Cs601 final term solve paper spring July 2011

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Question No: 1 ( Marks: 1 ) - Please choose one







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



► geometrical



► logical



► physical






Question No: 2 ( Marks: 1 ) - Please choose one







The internet model consists of _________ layers.



► three



► two



► five



► seven






Question No: 3 ( Marks: 1 ) - Please choose one







Encryption and encoding are the same terms.



► True



► False






Question No: 4 ( Marks: 1 ) - Please choose one







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



► phase



► voltage



► wavelength






Question No: 5 ( Marks: 1 ) - Please choose one







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



► NRZ-L



► NRZ-I



► RZ






Question No: 6 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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



► 100 Mbps



► 200 Mbps



► 250 Mbps



► 400 Mbps






Question No: 9 ( Marks: 1 ) - Please choose one







The RG number gives us information about ________.



► Twisted pairs



► Coaxial cables



► Optical fibers



► all of the given






Question No: 10 ( Marks: 1 ) - Please choose one







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



► Glass plastic



► Copper



► Bimetallic



► Liquid






Question No: 11 ( Marks: 1 ) - Please choose one







All of popular Fiber optic connectors are ___________ shaped.



► Conical



► Barrel



► Circular



► Rectangular






Question No: 12 ( Marks: 1 ) - Please choose one







The VLF and LF bands use _________ propagation for communications.



► Ground



► Sky



► Line of sight



► Space






Question No: 13 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







Which error detection method involves polynomials?



► Checksum



► Two-dimensional parity check



► CRC



► Simple parity check






Question No: 16 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







The Hamming code is a method of __________



► Error detection



► Error correction



► Error ecapsulation



► Error detection & Error encapsulation






Question No: 20 ( Marks: 1 ) - Please choose one







Sliding window requires that data frames be transmitted _______________



► Sequentially



► Frequently



► Synchronously



► Asynchronously






Question No: 21 ( Marks: 1 ) - Please choose one







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



► Retransmitted



► Forwarded



► Selected



► Rejected






Question No: 22 ( Marks: 1 ) - Please choose one







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



► MAC



► LLC



► Physical






Question No: 23 ( Marks: 1 ) - Please choose one







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



► TRUE



► FALSE






Question No: 24 ( Marks: 1 ) - Please choose one







IEEE divides the base band category into ______ standards.



► 5



► 4



► 3



► 6






Question No: 25 ( Marks: 1 ) - Please choose one







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



► Ring



► Mesh



► Star



► Bus






Question No: 26 ( Marks: 1 ) - Please choose one







Check sum method is used for _______________ layers.



► Physical



► Application



► Transport



► Datalink






Question No: 27 ( Marks: 1 ) - Please choose one







Repeater works on __________ layer.



► Data Link



► Physical



► Network



► Application






Question No: 28 ( Marks: 1 ) - Please choose one







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 ( Marks: 1 ) - Please choose one







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



► MUX



► DEMUX



► Switch



► Bridge






Question No: 30 ( Marks: 1 ) - Please choose one







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



► Increasing



► Decreasing



► Keeping Constant



► None of the given






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?



Answer:

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

Control Frames and
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:

ENQ/ACK (Enquiry Acknowledgement)





This is used for peer to peer communication.

Poll/ Select

This method is used for primary secondary communication.









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.















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.





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?





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.