CS601 FinalTerm Solved Paper 2010

FINALTERM  EXAMINATION
Spring 2010
CS601- Data Communication
Time: 90 min
M a r k s: 60

Question No: 1      ( M a r k s: 1 ) .



__________representation of links that connect nodes is called as physical topology.

       geometrical

       logical

       physical




Question No: 2      ( M a r k s: 1 ) .



The internet model consists of _________ layers.

       three

       two

       five

       seven




Question No: 3      ( M a r k s: 1 ) .



Encryption and encoding are the same terms.

       True

       False




Question No: 4      ( M a r k s: 1 ) .



The amplitude of a digital signal depends upon the________ to represent a bit.

       phase

       voltage

       wavelength




Question No: 5      ( M a r k s: 1 ) .



The inversion of the level at 1 bit is called as __________

       NRZ-L

       NRZ-I

       RZ




Question No: 6      ( M a r k s: 1 ) .



Modulation of an analog signal can be accomplished through changing the ___________ of the carrier signal.

       amplitude

       frequency

       phase

       all of the given




Question No: 7      ( M a r k s: 1 ) .



If FCC regulations are followed, the carrier frequencies of adjacent AM radio stations are ____________ apart.

       5 KHz

        10 KHz

       200 KHz

       530 KHz




Question No: 8      ( M a r k s: 1 ) .



Category 5 UTP cable is used for data transmission of upto__________.

       100 Mbps

       200 Mbps

       250 Mbps

       400 Mbps




Question No: 9      ( M a r k s: 1 ) .



The RG number gives us information about ________.

       Twisted pairs

       Coaxial cables

       Optical fibers

       all of the given




Question No: 10      ( M a r k s: 1 ) .



The inner core of an optical fiber is __________ in composition.

       Glass plastic

       Copper

       Bimetallic

       Liquid




Question No: 11      ( M a r k s: 1 ) .



All of popular Fiber optic connectors are ___________ shaped.

       Conical

       Barrel

       Circular

       Rectangular




Question No: 12      ( M a r k s: 1 ) .



The VLF and LF bands use _________ propagation for communications.

       Ground

       Sky

       Line of sight

       Space




Question No: 13      ( M a r k s: 1 ) .



Multiplexing is the set of techniques that allows simultaneous TX of multiple signals across  ___________ data link




       Single

       Multi

       Single and Multi

       none of the given




Question No: 14      ( M a r k s: 1 ) .



A portion of the path that carries TX b/w a given pair of devices is known as __________.

       Node

       Bridge

       Channel

       Gateway




Question No: 15      ( M a r k s: 1 ) .



Which error detection method involves polynomials?

       Checksum

       Two-dimensional parity check

       CRC

       Simple parity check




Question No: 16      ( M a r k s: 1 ) .



If the ASCII character G is sent and the character D is received, what type of error is this?

       Single-bit

       Multiple-bit

       Burst

       Recoverable




Question No: 17      ( M a r k s: 1 ) .



Which error detection method involves the use of parity bits?

       Simple parity check & two dimensional parity check

       CRC

       Two-dimensional parity check

       Simple parity check

Question No: 18      ( M a r k s: 1 ) .



Which error detection method can detect a single-bit error?

       Simple parity check

       Two-dimensional parity check

       CRC

       All of the given




Question No: 19      ( M a r k s: 1 ) .



The Hamming code is a method of __________

       Error detection

       Error correction

       Error ecapsulation

       Error detection & Error encapsulation




Question No: 20      ( M a r k s: 1 ) .



Sliding window requires that data frames be transmitted _______________

       Sequentially

       Frequently

       Synchronously

       Asynchronously




Question No: 21      ( M a r k s: 1 ) .



In selective-reject ARQ, only the specific damaged or lost frame is_____________.

       Retransmitted

       Forwarded

        Selected

       Rejected




Question No: 22      ( M a r k s: 1 ) .



Which of the following sublyer, resolves the contention for the shared media

       MAC

        LLC

       Physical




Question No: 23      ( M a r k s: 1 ) .



The PDU  has no flag fields, no CRC, and no station address

       TRUE

       FALSE




Question No: 24      ( M a r k s: 1 ) .



IEEE divides the base band category into  ______ standards.

       5

       4

       3

       6




Question No: 25      ( M a r k s: 1 ) .



Like 10 Base 5, 10 Base 2 is a _________ topology LAN

       Ring

       Mesh

       Star

       Bus




Question No: 26      ( M a r k s: 1 ) .



Check sum method is used for _______________ layers.

       Physical

       Application

       Transport

       Datalink




Question No: 27      ( M a r k s: 1 ) .



Repeater works on __________     layer.

       Data Link

       Physical

       Network

       Application




Question No: 28      ( M a r k s: 1 ) .



Trunks are transmission media such as _________ that handle the telephone to the nearest end office.

       Satellite links

       Twisted-pair & Fiber-optic

       Twisted-pair

       Fiber-optic




Question No: 29      ( M a r k s: 1 ) .



Which of the following ___________ uses a series of filters to decompose multiplexed signal into its constituent signals.

       MUX

       DEMUX

       Switch

       Bridge




Question No: 30      ( M a r k s: 1 ) .



In Fast Ethernet, data rate can be increased by _______________ collisions.

       Increasing

       Decreasing

       Keeping Constant

       None of the given




Question No: 31      ( M a r k s: 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      ( M a r k s: 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      ( M a r k s: 2 )



Which type of frames is present in BSC frames?

Answer:
There are two types of frames that are present in BSC.
  1. Control Frames and
  2. Data Frames

Question No: 34      ( M a r k s: 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.
  1. Poll/ Select
This method is used for primary secondary communication.
Question No: 35      ( M a r k s: 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.
  
Question No: 36      ( M a r k s: 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      ( M a r k s: 3 )

Write down disadvantages of Ring Topology.

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      ( M a r k s: 3 )



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


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      ( M a r k s: 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.


Question No: 40      ( M a r k s: 5 )

Explain Protocol Data Unit (PDU)?

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.

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