e"tit* that i$ #reate% he"e"#a$)ati"- %ata i"$i%e %ata )i"/ )a*er hea%er$ a"% trai)er$'

a. $ata

'. 0un&

c. Sementd. 6rame

e. ac&et

!. None o! tese=tere is no encapsulation '" te data lin& la"er

1 Differe"tiate 6etee" Li"/ $tate a"% Di$ta"#e >e#t!r R!ti"- a)-!rithm.

( 2e3ne 4outers

outers relay packets amon" multiple interconnected networks. They outepackets rom one network to any o a num2er o potential destinationnetworks oninternet routers operate in the physical, data link and network layer o 8/Imodel.

5 ,hat is meant #y hop count? The pathway re9uirin" the smallest num2er o relays, it is called hop-countroutin", in which every link is considered to 2e o e9ual len"th and "iven thevalue one. 5'. ow can the routin" 2e classi4ed:

The routin" can 2e classi4ed as, daptive routin"

!on-adaptive routin".

6 ,hat is time%to%li.e or packet lifetime?s the time-to-live 4eld is "enerated, each packet is marked with a lietime,usually the num2er o hops that are allowed 2eore a packet is considered


7/17

lost andaccordin"ly, destroyed. The time-to-live determines the lietime o a packet.

,hat is meant #y #router? 2router is a sin"le protocol or multiprotocol router that sometimes act as arouter and sometimes act as a 2rid"e.

$ ,rite the keys for understanding the distance .ector routing The three keys or understandin" the al"orithm are

;nowled"e a2out the whole networks

outin" only to nei"h2ors

Inormation sharin" at re"ular intervals

7 ,rite the keys for understanding the link state routing The three keys or understandin" the al"orithm are

;nowled"e a2out the nei"h2orhood. outin" to all nei"h2ors.

Inormation sharin" when there is a ran"e.

&0 "ow the packet cost referred in distance .ector and link staterouting?In distance vector routin", cost reer to hop count while in case o link stateroutin", cost is a wei"hted value 2ased on a variety o actors such assecurity levels, tra#cor the state o the link.

&& "ow the routers get the information a#out neigh#or? router "ets its inormation a2out its nei"h2ors 2y periodically sendin" themashort "reetin" packets. I the nei"h2orhood responds to the "reetin" ase


8/17

13.Identify whether each of the following IP addresses can be assigned to a

computer.

A. 192.168.5.6

B. 10.1.0.1

C. 172.20.0.0

D. 192.168. 200.255

E. 1.1.1.1

F. 10.0.0.0

G. 172.10.255.255

H. 172.61.9.8

I. 16.16.55.9

J. 127.6.1.3

Answer:

A. Yes.

B. Yes.

C. No. This is the network ID for the 172.20.0.0 network.

D. No. This is the broadcast address for the 192.168.200.0 network.

E. Yes.

F. No. This is the network ID for the 10.0.0.0 network.

G. No. This is the broadcast address for the 172.10.0.0 network.

H. Yes.

I. Yes.

J. No. The 127.0.0.0 network is reserved for network diagnostics.

14.For each of the following IP addresses, identify the network ID of the

network in which it lies.

A. 10.99.57.15/24

B. 192.168.6.9/30

C. 17.69.48.3/15

D. 56.57.58.60/22

E. 172.156.30.14/27

Answer:

A. 10.99.57.0/24

B. 192.168.6.8/30

C. 17.68.0.0/15

D. 56.57.56.0/22

E. 172.156.30.0/27


9/17

16.2e3ne+asking?7askin" is the process that e


10/17

*nd-to- end deliveryddressin"elia2le deliverylow control7ultiple. ,hat are the four aspects related to the relia#le deli.ery of data?*he four aspects are!*rror control/e9uence controlLoss controluplication control

'. ,hat is meant #y segment?t the sendin" and receivin" end o the transmission, T$+ divides lon"transmissions into smaller data units and packa"es each into a rame calleda

se"ment.

?. ,hat is meant #y segmentation?@hen the siAe o the data unit received rom the upper layer is too lon" orthe network layer data"ram or data link layer rame to handle, the transportprotocol divides it into smaller usa2le 2locks. The dividin" process is calledse"mentation.

>. S!$e a fi)e !f 17,777 6*te$ i$ t! 6e $e"t !ver a )i"e at (77 6$.

?a@ Ca)#)ate the !verhea% i" 6it$ a"% time i" $i"- a$*"#hr!"!$ #!mm"i#ati!". A$$me !"e

$tart

6it a"% a $t! e)eme"t !f )e"-th !"e 6it, a"% 3 6it$ t! $e"% the 6*te it$e)f f!r ea#h #hara#ter. The 3=

6it #hara#ter #!"$i$t$ !f a)) %ata 6it$, ith "! arit* 6it.


11/17

?6@ Ca)#)ate the !verhea% i" 6it$ a"% time i" $i"- $*"#hr!"!$ #!mm"i#ati!". A$$me thatthe %ata are $e"t i" frame$. Ea#h frame #!"$i$t !f 1777 #hara#ter$ 3777 6it$ a"% a" !verhea% !f

3 #!"tr!) 6it$ er frame.

?#@ What !)% 6e the a"$er$ t! art$ ?a@ a"% ?6@ 6e f!r a fi)e !f 177,777 #hara#ter$'

?%@ What !)% 6e a"$er$ t! art$ ?a@ a"% ?6@ 6e f!r the !ri-i"a) fi)e !f 17,777 #hara#ter$

e4#et at a %ata rate !f 577 6$'

Answer:

a) For each character(8 bits), we add 2 bits (start and stop).

So, each character has 25% overhead. For 10,000 characters, there are 20,000 extra

bits. This would take an extra 20,000/2400 = 8.33 seconds

b) The file takes 10 frames or 480 additional bits. The transmission time for the

additional bits is 480/2400 = 0.2 seconds.

BIG QUESTIONS

Question 1: (7 Marks)

Design (show only the network) a LAN system for small five-story building. One floor

is dedicated to two mail servers and separated three database servers. Each of

remaining floor has four computers with broadband access. Your design should meet

the following restrictions and conditions: three-input hubs, one bridge, and unlimited

Ethernet buses. The incoming broadband Internet access must be connected to a six-

repeater ring, no bus LAN is allowed outside of a floor, and a traffic analyzer must be

attached to the network.

Answer:


12/17

____________________________________________________________________________

Question 2 : (5 Marks)

What is the length of contention slot in CSMA/CD for (a) a 2-km twin-lead cable

(signal propagation speed is 82% of the signal propagation speed in vacuum)?, and

(b) a 40-km multimode fiber optic cable (signal propagation speed is 65% of the signal

propagation speed in vacuum)?

Answer:

(a) Signal propagation speed in twin lead is 2.46 × 108 m/sec. Signal propagation

time for 2 km is 8.13 μsec. So, the length of contention slot is 16.26 μsec. (b) Signal

propagation speed in multimode fiber is 1.95 × 108 m/s. Signal propagation time for

40 km is 205.13 μsec. So, the length of contention

slot is 410.26 μsec.

3. Consider a wide area network in which two hosts, A and B, are connected

through a 100 km communication link with the data speed of 1 Gb/s. Host A wants to transfer the content of a CD-ROM with 200 Kb of music data

while host B reserves portions of its 10 parallel buffers, each with the capacity

of 10,000 bits. The host A sends a SYN segment, where ISN = 2,000 and MSS =

2,000 and that host B sends ISN = 4,000 and MSS = 1,000. Sketch the

sequence of segment exchange, starting with host A sending data at time t = 0.

Assume host B sends ACK every five frames.


13/17

Answer:

Figure shows the operation.

ach packet size is 1000 bytes since that’s the MSS of host B. Data per packet is 1000 –

0 – 20 = 960 bytes. (i.e., 20 bytes for TCP header and 20 bytes for IP header).or Host A: MSS = 2000, ISN = 2000, File Size = 200 Kb = 25 KB.

or Host B: MSS = 1000, ISN = 4000.

Three stages in file transfer. Connection establishment, Segment transfer, and

Connection termination.

Question 3: (3 or 4 Marks)

Why do LANs tend to use broadcast networks? Why not use networks consisting of

multiplexers and switches?

Answer:

The computers in a LAN are separated by a short distance (typically


14/17

network interface cards in each computer. In addition, the LAN users usually belong

to the same group where all users are generally trusted, so broadcast does not pose

much security danger.

The original reason for avoiding a multiplexer and switch approach to LANs is that

centralized, a expensive “box” is required. The availability of Application Specific

Integrated Circuits (ASICs) has reduced the cost of switching boxes and made switch-

based LANs feasible, and in some environments the dominant approach.______________________________________________________________________________


Suppose that the ALOHA protocol is used to share a 56 kbps satellite channel.

Suppose that frames are 1000 bits long. Find the maximum throughput of the

system in frames/second.

Answer:

Maximum throughput for ALOHA = 0.184

Max. throughput in frames/sec =

(56000 bits/sec) x (1 frame/1000 bits) x 0.184 = 10.304

The maximum throughput is approximately 10 frames/sec.


You need to create 652 networks with the class B address 150.150.0.0.

a. What is the subnet mask?

b. List the first three valid network numbers.

c. List the range of host IP addresses on those three networks.

d. List the last valid network and range of IP addresses.

e. How many subnets does this solution allow?

f. How many host addresses can be on each subnet? Answer:

a. 255.255.255.192

b. 150.150.0.64, 150.150.0.128, 150.150.0.192

c. 150.150.0.65 – 150.150.0.126

150.150.0.129 – 150.150.0.190

150.150.0.193 – 150.150.0.254

d. Network: 150.150.255.128

Range of IP addresses: 150.150.255.129 – 150.150.255.190

______________________________________________________________________________

Question 5:

11. Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps.

(a) Calculate the overhead in bits and time in using asynchronous communication.

Assume one start

bit and a stop element of length one bit, and 8 bits to send the byte itself for each

character. The 8-bit character consists of all data bits, with no parity bit.


15/17

(b) Calculate the overhead in bits and time in using synchronous communication.

Assume that the data are sent in frames. Each frame consist of 1000 characters =

8000 bits and an overhead of 48 control bits per frame.

(c) What would be the answers to parts (a) and (b) be for a file of 100,000

characters?

(d) What would be answers to parts (a) and (b) be for the original file of 10,000

characters except at a data rate of 9600 bps? Answer:

a) For each character(8 bits), we add 2 bits (start and stop).

So, each character has 25% overhead.

For 10,000 characters, there are 20,000 extra bits.

This would take an extra 20,000/2400 = 8.33 seconds

b) The file takes 10 frames or 480 additional bits. The transmission time for the

additional bits is 480/2400 = 0.2 seconds.

c) Ten times as many extra bits & ten times as long for both.

d) The number of overhead bits would be the same, and the time would be decreased

by a factor of 4=9600/2400.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

uestion 6:

(a) In a CRC error-detecting scheme, choose P(x) = x4 + x + 1. Encode the bits

10010011011.

(b) Suppose the channel introduces an error pattern 100010000000000 (i.e., a

flip from 1 to 0 or from 0 to 1 in position 1 and 5). What is received? Can the error be

detected?

Answer:) Divide x10 + x7 + x4 + x3 + x + 1 by x4 + x + 1. The remainder is x3 + x2.

he CRC bits are 1100. The string 100100110111100 is sent.

) 100100110111100⊕ 100010000000000 = 000110110111100.

The string 000110110111100 is received, corresponding to x11+x10+x8+x7+x5+x4+x3+x2.

The remainder after division by x4+x+1 is x3+x2+x, which is nonzero. The errors are

detected.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


16/17


17/17

Documents

2016 question bank Computer Networks.doc