VTU 5TH SEM CSE COMPUTER NETWORKS-1 (DATA COMMUNICATION) SOLVED PAPERS

COMPUTER NETWORKS-1 SOLVED PAPER JUNE - 2013
1-1
1a. Explain OSI reference model. (10 Marks)
Ans:
LAYERS IN THE OSI MODEL
Physical Layer
• Main Responsibility:
Physical-layer (PL) is responsible for movements of individual bits from one node to
another node.
• Other responsibilities:
1) PL defines mechanical/electrical characteristics of the interface & transmission-medium
2) PL defines the type of transmission-medium: Wired or wireless.
3) PL defines the type of encoding i.e. how 0s and 1s are changed to signals.
4) PL defines the transmission-rate.
5) PL deals with the synchronization of the transmitter and receiver.
6) PL defines the nature of the connection ie point-to-point or multipoint configuration
7) PL defines the type of topology: mesh, star, ring or bus.
8) PL defines direction of data-transfer b/w 2 devices: Simplex, Half-duplex or Full-duplex
Data Link Layer
Data-link-layer (DLL) is responsible for moving frames from one node to another node.
1) DLL receives & divides the stream of bits from network-layer into frames.
2) DLL provides Physical-addressing
3) DLL provides flow-control.
4) DLL provides error-control.
5) DLL provides access-control.
Network Layer
Network-layer (NL) is responsible for source-to-destination delivery of a packet, possibly
across multiple-networks.
1) NL provides Logical-addressing
2) NL provides routing of packets.
Transport Layer
Transport-layer (TL) is responsible for process-to-process delivery of the entire message.
1) TL provides Service-point Addressing
2) TL is responsible for Segmentation & Reassembly
3) TL provides 2 services: i) connectionless or ii) connection-oriented.
4) TL is responsible for flow-control & error-control.

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Session Layer
Session-layer (SL) establishes, maintains, and synchronizes the interaction between 2
systems.
1) SL allows 2 systems to start communication with each other in half- or full-duplex.
2) SL provides synchronization
Presentation Layer
Presentation-layer (PL) is concerned with syntax & semantics of the info. exchanged b/w 2
systems.
1) PL is responsible for interoperability b/w encoding methods for different computers.
2) PL is responsible for encryption & decryption.
3) PL carries out data compression to reduce the size of the data to be transmitted.
Application Layer
The application-layer (AL)
→ provides services to the user
→ enables the user to access the network.
1) Mail Services
2) Directory Services
3) File Transfer, Access, and Management

1-3
1b. Explain categories of network and differentiate between them. (10 Marks)
Ans:
Network Types
• Two categories of networks: 1) LAN (Local Area Network) & 2) WAN (Wide Area Network)
LAN
• LAN is used to connect computers in a single office, building or campus (Figure 1.8).
• LAN is usually privately owned network.
• Each host in a LAN has an address that uniquely defines the host in the LAN.
• A packet sent by a host to another host carries both source- & destination-host’s addresses.
• LANs use a smart connecting switch.
• The switch is able to
→ recognize the destination address of the packet &
→ guide the packet to its destination.
• The switch
→ reduces the traffic in the LAN &
→ allows more than one pair to communicate with each other at the same time.
WAN
• WAN is used to connect computers anywhere in the world.
• WAN can cover larger geographical area. It can cover cities, countries and even continents.
• WAN interconnects connecting devices such as switches, routers, or modems.
• Normally, WAN is
→ created & run by communication companies (Ex: BSNL, Airtel)
→ leased by an organization that uses it.
• A WAN can be of 2 types:
1) Point-to-point WAN
 A point-to-point WAN is a network that connects 2 communicating devices through a
transmission media (Figure 1.9).
2) Switched WAN
 A switched WAN is a network with more than two ends.
 The switched WAN can be the backbones that connect the Internet.
 A switched WAN is a combination of several point-to-point WANs that are connected by
switches (Figure 1.10).

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Parameters LAN WAN
Expands to Local Area Network Wide Area Network
Meaning LAN is used to connect
computers in a single office,
building or campus
WAN is used to connect
computers in a large
geographical area such as
countries
Ownership of network Private Private or public
Range Small: up to 10 km Large: Beyond 100 km
Speed High: Typically 10, 100 and
1000 Mbps
Low: Typically 1.5 Mbps
Propagation Delay Short Long
Cost Low High
Congestion Less More
Design & maintenance Easy Difficult
Fault Tolerance More Tolerant Less Tolerant
Media used Twisted pair Optical fiber or radio waves
Used for College, Hospital Internet
Interconnects LAN interconnects hosts WAN interconnects connecting
devices such as switches,
routers, or modems

1-5
2a. Represent the given sequence 01001110 in unipolar. Manchesters, AMI,
Pseudoternary? (06 Marks)
Ans:
2b. The loss in a cable defined in debels/km (dB/km). If signal at beginning of a cable
with -3 dB/km has a power of 3 mW. What is power of the signal at 5 km? (06 Marks)
Ans:

1-6
2c. Explain the PCM encoder. (08 Marks)
Ans:
PCM
• PCM is a technique used to change an analog signal to digital data (digitization).
• PCM encoder has 3 processes (Figure 4.21): 1) Sampling 2) Quantization & 3) Encoding.
1) Sampling
• We convert the continuous time signal (analog) into the discrete time signal (digital).
• Pulses from the analog-signal are sampled every Ts sec
where Ts is the sample-interval or period.
• The inverse of the sampling-interval is called the sampling-frequency (or sampling-rate).
• Sampling-frequency is given by
• According to Nyquist theorem,
“The sampling-rate must be at least 2 times the highest frequency, not the bandwidth“.
i) If the analog-signal is low-pass, the bandwidth and the highest frequency are
the same value (Figure 4.23a).
ii) If the analog-signal is bandpass, the bandwidth value is lower than the value of
the maximum frequency (Figure 4.23b).
2) Quantization
• The sampled-signal is quantized.
• Result of sampling is a set of pulses with amplitude-values b/w max & min amplitudes of the
signal.
Quantization Level
 Let L = number of levels.
 The choice of L depends on
→ range of the amplitudes of the analog-signal and
→ how accurately we need to recover the signal.
 If the signal has only 2 amplitude values, we need only 2 quantization-levels.
If signal (like voice) has many amplitude values, we need more quantization-levels.
 In audio digitizing, L is normally chosen to be 256.
In video digitizing, L is normally thousands.
 Choosing lower values of L increases the quantization-error.
 Quantization-error is the difference b/w normalized PAM value & quantized values
3) Encoding
• The quantized values are encoded as n-bit code word.
• Relationship between number of quantization-levels (L) & number of bits (n) is given by
n=log2L or 2n
=L
• The bit-rate is given by:

1-7
3a. Explain frequency hopping spread spectrum (FHSS). (10 Marks)
Ans:
Frequency Hopping Spread-Spectrum (FHSS)
• This technique uses ‘M’ different carrier-frequencies that are modulated by the source-signal.
• At one moment, the signal modulates one carrier-frequency.
At the next moment, the signal modulates another carrier-frequency.
• Although the modulation is done using one carrier-frequency at a time, 'M’ frequencies are used
in the long run.
• The bandwidth occupied by a source is given by
• As shown in Figure 6.28.
A pseudorandom code generator (PN) creates a k-bit pattern for every hopping period Th
 The frequency-table
→ uses the pattern to find the frequency to be used for this hopping period and
→ passes the frequency to the frequency-synthesizer.
 The frequency-synthesizer creates a carrier-signal of that frequency.
 The source-signal modulates the carrier-signal.
• As shown in Figure 6.29, assume we have 8 hopping frequencies.
 Here, M = 8 and k = 3.
 The pseudorandom code generator will create 8 different 3-bit patterns.
 These are mapped to 8 different frequencies in the frequency table.
 The pattern for this station is 101, 111, 001, 000, 010, 111 & 100.
1) At hopping-period 1, the pattern is 101.
The frequency selected is 700 kHz; the source-signal modulates this carrier-
frequency.
2) At hopping-period 2, the pattern is 111.
The frequency selected is 900 kHz; the source-signal modulates this carrier-
frequency.

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• If there are many k-bit patterns & the hopping period is short, a sender & receiver can have
privacy.
If an attacker tries to intercept the transmitted signal, he can only access a small piece of
data because he does not know the spreading sequence to quickly adapt himself to the
next hop.
• The scheme has also an anti-jamming effect.
A malicious sender may be able to send noise to jam the signal for one hopping period
(randomly), but not for the whole period.
3b. Four 1 Kbps connections are multiplexed together. A unit is 1 bit. Find
i) The duration of 1 bit before multiplexing.
ii) The transmission rate the link
iii) The duration of a time slot.
iv) The duration of a frame. (05 Marks)
Ans:

1-9
3c. Differentiate between circuit switched, datagram networks and virtual circuit
networks. (05 Marks)
Ans:

1-10
4a. Explain structure of encoder and decoder for hamming code. (08 Marks)
Ans:
Hamming Distance
• Hamming distance b/w 2 words is the number of differences between the corresponding bits.
• For example: The Hamming distance d(000, 011) is 2 because 000⊕011= 011 (two 1s).
Parity Check Code C(5, 4)
• This code is a linear block code. This code can detect an odd number of errors.
• A 4-bit data-word is changed to an 5-bit code-word. One extra bit is called the parity-bit.
• The parity-bit is selected to make the total number of 1s in the code-word even.
• Minimum hamming distance dmin = 2. This means the code is a single-bit error-detecting code.
• Here is how it works (Figure 10.4):
1) At Sender
 The encoder uses a generator that takes a copy of a 4-bit data-word (a0, a1, a2, and a3)
and generates a parity-bit r0.
 The encoder
→ accepts a copy of a 4-bit data-word (a0, a1, a2, and a3) and
→ generates a parity-bit r0 using a generator
→ generates a 5-bit code-word
 The parity-bit & 4-bit data-word are added to make the no. of 1s in code-word even.
 The addition is done by using the following:
 The result of addition is the parity-bit.
1) If the no. of 1s in data-word = even, result = 0. (r0=0)
2) If the no. of 1s in data-word = odd, result = 1. (r0=1)
3) In both cases, the total number of 1s in the code-word is even.
 The sender sends the code-word, which may be corrupted during transmission.
2) At Receiver
 The receiver receives a 5-bit word.
 The checker performs the same operation as the generator with one exception:
The addition is done over all 5 bits.
 The result is called the syndrome bit (so).
 Syndrome bit = 0 when the no. of 1s in received code-word is even; otherwise, it is 1.
 The syndrome is passed to the decision logic analyzer.
1) If s0=0, there is no error in the received code-word. The data portion of the
received code-word is accepted as the data-word.
2) If s0=1, there is error in the received code-word. The data portion of the
received code-word is discarded. The data-word is not created.

1-11
4b. Find the codeword using CRC given dataword 1001 and generator 1011 (06 Marks)
Ans:

1-12
4c. What is Internet checksum? With an example, list the steps undertaken by the
sender and receiver for error detection. (06 Marks)
Ans:
Internet Checksum
• Internet Checksum is an error-detecting technique.
• This is based on the concept of redundancy.
• Traditionally, the Internet has been using a 16-bit checksum.
• The sender or the receiver uses five steps.
Algorithm
Example

1-13
5a. Explain stop-and-wait ARQ protocol with neat diagram. (08 Marks)
Ans:
Stop & Wait Protocol
• This uses both flow and error control.
• Normally, the receiver has limited storage-space.
• If the receiver is receiving data from many sources, the receiver may
→ be overloaded with frames &
→ discard the frames.
• To prevent the receiver from being overloaded with frames, we need to tell the sender to slow
down.
Design
1) At Sender
 The sender
→ sends one frame & starts a timer
→ keeps a copy of the sent-frame and
→ waits for ACK-frame from the receiver (okay to go ahead).
 Then,
1) If an ACK-frame arrives before the timer expires, the timer is stopped and the
sender sends the next frame.
Also, the sender discards the copy of the previous frame.
2) If the timer expires before ACK-frame arrives, the sender resends the previous
frame and restarts the timer
2) At Receiver
 To detect corrupted frames, a CRC is added to each data frame.
 When a frame arrives at the receiver-site, the frame is checked.
 If frame’s CRC is incorrect, the frame is corrupted and discarded.
 The silence of the receiver is a signal for the sender that a frame was either corrupted
or lost.
FSMs

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1) Sender States
 Sender is initially in the ready state, but it can move between the ready and blocking
state.
i) Ready State: When the sender is in this state, it is only waiting for a packet
from the network layer.
If a packet comes from the network layer, the sender creates a frame,
saves a copy of the frame, starts the only timer and sends the frame. The
sender then moves to the blocking state.
ii) Blocking State: When the sender is in this state, three events can occur:
a) If a time-out occurs, the sender resends the saved copy of the frame and
restarts the timer.
b) If a corrupted ACK arrives, it is discarded.
c) If an error-free ACK arrives, the sender stops the timer and discards the
saved copy of the frame. It then moves to the ready state.
2) Receiver
 The receiver is always in the ready state. Two events may occur:
a) If an error-free frame arrives, the message in the frame is delivered to the
network layer and an ACK is sent.
b) If a corrupted frame arrives, the frame is discarded.

1-15
5b. What is framing? Explain bit and character stuffing with an example. (04 Marks)
Ans:
Framing
• A frame is a group of bits.
• Framing means organizing the bits into a frame that are carried by the physical layer.
• The data-link-layer needs to form frames, so that each frame is distinguishable from another.
Byte Stuffing
• In byte stuffing, a special byte is added to the data-section of the frame when there is a
character with the same pattern as the flag.
• The data-section is stuffed with an extra byte. This byte is called the escape character (ESC),
which has a predefined bit pattern.
• When a receiver encounters the ESC character, the receiver
→ removes ESC character from the data-section and
→ treats the next character as data, not a delimiting flag. (Figure 11.2).
Bit Stuffing
• In bit stuffing, if a 0 and five consecutive 1 bits are encountered, an extra 0 is added. This
extra stuffed bit is eventually removed from the data by the receiver. (Figure 11.4).
• This guarantees that the flag field sequence does not inadvertently appear in the frame.

1-16
5c. Write short notes on HDLC. (08 Marks)
Ans:
High-Level Data Link Control (HDLC)
• HDLC is a bit-oriented protocol for communication over point-to-point and multipoint links.
Configurations and Transfer Modes
• HDLC provides 2 common transfer modes:
1) Normal response mode (NRM) & 2) Asynchronous balanced mode (ABM).
1) NRM
 The station configuration is unbalanced (Figure 11.14).
 We have one primary station and multiple secondary stations.
 A primary station can send commands, a secondary station can only respond.
 The NRM is used for both point-to-point and multiple-point links.
2) ABM
 The configuration is balanced (Figure 11.15).
 Link is point-to-point, and each station can function as a primary and a secondary
(acting as peers).
 This is the common mode today.
Framing
• HDLC defines three types of frames:
1) Information frames (I-frames): are used to transport user data and control information
relating to user data (piggybacking).
2) Supervisory frames (S-frames): are used only to transport control information.
3) Unnumbered frames (U-frames): are reserved for system management.
Information carried by U-frames is intended for managing the link itself.
• Each type of frame serves as an envelope for the transmission of a different type of message.
Frame Format
• Various fields of HDLC frame (Figure 11.16):
1) Flag Field
 This field has a synchronization pattern 01111110.
 This field identifies both the beginning and the end of a frame.

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2) Address Field
 This field contains the address of the secondary station.
 If a primary station created the frame, it contains a to-address.
 If a secondary creates the frame, it contains a from-address.
 This field can be 1 byte or several bytes long, depending on the needs of the network.
3) Control Field
 This field is one or two bytes used for flow and error control.
4) Information Field
 This field contains the user's data from the network-layer or management information.
 Its length can vary from one network to another.
5) FCS Field
 This field is the error-detection field. (FCS  Frame Check Sequence)
 This field can contain either a 2- or 4-byte standard CRC.

1-18
6a. Explain CDMA. (06 Marks)
Ans:
CHANNELIZATION PROTOCOLS
• Channelization is a multiple-access method.
• The available bandwidth of a link is shared b/w different stations in time, frequency, or through
code.
CDMA
• CDMA simply means communication with different codes.
• CDMA differs from TDMA because
→ only one channel occupies the entire bandwidth of the link.
• CDMA differs from FDMA because
→ all stations can send data simultaneously; there is no timesharing.
Implementation
• Let us assume we have four stations 1, 2, 3, and 4 connected to the same channel.
• The data from station-1 are d1, from station-2 are d2, and so on.
• The code assigned to the first station is c1, to the second is c2, and so on.
• We assume that the assigned codes have 2 properties.
1) If we multiply each code by another, we get 0.
2) If we multiply each code by itself, we get 4 (the number of stations).
 Station-1 multiplies the data by the code to get d1.c1.
 Station-2 multiplies the data by the code to get d2.c2. And so on.
 The data that go on the channel are the sum of all these terms.
 The receiver multiplies the data on the channel by the code of the sender.
 For example, suppose stations 1 and 2 are talking to each other.
 Station-2 wants to hear what station-1 is saying.
 Station-2 multiplies the data on the channel by c1 the code of station-1.
(c1.c1)=4, (c2.c1)=0, (c3.c1)=0, and (c4.c1)=0,
Therefore, station-2 divides the result by 4 to get the data from station-1.
Data Representation
• We follow the following rules for encoding:
1) To send a 0 bit, a station encodes the bit as -1
2) To send a 1 bit, a station encodes the bit as +1
3) When a station is idle, it sends no signal, which is interpreted as a 0.

1-19
Encoding and Decoding
 Each station multiplies the corresponding number by its chip (its orthogonal sequence).
 The result is a new sequence which is sent to the channel.
 The sequence on the channel is the sum of all 4 sequences.
 Now imagine station-3, which is silent, is listening to station-2.
 Station-3 multiplies the total data on the channel by the code for station-2, which is
[+1 -1 +1-1], to get
6b. A slotted ALOHA network transmits 200 bit frames using a shared channel with 200
Kbits/sec bandwidth. Find throughput if system produces
i) 1000 frames/sec
ii) 500 frames/sec
iii) 250 frames/sec. (06 Marks)
Ans:

1-20
6c. Explain 802.3 MAC frame format. (08 Marks)
Ans:
802.3 MAC Frame Format
• The Ethernet frame contains 7 fields (Figure 13.3):
1) Preamble
 This field contains 7 bytes (56 bits) of alternating 0s and 1s.
 This field
→ alerts the receiving-system to the coming frame and
→ enables the receiving-system to synchronize its input timing.
 The preamble is actually added at the physical-layer and is not (formally) part of the
frame.
2) Start frame delimiter (SFD)
 This field signals the beginning of the frame.
 The SFD warns the stations that this is the last chance for synchronization.
 This field contains the value: 10101011.
 The last 2 bits (11) alerts the receiver that the next field is the destination-address.
3) Destination-address (DA)
 This field contains the physical-address of the destination-station.
4) Source-address (SA)
 This field contains the physical-address of the sender-station.
5) Length or type
 This field is defined as a i) type field or ii) length field.
i) In original Ethernet, this field is used as the type field.
¤ Type field defines the upper-layer protocol using the MAC frame.
ii) In IEEE standard, this field is used as the length field.
¤ Length field defines the number of bytes in the data-field.
6) Data
 This field carries data encapsulated from the upper-layer protocols.
 Minimum data size = 46 bytes. Maximum data size = 1500 bytes.
7) CRC
 This field contains error detection information such as a CRC-32.

1-21
7a. Explain the architecture of IEEE 802.11. (10 Marks)
Ans:
IEEE 802.11 Architecture
• Two kinds of services defined are: 1) Basic service set (BSS) and 2) Extended service set (ESS)
1) BSS
• IEEE 802.11 defines the basic service set (BSS) as the building block of a wireless-LAN.
• A basic service set is made of (Figure 15.4):
→ stationary or mobile wireless stations and
→ optional central base station, known as the access point (AP).
• There are 2 types of architecture:
1) Ad hoc Architecture
 The BSS without an AP is a stand-alone network and cannot send data to other BSSs.
 Stations can form a network without the need of an AP.
 Stations can locate one another and agree to be part of a BSS.
2) Infrastructure Network
 A BSS with an AP is referred to as an infrastructure network.
2) ESS
• The ESS is made up of 2 or more BSSs with APs (Figure 15.5).
• The BSSs are connected through a distribution-system, which is usually a wired LAN.
• The distribution-system connects the APs in the BSSs.
• IEEE 802.11 does not restrict the distribution-system;
The distribution-system can be any IEEE LAN such as an Ethernet.
• The ESS uses 2 types of stations:
1) Mobile stations are normal stations inside a BSS.
2) Stationary stations are AP stations that are part of a wired LAN.
• When BSSs are connected, the stations within reach of one another can communicate without
the use of an AP.
• However, communication b/w two stations in two different BSSs usually occurs via two APs.
3) Station Types
• IEEE 802.11 defines three types of stations based on their mobility in a wireless-LAN:
1) No-transition 2) BSS-transition 3) ESS-transition mobility
1) A station with no-transition mobility is either
→ stationary (not moving) or
→ moving only inside a BSS.
2) A station with BSS-transition mobility can move from one BSS to another, but the
movement is confined inside one ESS.
3) A station with ESS-transition mobility can move from one ESS to another.

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7b. Explain connecting devices. (10 Marks)
Ans:
CONNECTING DEVICES
Passive Hubs
• A passive hub is just a connector.
• It connects the wires coming from different branches.
• In a star-topology Ethernet LAN, a passive hub is just a point where the signals coming from
different stations collide; the hub is the collision point.
• This type of a hub is part of the media; its location in the Internet model is below the physical
layer.
Repeater
• A repeater is a device that operates only in the physical layer (Figure 15.2).
• Signals that carry information within a network can travel a fixed distance before attenuation
endangers the integrity of the data.
• A repeater receives a signal and, before it becomes too weak or corrupted, regenerates and
retimes the original bit pattern.
• The repeater then sends the refreshed signal.
• A repeater can extend the physical length of a LAN.
• A repeater forwards every frame; it has no filtering capability
• A repeater is a regenerator, not an amplifier
Active Hubs
• An active hub is actually a multipart repeater.
• It is normally used to create connections between stations in a physical star topology.
• Hubs can also be used to create multiple levels of hierarchy (Figure 15.4).

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Bridges
• A bridge operates in both the physical and the data link layer (Figure 15.5).
• As a physical layer device, it regenerates the signal it receives.
• As a data link layer device, the bridge can check the physical (MAC) addresses (source and
destination) contained in the frame.
• A bridge has filtering capability.
• It can check the destination address of a frame and decide if the frame should be forwarded or
dropped.
• If the frame is to be forwarded, the decision must specify the port.
• A bridge has a table that maps addresses to ports.
• A bridge has a table used in filtering decisions.
• A bridge does not change the physical (MAC) addresses in a frame
Routers
• A router is a three-layer device that routes packets based on their logical addresses(host-to-
host addressing).
• A router normally connects LANs and WANs in the Internet and has a routing table that is used
for making decisions about the route (Figure 15.11).
• The routing tables are normally dynamic and are updated using routing protocols.
Gateway
• A gateway is normally a computer that operates in all five layers of the Internet or seven layers
of OSI model.
• A gateway takes an application message, reads it, and interprets it.
• This means that it can be used as a connecting device between two internetworks that use
different models.
• For example, a network designed to use the OSI model can be connected to another network
using the Internet model.
• The gateway connecting the two systems can take a frame as it arrives from the first system,
move it up to the OSI application layer, and remove the message.
• Gateways can provide security.

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8a. Draw IPV4 header format and explain. (08 Marks)
Ans:
IP Header Format
• IP uses the packets called datagrams.
• A datagram consist of 2 parts (Figure 19.2): 1) Payload 2) Header.
1) Payload
• Payload (or Data) is the main reason for creating a datagram.
• Payload is the packet coming from other protocols that use the service of IP.
2) Header
• Header contains information essential to routing and delivery.
• IP header contains following fields:
1) Version Number (VER)
 This field indicates version number used by the packet. Current version=4
2) Header Length (HLEN)
 This field specifies length of header.
 When a device receives a datagram, the device needs to know
→ when the header stops and
→ when the data starts.
3) Service Type
 This field specifies priority of packet based on delay, throughput, reliability & cost
requirements.
4) Total Length
 This field specifies the total length of the datagram (header plus data).
 Maximum length=65535 bytes.
5) Identification, Flags, and Fragmentation Offset
 These 3 fields are used for fragmentation and reassembly of the datagram.
 Fragmentation occurs when the size of the datagram is larger than the MTU of the
network.
6) Time-to-Live (TTL)
 This field is indicates amount of time, the packet is allowed to remain in the network.
 If TTL becomes 0 before packet reaches destination, the router
→ discards packet and
→ sends an error-message back to the source.
7) Protocol
 This field specifies upper-layer protocol that is to receive the packet at the destination-
host.
 For example: For TCP, protocol = 6 For UDP, protocol = 17
8) Header Checksum
 This field is used to verify integrity of header only.
 If the verification process fails, packet is discarded.
9) Source and Destination Addresses
 These 2 fields contain the IP addresses of source and destination hosts.
10) Options
 This field allows the packet to request special features such as
→ security level
→ route to be taken by packet and
→ timestamp at each router.
 This field can also be used for network testing and debugging.
11) Padding
 This field is used to make the header a multiple of 32-bit words.

1-25
8b. A ISP is granted a block of address starting with 190.100.0.0/16 (655.536
address). The ISP needs to distribute these addressing to 3 groups of customers.
i) First group has 64 customers each needs 256 address.
ii) Second group has 128 customers each needs 128 address.
iii) The third group has 128 customers each needs 64 address.
Design the subblock and findout. How many addresses are still available after their
allocations? (07 Marks)
Ans:
Solution:
Group 1
 For this group, each customer needs 256 addresses. This means the suffix length is 8
(28
=256). The prefix length is then 32−8 = 24.
01: 190.100.0.0/24  190.100.0.255/24
02: 190.100.1.0/24 190.100.1.255/24
.........................................
64: 190.100.63.0/24  190.100.63.255/24
Total = 64 ×256 = 16,384
Group 2
(27
=128). The prefix length is then 32 −7 = 25.
The addresses are:
001 : 190.100.64.0/25  190.100.64.127/25
002 : 190.100.64.128/25  190.100.64.255/25
003: 190.100.127.128/25  190.100.127.255/25
Total =128×128 = 16,384
Group 3
(26
=64). The prefix length is then 32−6 =26.
001:190.100.128.0/26  190.100.128.63/26
002:190.100.128.64/26  190.100.128.127/26
..............................
128 :190.100.159.192/26  190.100.159.255/26
Total =128 ×64 = 8,192
Number of granted addresses : 65,536
Number of allocated addresses: 40,960
Number of available addresses: 24,576
8c. Compare between IPV4 & IPV6. (05 Marks)
Ans:

COMPUTER NETWORKS-1 SOLVED PAPER DEC - 2013
2-1
1a. What is data communication? List and explain the five components of data
communication system. (06 Marks)
Ans:
DATA COMMUNICATIONS
• Data Communication is defined as exchange of data between 2 devices over a transmission-
medium.
• A communication-system is made up of
→ hardware (physical equipment) and
→ software (programs)
• For data-communication, the communicating-devices must be part of a communication-system.
• Four attributes of a communication-system:
1) Delivery
 The system must deliver data to the correct destination.
2) Accuracy
 The system must deliver the data accurately.
 Normally, the corrupted-data are unusable.
3) Timeliness
 The system must deliver audio/video data in a timely manner.
 This kind of delivery is called real-time transmission.
 Data delivered late are useless.
4) Jitter
 Jitter refers to the variation in the packet arrival-time.
 In other words, jitter is the uneven delay in the delivery of audio/video packets.
Components of Communication System
• Five components of a communication-system (Figure 1.1):
1) Message
2) Sender
3) Receiver
4) Transmission-Medium
5) Protocol
1) Message
 Message is the information (or data) to be communicated.
 Message may consist of
→ number/text
→ picture or
→ audio/video
2) Sender
 Sender is the device that sends the data-message.
 Sender can be
→ computer and
→ mobile phone
3) Receiver
 Receiver is the device that receives the message.
 Receiver can be
→ computer and
→ mobile phone
4) Transmission Medium
 Transmission-medium is physical-path by which a message travels from sender to
receiver.

2-2
 Transmission-medium can be wired or wireless.
 Examples of wired medium:
→ twisted-pair wire (used in landline telephone)
→ coaxial cable (used in cable TV network)
→ fiber-optic cable
 Examples of wireless medium:
→ radio waves
→ microwaves
→ infrared waves (ex: operating TV using remote control)
5) Protocol
 A protocol is a set of rules that govern data-communications.
 In other words, a protocol represents an agreement between the communicating-
devices.
 Without a protocol, 2 devices may be connected but not communicating.
1b. Discuss OSl layered model, bringing out functionalities of each layer.(10 Marks)
Ans: For answer, refer Solved Paper June-2013 Q.No.1a.
1c. Differentiate between
i) ARP and RARP
ii) UDP and TCP (04 Marks)
Ans:
i) ARP
 ARP is used to associate a Internet-address with a physical-address.
 In other words, ARP is used to find the physical-address of the node when its Internet-address
is known.
 On a typical physical-network, each device on a link is identified by a physical-address.
 Usually, physical-address is imprinted on the NIC (network interface card).
RARP
 RARP is used to find the Internet-address of the node when its physical-address is known.
 RARP is used
→ when a computer is connected to a network for the first time or
→ when a diskless computer is booted.
ii) TCP
 TCP is a reliable connection-oriented protocol.
 A connection is established b/w the sender and receiver before the data can be transmitted.
 TCP provides
→ flow control
→ error control and
→ congestion control
UDP
 UDP is the simplest of the 3 transport protocols.
 It is an unreliable, connectionless protocol.
 It does not provide flow, error, or congestion control.
 Each datagram is transported separately & independently.
 It is suitable for application program that
→ needs to send short messages &
→ cannot afford the retransmission.

2-3
2a. Write a descriptive note on three causes of transmission impairment. (08 Marks)
Ans:
THREE CAUSES OF TRANSMISSION IMPAIRMENT
1) Attenuation
• As signal travels through the medium, its strength decreases as distance increases. This is
called attenuation (Figure 3.27).
• As the distance increases, attenuation also increases.
• For example:
Voice-data becomes weak over the distance & loses its contents beyond a certain distance.
• To compensate for this loss, amplifiers are used to amplify the signal.
Decibel
 The decibel (dB) measures the relative strengths of
→ 2 signals or
→ one signal at 2 different points.
 The decibel is negative if a signal is attenuated.
The decibel is positive if a signal is amplified.
where P1 and P2 = powers of a signal at points 1 and 2, respectively.
2) Distortion
• Distortion means that the signal changes its form or shape (Figure 3.29).
• Distortion can occur in a composite signal made of different frequencies.
• Different signal-components
→ have different propagation speed through a medium.
→ have different delays in arriving at the final destination.
• Differences in delay create a difference in phase if delay is not same as the period-duration.
• Signal-components at the receiver have phases different from what they had at the sender.
• The shape of the composite signal is therefore not the same.

2-4
3) Noise
• Noise is defined as an unwanted data (Figure 3.30).
• In other words, noise is the external energy that corrupts a signal.
• Due to noise, it is difficult to retrieve the original data/information.
• Four types of noise:
i) Thermal Noise
 It is random motion of electrons in wire which creates extra signal not originally sent by
transmitter.
ii) Induced Noise
 Induced noise comes from sources such as motors & appliances.
 These devices act as a sending-antenna.
The transmission-medium acts as the receiving-antenna.
iii) Crosstalk
 Crosstalk is the effect of one wire on the other.
 One wire acts as a sending-antenna and the other as the receiving-antenna.
iv) Impulse Noise
 Impulse Noise is a spike that comes from power-lines, lightning, and so on.
Signal-to-Noise Ratio (SNR)
¤ SNR is used to find the theoretical bit-rate limit.
¤ SNR is defined as
¤ A high-SNR means the signal is less corrupted by noise.
A low-SNR means the signal is more corrupted by noise.

2-5
2b. Explain the transmission modes? (06 Marks)
Ans:
TRANSMISSION MODES
1) PARALLEL TRANSMISSION
• Multiple bits are sent with each clock-tick (Figure 4.32).
• „n‟ bits in a group are sent simultaneously.
• „n‟ wires are used to send „n‟ bits at one time.
• Each bit has its own wire.
• Typically, the 8 wires are bundled in a cable with a connector at each end.
• Advantage:
1) Speed: Parallel transmission can increase the transfer speed by a factor of n over serial
transmission.
• Disadvantage:
1) Cost: Parallel transmission requires n communication lines just to transmit the data-
stream.
2) SERIAL TRANSMISSION
• One bit is sent with each clock-tick using only a single link (Figure 4.33).
• Advantage:
1) Cost: Serial transmission reduces cost of transmission over parallel by a factor of n.
• Disadvantage:
1) Since communication within devices is parallel, following 2 converters are required at
interface:
i) Parallel-to-serial converter
ii) Serial-to-parallel converter
• Three types of serial transmission: asynchronous, synchronous, and isochronous.

2-6
i) Asynchronous Transmission
• Asynchronous transmission is so named because the timing of signal is not important (Fig 4.34)
• Prior to data transfer, both sender & receiver agree on pattern of information to be exchanged.
• Normally, patterns are based on grouping the bit-stream into bytes.
• The sender transmits each group to the link without regard to a timer.
• As long as those patterns are followed, the receiver can retrieve the info. w/o regard to a timer.
• There may be a gap between bytes.
• We send
→ 1 start bit (0) at the beginning of each byte
→ 1 stop bit (1) at the end of each byte.
• Start bit alerts the receiver to the arrival of a new group.
Stop bit lets the receiver know that the byte is finished.
• Here, the term asynchronous means “asynchronous at the byte level”.
However, the bits are still synchronized & bit-durations are the same.
ii) Synchronous Transmission
• We send bits one after another without start or stop bits or gaps (Figure 4.35).
• The receiver is responsible for grouping the bits.
• The bit-stream is combined into longer "frames," which may contain multiple bytes.
• If the sender wants to send data in separate bursts, the gaps b/w bursts must be filled with a
special sequence of 0s & 1s (that means idle).
iii) Isochronous
• Synchronization between characters is not enough; the entire stream of bits must be
synchronized.
• The isochronous transmission guarantees that the data arrive at a fixed rate.
• In real-time audio/video, jitter is not acceptable. Therefore, synchronous transmission fails.
• For example: TV images are broadcast at the rate of 30 images per second. The images must
be viewed at the same rate.

2-7
2c. Explain delta modulation. (06 Marks)
Ans:
Delta Modulation (DM)
• PCM finds the value of the signal amplitude for each sample; DM finds the change from the
previous sample.
Modulator
 The modulator is used at the sender site to create a stream of bits from an analog
signal.
 The process records the small positive or negative changes, called delta δ.
 If the delta is positive, the process records a 1; if it is negative, the process records a 0.
Demodulator
 The demodulator takes the digital data and, using the staircase maker and the delay
unit, creates the analog signal.
 The created analog signal, however, needs to pass through a low-pass filter for
smoothing.
Quantization Error
 It is obvious that DM is not perfect.
 Quantization error is always introduced in the process.
 The quantization error of DM, however, is much less than that for PCM.

2-8
3a. What is FDM? Briefly explain its multiplexing & demultiplexing process. (06 Marks)
Ans:
Frequency Division Multiplexing (FDM)
• FDM is an analog multiplexing technique that combines analog signals (Figure 6.3).
• FDM can be used when the bandwidth of a link is greater than the combined bandwidths of the
signals to be transmitted..
1) Multiplexing Process
1) Each sending-device generates modulated-signals with different carrier-frequencies (f1,
f2, & f3).
2) Then, these modulated-signals are combined into a single multiplexed-signal.
3) Finally, the multiplexed-signal is transported by the link.
• Carrier-frequencies are separated by sufficient bandwidth to accommodate the modulated-
signal.
• Channels can be separated by strips of unused bandwidth called guard bands.
• Guard bands prevent signals from overlapping.
• In addition, carrier-frequencies must not interfere with the original data frequencies.
• Although FDM is considered as analog multiplexing technique, the sources can produce digital-
signal.
• The digital-signal can be sampled, changed to analog-signal, and then multiplexed by using
FDM.
2) Demultiplexing Process
1) The demultiplexer uses filters to divide the multiplexed-signal into individual-signals.
2) Then, the individual signals are passed to a demodulator.
3) Finally, the demodulator
→ separates the individual signals from the carrier signals and
→ passes the individual signals to the output-lines.

2-9
3b. Four sources create 250 characters per second. The frame contain one character
from each source and one extra bit for synchronization. Find
i) The data rate of each source
ii) Duration of each character in each source
iii) Frame rate
iv) Duration of output frame
v) Frame size in bits
vi) Data rate of link (06 Marks)
Ans:

2-10
3c. What is TDM? Explain how TDM overcomes the disadvantages of synchrononous
TDM. (08 Marks)
Ans:
Time-Division Multiplexing (TDM)
• TDM is a digital multiplexing technique that combines digital signals (Figure 6.12).
• TDM combines several low-rate channels into one high-rate one.
• Each connection occupies a portion of time in the link.
• Two types of TDM:
1) Synchronous and 2) Statistical.
Statistical TDM
• Problem: Synchronous TDM is not efficient.
For ex: If a source does not have data to send, the corresponding slot in the output-frame
is empty (Figure 6.26a).
Solution: Use statistical TDM (Figure 6.26b).
Slots are dynamically allocated to improve bandwidth-efficiency.
Only when an input-line has data to send, the input-line is given a slot in the
output-frame.
• The number of slots in each frame is less than the number of input-lines.
• The multiplexer checks each input-line in round robin fashion.
If the line has data to send;
Then, multiplexer allocates a slot for an input-line;
Otherwise, multiplexer skips the line and checks the next line.
1) Addressing
Synchronous TDM Statistical TDM
An output-slot needs to carry only data of
the destination (Figure 6.26a).
An output-slot needs to carry both data &
address of the destination (Figure 6.26b).
There is no need for addressing.
Synchronization and pre-assigned
relationships between the inputs and
outputs serve as an address.
There is no fixed relationship between the
inputs and outputs because there are no
pre-assigned or reserved slots.
We need to include the address of the
receiver inside each slot to show where it is
to be delivered.

2-11
2) Slot Size
 Usually, a block of data is many bytes while the address is just a few bytes.
 A slot carries both data and address.
 Therefore, address-size must be very small when compared to data-size.
This results in efficient transmission.
3) No Synchronization Bit
 In statistical TDM, the frames need not be synchronized, so synchronization-bits are not
needed.
4) Bandwidth
 Normally, the capacity of the link is less than the sum of the capacities of each channel.
 The designers define the capacity of the link based on the statistics of the load for each
channel.
4a. Describe different types of errors. (03 Marks)
Ans:
Types of Errors
• When bits flow from 1 point to another, they are subject to unpredictable-changes „.‟ of
interference.
• The interference can change the shape of the signal.
• Two types of errors: 1) Single-bit error 2) Burst-error.
1) Single-Bit Error
 Only 1 bit of a given data is changed
→ from 1 to 0 or
→ from 0 to 1 (Figure 10.1a).
2) Burst Error
 Two or more bits in the data have changed
→ from 1 to 0 or
→ from 0 to 1 (Figure 10.1b).
 A burst-error occurs more than a single-bit error. This is because:
Normally, the duration of noise is longer than the duration of 1-bit.
 When noise affects data, the noise also affects the bits.
 The no. of corrupted-bits depends on
→ data-rate and
→ duration of noise.

2-12
4b. Explain error detection and error correction wrt block coding. (08 Marks)
Ans:
Block Coding
• The message is divided into k-bit blocks. These blocks are called data-words.
• Here, r-redundant-bits are added to each block to make the length n=k+r.
• The resulting n-bit blocks are called code-words.
• Block-coding process is 1-to-1; the same data-word is always encoded as the same code-word.
Error Detection
• If the following 2 conditions are met, the receiver can detect a change in the original code-
word:
1) The receiver has a list of valid code-words.
2) The original code-word has changed to an invalid code-words.
1) At Sender
i) The sender creates code-words out of data-words by using a generator.
The generator applies the rules and procedures of encoding.
ii) During transmission, each code-word sent to the receiver may change.
2) At Receiver
i) a) If the received code-word is the same as one of the valid code-words,
the code-word is accepted;
the corresponding data-word is extracted for use.
b) If the received code-word is invalid, the code-word is discarded.
ii) However, if the code-word is corrupted but the received code-word still matches a valid
code-word, the error remains undetected.
• An error-detecting code can detect only the types of errors for which it is designed; other types
of errors may remain undetected.
Example
4c. Find code word, using CRC given data word ‘1001’ and generator ‘1011’. (09 Marks)
Ans: For answer, refer Solved Paper June-2013 Q.No.4b.

2-13
5a. Explain briefly, with figure i) Stop and Wait ARQ and ii) Go Back N ARQ. (12 Marks)
Ans (i): For answer, refer Solved Paper June-2013 Q.No.5a.
Ans (ii):
Go-Back-N ARQ Protocol
• We send several frames before receiving acknowledgments; we keep a copy of these frames
until the acknowledgments arrive.
Sequence-numbers
 Frames from a sending station are numbered sequentially.
 If m bits are used for sequence-number, the sequence-numbers range from 0 to 2m - 1
.
For example, if m=4, the only sequence-numbers are 0 to 15.
 However, we can repeat the sequence.
 So the sequence-numbers are
0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,0,1,2,3,4,5,6,7,8,9,10,11,...
Sliding Window
 Sliding window defines the range of sequence-numbers that can be used for the sender
& receiver.
 The range which is the used for the sender is called the send sliding window.
The range that is the used for the receiver is called the receiver sliding window.
 The send window is an imaginary box covering the sequence-numbers of the data-
frames.
 In each window position,
Some sequence-numbers define the frames that have been sent.
Some sequence-numbers define the frames that can be sent.
 The maximum size of the window is 2m
– 1.
 The size can be fixed and set to the maximum value.
 Figure 11.12 shows a sliding window of size 15 (m =4).

2-14
5b. Explain frame format and transitional phases of point to point protocol. (08 Marks)
Ans:
PPP Frame Format (Point to Point Protocol)
• PPP uses a character-oriented (or byte-oriented) frame (Figure 11.20).
• Various fields of PPP frame are:
1) Flag
 This field has a synchronization pattern 01111110.
 This field identifies both the beginning and the end of a frame.
2) Address
 This field is set to the constant value 11111111 (broadcast address).
3) Control
 This field is set to the constant value 00000011.
 PPP does not provide any flow control.
 Error control is also limited to error detection.
4) Protocol
 This field defines what is being carried in the payload field.
 Payload field carries either i) user data or ii) other control information.
 By default, size of this field = 2 bytes.
5) Payload field
 This field carries either i) user data or ii) other control information.
 By default, maximum size of this field = 1500 bytes.
 This field is byte-stuffed if the flag-byte pattern appears in this field.
 Padding is needed if the payload-size is less than the maximum size.
6) FCS
 This field is the PPP error-detection field.
 This field can contain either a 2- or 4-byte standard CRC.
PPP Transition Phases
• The transition diagram starts with the dead state (Figure 11.21).
1) Dead State
 In dead state, there is no active carrier and the line is quiet.
2) Establish State
 When 1 of the 2 nodes starts communication, connection goes into the establish state.
 In establish state, options are negotiated between the two parties.
3) Authenticate State
 If the 2 parties agree that they need authentication,
Then the system needs to do authentication;
Otherwise, the parties can simply start communication.
4) Open State
 Data transfer takes place in the open state.
5) Terminate State
 When 1 of the endpoints wants to terminate connection, system goes to terminate state

2-15
6a. Explain i) CSMA ii) CSMA/CD. (12 Marks)
Ans (i):
CSMA
• CSMA is based on the principle “sense before transmit” or “listen before talk.”
• Here is how it works:
1) Each station checks the state of the medium: idle or busy.
2) i) If the medium is idle, the station sends the data.
ii) If the medium is busy, the station defers sending.
• CSMA can reduce the possibility of collision, but it cannot eliminate it.
Vulnerable Time
 The vulnerable time is the propagation time Tp. (Figure 12.7).
 Collision occurs when
→ a station sends a frame, and
→ other station also sends a frame during propagation time.
Persistence Methods
 Q: What should a station do if the channel is busy or idle?
Three methods can be used to answer this question:
1) 1-persistent method 2) Non-persistent method and 3) p-persistent method

2-16
1) 1-Persistent
¤ Before sending a frame, a station senses the line (Figure 12.10a).
i) If the line is idle, the station sends immediately (with probability = 1).
ii) If the line is busy, the station continues sensing the line.
2) Non-persistent
¤ Before sending a frame, a station senses the line (Figure 12.10b).
i) If the line is idle, the station sends immediately.
ii) If the line is busy, the station waits a random amount of time and then
senses the line again.
3) P-Persistent
¤ After the station finds the line idle, it follows these steps(Figure 12.10c):
1) With probability p, the station sends the frame.
2) With probability q=1-p, the station waits for the beginning of the next
time-slot and checks the line again.
i) If line is idle, it goes to step 1.
ii) If line is busy, it assumes that collision has occurred and uses the
back off procedure.
Ans (ii):
CSMA/CD
• Disadvantage of CSMA: CSMA does not specify the procedure after a collision has occurred.
Solution: CSMA/CD enhances the CSMA to handle the collision.
1) A station
→ sends the frame &
→ then monitors the medium to see if the transmission was successful or not.
2) If the transmission was unsuccessful (i.e. there is a collision), the frame is sent again.
• In the Figure 12.12,
 At time t1, station A has executed its procedure and starts sending the bits of its frame.
 At time t2, station C has executed its procedure and starts sending the bits of its frame.
 The collision occurs sometime after time t2.
 Station C detects a collision at time t3 when it receives the first bit of A's frame.
Station C immediately aborts transmission.
 Station A detects collision at time t4 when it receives the first bit of C's frame.
Station A also immediately aborts transmission.
• Station A transmits for the duration t4–t1.
Station C transmits for the duration t3-t2.
• For the protocol to work:
Length of any frame divided by the bit rate must be more than either of these durations.
Minimum Frame Size
• For CSMA/CD to work, we need to restrict the frame-size.
• Before sending the last bit of the frame, the sender must
→ detect a collision and
→ abort the transmission.
• This is so because the sender
→ does not keep a copy of the frame and
→ does not monitor the line for collision-detection.
• Frame transmission time Tfr is given by
Tfr=2Tp where Tp=maximum propagation time

2-17
Procedure
• CSMA/CD is similar to ALOHA with 2 differences (Figure 12.13):
1) Addition of the persistence process.
¤ We need to sense the channel before sending the frame by using non-persistent,
1-persistent or p-persistent.
2) Frame transmission.
i) In ALOHA, first the entire frame is transmitted and then acknowledgment is
waited for.
ii) In CSMA/CD, transmission and collision-detection is a continuous process.
Throughput
• The throughput of CSMA/CD is greater than pure or slotted ALOHA.
• The maximum throughput is based on
→ different value of G
→ persistence method used (non-persistent, 1-persistent, or p-persistent) and
→ „p‟ value in the p-persistent method.
• For 1-persistent method, the maximum throughput is 50% when G =1.
• For non-persistent method, the maximum throughput is 90% when G is between 3 and 8.
6b. Describe 802.3 MAC frame. (08 Marks)
Ans: For answer, refer Solved Paper June-2013 Q.No.6c.
7a. Explain IEEE 802.11 architecture. (10 Marks)
7b. Bring out the differences b/w repeaters, bridges, routers and gateways. (10 Marks)

2-18
8a. Explain wrt IPV4, classful addressing and classless addressing. (10 Marks)
Ans:
Classful Addressing
• IPv4 addressing used the concept of classes.
• This architecture is called classful addressing.
• In classful addressing, the address space is divided into five classes: A, B, C, D, and E.
• Each class occupies some part of the address space.
Classes and Blocks
 Class A addresses were designed for large organizations with a large number of
attached hosts or routers.
 Class B addresses were designed for midsize organizations with tens of thousands of
 Class C addresses were designed for small organizations with a small number of
 In classfnl addressing, a large part of the available addresses were wasted.
 In classful addressing, an IP address in class A, B, or C is divided into netid and hostid.
 These parts are of varying lengths, depending on the class of the address.
Mask
 The mask can help us to find the netid and the hostid.
 For example, the mask for a class A address has eight 1s, which means the first 8 bits
of any address in class A define the netid; the next 24 bits define the hostid.

2-19
Classless Addressing
• To overcome address depletion and give more organizations access to the Internet, classless
addressing was designed and implemented.
• In this scheme, there are no classes, but the addresses are still granted in blocks
Address Blocks
 In classless addressing, when an entity, small or large, needs to be connected to the
Internet, it is granted a block (range) of addresses.
 The size of the block (the number of addresses) varies based on the nature and size of
the entity.
 For example,
A householdmay be given only two addresses;
A large organization may be given thousands of addresses.
Restriction
• To simplify the handling of addresses, the Internet authorities impose three restrictions
on classless address blocks:
1) The addresses in a block must be contiguous, one after another.
2) The number of addresses in a block must be a power of 2 (I, 2, 4, 8, ... ).
3) The first address must be evenly divisible by the number of addresses.
Mask
 In classless addressing the mask for a block can take any value from 0 to 32. It is very
convenient to give just the value of n preceded by a slash (CIDR notation).
 In IPv4 addressing, a block of addresses can be defined as x.y.z.t/n in which x.y.z.t
defines one of the addresses and the /n defines the mask.

2-20
8c. Explain in detail IPV6 packet format. (10 Marks)
Ans:
IPV6 PACKET FORMAT
• IP header contains following fields (Figure 22.6):
1) Version
 This specifies version number of protocol. For IPv6, version=6.
2) Traffic Class
 This field is used to distinguish different payloads with different delivery requirements.
(Traffic class replaces the type-of-service field in IPv4).
3) Flow Label
 This field is designed to provide special handling for a particular flow of data.
4) Payload Length
 This indicates length of data (excluding header). Maximum length=65535 bytes.
 The length of the base-header is fixed (40 bytes); only the length of the payload needs
to be defined.
5) Next Header
 This identifies type of extension header that follows the basic header.
6) Hop Limit
 This specifies number of hops the packet can travel before being dropped by a router.
(Hop limit serves the same purpose as the TTL field in IPv4).
7) Source and Destination Addresses
 These identify source host and destination host respectively.
8) Payload
 The payload contains zero or more extension headers (options) followed by the data
from other protocols (UDP, TCP, and so on).
 The payload can have many extension headers as required by the situation.
 Each extension header has 2 mandatory fields (Figure 22.7):
1) Next header and
2) Length
• Two mandatory fields are followed by information related to the particular option.

2-21
Concept of Flow & Priority in IPv6
• To a router, a flow is a sequence of packets that share the same characteristics such as
→ traveling the same path
→ using the same resources or
→ having the same kind of security
• A router that supports the handling of flow labels has a flow label table.
• The table has an entry for each active flow label.
Each entry defines the services required by the corresponding flow label.
• When a router receives a packet, the router consults its flow label table.
• Then, the router provides the packet with the services mentioned in the entry.
Fragmentation & Reassembly
• Fragmentation of the packet is done only by the source, but not by the routers.
The reassembling is done by the destination.
• At routers, the fragmentation is not allowed to speed up the processing in the router.
• Normally, the fragmentation of a packet in a router needs a lot of processing. This is because
1) The packets need to be fragmented.
2) All fields related to the fragmentation need to be recalculated.
• The source will
→ check the size of the packet and
→ make the decision to fragment the packet or not.
• If packet-size is greater than the MTU of the network, the router will drop the packet.
• Then, the router sends an error message to inform the source.

3-1
1a. What is data communication? What are its four fundamental characteristics? With a
neat diagram, explain the components of data communication system. (08 Marks)
Ans: For answer, refer Solved Paper Dec-2013 Q.No.1a.
1b. Assume that five devices are connected in mesh topology. How many cables are
needed? How many ports are needed for each device? (02 Marks)
Ans:
Solution:
 For ‘n’ nodes, there are n(n-1)/2 duplex-mode links.
Here n=5, therefore number of cables needed = 5(5-1)/2 = 10
 Every device must have (n–1) I/O ports to be connected to the other (n-1) devices.
Here n=5, number of ports needed for each device = 5-1 = 4
1c. With a diagram, explain the functionalies of each layer of OSI model (10 Marks)
2a. Explain the different causes for transmission impairments during signal
transmission through media. (06 Marks)
2b. Define bandwidth: A peroidic signal has bandwidth 20 Hz.The highest frequency is
60 Hz. What is the lowest frequency? Draw the spectrum, if signal contains all
frequencies of same amplitude. (04 Marks)
Ans:
Bandwidth
• Bandwidth is defined as the range of frequencies that the channel can carry. (Figure 3.13).
• Bandwidth of a signal is expressed in terms of its frequencies.
• Bandwidth is calculated by the difference b/w the max. frequency and the min. frequency.

3-2
2c. What is line coding? Describe and represent the information sequence ‘101000110’
using Biphase and Bipolar schemes. (10 Marks)
Ans:
LINE CODING
• Line-coding is the process of converting digital-data to digital-signals (Figure 4.1).
• The data are stored in computer memory as sequences of bits (0s or 1s).
• Line-coding converts a sequence of bits to a digital-signal.
Solution:

3-3
3a. What is multiplexing? With neat diagram, explain FDM. (06 Marks)
Ans:
MULTIPLEXING
• Multiplexing allows simultaneous transmission of multiple signals across a single data-link.
Frequency Division Multiplexing (FDM)
• FDM is an analog multiplexing technique that combines analog signals.
• FDM can be used when the bandwidth of a link is greater than the combined bandwidths of the
signals to be transmitted.
1) Multiplexing Process
1) Each sending-device generates modulated-signals with different carrier-frequencies (f1,
f2, & f3).
2) Then, these modulated-signals are combined into a single multiplexed-signal.
3) Finally, the multiplexed-signal is transported by the link.
• Carrier-frequencies are separated by sufficient bandwidth to accommodate modulated-signal.
• Channels can be separated by strips of unused bandwidth called guard bands.
• Guard bands prevent signals from overlapping.
• In addition, carrier-frequencies must not interfere with the original data frequencies.
• Although FDM is considered as analog multiplexing technique,sources can produce digital-signal
• Digital-signal can be sampled, changed to analog-signal, and then multiplexed by using FDM.
2) Demultiplexing Process
1) The demultiplexer uses filters to divide the multiplexed-signal into individual-signals.
2) Then, the individual signals are passed to a demodulator.
3) Finally, the demodulator
→ separates the individual signals from the carrier signals and
→ passes the individual signals to the output-lines.

3-4
3b. What is spread spectrum? Explain with an example DSSS. (06 Marks)
Ans:
SPREAD SPECTRUM
• Spread-spectrum is used in wireless applications. (Figure 6.27).
• Spread-spectrum is used to provide secure communication by spreading the original spectrum
needed for each station.
Where B = required bandwidth
Bss= expanded bandwidth
• Two types of spread-spectrum:
1) Frequency hopping spread-spectrum (FHSS) and
2) Direct sequence spread-spectrum (DSSS).
Direct Sequence Spread-Spectrum (DSSS)
• This technique expands the bandwidth of the original signal.
• Each data-bit is replaced with ‘n’ bits using a spreading-code.
• Each bit is assigned a code of ‘n’ bits called chips.
• The chip-rate is ‘n’ times that of the data-bit (Figure 6.32).
• For example (Figure 6.33):
 Consider the Barker sequence used in a wireless LAN. Here n =11.
 Assume: The original signal and the chips in the chip-generator use polar NRZ encoding.
 The spreading-code is 11 chips having the pattern 10110111000.
 If the original signal-rate is N, the rate of the spread signal is 1/N.
 This means that the required bandwidth for the spread signal is 11 times larger than the
bandwidth of the original signal.
• The spread signal can provide privacy if the attacker does not know the code.
• It can also provide immunity against interference if each station uses a different code.

3-5
3c. With a neat diagram, explain how message can be sent from one system to another
using datagram networks. (08 Marks)
Ans:
Datagram Networks
• This is analogous to postal system.
• The message is divided into packets of fixed or variable size (Figure 8.7).
• Each packet is routed independently through the network.
• Each packet has a header that contains source and destination addresses.
• Each switch examines the header to determine the next hop in the path to the destination.
• If the transmission line is busy
then the packet is placed in the queue until the line becomes free.
• Advantage:
1) High utilization of transmission-line can be achieved by sharing among multiple
packets.
2) There is no resource reservation; resources are allocated on-demand.
• Disadvantages:
1) Packets may arrive out-of-order, and re-sequencing may be required at the destination
2) Loss of packets may occur when a switch has insufficient buffer
• It is the responsibility of an upper-layer protocol to
→ reorder the datagrams or
→ ask for lost datagrams.
• The datagram-networks are referred to as connectionless networks. This is because
1) The switch does not keep information about the connection state.
2) There are no setup or teardown phases.
3) Each packet is treated the same by a switch regardless of its source or destination.
Routing Table
 Each switch has a routing-table which is based on the destination-address.
 The routing-tables are dynamic & updated periodically. (Figure 8.8).
 The destination-addresses and the corresponding forwarding output-ports are recorded
in the tables.

3-6
Efficiency
¤ Datagram-networks are more efficient when compared to circuit-switched-
network. This is because
1) Resources are allocated only when there are packets to be transferred.
2) If there is a delay, the resources can be re-allocated.
Delay
¤ Datagram-networks may have greater delay when compared to circuit-switched-
network. This is because each packet may experience a wait at a switch before it is
forwarded.
¤ In Figure 8.9, there are three transmission times (3T), three propagation delays,
and two waiting times (W1+ W2).
4a. Define hamming distance. Explain simple parity check code C(5, 4) with dmin = 2.
How many bits can be corrected. (06 Marks)
4b. Find code word c(x), using CRC for the information d(x) = x3
+1 with generator
polynomial t(x) = x3
+x+ 1. (08 Marks)
4c. Explain with an example, the computation of Internet checksum. List the steps
undertaken by the sender and receiver for error detection. (06 Marks)
5a. Why bit stuffing and byte stuffing are needed? Explain with examples. (06 Marks)

3-7
5b. With neat figures, explain briefly:
i) Go-back N
ii) Selective repeat ARQ protocols. (10 Marks)
Ans (i): For answer, refer Solved Paper Dec-2013 Q.No.5a.(ii).
Ans (ii):
Selective Repeat ARQ Protocol
• In a noisy link, a frame has a higher probability of damage, which means the resending of
multiple frames.
• This resending uses up the bandwidth and slows down the transmission.
• For noisy links, there is another mechanism that does not resend N frames when just one frame
is damaged; only the damaged frame is resent.
• This mechanism is called Selective Repeat ARQ.
• It is more efficient for noisy links, but the processing at the receiver is more complex.
• The Selective Repeat Protocol also uses two windows: 1) a send window and 2) a receive
window. (Figure 11.18).
• The size of the send window is much smaller; it is 2m-1
.
• The smaller window size means less efficiency in filling the pipe, but the fact that there are
fewer duplicate frames can compensate for this.
• The Selective Repeat Protocol allows as many frames as the size of the receive window to arrive
out of order and be kept until there is a set of in-order frames to be delivered to the network-
layer. (Figure 11.19).
• Because the sizes of the send window and receive window are the same, all the frames in the
send frame can arrive out of order and be stored until they can be delivered.
• We need, however, to mention that the receiver never delivers packets out of order to the
network-layer.
5c. Explain the frame format of PPP protocol. (04 Marks)
Ans: For answer, refer Solved Paper Dec-2013 Q.No.5b.
6a. Describe CSMA/CD protocol, with neat flow diagram. (06 Marks)
Ans: For answer, refer Solved Paper Dec-2013 Q.No.6a.(ii)
6b. What is channelization? Explain CDMA, with an example. (08 Marks)

3-8
6c. Discuss i) Frame Format and ii) Frame Length of 802.3 MAC (06 Marks)
Ans (i): For answer, refer Solved Paper June-2013 Q.No.6c.
Ans (ii):
Frame Length
• Ethernet has imposed restrictions on both minimum & maximum lengths of a frame (Fig 13.5).
• The minimum length restriction is required for the correct operation of CSMA/CD.
• Minimum length of frame = 64 bytes.
1) Minimum data size = 46 bytes.
2) Header size + Trailer size = 14 + 4 = 18 bytes.
• The minimum length of data from the upper layer = 46 bytes.
• If the upper-layer packet is less than 46 bytes, padding is added to make up the difference.
• Maximum length of frame =1518 bytes.
1) Maximum data size = 1500 bytes.
2) Header size + trailer size = 14 + 4 = 18 bytes.
• The maximum length restriction has 2 reasons:
1) Memory was very expensive when Ethernet was designed.
A maximum length restriction helped to reduce the size of the buffer.
2) This restriction prevents one station from
→ monopolizing the shared medium
→ blocking other stations that have data to send.

3-9
7a. Explain the different types of addressing mechanisms in IEEE 802.11. (06 Marks)
Ans:
IEEE 802.11 Addressing
• In an Ethernet-network, each station has its own NIC (6-byte  48 bits).
• The NIC provides the station with a 6-byte physical-address (or Ethernet-address).
• For example, the following shows an Ethernet MAC address:
(NIC  network interface card)
Unicast, Multicast, and Broadcast Addresses
• A source-address is always a unicast address i.e. the frame comes from only one station.
• However, the destination-address can be 1) Unicast 2) Multicast or 3) Broadcast.
• As shown in Figure 13.4,
If LSB of first byte in a destination-address is 0,
Then, the address is unicast;
Otherwise, the address is multicast.
1) A unicast destination-address defines only one recipient.
¤ The relationship between the sender and the receiver is one-to-one.
2) A multicast destination-address defines a group of addresses.
¤ The relationship between the sender and the receivers is one-to-many.
3) The broadcast address is a special case of the multicast address.
¤ The recipients are all the stations on the LAN.
¤ A broadcast destination-address is 48 1s (6-byte  48 bits).
• Standard Ethernet uses a coaxial cable (bus topology) or a set of twisted-pair cables with a hub
(star topology) (Figure 13.5).

3-10
7b. With neat diagram, explain layers of Bluetooth. (06 Marks)
Ans:
Bluetooth Layers
• Bluetooth uses several layers that do not exactly match those of the Internet model (Fig 15.19)
1) Radio Layer
• The radio layer is roughly equivalent to the physical layer of the Internet model.
• Bluetooth devices are low-power and have a range of 10 m.
1) Band
 Bluetooth uses a 2.4-GHz ISM band divided into 79 channels of 1 MHz each.
2) FHSS
 Bluetooth uses the FHSS in the physical layer to avoid interference from other devices
or other networks.
3) Modulation
 To transform bits to a signal, Bluetooth uses a sophisticated version of FSK, called GFSK
(FSK with Gaussian bandwidth filtering).
 GFSK has a carrier frequency.
2) Baseband Layer
• The baseband layer is roughly equivalent to the MAC sublayer in LANs.
• The access method is TDMA.
• The primary and secondary communicate with each other using time slots.
• The communication is only between the primary and a secondary; secondaries cannot
communicate directly with one another.
TDMA
 Bluetooth uses a form of TDMA that is called TDD-TDMA (timedivision duplex TDMA).
 TDD-TDMA is a kind of half-duplex communication in which the secondary and receiver
send and receive data, but not at the same time (halfduplex);
Links
 Two types of links can be created between a primary and a secondary:
1) SCA link (Synchronous Connection-oriented Link)
¤ This link is used when avoiding latency is more important than data-integrity.
(Latency  delay in data delivery Integrity  error-free delivery)
2) ACL link (Asynchronous Connectionless Link)
¤ This link is used when data-integrity is more important than avoiding latency.
3) L2CAP
• The L2CAP is roughly equivalent to the LLC sublayer in LANs.
• It is used for data exchange on an ACL link. (L2CAP  Logical Link Control and Adaptation
Protocol)
• The L2CAP has specific duties: 1) Multiplexing
2) Segmentation and reassembly
3) QoS (quality of service) and
4) Group management.

3-11
7c. What is a bridge? Explain with an example the bridge learning and forwarding
process of transparent bridge. (08 Marks)
Ans:
Bridges
• A bridge operates in both the physical and the data link layer (Figure 15.5).
• As a physical layer device, it regenerates the signal it receives.
• As a data link layer device, the bridge can check the physical (MAC) addresses (source and
destination) contained in the frame.
• A bridge has filtering capability.
• It can check the destination address of a frame and decide if the frame should be forwarded or
dropped.
• If the frame is to be forwarded, the decision must specify the port.
• A bridge has a table that maps addresses to ports.
• A bridge has a table used in filtering decisions.
• A bridge does not change the physical (MAC) addresses in a frame
Transparent Bridges
• A transparent bridge is a bridge in which the stations are completely unaware of the bridge's
existence.
• If a bridge is added or deleted from the system, reconfiguration of the stations is unnecessary.
• According to the IEEE 802.1 d specification, a system equipped with transparent bridges must
meet three criteria:
1) Frames must be forwarded from one station to another.
2) The forwarding table is automatically made by learning frame movements in the
network.
3) Loops in the system must be prevented.
Forwarding
• A transparent bridge must correctly forward the frames. (Figure 15.6).
Learning
• The earliest bridges had forwarding tables that were static.
• The systems administrator would manually enter each table entry during bridge setup.
• If a station was added or deleted, the table had to be modified manually

3-12
8a. Explain the following fields in IPV4 packet header: (06 Marks)
i) Identification ii) flags iii) fragmentation offset.
Ans:
Fragmentation
Maximum Transfer Unit (MTU)
• Each network imposes a restriction on maximum size of packet that can be carried. This is
called the MTU (maximum transmission unit).
• For example:
For Ethernet, MTU = 1500 bytes
• When IP wants send a packet that is larger than MTU of physical-network, IP breaks packet into
smaller fragments. This is called fragmentation (Figure 19.5).
• Source host or router is responsible for fragmentation.
Destination host is responsible for reassembly.
Figure 19.5: Packet Fragmentation
Fields Related to Fragmentation & Reassembly
• Three fields in the IP header are used to manage fragmentation and reassembly:
1) Identification 2) Flags 3) Fragmentation offset.
1) Identification
• This field is used to identify to which datagram a particular fragment belongs to (so that
fragments for different packets do not get mixed up).
• To guarantee uniqueness, the IP protocol uses an up-counter to label the datagrams.
• When the IP protocol sends a datagram, IP protocol
→ copies the current value of the counter to the identification field and
→ increments the up-counter by 1.
• When a datagram is fragmented, the value in the identification field is copied into all fragments.
• The identification number helps the destination in reassembling the datagram.
2) Flags
• This field has 3 bits.
1) The leftmost bit is not used.
2) DF bit (Don't Fragment):
i) If DF=1, the router should not fragment the datagram. Then, the router
→ discards the datagram and
→ sends an error-message to the source host.
ii) If DF=0, the router can fragment the datagram if necessary.
3) MF bit (More Fragment):
i) If MF=1, there are some more fragments to come.
ii) If MF=0, this is last fragment.
3) Fragmentation Offset
• This field identifies location of a fragment in a packet.
• This field is the offset of the data in the original datagram.

3-13
8b. What is NAT? How can NAT help in address depletion with a diagram? (06 Marks)
Ans:
NAT (Network address translation)
• NAT enables a user to have a large set of addresses internally and one address, or a small set
of addresses, externally. (Figure 19.10).
• The traffic inside can use the large set; the traffic outside, the small set.
• To separate the addresses used inside the home/business and the ones used for the Internet,
the Internet authorities have reserved 3 sets of addresses as private addresses (Table 19.3).
• Any organization can use an address out of this set without permission from the Internet
authorities.
• Everyone knows that these reserved addresses are for private networks.
• They are unique inside the organization, but they are not unique globally.
• No router will forward a packet that has one of these addresses as the destination address.
• The site must have only one single connection to the global Internet through a router that runs
the NAT software.
Address Translation
• All the outgoing packets go through the NAT router, which replaces the source address in the
packet with the global NAT address. (Figure 19.11).
• All incoming packets also pass through the NAT router, which replaces the destination address
in the packet (the NAT router global address) with the appropriate private address (Fig19.11)

3-14
8c. i) What is the need to change from IPV4 to 1PV6? ii) Write IPV6 header and
describe its fields. (08 Marks)
Ans (i):
Need to Change from IPv4 to IPv6
1) Header Format
 IPv6 uses a new header format.
 Options are
→ separated from the base-header and
→ inserted between the base-header and the data.
 This speeds up the routing process (because most of the options do not need to be
checked by routers).
2) New Options
 IPv6 has new options to allow for additional functionalities.
3) Extension
 IPv6 is designed to allow the extension of the protocol if required by new technologies
or applications.
4) Resource Allocation
 In IPv6,
→ type-of-service (TOS) field has been removed
→ two new fields: 1) traffic class and 2) flow label, are added to enable the source
to request special handling of the packet.
 This mechanism can be used to support real-time audio and video.
5) Security
 The encryption option provides confidentiality of the packet.
 The authentication option provides integrity of the packet.
Ans (ii): For answer, refer Solved Paper Dec-2013 Q.No.8b.

4-1
1a. What are the components of data communication system? Explain. (05 Marks)
1b. With a neat diagram, explain interaction b/w layers in the OSI model. (10 Marks)
Ans:
OSI MODEL
• The OSI model describes an abstract reference-model for network-architecture.
• Purpose of reference model: To provide a framework for the development of protocols.
• OSI consists of 7 separate layers.
• Each layer defines a part of the process of moving information across a network.
Layered Architecture
• The OSI model consists of 7 layers:
1) Physical (layer 1)
2) Data link (layer 2)
3) Network (layer 3)
4) Transport (layer 4)
5) Session (layer 5)
6) Presentation (layer 6) and
7) Application (layer 7).
• Figure 2.3 shows the layers involved when a message is sent from device-A to device-B.
• As the message travels from device-A to device-B, the message may pass through many
intermediate-nodes.
• Usually, the intermediate-nodes uses only first 3 layers of the OSI model.
• Each layer calls upon the services of the layer just below it.
For example: Layer-3 calls upon the services of Layer-2.
Peer-to-Peer Processes
• The processes on each machine that communicate at a given layer are called peer-to-peer
processes.
• At the physical-layer, communication is direct: In figure 2.3, device-A sends a stream of bits to
device-B (through intermediate-nodes).
• At the higher layers,
→ firstly, data must move down through the layers on device-A
→ then, data must move over to intermediate-nodes
→ finally, data must move back up through the layers on device-B.

4-2
• Here is how data transmission occurs:
1) In the sending-device, each layer
→ adds a header to the message received from the above layer and
→ passes the whole package to the below layer.
2) At Layer-1, the entire package is
→ converted to an electrical signal &
→ transmitted to the receiving-device.
3) At the receiving-device,
→ the message is unwrapped layer by layer
→ each layer receives and removes the data meant for it.
For example:
i) Layer-2 removes the data meant for it & passes the remaining data to Layer-3.
ii) Then, Layer-3 removes data meant for it & passes remaining data to Layer-4.
Interfaces Between Layers
 An interface between layers makes it possible for the passing of the message
→ down through the layers of the sending-device and
→ back up through the layers of the receiving-device.
 Each interface defines the services a layer must provide for the layer above it.
 Well-defined interfaces & layer-functions provide modularity to a network.
1c. What is the difference between a physical and logical address? Explain with
example. (05 Marks)
Ans:
Logical address
• At the network-layer, addresses are called IP addresses.
• IP address uniquely defines the connection of a device to the Internet.
• IP address is assigned to each system in the Internet.
• The IP addresses are global, with the whole Internet as the scope.
• Address size = 32 bits.
Eg: 192.12.41.91 in hexadecimal format
Physical Address
• At the data link-layer, addresses are called MAC addresses
• The MAC address defines a specific host or router in a network (LAN or WAN).
• The MAC address is burned into the ROM of the NIC card.
• The MAC address is unique for each system and cannot be changed.
• The MAC addresses are locally defined addresses.
• Address size = 48 bits.
Eg: 12.14.14.19.25.37 in hexadecimal format

4-3
2a. Distinguish between low pass channel and a band pass channel. (06 Marks)
Ans:
Transmission of Digital Signals
• Two methods for transmitting a digital signal:
1) Baseband transmission 2) Broadband transmission (using modulation).
1) Baseband Transmission
• Baseband transmission means sending a digital signal over a channel without changing the
digital signal to an analog signal (Figure 3.19).
• Baseband transmission requires that we have a low-pass channel.
• Low-pass channel means a channel with a bandwidth that starts from zero.
• For example, we can have a dedicated medium with a bandwidth constituting only one channel.
• Two cases of a baseband communication:
Case 1: Low-pass channel with a wide bandwidth (Figure 3.20a).
Case 2: Low-pass channel with a limited bandwidth (Figure 3.20b).
Low-Pass Channel with Wide Bandwidth
 To preserve the shape of a digital signal, we need to send the entire spectrum i.e. the
continuous range of frequencies between zero and infinity.
 This is possible if we have a dedicated medium with an infinite bandwidth between the
sender and receiver.
 If we have a medium with a very wide bandwidth, 2 stations can communicate by using
digital signals with very good accuracy (Figure 3.21).

4-4
2) Broadband Transmission (Using Modulation)
• Broadband transmission or modulation means changing the digital signal to an analog signal for
transmission.
• Modulation allows us to use a bandpass channel (Figure 3.24).
• Bandpass channel means a channel with a bandwidth that does not start from zero.
• This type of channel is more available than a low-pass channel.
• If the available channel is a bandpass channel,
We cannot send the digital signal directly to the channel;
We need to convert digital signal to an analog signal before transmission (Figure 3.25).
2b. A network with bandwidth of 10Mbps can pass only an average of 18,000 frames
per minute with each frame carrying an average of 10,000 bits. What is the throughput
of this network? (04 Marks)
Ans:
2c. Compare and contrast between i) PCM and ii) DM. (06 Marks)
Ans(i): For answer, refer Solved Paper June-2013 Q.No.2c.
Ans(ii): For answer, refer Solved Paper Dec-2013 Q.No.2c.

4-5
2d. Explain polar biphase Manchester and differential Manchester encoding schemes
with example. (04 Marks)
Ans:
1) Manchester Encoding
• This is a combination of NRZ-L & RZ schemes.
• There is always a transition at the middle of the bit. Either
i) from high to low (for 0) or
ii) from low to high (for 1).
• It uses only two voltage levels (Figure 4.8).
• The duration of the bit is divided into 2 halves.
• The voltage
→ remains at one level during the first half &
→ moves to the other level in the second half.
• The transition at the middle of the bit provides synchronization.
2) Differential Manchester
• This is a combination of NRZ-I and RZ schemes.
• There is always a transition at the middle of the bit, but the bit-values are determined at the
beginning of the bit.
• If the next bit is 0, there is a transition.
If the next bit is 1, there is none.
3a. A multiplexer combines four 100 kbps channels using a time slot of 2 bits. Show the
output with arbitarary inputs. What is the frame rate? What is the frame duration?
What is the bit rate? Whart is the bit duration? (04 Marks)
Ans:

4-6
3b. Explain following modulation techniques (06 Marks)
i) Amplitude modulation ii) Frequency modulation
Ans(i):
Amplitude Modulation
• In AM transmission, the carrier signal is modulated so that its amplitude varies with the
changing amplitudes of the modulating signal (Figure 5.16).
• The frequency and phase of the carrier remain the same; only the amplitude changes to follow
variations in the information.
AM Bandwidth
 The modulation creates a bandwidth that is twice the bandwidth of the modulating
signal and covers a range centered on the carrier frequency.
 However, the signal components above and below the carrier frequency carry exactly
the same information.
 For this reason, some implementations discard one-half of the signals and cut the
bandwidth in half.
Ans(ii):
Frequency Modulation
•In FM transmission, the frequency of the carrier signal is modulated to follow the changing
voltage level (amplitude) of the modulating signal (Figure 5.18).
• The peak amplitude and phase of the carrier signal remain constant, but as the amplitude of
the information signal changes, the frequency of the carrier changes correspondingly
FM Bandwidth
 The actual bandwidth is difficult to determine exactly, but it can be shown empirically
that it is several times that of the analog signal or 2(1 + β)B where β is a factor that
depends on modulation technique with a common value of 4.

4-7
3c. With diagram, explain data transfer phase in virtual circuit network. (10 Marks)
Ans:
Virtual Circuit Network (VCN)
• This is similar to telephone system.
• A virtual-circuit network is a combination of circuit-switched-network and datagram-network.
• The network has switches that allow traffic from sources to destinations.
Three Phases
• A source and destination need to go through 3 phases: setup, data-transfer, and teardown.
1) In setup phase, the source and destination use their global addresses to help
switches make table entries for the connection.
2) Data-transfer occurs between these 2 phases.
3) In the teardown phase, the source and destination inform the switches to delete
the corresponding entry.
1) Data Transfer Phase
• To transfer a frame from a source to its destination, all switches need to have a table-entry for
this virtual-circuit.
• The table has four columns.
• The switch holds 4 pieces of information for each virtual-circuit that is already set up.
 As shown in Figure 8.12, a frame arrives at port 1 with a VCI of 14.
 When the frame arrives, the switch looks in its table to find port 1 and a VCI of 14.
 When it is found, the switch knows to change the VCI to 22 & send out the frame from
port 3.
 As shown in Figure 8.13, each switch changes the VCI and routes the frame.
 The data-transfer phase is active until the source sends all its frames to the destination.
 The procedure at the switch is the same for each frame of a message.
 The process creates a virtual circuit, not a real circuit, b/w the source and destination.

4-8
2) Setup Phase
• A switch creates an entry for a virtual-circuit.
• For example, suppose source A needs to create a virtual-circuit to B.
• Two steps are required: i) Setup-request and ii) Acknowledgment.
i) Setup Request
 A setup-request frame is sent from the source to the destination (Figure 8.14).
ii) Acknowledgment
 A special frame, called the acknowledgment-frame, completes the entries in the
switching-tables (Figure 8.15).
3) Teardown Phase
• Source-A, after sending all frames to B, sends a special frame called a teardown request.
• Destination-B responds with a teardown confirmation frame.
• All switches delete the corresponding entry from their tables.

4-9
4a. Explain CRC error detection method with an example (06 Marks)
Ans:
Cyclic Redundancy Check (CRC)
• CRC is a cyclic code that is used in networks such as LANs and WANs.
• Let Size of data-word = k bits (here k=4). Size of code-word = n bits (here n=7).
Size of divisor = n-k+1 bits (here n-k+1=4). (Augmented  increased)
1) At Sender
 n-k 0s is appended to the data-word to create augmented data-word. (here n-k=3).
 The augmented data-word is fed into the generator (Figure 10.6).
 The generator divides the augmented data-word by the divisor.
 The remainder is called check-bits (r2r1r0).
 The check-bits (r2r1r0) are appended to the data-word to create the code-word.
2) At Receiver
 The possibly corrupted code-word is fed into the checker.
 The checker is a replica of the generator.
 The checker divides the code-word by the divisor.
 The remainder is called syndrome bits (r2r1r0).
 The syndrome bits are fed to the decision-logic-analyzer.
 The decision-logic-analyzer performs following functions:
i) For No Error
¤ If all syndrome-bits are 0s, the received code-word is accepted.
¤ Data-word is extracted from received code-word.
ii) For Error
¤ If all syndrome-bits are not 0s, the received code-word is discarded (Fig 10.7b).

4-10
4b. Explain the structure of encoder and decoder for a Hamming code. (04 Marks)
4c. What is internet checksum? If a sender needs to send four data items 0X3456,
0XABCC, 0X02BC, 0XEEEE, answer the following
i) Find the checksum at sender's site.
ii) Find the checksum at receiver's site if there is no error. (10 Marks)
Ans:
Solution:
5a. Explain i) GO-BACK-N ARQ and ii) selective-repeat-ARQ. List the differences
between them. (10 Marks)
Ans (i): For answer, refer Solved Paper Dec-2013 Q.No.5a.(ii).
Ans (ii): For answer, refer Solved Paper June-2014 Q.No.5b.(ii).

4-11
5b. Explain the different frame types in HDLC. (06 Marks)
Ans:
Three Types of Frames
1) Information frames (I-frames): are used to transport user data and control information
relating to user data (piggybacking).
2) Supervisory frames (S-frames): are used only to transport control information.
3) Unnumbered frames (U-frames): are reserved for system management.
Information carried by U-frames is intended for managing the link itself.
• Each type of frame serves as an envelope for the transmission of a different type of message.
Control Fields of HDLC Frames
• The control field determines the type of frame and defines its functionality (Figure 11.17).
1) Control Field for I-Frames
• The subfields in the control field are:
1) The first bit defines the type.
If the first bit of the control field is 0, this means the frame is an I-frame.
2) The next 3 bits N(S) define the sequence-number of the frame.
With 3 bits, we can define a sequence-number between 0 and 7
3) Last 3 bits N(R) correspond to the acknowledgment-number when piggybacking is used
4) The single bit between N(S) and N(R) is called the P/F bit.
The P/F field is a single bit with a dual purpose. It can mean poll or final.
i) It means poll when the frame is sent by a primary station to a secondary (when
the address field contains the address of the receiver).
ii) It means final when the frame is sent by a secondary to a primary (when the
address field contains the address of the sender).
2) Control Field for S-Frames
• The subfields in the control field are:
1) If the first 2 bits of the control field is 10, this means the frame is an S-frame.
2) The last 3 bits N(R) corresponds to the acknowledgment-number (ACK) or negative
acknowledgment-number (NAK).
3) The 2 bits called code is used to define the type of S-frame itself.
With 2 bits, we can have four types of S-frames:
1) Receive ready (RR) = 00
¤ This acknowledges the receipt of frame or group of frames.
¤ The value of N(R) is the acknowledgment-number.
2) Receive not ready (RNR) = 10
¤ This is an RR frame with 1 additional function:
i) It announces that the receiver is busy and cannot receive more frames.
¤ It acts as congestion control mechanism by asking the sender to slow down.
¤ The value of N(R) is the acknowledgment-number.
3) Reject (REJ) = 01
¤ It is a NAK frame used in Go-Back-N ARQ to improve the efficiency of the
process.
¤ It informs the sender, before the sender time expires, that the last frame is lost
or damaged.
¤ The value of N(R) is the negative acknowledgment-number.
4) Selective reject (SREJ) = 11
¤ This is a NAK frame used in Selective Repeat ARQ.
¤ The value of N(R) is the negative acknowledgment-number.
3) Control Field for U-Frames
• U-frame codes are divided into 2 sections:
i) A 2-bit prefix before the P/F bit
ii) A 3-bit suffix after the P/F bit.
• Together, these two segments (5 bits) can be used to create up to 32 different types of U-
frames.

4-12
5c. Write a short note on piggybacking. (04 Marks)
Ans:
Piggybacking
• The technique of temporary delaying outgoing acknowledgements so that they can be hooked
onto next outgoing data frame is known as Piggybacking.
• Piggybacking technique is used to improve the efficiency of the bidirectional protocols.
• As shown in Figure 5.1,
 When a frame is carrying data from A to B, it can also carry control information about
arrived (or lost) frames from B;
 When a frame is carrying data from B to A, it can also carry control information about
the arrived (or lost) frames from A
Figure 5.1: Piggybacking
6a. With a flow diagram, explain the working of CSMA/CD. (10 Marks)
Ans: For answer, refer Solved Paper Dec-2013 Q.No.6a.(ii)

4-13
6b. Explain the following channelization techniques: i) CDMA ii) TDMA. (10 Marks)
Ans(i): For answer, refer Solved Paper June-2013 Q.No.6a.
Ans (ii):
TDMA
• The stations share the bandwidth of the channel in time (Figure 12.22).
• Each time-slot is reserved for a specific station.
• Each station can send the data in the allocated time-slot.
• Main problem: Achieving synchronization between the different stations.
i.e. each station needs to know the beginning of its slot and the location of its slot.
This may be difficult because of propagation delays introduced in the system.
• To compensate for the delays, we can insert guard-times.
• Normally, synchronization is accomplished by having some synchronization bits at the
beginning of each slot.
• TDMA vs. TDM
1) TDM
 TDM is a multiplexing method in the physical layer.
 TDM
→ combines the individual-data from slower channels and
→ transmits the aggregated- data by using a faster channel.
 The multiplexer interleaves data units from each channel.
2) TDMA
 TDMA is an access method in the data link layer.
 In each station, the data link layer tells the physical layer to use the allocated time-slot.
 There is no physical multiplexer at the physical layer.

4-14
7a. What do you mean by hidden and exposed station problems in IEEE 802.11
protocol. Explain in detail. (06 Marks)
Ans:
Hidden Station Problem
• Figure 15.3 shows an example of the hidden station problem.
• Every station in transmission range of Station B can hear any signal transmitted by station B.
• Every station in transmission range of Station C can hear any signal transmitted by station C.
• Station C is outside the transmission range of B;
Likewise, station B is outside the transmission range of C.
• However, Station A is in the area covered by both B and C;
Therefore, Station A can hear any signal transmitted by B or C.
Exposed Station Problem
• In this problem, a station refrains from using a channel even when the channel is available for
use.
• In the figure 14.12, station A is transmitting to station B.
• Station C has some data to send to station D, which can be sent without interfering with the
transmission from A to B.
• However, station C is exposed to transmission from A i.e. station C hears what A is sending and
thus refrains from sending.
• In other words, C is too conservative and wastes the capacity of the channel.

4-15
7b. With neat diagram, explain the architecture of Piconet and Scatternet Bluetooth
networks. (06 Marks)
Ans:
BLUETOOTH
• Bluetooth is a wireless-LAN technology designed to connect devices of different functions such
as telephones, notebooks, computers, cameras, printers, coffee makers, and so on.
• The devices
→ find each other and
→ make a network called a piconet (Usually, devices are called gadgets)
Architecture
• Bluetooth defines 2 types of networks: 1) Piconet and 2) Scatternet.
1) Piconets
• A Bluetooth network is called a piconet, or a small net. (Figure 15.17).
• A piconet can have up to 8 stations. Out of which
i) One of station is called the primary.
ii) The remaining stations are called secondaries.
• All the secondary-stations synchronize their clocks and hopping sequence with the primary
station.
• A piconet can have only one primary station.
• The communication between the primary and the secondary can be one-to-one or one-to-
many.
• Although a piconet can have a maximum of 7 secondaries, an additional 8 secondaries can be
in the parked state.
• A secondary in a parked state is synchronized with the primary, but cannot take part in
communication until it is moved from the parked state.
• Because only 8 stations can be active in a piconet, activating a station from the parked state
means that an active station must go to the parked state.
2) Scatternet
• Piconets can be combined to form a scatternet (Figure 15.18).
• A station can be a member of 2 piconets.
• A secondary station in one piconet can be the primary in another piconet. This is called
mediator station.
1) Acting as a secondary, mediator station can receive messages from the primary in the
first piconet.
2) Acting as a primary, mediator station can deliver the message to secondaries in the
second piconet.

4-16
7c. Explain the working of global system for mobile (GSM) in detail. (08 Marks)
Ans:
GSM
• Aim of GSM: to replace a number of incompatible 1G technologies.
• Here we discuss, two issues: 1) Bands 2) Transmission
1) Bands
 The system uses two bands for duplex communication (Figure 16.11).
 Each band is 25 MHz in width.
 Each band is divided into 124 channels of 200 kHz.
2) Transmission
 Each voice channel is digitized and compressed to a 13-kbps digital signal (Fig 16.12).
 Each slot carries 156.25 bits.
 Eight slots share a frame (TDMA).
 26 frames also share a multiframe (TDMA).
 We can calculate the bit rate of each channel as follows.
 Each 270.8-kbps digital channel modulates a carrier using GMSK (a form of FSK); the
result is a 200-kHz analog signal.
 Finally, 124 analog channels of 200 kHz are combined using FDMA. The result is a 25-
MHz band (Figure 16.13).

4-17
8a. Explain i) Header format & ii) Extension headers of IPV6 (10 Marks)
Ans (i): For answer, refer Solved Paper Dec-2013 Q.No.8c.
Ans (ii):
Extension Header
• An IP packet is made of
→ base-header &
→ some extension headers.
• To support extra functionalities, extension headers can be placed b/w base header and payload.
• There are Six types of extension headers (Figure 22.8).
1) Hop-by-Hop Option
• This option is used when the source needs to pass information to all routers visited by the
datagram.
• Three options are defined: i) Pad1, ii) PadN, and iii) Jumbo payload.
i) Pad1
 This option is designed for alignment purposes.
 Some options need to start at a specific bit of the 32-bit word.
 Pad1 is added, if one byte is needed for alignment.
ii) PadN
 PadN is similar in concept to Pad1.
 The difference is that PadN is used when 2 or more bytes are needed for alignment.
iii) Jumbo Payload
 This option is used when larger packet has to be sent. (> 65,535 bytes)
 Large packets are referred to as jumbo packets.
 Maximum length of payload = 65,535 bytes.
2) Destination Option
• This option is used when the source needs to pass information to the destination only.
• Intermediate routers are not allowed to access this information.
• Two options are defined: i) Pad1 & ii) PadN
3) Source Routing
• This option combines the concepts of
→ strict source routing and
→ loose source routing.
4) Fragmentation
• In IPv6, only the original source can fragment.
• A source must use a “Path MTU Discovery technique” to find the smallest MTU along the path
from the source to the destination.
• Minimum size of MTU = 1280 bytes. This value is required for each network connected to the
Internet.
• If a source does not use a Path MTU Discovery technique, the source fragments the datagram
to a size of 1280 bytes.
5) Authentication
• This option has a dual purpose:
i) Validates the message sender: This is needed so the receiver can be sure that a
message is from the genuine sender and not from an attacker.
ii) Ensures the integrity of data: This is needed to check that the data is not altered in
transition by some attacker.
6) Encrypted Security Payload (ESP)
• This option provides confidentiality and guards against attacker.

4-18
8b. Write short note for following: (10 Marks)
i) FHSS ii) Gigabit Ethernet
iii) Polling iv) Token passing
Ans (i): For answer, refer Solved Paper June-2013 Q.No.3a.
Ans (ii):
GIGABIT ETHERNET
• Goals of Gigabit-Ethernet:
1) Upgrade the data-rate to 1 Gbps.
2) Make it compatible with Standard or Fast-Ethernet.
3) Use the same 48-bit address.
4) Use the same frame format.
5) Keep the same minimum and maximum frame-lengths.
6) To support auto-negotiation as defined in Fast-Ethernet.
MAC Sublayer
• Gigabit-Ethernet has 2 distinctive approaches for medium access: half-duplex and full-duplex.
1) Full-Duplex Mode
• There is a central switch connected to all computers or other switches.
• Each switch has buffers for each input-port in which data are stored until they are transmitted.
• There is no collision. This means that CSMA/CD is not used.
2) Half-Duplex Mode
• A switch is replaced by a hub, which acts as the common cable in which a collision might occur.
• CSMA/CD is used.
• The maximum length of the network is totally dependent on the minimum frame size.
• Three methods have been defined:
i) Traditional
 Like traditional Ethernet, the minimum length of a frame is 512 bits.
 The maximum length of the network is 25 m.
ii) Carrier Extension
 To allow for a longer network, we increase the minimum frame-length.
 Minimum length of frame is 512 bytes (4096 bits).
 The maximum length of the network is 200 m.
iii) Frame Bursting
 Instead of adding an extension to each frame, multiple frames are sent.
Ans (iii):
Polling
• In a network,
One device is designated as a primary station and
Other devices are designated as secondary stations.
• Functions of primary-device:
1) Primary-device controls the link.
2) Primary-device is always the initiator of a session.
3) Primary-device determines which device is allowed to use the channel at a given time.
4) All data exchanges must be made through the primary-device.
• The secondary devices follow instructions of primary-device.
• Disadvantage: If the primary station fails, the system goes down.
• Poll and select functions are used to prevent collisions (Figure 12.19).
1) Select
 If the primary wants to send data, it tells the secondary to get ready to receive; this is
called select function.
2) Poll
 If the primary wants to receive data, it asks the secondaries if they have anything to
send; this is called poll function.

4-19
Ans (iv):
Token Passing
• In a network, the stations are organized in a ring fashion i.e. for each station; there is a
predecessor and a successor.
1) The predecessor is the station which is logically before the station in the ring.
2) The successor is the station which is after the station in the ring.
• The current station is the one that is accessing the channel now.
• A token is a special packet that circulates through the ring.
• Here is how it works:
 A station can send the data only if it has the token.
 When a station wants to send the data, it waits until it receives the token from its
predecessor.
 Then, the station holds the token and sends its data.
 When the station finishes sending the data, the station
→ releases the token
→ passes the token to the successor.
• There are four physical topologies to create a logical ring (Figure 12.20).

5-1
1a. What is data communication? What are the five components of data communication
system? (06 Marks)
1b. Explain OSI reference model, listing the functions of each layer in brief. (10 Marks)
1c. What are the four level of addresses used in internet employing TCP/IP. (04 Marks)
Ans:
Addressing
• Any communication that involves 2 parties needs 2 addresses: source address and destination
address.
• We need 4 pairs of addresses (Figure 2.9):
1) At the application layer, we normally use names to define
→ site that provides services, such as vtunotesbysri.com, or
→ e-mail address, such as vtunotesbysree@gmail.com.
2) At the transport layer, addresses are called port numbers.
 Port numbers define the application-layer programs at the source and destination.
 Port numbers are local addresses that distinguish between several programs running at
the same time.
3) At the network-layer, addresses are called IP addresses.
 IP address uniquely defines the connection of a device to the Internet.
 The IP addresses are global, with the whole Internet as the scope.
4) At the data link-layer, addresses are called MAC addresses
 The MAC addresses defines a specific host or router in a network (LAN or WAN).
 The MAC addresses are locally defined addresses.
2a. Using Shannon's theorem, compute the maximum bit rate for a channal having a
band width of 3100 Hz and signal to noise ratio of 20 db. (06 Marks)
Ans:
Solution:
SNRdB = 10 log10 SNR  SNR = 1OSNRdB/10
 SNR= 102
=100
C =B log2 (1 + SNR) =3100 log2 (1 + 100) = 3100 log2 101 =3100 x 6.65= 20615 bps
2b. Sketch the signal waveform when 01001110 is transmit using following line coding
schemes i) RZ ii) NRZ—L iii) Manchester coding. (06 Marks)
2c. Explain different types of transmission modes. (08Marks)
3 a. Four 1-kbps connections are muhiplexed together a unit is 1 bit. Find :
i) the duration of 1-bit before multiplexing
ii) the duration of a timeslot
iii) the duration a frame. (06 Marks)

5-2
3b. Define direct sequence spread spectrum (DSSS) and explain how it achieves
bandwith spread using relevant sketch. (08 Marks)
3c. What is virtual circuit network? List the five characteristics of the same. (06 Marks)
Ans:
Virtual Circuit Network (VCN)
• This is similar to telephone system.
• A virtual-circuit network is a combination of circuit-switched-network and datagram-network.
• Five characteristics of VCN:
1) As in a circuit-switched-network, there are setup & teardown phases in addition to the
data transfer phase.
2) As in a circuit-switched-network, resources can be allocated during the setup phase.
As in a datagram-network, resources can also be allocated on-demand.
3) As in a datagram-network, data is divided into packets.
Each packet carries an address in the header.
However, the address in the header has local jurisdiction, not end-to-end jurisdiction.
4) As in a circuit-switched-network, all packets follow the same path established during
the connection.
5) A virtual-circuit network is implemented in the data link layer.
A circuit-switched-network is implemented in the physical layer.
A datagram-network is implemented in the network layer.
 The Figure 8.10 is an example of a virtual-circuit network.
 The network has switches that allow traffic from sources to destinations.
 A source or destination can be a computer, packet switch, bridge, or any other device
that connects other networks.
4a. Given the data word 1001 and divisor 1011 : (10 Marks)
i) Show the generation code word at the sender site
ii) Show the checking of code word at receiver site (assume no error).
4b. Explain process of error detection using block coding. (06 Marks)
4c. What is internet checksum? List the steps under taken by sender to calculate check
sum. (04 Marks)
5a. With neat diagram of point-to-point protocol (PPP) frame format, explain each of
the fields. (08 Marks)
5b. Explain stop and wait automatic repeat request protocol. (06 Marks)

5-3
5c. What is framing? With sketches explain bit stuffing and unstuffing. (06 Marks)
6a. With neat diagram explain TDMA. (06 Marks)
Ans: For answer, refer Solved Paper Dec-2014 Q.No.6b.(ii)
6b. Mention different categories of standard Ethernet and explain implementation of 10
base 5-thick Ethernet. (08 Marks)
Ans:
Implementation
• The Standard-Ethernet defines several physical-layer implementations (Table 13.1).
Encoding and Decoding
• All standard implementations use digital-signaling (baseband) at 10 Mbps (Figure 13.6).
1) At the sender, data are converted to a digital-signal using the Manchester scheme.
2) At the receiver, the received-signal is
→ interpreted as Manchester and
→ decoded into data.
10Base5: Thick Ethernet
 10Base5 uses a bus topology (Figure 13.7).
 A external transceiver is connected to a thick coaxial-cable.
(transceiver  transmitter/receiver)
 The transceiver is responsible for
→ transmitting
→ receiving and
→ detecting collisions.
 The transceiver is connected to the station via a coaxial-cable.
The cable provides separate paths for sending and receiving.
The collision can only happen in the coaxial cable.
 The maximum-length of the cable must not exceed 500m.
If maximum-length is exceeded, then there will be excessive degradation of the
signal.
 If a cable-length of more than 500 m is needed, the total cable-length can be divided
into up to 5 segments.
 Each segment of maximum length 500-meter, can be connected using repeaters.

5-4
6c. Mention the five goals of fast Ethernet. And give the importance of
“AUTONEGOTIATION”. (06 Marks)
Ans:
FAST ETHERNET (100 MBPS)
• IEEE created Fast-Ethernet under the name 802.3u.
• Goals of Fast-Ethernet:
1) Upgrade the data-rate to 100 Mbps.
2) Make it compatible with Standard-Ethernet.
3) Keep the same 48-bit address.
4) Keep the same frame format.
5) Keep the same minimum and maximum frame-lengths.
Access Method
• Only the star topology is used.
• For the star topology, there are 2 choices:
1) In half-duplex approach, the stations are connected via a hub.
CSMA/CD was used as access-method.
2) In full-duplex approach, the connection is made via a switch with buffers at each port.
There is no need for CSMA/CD.
Autonegotiation
 A new feature added to Fast-Ethernet is called autonegotiation.
 It provides a station/hub with a range of capabilities.
 It was used for the following purposes:
1) To allow 2 devices to negotiate the mode or data-rate of operation.
2) To allow incompatible devices to connect to one another.
For example: a device with a maximum capacity of 10 Mbps can
communicate with a device with a 100 Mbps capacity.
3) To allow one device to have multiple capabilities.
4) To allow a station to check a hub's capabilities.
7a. What is blue tooth? Explain its architecture. (06 Marks)
7b. Explain the following connecting devices (08 Marks)
i) Hub ii) Bridge iii) Router iv) Gateway.

5-5
7c. Discuss cellular telephone in brief. (06 Marks)
Ans:
CELLULAR TELEPHONY
• Cellular telephony is designed to provide communications
→ between two moving units called mobile-stations (MSs) or
→ between one mobile-station and one stationary unit called a land unit (Figure 16.6).
• A service-provider is responsible for
→ locating & tracking a caller
→ assigning a channel to the call and
→ transferring channel from base-station to base-station as caller moves out-of-range.
• Each cellular service-area is divided into small regions called cells.
• Each cell contains an antenna.
• Each cell is controlled by AC powered network-station called the base-station (BS).
• Each base-station is controlled by a switching office called a mobile-switching-center (MSC).
• MSC coordinates communication between all the base-stations and the telephone central office.
• MSC is a computerized center that is responsible for
→ connecting calls
→ recording call information and
→ billing.
• Cell-size is not fixed; Cell-size can be increased or decreased depending on population of the
area.
• Cell-radius = 1 to 12 mi.
• Compared to low-density areas, high-density areas require many smaller cells to meet traffic
demands.
• Cell-size is optimized to prevent the interference of adjacent cell-signals.
8a. List i) deficiencies of IPV4 and ii) advantages of IPV6 over IPV4. (10 Marks)
Ans(i):
Deficiencies of IPV4
1) Scarcity of IPv4 Addresses: The recent exponential growth of the Internet and the exhaustion
of the IPv4 address space
2) Security Related Issues: The requirement for security at the Internet layer
3) Absence of support of QoS: The need for better support for prioritized and real-time delivery
of data.
4) Absence of the autoconfiguration addresses mechanism: Internet is expanding and many new
devices are using IP. The configuration of IP addresses should be simple.
5) The problems related to fragmentation: The size of the maximal block of data transmission
on each concrete way is not defined;
Ans(ii): For answer, refer Solved Paper June-2014 Q.No.8c.(i)
8b. Draw format of an IPV6 datagram and explain. (10 Marks)

VTU 5TH SEM CSE COMPUTER NETWORKS-1 (DATA COMMUNICATION) SOLVED PAPERS

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