CS2302-COMPUTERNETWORKS

29. What is the purpose of hamming code?

A hamming code can be designed to correct burst errors of certain lengths. So the simple strategy used by the hamming code to correct single bit errors must be redesigned to be applicable for multiple bit correction.

30. What is redundancy?

It is the error detecting mechanism, which means a shorter group of bits or extra bits may be appended at the destination of each unit.

31. Define flow control?

Flow control refers to a set of procedures used to restrict the amount of data. The sender can send before waiting for acknowledgment.

32. Mention the categories of flow control?

There are 2 methods have been developed to control flow of data across communication links. a) Stop and wait- send one from at a time. b) Sliding window- send several frames at a time.

33. What is a buffer?

Each receiving device has a block of memory called a buffer, reserved for storing incoming data until they are processed.

34.What is the difference between a passive and an active hub?

An active hub contains a repeater that regenerates the received bit patterns before sending them out. A passive hub provides a simple physical connection between the attached devices.

35.  For n devices in a network, what is the number of cable links required for a

mesh and ring topology?

·         Mesh topology – n (n-1)/2

·         Ring topology – n

36. Group the OSI layers by function. (MAY/JUNE2007)

The seven layers of the OSI model belonging to three subgroups. Physical, data link and network layers are the network support layers; they deal with the physical aspects of moving data from one device to another. Session, presentation and application layers are the user support layers; they allow interoperability among unrelated software systems. The transport layer ensures end-to-end reliable data transmission.

37.We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?

First, we use the Shannon formula to find our upper limit.

C = B log₂ (1 + SNR) = 10⁶ log₂ (1 + 63) = 10⁶ log₂ (64) = 6 Mbps

Then we use the Nyquist formula to find the

number of signal levels.

4 Mbps = 2 ´ 1 MHz ´ log₂ L  è  L = 4

= B log₂ (1) = B ´ 0 = 0

38.List the Channelization Protocols

n  Frequency Division Multiple Access (FDMA)

n  The total bandwidth is divided into channels.

n  Time Division Multiple Access (TDMA)

n  The band is divided into one channel that is time shared

n  Code Division Multiple Access (CDMA)

n  One channel carries all transmission simultaneously

39.What is protocol?What are its key elements?(NOV/DEC 2007)

            Set of rules that govern the data communication is protocol. The key elements are

                        i)Syntax  ii)Semantics iii)Timing

40.We have a channel with a 1 MHz bandwidth. The SNR for this channel is 63; what is the appropriate bit rate and signal level?

First, we use the Shannon formula to find our upper limit.

C = B log₂ (1 + SNR) = 10⁶ log₂ (1 + 63) = 10⁶ log₂ (64) = 6 Mbps

Then we use the Nyquist formula to find the

number of signal levels.

4 Mbps = 2 ´ 1 MHz ´ log₂ L  è  L = 4

= B log₂ (1) = B ´ 0 = 0

41.List the Channelization Protocols

n  Frequency Division Multiple Access (FDMA)

n  The total bandwidth is divided into channels.

n  Time Division Multiple Access (TDMA)

n  The band is divided into one channel that is time shared

n  Code Division Multiple Access (CDMA)

n  One channel carries all transmission simultaneously

42.What is protocol?What are its key elements?(NOV/DEC 2007)

            Set of rules that govern the data communication is protocol.The key elements are

                        i)Syntax  ii)Semantics iii)Timing