CS601-Data Communication Subjective Questions Solved For FinalTerm Papers

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Question No: 41 ( Marks: 2 )
What are the conditions for the polynomial used by the CRC generator?
CRC generator:
CRC generator (the divisor) is most often represented not as a1’s and 0’s but as an algebraic polynomial.
conditions for the polynomial:
it should have following properties:
It should not be divisible by “x”.
It should not be divisible by “x+1”.
The first condition guarantees that all burst error of a length equal to degree of the polynomial is detected.
The 2nd condition guarantees that all burst error affecting an odd number of bits are detected.

Question No: 42 ( Marks: 2 )
What are intelligent modems?
Intelligent modems:
A modem that responds to commands and can accept new instructions during online transmission. It was originally developed by Hayes.
Example:
• Automatic answering,
• Dialing etc.
Question No: 43 ( Marks: 2 )
What is the basic purpose of Router?
Basic purpose of Router:
"A router is a device that extracts the destination of a packet it receives, selects the best path to that destination, and forwards data packets to the next device along this path. They connect networks together;
a LAN to a WAN for example, to access the Internet. 
"A more precise definition of a router is a computer networking device that interconnects separate logical subnets."
Question No: 44 ( Marks: 3 )
What are the fractional T Lines?
http://www.allvupastpapers.blogspot.com/The fractional T Lines:
Many subscribers don’t need the entire capacity of the T-line. 
For example,
A small business may need only one-fourth of the capacity of T-line. if four business of same size lie in the same building, they can share T-line.DSU/CSU allow the capacity of T-line to be interleaved in to four channels

Question No: 45 ( Marks: 3 )
What are the light sources used for optic fiber?
• light sources used for optic fiber:
• The light source can weather be an LED or ILD 
• LED (Light emitting diode) cheaper but provide unfocused light that strikes the boundaries of channel at uncontrollable angles.
• Limited to short distance use.
• LASSER 
• Can be focused to a narrow range allowing control over angle of incidence.

Question No: 46 ( Marks: 3 )
What is Multi Access Unit (MAU) in Token Ring?
Multi Access Unit (MAU) in Token Ring:
• Individual automatic switches are combined in to a hub
• One MAU can support up to 8 stations.
• Although it looks like a star, it is in fact a ring.


Question No: 47 ( Marks: 5 )
Give characteristics of Dual Ring, if necessary then draw the diagram. [5]
Characteristics of Dual Ring:
A network topology in which two concentric rings connect each node on a network instead of one network ring that is used in a ring topology. Typically, the secondary ring in a dual-ring topology is redundant. It is used as a backup in case the primary ring fails. In these configurations, data moves in opposite directions around the rings. Each ring is independent of the other until the primary ring fails and the two rings are connected to continue the flow of data traffic.


Question No: 48 ( Marks: 5 )
What the receiver will receive if the checksum method is applied to the following 
bit.
10101001 00111001
Ans:
the receiver will receive the checksum method is applied to the following bit.
10101001 00111001
10101001 00111001
Sum of 2 bits are
10101001
00111001
-------------------
11100010
00011101 1's complement
1
--------------
00011110 2's complement
-----------
10101001 00111001 ==> 00011110

www.allvupastpapers.blogspot.comSo the data transmitted which will receiver get:
10101001 00111001 00011110
Question No: 49 ( Marks: 5 )
What is rafraction in terms of optic fiber? Give one example.
Refraction:
Light travels in a straight line as long as it is moving through a single uniform structure If a ray of light traveling through one substance enters another (more or less dense) substance, its speed changes abruptly causing the ray to change direction. This phenomenon is called Refraction.
Refraction in terms of optic fiber:
the propagation of light in an optical fiber which in its simplest form consists of a circular core of uniform refractive index surrounded by a cladding of slightly lower refractive index. The light is launched into the entrance face of the fiber. 
The light is propagated by the total internal reflection at the interface between core and cladding. However the rays incident at angles larger than a certain angle, called the cut-off angle, suffer both refraction and reflection at the interface between the core and the cladding. 
They, therefore, are not guided. Due to this the optical fiber has a numerical aperture. The numerical aperture is given by the square root of (n12-n22). Typical values of numerical aperture lie between 0.1 and 0.3.
The refractive indices of the core and the cladding are n1 and n 2 respectively. The fiber is normally in air (n0=1) but could also be in a medium of refractive index n0.

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Question No: 50 ( Marks: 10 )
What are the asynchronous protocols in data link layer? Discuss in detail with examples. [10 marks]
Asynchronous protocols in data link layer:
Asynchronous communication at the data link layer or higher protocol layers is known as statistical multiplexing or packet mode communication,
For example :
Asynchronous transfer mode (ATM). In this case the asynchronously transferred blocks are called data packets, 
Async protocols in Data link layer is called statistical multiplexing. for example ATM cells.
The opposite is circuit switched communication, which provides constant bit rate, for example ISDN and SONET/SDH.
The packets may be encapsulated in a data frame, with a frame synchronization bit sequence indicating the start of the frame, and sometimes also a bit synchronization bit sequence, typically 01010101, for identification of the bit transition times. Note that at the physical layer, this is considered as synchronous serial communication.
Examples of packet mode data link protocols that can be/are transferred using synchronous serial communication are the 
• HDLC,
• Ethernet,
• PPP and 
• USB protocols.

Question No: 51      ( Marks: 2 )

What is the formula to calculate the number of redundancy bits required to correct a bit error in a given number of data bits? [2]

Messages(frames) consist of m data (message) bits, yielding an n=(m+r)-bit codeword.

Question No: 52      ( Marks: 2 )

What is R G rating of coaxial cable?

Different coaxial cable designs are categorized by their Radio government (

RG ) ratings

Each cable defined by RG rating is adapted for a specialized function:

RG-8

www.allvupastpapers.blogspot.com·        Used in Thick Ethernet

RG-9

·        Used in Thick Ethernet

RG-11

·        Used in Thick Ethernet

RG-58

·        Used in Thin Ethernet

RG-59

·        Used for TV

Question No: 53      ( Marks: 2 )

What are the advantages of thin ethernet?

The advantages of thin Ethernet are :

·        reduced cost and

·        ease of installation

Because the cable is lighter weight and more flexible than that used in Thicknet

Question No: 54      ( Marks: 3 )

What is the difference between a unicast, multicast, and broadcast address? [3]

Three methods can be used to transmit packets over a network: unicast, multicast, and broadcast.

Uhttp://www.allvupastpapers.blogspot.com/nicast involves communication between a single sender and a single receiver. This is a type of point-to-point transmission; since the packet is transmitted to one destination at a time.

Multicast is used to send packets to a group of addresses, represented by a "group address." In this case, packets are transmitted from a single sender to multiple receivers. Since the same data packet can be sent to multiple nodes by sending just one copy of the data, the load of the sender and the overall load of the network are both reduced.

Broadcast involves sending packets to all nodes on a network simultaneously. This type of transmission is used to establish communication with another host, and for DHCP type methods of assigning IP addresses.

Question No: 55      ( Marks: 3 )

T lines are designed for Digital data how they can be used for Analog Transmission ?

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T Lines are digital lines designed for digital data however; they can also be used for analog transmission (Telephone connections). Analog signals are first sampled and the Time Multiplexed.

Question No: 56      ( Marks: 3 )

What are the three types of Guided Media?

Guided Media, are those media that provide a conduit from one device to another. Three types are

1.      Twisted pair cable

2.      Coaxial cable

3.      Fiber-optic Cable

Question No: 57      ( Marks: 5 )

Why do we need Inverse Multiplexing? [5]

Data & Video can be broken into smaller portions using Inverse Multiplexing and TX. An inverse multiplexer (often abbreviated to "inverse mux" or "imux") allows a data stream to be broken into multiple lower data rate communication links. An inverse multiplexer differs from a demultiplexer in that each of the low rate links coming from it is related to the others and they all work together to carry their respective parts of the same higher rate data stream. By contrast, the output streams from a demultiplexer may be completely independent from each other and the demultiplexer does not have to understand them in any way.

This is the opposite of a multiplexer which creates one high speed link from multiple low speed ones.

It can lease a 1.544 Mbps line from a common carrier and only use it fully for

sometime

Or it can lease several separate channels of lower data rates

Voice can be sent over any of these channels

Question No: 58      ( Marks: 5 )

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Describe method of checksum briefly?

The sender subdivides data units into equal segments of ‘n’ bits(16 bits).These segments are added together using one’s complement. The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM. The extended data unit is transmitted across the network. The receiver subdivides data unit as above and adds all segments together and complement the result. If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero. If the result is not zero, the packet contains an error & the receiver rejects it

Question No: 59      ( Marks: 10 )

Explain Asynchronous Time Division Multiplexing in detail? Also discuss its advantages over synchronous TDM?

Asynchronous time-division multiplexing (ATDM) is a method of sending information that resembles normal TDM, except that time slots are allocated as needed dynamically rather than preassigned to specific transmitters. ATDM is more intelligent and has better bandwidth efficiency than TDM.

Time-division multiplexing (TDM) is a type of digital or (rarely) analog multiplexing in which two or more signals or bit streams are transferred apparently simultaneously as sub-channels in one communication channel, but are physically taking turns on the channel. The time domain is divided into several recurrent timeslots of fixed length, one for each sub-channel. A sample byte or data block of sub-channel 1 is transmitted during timeslot 1, sub-channel 2 during timeslot 2, etc. One TDM frame consists of one timeslot per sub-channel. After the last sub-channel the cycle starts all over again with a new frame, starting with the second sample, byte or data block from sub-channel 1, etc.

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asynchronous time-division multiplexing comprising receive circuits (CRl/i) supplying cells received via input links, transmit circuits (CTl/j) transmitting retransmitted cells on output links, a buffer memory (MT) storing the received cells and delivering the cells to be retransmitted and a buffer memory addressing device (SMT) including a write address source (SAE) and a read address source (fsl/j).

The switching unit further comprises a write disabling circuit (pi) conditioned by a signal (adl) derived from the content of at least one received cell or a signal (tle) derived from the absence of any received cell and supplying a disabling signal (spi) and the address source includes a disabling device (pac, pal) influenced by the disabling signal (spi) so that no memory location is then occupied in the buffer memory (MT).

Advantages asynchronous TDM:

In asynchronous TDM, the timeslots are not fixed. They are assigned dynamically as needed.

In order to reduce the communications costs in time-sharing systems and multicomputer communication systems, multiplexing techniques have been introduced to increase channel utilization. A commonly used technique is Synchronous Time Division Multiplexing (STDM). In Synchronous Time Division Multiplexing, for example, consider the transmission of messages from terminals to computer, each terminal is assigned a fixed time duration. After one user's time duration has elapsed, the channel is switched to another user. With synchronous operation, buffering is limited to one character per user line, and addressing is usually not required. The STDM technique, however, has certain disadvantages. As shown in Figure 1, it is inefficient in capacity and cost to permanently assign a segment of bandwidth that is utilized only for a portion of the time. A more flexible system that efficiently uses the transmission facility on an "instantaneous time-shared" basis could be used instead. The objective would be to switch from one user to another user whenever the one user is idle, and to asynchronously time multiplex the data. With such an arrangement, each user would be granted access to the channel only when he has a message to transmit. This is known as an Asynchronous Time Division Multiplexing System (ATDM). A segment of a typical ATDM data stream is shown in Figure 2. The crucial attributes of such a multiplexing technique are:
1. An address is required for each transmitted message, and
2. Buffering is required to handle the random message arrivals.

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Question No: 31      ( Marks: 2 )

What are the advantages of a multipoint connection over a point-to-point connection?

Answer:

Point-to-point connection is limited to two devices, where else more than two devices share a single link in multipoint connection. Multipoint connection can be used for fail-over and reliability.

Question No: 32      ( Marks: 2 )

What's the name of the telephone service in which there is no need of dialing.

Answer:

DSS (digital data service)   is the telephone service in which there is no need of dialing.

 

Question No: 33      ( Marks: 2 )

Which type of frames are present in BSC frames?

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Answer:

There are two types of frames that are present in BSC.

1.      Control Frames and

2.      Data Frames

Question No: 34      ( Marks: 2 )

What methods of line discipline are used for peer to peer and  primary secondary communication?

Answer:

Line discipline is done in two ways:

1.      ENQ/ACK      (Enquiry Acknowledgement)

         This is used for peer to peer communication.

2.      Poll/ Select

This method is used for primary secondary communication.

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Question No: 35      ( Marks: 3 )

How does the checksum checker know that the received data unit is undamaged? [3]

Answer:

Checksum Checker or generator:

The sender subdivides data units into equal segments of ‘n’ bits(16 bits)

1.      These segments are added together using one’s complement.

2.      The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM.

3.      The extended data unit is transmitted across the network.

4.      The receiver subdivides data unit and adds all segments together and complement the result.

5.      If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero.

6.      If the result is not zero, the packet contains an error & the receiver rejects it.

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Question No: 36      ( Marks: 3 )

Which one has more overhead, a repeater or a bridge? Explain your answer. [3]

Answer:

A bridge has more overhead than a repeater. A bridge processes the packet at two

layers ; a repeater processes a frame at only one layer. A bridge needs to search a

table and find the forwarding port as well as to regenerate the signal; a repeater

only regenerates the signal. In other words, a bridge is also a repeater (and more); a

repeater is not a bridge.

Question No: 37      ( Marks: 3 )

Write down disadvantages of Ring Topology.

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Answer:

Disadvantages of Ring Topology

       Unidirectional Traffic

       A break in a ring that is a disabled station can disable the entire network

      Can be solved by using:

       Dual Ring or

       A switch capable of closing off the Break

Question No: 38      ( Marks: 3 )

How parity bits are counted in VRC error detection method technique in case of odd parity generator?

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Answer:

For example:

We want to TX the binary data unit 1100001

Adding together the number of 1’s gives us 3, an odd number

Before TX, we pass the data unit through a parity generator, which counts the 1’s and appends the parity bit (1) to the end

The total number of 1’s is now 4, an even number

The system now transfers the entire expanded across the network link

When it reaches its destination, the RX puts all 8 bits through an even parity checking function

If the RX sees 11100001, it counts four ones, an even number and the data unit passes

 When the parity checker counts the 1’s, it gets 5 an odd number

The receiver knows that an error has occurred somewhere and therefore rejects the whole unit

Some systems may also use ODD parity checking

The principal is the same as even parity

Question No: 39      ( Marks: 5 )

How are lost acknowledgment and a lost frame handled at the sender site? [5]

Answer:

At some error rates (16%-20%) the protocol hung up in an infinite loop, while it worked fine for other error rates. On examining the code it was determined that this problem resulted from improper variable initialization. On these certain error rates the pseudo-random number generator caused the very first frame sent to be lost or damaged. The receiver used a variable to keep track of the last in sequence frame received. This was erroneously initialized to 0. Therefore if the first frame got lost (sequence no 0), when the receiver received the second frame (sequence number 1) it sent an acknowledgment for the last in sequence frame  received, which had been initialized to 0. Therefore the sender received an acknowledgement for sequence number 0 and moved its window up accordingly. It caused everything to get out of synch, and caused the protocol to go into infinite loop.  This was resolved by initializing the variable to remember the last in sequence frame received to an out of range sequence number.

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Question No: 40      ( Marks: 5 )

Explain Protocol Data Unit (PDU)?

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Answer:

Protocol data unit (PDU) is an OSI term that refers generically to a group of information added or removed by a particular layer of the OSI model. In specific terms, an LxPDU implies the data and headers defined by layer x. Each layer uses the PDU to communicate and exchange information. The PDU information is only read by the peer layer on the receiving device and then stripped off, and data is handed over to the next upper layer.

Question No: 39      ( Marks: 5 )

Compare line decipline methods ENQ/ACK and Poll/ Select?

=>ENQ/ACK coordinates which device may start a transmission and whether or not the intended recipient is ready and enabled.

=> Using ENQ/ACK, a session can be initiated by either station on a link as long as both are of equal rank.

=> In both half-duplex and full-duplex transmission, the initiating device establishes the session.

=> In half duplex, the initiator then sends its data while the responder waits. The responder may take over the link when the initiator is finished or has requested a response.

=> In full duplex, both devices can transmit simultaneously once the session has been established. 

POLL/SELECT:

=> The poll/select method of line discipline works with topologies where one device is designated as a primary station and the other devices are secondary stations.

www.allvupastpapers.blogspot.com=> Multipoint systems must coordinate several nodes, not just two.

=> The question to be determined in these cases, therefore, is more than just, are you ready? It is also, which of the several nodes has the right to use the channel?