data communication sample tutorial& answer

8/10/2019 data communication Sample Tutorial& answer

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INSTRUCTION(S) TO CANDIDATES

DO NOT OPEN UNTIL YOU ARE ASKED TO DO SO

Total marks of this examination is 100.

This examination is worth 50%of the total assessment.

Answer any Five (5)questions out of Six (6)questions.

State all your assumptions clearly. Show all your work.

Any form of cheating or attempt to cheat is a serious

offence which may lead to dismissal.


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Data Communications ECE 4111

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Q.1 [20 marks]

(a) Why are protocols needed in data communication? (3 marks)

Protocols are needed since they provide a set of rules that govern data

communication. A protocol defines what is communicated, in what way and when.

This provides accurate and timely transfer of information between differentdevices on a network.

(b) For each of the following four networks, discuss the consequences if a connection

fails.

i. Five devices arranged in a mesh topology (1 marks)

ii. Five devices arranged in a star topology (not counting the hub) (1 marks)iii. Five devices arranged in a bus topology (1 marks)

iv. Five devices arranged in a ring topology (1 marks)

i. Mesh topology: If one connection fails, the other connections will still beworking.

ii. Star topology: The other devices will still be able to send data through thehub; there will be no access to the device which has the failed connection

to the hub.

iii. Bus Topology: All transmission stops if the failure is in the bus. If thedrop-line fails, only the corresponding device cannot operate.

iv. Ring Topology: The failed connection may disable the whole networkunless it is a dual ring or there is a by-pass mechanism.

(c) Find the 8-bit data stream for each case shown in Fig. 1. (8 marks)


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Fig. 1The data stream can be found as

a. NRZ-I: 01110011

b. Differential Manchester: 10101110

c. AMI: 01100101

d. Manchester: 10110011

(for each case: 0.5 mark deduction for 1 bit error, for more errors (consider the

result totally wrong) deduct 2 marks)

(d) Identify different layers of TCP/IP reference model and list at least 1 protocol

used in each layer excluding TCP and IP protocol. (5 marks)

Layer # Layer Name Protocol used

5 Application HTTP, SMPT, FTP

4 Transport UDP, SCTP

3 Network ICMP, IGMP, ARP,

BGP, OSPF, RIP2 Data link PPP, HDLC, Frame

Relay, ATM, DSL

1 Physical EIA/TIA-232, V.24, V.35

a. NZR-I code

b. Differential Manchester code

c. AMI code

c. Manchester code


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Q.2 [20 marks]

(a) We have a channel with a 1-MHz bandwidth. The SNR for this channel is 1023. i. What is the maximum bit rate supported by this channel (4 marks)

ii. How many signal levels are needed to be used to achieve 80% of the

maximumbit rate? (4 marks)

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

C = B log2(1 + SNR) = 106log2(1 + 1023) = 10 Mbps

ii. 80% of the maximum bit rate = 10*.8 = 8Mbps

Then we use the Nyquist formula to find the number of signal levels.

8 Mbps = 2*1MHz* log2L

L = 24= 16 levels

(b) Describe the goal of multiplexing. (2 marks)

Multiplexing is the set of techniques that allows the simultaneous transmission of

multiple signals across a single data link.

(c) We need to use synchronous TDM and combine 20 digital sources, each of 100

Kbps. Each output slot carries 1 bit from each digital source, but one extra bit is

added to each frame for synchronization. Answer the following questions:

i. What is the size of an output frame in bits? (2 marks)

ii. What is the output frame rate? (2 marks)iii. What is the duration of an output frame? (2 marks)

iv. What is the output data rate? (2 marks)

v. What is the efficiency of the system (ratio of useful bits to the total bits).

(2 marks)

i. Each output frame carries 1 bit from each source plus one extra bit forsynchronization. Frame size = 20 1 + 1 = 21 bits.

ii. Each frame carries 1 bit from each source. Frame rate = 100,000frames/s.

iii. Frame duration = 1 /(frame rate) = 1 /100,000 = 10 s.iv. Data rate = (100,000 frames/s) (21 bits/frame) = 2.1 Mbpsv. In each frame 20 bits out of 21 are useful. Efficiency = 20/21= 95%

Q.3 [20 marks]

(a) Identify the differences between circuit switching and virtual circuit switching.

(6 marks)

Circuit switching is the transmission technology that has been used since the first

communication networks in the nineteenth century. In circuit switching, a caller

must first establish a connection to a callee before any communication is possible.


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During the connection establishment, resources are allocated between the caller

and the callee. Generally, resources are frequency intervals in a Frequency

Division Multiplexing (FDM) scheme or more recently time slots in a Time

Division Multiplexing (TDM) scheme. The set of resources allocated for a

connection is called a circuit. A path is a sequence of links located between nodes

called switches. The path taken by data between its source and destination isdetermined by the circuit on which it is flowing, and does not change during the

lifetime of the connection. The circuit is terminated when the connection is closed.

In circuit switching, resources remain allocated during the full length of a

communication, after a circuit is established and until the circuit is terminated

and the allocated resources are freed. Resources remain allocated even if no data

is flowing on a circuit, hereby wasting link capacity when a circuit does not carry

as much traffic as the allocation permits. This is a major issue since frequencies

(in FDM) or time slots (in TDM) are available in finite quantity on each link, and

establishing a circuit consumes one of these frequencies or slots on each link of

the circuit. As a result, establishing circuits for communications that carry lesstraffic than allocation permits can lead to resource exhaustion and network

saturation, preventing further connections from being established. If no circuit

can be established between a sender and a receiver because of a lack of

resources, the connection is blocked.

Virtual circuit packet switching (VC-switching) is a packet switching technique

which merges datagram packet switching and circuit switching to extract both of

their advantages. VC-switching is a variation of datagram packet switching

where packets flow on so-called logical circuits for which no physical resources

like frequencies or time slots are allocated. Each packet carries a circuit

identifier which is local to a link and updated by each switch on the path of the

packet from its source to its destination. A virtual circuit is defined by the

sequence of the mappings between a link taken by packets and the circuit

identifier packets carry on this link. This sequence is set up at connection

establishment time and identifiers are reclaimed during the circuit termination.

There is a trade-off between connection establishment and forwarding time costs

that exists in circuit switching and datagram packet switching. In VC-switching,

routing is performed at circuit establishment time to keep packet forwarding fast.

Other advantages of VC-switching include the traffic engineering capability of

circuit switching, and the resources usage efficiency of datagram packet

switching. Nevertheless, a main issue of VC-Switched networks is the behavior on

a topology change. As opposed to Datagram Packet Switched networks which

automatically recompute routing tables on a topology change like a link failure,

in VC-switching all virtual circuits that pass through a failed link are interrupted.

Hence, rerouting in VC-switching relies on traffic engineering techniques.

(b) Explain with proper diagram the steps needed to create a virtual circuit between

two computers assuming there are 3 switches in between. (8 marks)


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Steps to create a Switched Virtual Circuits (SVC)

Setup request

Acknowledgement

VC setup request

VC setup acknowledgement

(c) A sender needs to send the four data items0x3456

,OxABCC

,0x02BC

, andOxEEEE. Answer the following:

i. Find the checksum at the sender site. (3 marks)ii. Find the checksum at the receiver site if the second data item is changed to

OxABCE. (3 marks)


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Q.4 [20 marks]

(a) Define framing and the reason for its need. (4 marks)

The data link layer needs to pack bits into frames.

Framing divides a message into smaller entities to make flow and error control

more manageable.

(b) Explain what is meant by the term data transparency and how it is achieved inHDLC. (4 marks)

HDLC is a data link protocol which uses a unique bit sequence to delimit the start

and end of each PDU transported by the data link layer service. In HDLC, frames

are delimited by a sequence of bits known as a "flag". The flag sequence is a

unique 8-bit sequence of the form 0111 1110. The way in which this is performed

is described in the text and diagrams which follow.


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The flags before and after an HDLC frame indicate the start and end of the frame

Transparency: The flag sequence must never occur within the content of a frame

otherwise it could be confused with an intentionally sent flag. A technique known

as 0-bit insertion or bit stuffing is used to prevent random data synthesising a

flag. The technique is said to make HDLC transparent, since any stream of bitsmay be present between the open and closing flag of a frame. The transparency is

achieved by encoding the data by inserting a 0-bit after any sequence of 5

consecutive 1's within the payload, as shown,

Insertion of a "zero-bit" into the content of a frame to ensure transparency

Bit stuffing used to avoid confusion with data containing 01111110:

0 inserted after every sequence of five 1s.

If the receiver detects five 1s it checks the next bit.

If it is 0, it is deleted.

If it is 1 and seventh bit is 0, accept as a flag.

If sixth and seventh bits are 1, the sender is indicating an abort condition.

(c) Data is transmitted over a half-duplex radio link at a rate of 28.8 kbps using a

stop-and-wait ARQ strategy. Frames have a block length of 256 bytes of which 5are non-information bytes. If the propagation delay is 1 ms and processing delays

and acknowledgement transmission time can be neglected, determine:i. The throughput in the absence of errors, (2 marks)

ii. The throughput in the presence of a bit error rate of 104

(4 marks)

Frame transmission time, 071.0108.28

82563 =

=ft sec = 71 msec

propagation delay, 1=dt msec

no. of information bits, 251=k kbps


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(a) in the absence of errors:

throughput = 5.27271

8251

2=

+

=

+ df tt

kkbps

(b) bit error rate = 104= 0.0001probability of an error-free bit = 1 0.0001 = 0.9999

probability of an error-free frame = 0.99992048

= 0.815

frame error rate, P = 1 0.815 = 0.185

average number of times a frame is retransmitted = 227.1815.0

185.01

11 =+=

+

P

P

The transmission and delay times will increase by this amount, giving a

throughput of:

( )4.22

73227.1

2008

271227.1

8251=

=

+

kbps

(d) An Ethernet MAC sublayer receives 3020 bytes of data from the upper layer.

i. How many frames need to be generated and sent to carry the data?(3 marks)

ii. What is the size of data in each frame and why? (3 marks)

i. The maximum data size in the Standard Ethernet is 1500 bytes. The data of 3020

bytes, therefore, must be split between three frames. The standard dictates that

the all frames must carry the maximum possible number of bytes (1500) except

the last; the last (third) frame then needs to carry only 20 bytes of data (it

requires padding).

ii. The follow-ing shows the breakdown:

Data size for the first frame: 1500 bytes

Data size for the second frame: 1500 bytes

Data size for the third frame: 46 bytes (with padding) (20 bytes data + 26 bytes of

padding)

Q.5 [20 marks]

(a) i. List the main categories of protocols that have been devised to handle

access to a shared link. (2 marks)

Random access protocolControlled access protocol

Channelization protocol

ii. Highlight the key functional differences of the categories that have been

identified in Q.5 (a) i. (6 marks)


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Random access protocol:

In random access or contention methods, no station is superior to another

station and none is assigned the control over another. No station permits,

or does not permit, another station to send. At each instance, a station that

has data to send uses a procedure defined by the protocol to make a

decision on whether or not to send. This decision depends on the state ofthe medium (idle or busy). In other words, each station can transmit when

it desires on the condition that it follows the predefined procedure,

including the testing of the state of the medium.

Two features give this method its name. First, there is no scheduled time

for a station to transmit. Transmission is random among the stations. That

is why these methods are called random access. Second, no rules specify

which station should send next. Stations compete with one another to

access the medium. That is why these methods are also called contention

methods.

Controlled access protocol:In controlled access, the stations consult one another to find which station

has the right to send. A station cannot send unless it has been authorized

by other stations. Three popular controlled-access methods are:

Reservation, Polling, and Token Passing.

Channelization protocol:

Channelization is a multiple-access method in which the available

bandwidth of a link is shared in time, frequency, or through code, between

different stations. Three well knownchannelization protocols: FDMA,

TDMA, and CDMA.

(b) Explain CSMA/CA method with suitable process flowchart. (7 marks)


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(c) In a CDMA/CD network with a data rate of 10 Mbps, the minimum frame size is

found to be 512 bits for the correct operation of the collision detection process.

What should be the minimum frame size if we increase the data rate to 100 Mbps?

(5 marks)The relationship between the minimum frame size and the data rate can be found

as follows.

We know that

Tfr = (frame size) / (data rate) = 2 Tp = 2 distance / (propagation

speed)

or

(frame size) = [2 (distance) / (propagation speed)] (data rate)]

or

(frame size) = K (data rate)

This means that minimum frame size is proportional to the data rate (K is a con-stant). When the data rate is increased, the frame size must be increased in a net-

work with a fixed length to continue the proper operation of the CSMA/CD. In

Example 12.5, we mentioned that the minimum frame size for a data rate of 10

Mbps is 512 bits. We calculate the minimum frame size based on the above pro-

portionality relationship

Data rate = 10 Mbps minimum frame size = 512 bits

Data rate = 100 Mbps minimum frame size = 5120 bits


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Q.6 [20 marks]

(a) According to IEEE specification bridges are required to use spanning tree

algorithm. What type of problem is solved by creating spanning tree in LAN

systems? (2 marks)

A spanning tree is a graph in which there is no loop thus spanning tree algorithm

helps to create loop free network topology.

(b) Find the Spanning tree for the system shown in Fig.1. Choose B1 as the root

bridge. You must demonstrate with diagrams the forwarding and the blocking

ports, after applying the spanning tree procedure. (6 marks)

LAN 1 LAN 2

LAN 3

B1

B3B2 B4

B5LAN 7

LAN 5LAN 4

1

1

1

1

1 1

1 2B6

B7

22

2

2

22

3

3

LAN 6


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A spanning tree is a graph in which there is no loop.

Process of creating a logical topology without loop (using Spanning tree)

Select the root bridge (as the root of the tree) with the smallest ID

Make one port of each bridge (except the root bridge) as the root port (having least cost

path from the bridge to the root bridge)

Choose a designated bridge for each LAN (having the least-cost path between the LAN

and the root bridge)

Make the corresponding port connecting the designated bridge and LAN for each LAN

the designated port.

Mark the root port and the designated port as forwarding ports, the others as blocking

ports.

(c) Explain the advantages of dividing an Ethernet LAN with a bridge? (4 marks)

A bridge can raise the bandwidth and separate collision domains.

(d) Compare and contrast X.25 with Frame Relay by identifying the limitations of

X.25 and highlighting the key features of Frame Relay (8 marks)

LAN 1 LAN 2

LAN 3

B1

B3B2 B4

B5LAN 7

LAN 5LAN 4

1

1

1

1

1 1

1 2B6

B7

22

2

2

22

3

3

LAN 6


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Limitations of X.25:

i. X.25 has a low 64-kbps data rate. By the 1990s, there was a need forhigher data-rate WANs.

ii. X.25 has extensive flow and error control at both the data link layer and

the network layer. This was so because X.25 was designed in the 1970s,when the available transmission media were more prone to errors. Flow

and error control at both layers create a large overhead and slow down

transmissions. X.25 requires acknowledgments for both data link layer

frames and network layer packets that are sent between nodes and

between source and destination.

iii. Originally X.25 was designed for private lise, not for the Internet. X.25has its own network layer. This means that the user's data are encapsulated

in the network layer packets of X.25. The Internet, however, has its own

network layer, which means if the Internet wants to use X.25, the Internet

must deliver its network layer packet, called a datagram, to X.25 forencapsulation in the X.25 packet. This doubles the overhead.

The key features of Frame Relay

i.

Frame relay operates at a higher speed (1.544 Mbps and recently 44.376

Mbps) hence it can easily be used instead of a mesh of T-1 or T-3 lines.

ii.

Frame relay operates in physical and data link layers. Therefore it can

easily be used as a backbone network to provide services to protocols that

already have a network layer protocol, such as Internet.

iii.

Frame relay allows bursty data.

iv.

Frame relay allows a frame size of 9000 bytes, which can easily

accommodate all LAN frame sizes.

v.

Frame relay is less expensive than other traditional WANs.

vi.

Frame relay has error detection at the data link layer only. There is no

flow control and error control (no retransmission policy). It was designed

this way to provide fast transmission capability for more reliable media

and for those protocols that have flow and error control at the higher

layers.

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