IGNOU MCS-042 Solved Assignment 2012-13

8/13/2019 IGNOU MCS-042 Solved Assignment 2012-13

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Course Code : MCS-042

Course Title : Data Communication and Computer Network

Assignment Number : MCA (4)/042/Assign/12

Q 1: Assume a data stream is made of 000010 Encode this stream using the

following encoding schemes.

(i) Manchester

(ii) Differential Manchester

(iii) UNI polar

(iv) Polar NR Z-1

(v) RZ

Also discuss the usefulness of each scheme.

Ans:(i)

A method of transmitting bits which enables the receiver to easilysynchronise with the sender.

A simple way of signalling bits might be to transmit a high voltagefor some period for a 1-

bit and a low voltage for a 0 bit:
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Bits Sent: 1 1 0 0

Signal: High ___________ Low |___________

Time: -> . . . . .

However, when several identical bits are sent in succession, thisprovides no information to the re

ceiver about when each bit startsand stops.

Manchester encoding splits each bit period into two, and ensuresthat there is always a transition

between the signal levels in themiddle of each bit. This allows the receiver to synchronise withth

e sender.

In normal Manchester encoding, a 1-

bit is transmitted with a highvoltage in the first period, and a low voltage in the second, andvice

verse for the 0 bit:

Bits Sent: 1 1 0 0

Signal: High __ __ __ __ Low |__| |_____| |__|

Time: -> . ' . ' . ' . ' .

In Differential Manchester encoding, a 1-bit is indicated by makingthe first half of the signal equal to the last half of the previousbit's sign

al and a 0-

bit is indicated by making the first half of thesignal opposite to the last half of the previous bit's s

ignal. That is,a zero bit is indicated by a transition at the beginning of the bit.

Like normal Manchester encoding, there is always a transition inthe middle of the transmission o

f the bit.

Differential Manchester Encoding

Bits Sent: 1 1 0 0

Signal: High ____ __ __ __ Low |_____| |__| |__|

Time: -> . ' . ' . ' . ' .With each bit period half as long, twice as muchbandwidth isrequired when using either of the

Manchester
http://dictionary.reference.com/browse/bandwidthhttp://dictionary.reference.com/browse/bandwidth


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(iii)

Unipolar Coding

The most basic transmission code is unipolar or unbalanced coding. In this scheme each discrete

variable is transmitted with a different assigned level, 0V and for example +2.5V. But this holds

a number of disadvantages:

The average power is two times other bipolar codes

The coded signal contains DC and low frequency components.

When long strings of zeros are present, a DC or baseline wander occurs.

This results in loss of timing and data because a receiver/repeater cannot optimally

discriminate ones and zeros.

Repeaters/receivers require a minimum pulse density for proper timing extraction. Longstrings of ones or zeros contain no timing information and lead to timing jitter (when a

clock recovery is used) and possible loss of synchronization.

There is no provision for line error rate monitoring.


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(iv)

1 = signal on

0 = signal off (no signal)

NRZ is used on low speed links, such as serial ports. Its problems are lack of clock recovery

during long string of 0 or 1 bits and it has a DC component resulting in baseline wander duringlong strings of 0 or 1 bits.

(v)

NRZI Nonreturn to Zero Inverted1 = change of signal level (on-off or off-on)

0 = no change of signal level

NRZI is a differential encoding used in 4B/5B on fast ethernet. It fixes problems in clocking

during long strings of 1 bits. The problems are the DC component and the lack of clock recovery

during long string of 0 bits.

Q 2: (i) Illustrate constellation diagram of 8- PSK and 8- QAM.

Ans:


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(ii) Given a 10 bit sequence frame: 100101001and a divisor (polynomial) of

10011, Find the CRC.

Ans:

A CRC has a generating polynomial of the form:

Gn= dn-1xn-1

+ dn-2xn-2

+ ... + d0x0

For example, for n = 9, a generating polynomial might be:


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G = 1x8+ 0x7 + 0x6 + 1x5 + 1x4 + 0x3 + 1x2 + 0x1+ 1x0

or

G = 100110101

The generating polynomial divides the binary message to determine the remainder which is then

subtracted from the message to get a message that is evenly divided by the generating

polynomial. At the receiving end, if the message divided by the generating polynomial yields a

remainder, it must be corrupted. Both sides have to use the same generating polynomial. Also, so

that no data in the message must be destroyed to subtract the remainder, n-1 zeros are added to

the original message before dividing.

All arithmetic is done in one's complement form, so that there are no carries or borrows. One's

complement arithmetic uses an exclusive or (XOR) to perform both addition and subtraction:

(0+1) = (0-1) = 1.

For example, if the message is 100101001 and the generating polynomial is 10011, then:

1. Append 4 zeros to the message (100101001,0000)

2. Do one's complement polynomial division:

3. 10001 1011

4. ----------------------------

5. 10011 | 100101001,0000

6. 10011

7. --------

8. 11001

9. 10011

10. --------

11. 10100

12. 10011

13. --------

14. 11100

15. 10011

16. --------17. 11110

18. 10011

19. --------

20. 1111 = remainder

21.

22.Subtract the remainder form the message and send the result.


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23.100101001,0000

24. 1111

25.---------------

26.1001010011111 = the message transmitted to the other end

You can check the result by dividing the transmitted message by the generating polynomial. It

will be zero.

Q 3: How are problems of Hidden Station and Exposed Station resolved in

wireless LAN? Explain in detail.

Ans:

There are two fundamental problems associated with a wireless network. Assume that there are

four nodes A, B, C and D. B and C are in the radio range of each other. Similarly A and B are in

the radio range of each other. But C is not in the radio range of A. Now, suppose that there is a

transmission going on between A and B. If C also wants to transmit to B, first, it will sense the

medium but will not listen to As transmission to B because, A is outside its range. Thus, C will

create garbage for the frame coming from A if, it transmits to B. This is called the hidden station

problem. The problem of a station not being able to detect another node because that node is too

far away is called hidden station problem. Now, let us consider the reverse situation called the

exposed station problem.

In this case, B is transmitting to A. Both are within radio range of each other. Now C wants

to transmit to D. As usual, it senses the channel and hears an ongoing transmission and

falsely concludes that it cannot transmit to D. But the fact is transmission between C and D

would not have caused any problems because, the intended receivers C and D are in a


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different range. This is called exposed station problem.

Q 4: Explain the 3-way handshake method. How is it different from 2-way

handshake method?

Ans:

3-Way Handshake

The TCP three-way handshake in Transmission Control Protocol (also called the TCP-

handshake; three message handshake and/or SYN-SYN-ACK) is the method used by TCP set

up a TCP/IP connection over an Internet Protocol based network. TCP's three way

handshaking technique is often referred to as "SYN-SYN-ACK" (or more accurately SYN,

SYN-ACK, ACK) because there are three messages transmitted by TCP to negotiate and start

a TCP session between two computers. The TCP handshaking mechanism is designed so that

two computers attempting to communicate can negotiate the parameters of the network TCPsocket connection before transmitting data such as SSH and HTTP web browser requests.

This 3-way handshake process is also designed so that both ends can initiate and negotiate

separate TCP socket connections at the same time. Being able to negotiate multiple TCP

socket connections in both directions at the same time allows a single physical network

interface, such as ethernet, to be multiplexed to transfer multiple streams of TCP data

simultaneously. Below is a (very) simplified diagram of the TCP 3-way handshake process.

Have a look at the diagram on the right as you examine the list of events on the left.

Synchronize and Acknowledge messages are indicated by a either the SYN bit, or the ACK

bit inside the TCP header, and the SYN-ACK message has both the SYN and the ACK bitsturned on (set to 1) in the TCP header. TCP knows whether the network TCP socket connection

is opening, synchronizing, established by using the Synchronize and Acknowledge messages

when establishing a network TCP socket connection.

When the communication between two computers ends, another 3-way communication is

performed to tear down the TCP socket connection. This setup and teardown of a TCP socket

connection is part of what qualifies TCP a reliable protocol. TCP also acknowledges that data

is successfully received and guarantees the data is reassembled in the correct order.

Note that UDP is connectionless. That means UDP doesn't establish connections as TCP

does, so UDP does not perform this 3-way handshake and for this reason, it is referred to as

an unreliable protocol. That doesn't mean UDP can't transfer data, it just doesn't negotiate

how the connection will work, UDP just transmits and hopes for the best.

Q 5: How does BGP work? How does it solve the Count to Infinity problem?

Ans:


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BGP: The Exterior Gateway Routing Protocol

The purpose of Border Gateway Protocol is to enable two different ASes to exchange routing

information so that, IP traffic can flow across the AS border. A different protocol is needed

between the ASes because the objectives of an interior gateway and exterior gateway routing

protocol are different. Exterior gateway routing protocol such as BGP is related to policy

matters. BGP is fundamentally a distance vector protocol but, it is more appropriately

characterised as path vector protocol.

Instead of maintaining just the cost to each destination, each BGP router keeps track of the path

used .Neighbouring BGP routers, known as BGP peers exchange detailed information along with

the list of ASes on a path to a given destination rather than record cost information.

The main advantage of using BGP is to solve the count to infinity problem which is illustrated.

there are A, B, C, D, E, F, G, H, I, J and K routers.

Now consider Gs routing table. G uses G C D K path to forward a packet to K. As discussed

earlier whenever a router gives any routing information, it provides a complete path.For ex. From A, the path used to send a packet to K is ABCDK

From B-the path used is BCDK

From C-the path used is CGJK

From E-EFGJK

From H-HIJK.

After receiving all the paths from the neighbours, G will find the best route available. It will

outright reject the path from C and E, since they pass through G itself. Therefore, the choice left

is between a route announced by B and H. BGP easily solves count to infinity problems. Now,


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z = (g^x % p)^x' % p = (g^x' % p)^x % p

All of these numbers are positive integers

x^y means: x is raised to the y power

x%y means: x is divided by y and the remainder is returned

Example:

Prime Number p = 353 and primitive root of 353, in this case is g = 3. Let A is sender B is

receives A & B select secret key as XA = 97 and XB = 233. Now A & B computes their public

keys. YA = 3233. Now A & B computes their public keys.

YA = 397 mod 353 = 40

YB = 3233 mod 353 = 248

After exchanging their public keys, A & B can compute the common secret key: A computes:

Z=(YB) mod 253

B computes = 24897 mod 353 = 160

Z = (YA)xbmod353 = 40233 mod 353 = 160

Q 7: What is the utility of digital certificate? How are these signatures

created?

Ans:

This is a certificate issued by the Certifying Authority (Figure 11) to the holder of the public key.

The contents of a digital certificate are issued by a CA as, a data message and are always

available online.

Sr. No of the Certificate

Applicants name, Place and Date of Birth, Name of the Company Applicants legal domicile and virtual domicile

Validity period of the certificate and the signature

CAs name, legal domicile and virtual domicile

Users public key

Information indicating how the recipient of a digitally signed document can verify the senders

public key


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CAs digital signature.

Uses of Digital Signature

Contracts: The next time you purchase a car, a home, or an insurance policy, you may never

need to meet with an agent or sales representative. You may be able to review and sign alldocuments online, and save secure backup copies to your own disk.

Checks and money orders: Buying online is now easy with a credit card, but digital checks or

money orders (authenticated by secure digital signatures) may be preferable for some

transactions,

especially when you dont want to face a large credit card bill.

Letters and memos: Businesses already transmit many letters and memos online, especially

those that are only distributed internally. But when a letter or memo needs the weight of a

managers signature, it must be printed, signed, duplicated, and distributed manually or through

the mail. Digital signatures will save companies the time and expense of this manual process.

Approvals: Many kinds of documents are collaborative works, such as legal briefs, contracts,

reports, and others. Using digital signatures, people can collaborate on documents online and

approve final drafts, before/prior to releasing them for use.

CREATION OF DIGITAL SIGNATURES

Ravi works at the companys certificate authority centre. Ravi can create a digital certificate for

Ram by using Rams public key as well as some information Ram . Rams co-workers can

verify Rams trusted certificate to make sure that his public key truly belongs to him. In fact, no

one at Rams company accepts a signature for which there does not exist a certificate generated

by Ravi. This gives Ravi the power to revoke signatures if private keys are compromised, or nolonger needed. There are even more widely accepted certificate authorities that certify Ravi.

If Ram sends a signed document to Shayam, to verify the signature on the document,

Shayams software first uses Ravis (the certificate authoritys) public key to check the

signature on Rams certificate. Successful de-encryption of the certificate proves that Ravi

created it. After the certificate is de-encrypted, Shayams software can check if Ram is in

good standing with the certificate authority and that the certificate information concerning Rams

identity has not been altered. Shayams software then takes Rams public key from the

certificate and uses it to check Rams signature. If Ram's public key de-encrypts the signature

successfully, then Shayam is assured that the signature was created using Rams private key, for

Ravi has certified the matching public key. And of course, if the signature is valid, then we knowthat Mohan didnt try to change the signed content.

Q 8: Differentiate between the following:

(i) Leaky Bucket Traffic Shaper and Token bucket Traffic Shaper


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Ans:

An important difference between two traffic shaping algorithms: token bucket throws away

tokens when the bucket is full but never discards packets while leaky bucket discards packets

when the bucket is full. Unlike leaky bucket, token bucket allows saving, up to maximum size of

bucket n. This means that bursts of up to n packets can be sent at once, giving faster response tosudden bursts of input. Leaky bucket forces bursty traffic to smooth out, token bucket permits

burstiness but bounds it. Token bucket has no discard or priority policy. Token bucket when

compared to leaky bucket, is easy to implement. Each flow needs just a counter to count tokens

and a timer to determine when to add new tokens to the counter.

A token bucket flow is defined by (r,b), r denotes the rate at which tokens(credits) are

accumulated and b is the depth of the token pool(in bytes).

As it shown above the new token are adding to the bucket at rate of r tokens/sec, the maximum

token can be accumulated is b bytes. If the bucket is full, the incoming tokens will be thrown

remove

token

b

r

tokens/sec

ackets

TokenTo network


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away. The Token Bucket(TB) profile contains three parameters: an average rate, a peak rate, and

burst size.

(ii) Distance vector routing and Link state routing

Ans:

Distance vector routing and Link state routing

"Distance Vector" and "Link State" are terms used to describe routing protocols which are

used by routers to forward packets between networks. The purpose of any routing

protocol is to dynamically communicate information about all network paths used to

reach a destination and to select the from those paths, the best path to reach a destination

network. The terms distance vector and link state are used to group routing protocols into

two broad categories based on whether the routing protocol selects the best routing path

based on a distance metric (the distance) and an interface (the vector), or selects the best

routing path by calculating the state of each link in a path and finding the path that has the

lowest total metric to reach the destination.

Distance is the cost of reaching a destination, usually based on the number of hosts the

path passes through, or the total of all the administrative metrics assigned to the links in

the path. From the standpoint of routing protocols, the vector is the interface traffic will be

forwarded out in order to reach an given destination network along a route or path

selected by the routing protocol as the best path to the destination network.

Distance vector protocols use a distance calculation plus an outgoing network interface (a

vector) to choose the best path to a destination network. The network protocol (IPX, SPX,IP, Appletalk, DECnet etc.) will forward data using the best paths selected.

Common distance vector routing protocols include:

Appletalk RTMP

IPX RIP

IP RIP

IGRP

Advantages of Distance Vector Protocols

Well Supported

Protocols such as RIP have been around a long time and most, if not all devices that

perform routing will understand RIP.

LINK STATE

Link State protocols track the status and connection type of each link and produces a

calculated metric based on these and other factors, including some set by the network

administrator. Link state protocols know whether a link is up or down and how fast it is

and calculates a cost to 'get there'. Since routers run routing protocols to figure out how to


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get to a destination, you can think of the 'link states' as being the status of the interfaces

on the router. Link State protocols will take a path which has more hops, but that uses a

faster medium over a path using a slower medium with fewer hops.

Because of their awareness of media types and other factors, link state protocols require

more processing power (more circuit logic in the case of ASICs) and memory. Distance

vector algorithms being simpler require simpler hardware.

A Comparison: Link State vs. Distance Vector

See Fig. 1-1 below. If all routers were running a Distance Vector protocol, the path or

'route' chosen would be from A B directly over the ISDN serial link, even though that

link is about 10 times slower than the indirect route from A C D B.

A Link State protocol would choose the A C D B path because it's using a faster

medium (100 Mb ethernet). In this example, it would be better to run a Link State routing

protocol, but if all the links in the network are the same speed, then a Distance Vector

protocol is better.

------------------------------------------------- THE END -----------------------------------------------------

IGNOU MCS-042 Solved Assignment 2012-13

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