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1. Question(1.0 mark):Identify and explain briefly the five components of a datacommunications system.

Five components of data communications system are:

Message: is the information (data) to be communicated. The message can be text,

numbers, pictures, audio and video.

Sender: is a device that sends the data message. It can be computer, workstation,

telephone handset, video camera, and so on.

Receiver: is a device that receives the data message. It can be computer, workstation,

telephone handset, television, and so on.

Transmission medium: is the physical path which a message travels from the sender to

the receiver. Some examples of transmission media include twisted-pair wire, coaxialcable, fiber-optic cable, and radio wave.

Protocol: is a set of rules that govern data communications. It presents an agreement

between the communicating devices.

2. Question(1.0 mark):Distinguish simplex, half-duplex and full-duplex transmissionmode.

Simplex: the communication is unidirectional (theo mt hng), as on a one-way street.

Only one of the two devices on a link can transmit; the other can only receive.

Half-Duplex: each station can both transmit and receive, but not at the same time. This

means, when one station is the transmitter, the other one must be the receiver and vice

versa.

Full-Duplex: both stations can transmit and receive simultaneously.In this mode,

signals going in one direction share the capacity of the link with signals going in the

other direction.

3. Question(1.0 mark):Distinguish unicast, multicast and broadcast communicationsUnicast: is communication where a piece of information is sent from one point to the

another point. In this case, there is just one sender, and one receiver.

Multicast: is communication where a piece of information is sent from one or more

point to a set of other points. In this case, there is one or more sender and the

information is distributed to a set of receivers.

Broadcast: is communication where a piece of information is sent from one point to all

the other points. In this case, there is just one sender, but the information is sent to allconnected receivers.


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4. Question(1.0 mark):Describe briefly the differences between three technicalterms: (channel) bandwidth, transmission rate, throughput ?

Bandwidth:

In networking, we use the term bandwidth in two context:

- The first, bandwidth in hertz, refers to the range of frequencies in a compositesignal or the range of frequencies that a channel can pass.

- The second, bandwidth in bits per second, refers to the speed of bit transmissionin a channel or link.In this case, bandwidth refers to the data-carrying capacity of

a network or data transmission medium. It indicates the maximum amount of

data that can pass from a point to another in a unit of time. The bandwidth of a

channel is unchanged. For example, we say we have a slow 9.6Kps dial-up line.

Throughput:

Throughput is the measure how fast we can actually send data through a network.

Throughput is never a constant, it fluctuates and is obviously lower than the bandwidth

of the channel.

Transmission rate:

Transmission Rate is the total amount of data that can be sent from one place to another

in a given period of time.

5. Question(1.0 mark):Name the four basic network topologies and illustrate themby figures.

Mesh: Ring

Star Bus


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6. Question(1.0 mark):What are the differences between physical, logical, port andspecific address?

Physical address:

- It is also known as link address, is the address of a node as defined by its LAN orWAN.It is used by Data Link layer to deliver data from one node to another within

the same network.It is generally the MAC address and is the lowest-level address.

- The size and format vary depending on the network (it changes from hop to hop).Most local-area networks use a 48-bit (6-byte) physical address written as 12

hexadecimal digits; every byte is seperated by a colon, as shown as:

07:01:02:01:2C:4B

Logical address:

- Is address at the Network layer and generally IP address.- A logical address in the Internet is currently a 32-bit address that can uniquely

define a host connected to the Internet. It remains the same from hop to hop. An

example of logical address is shown as: 200.100.50.100

Port address:

- Is address at Transport layer, used to identify the particular application running onthe destination machine.

- It remains the same from hop to hop.- A port address in TCP/IP is 16 bits in length.Specific address:

- Is address at the Application layer and is user-friendly address. For example, theemail address ([email protected])and the Universal Resource Locator (URL)

(vnu.edu.vn)

7. Question (1.0 mark): List the roles of the data link layer in the TCP/IP model.- Framing: the data layer divides the stream of bits received from the network layer

into manageable data units called frame.
mailto:[email protected]:[email protected]:[email protected]:[email protected]


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- Physical addressing: if frames are to be distributed to different systems on thenetwork, the data link layer adds a header to the frame to define the sender and/or

receiver of the frame.

- Flow control: if the rate at which the data are absorbed by the receiver is less thanthe rate at which data are produced in the sender, the data link layer imposes (tc

ng, cng ng)a flow control mechanisms to avoid overwhelming (trn, ln t)the receiver.

- Error control: the data link layer adds reliability to the physical layer by addingmechanisms to detect and retransmit damaged or lost frames. It also uses a

mechanism to recognize duplicate frames.

- Access control: when two or more devices are connected to the same link, data linklayer protocols are necessary to determine which device has control over the link at

any given time.

8. Question (1.0 mark)What are the responsibilities of the network layer in theTCP/IP model.

- Logical addressing: If a packet passes the network boundary, we need anotheraddressing system to help distinguish the source and destination systems. Address

in this case is logical address.

- Creating a connection between the source computer and the destination com-puter, and routing the packet through possible routes. The communication atthe Network layer is host-to-host. However, there are several routers from the

source to the destination, the routers in the path are responsible for choosing the

best route for each packet.

- Defining the format of the packet.The Network layer in the Internet include theInternet Protocol (IP), that defines the format of the packet, called a datagram.

9. Question (1.0 mark)What are the responsibilities of the transport layer in theTCP/IP model.

- Process-to-process delivery of the entire message. A process is an applicationprogram running on the host. The Transport layer ensures that the whole message

arrives intact (nguyn vn) and in order, overseeing both error control and flowcontrol at the source-to-destination level.

- Service-point addressing. Computers often run several programs at the same time.For this reason, source-to-destination delivery means delivery from a specific

process (running program) on one computer to a specific process on the other. The

transport layer must therefore include a type of address called a service-point


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address (or port address). The transport layer gets the entire message to the correct

process on that computer.

- Segmentation and reassembly. A message is divided into transmittable segmentswith each segment containing a sequence of number. These numbers enable the

transport layer to resemble the message correctly.

- Connection control. Transport layer can control the delivery segments to thetransport layer at the destination machine through either connectionless or

connection oriented.

- Flow control. Unlike the data link layer, flow control at transport layer is per-formed end to end rather than across a single link.

- Error control. Unlike that data link layer, error control at transport layer is per-formed process-to-process rather than across a single link. Error correction is

usually achived through retransmission.

10. Question (1.0 mark)What are the responsibilities of the application layer in theTCP/IP model.

- Network virtual terminal.A network virtual terminal is a software version of aphysical terminal, and it allows a user to log on to a remote host.

- File transfer, access, and management. This application allows a user to accessfile in a remote host, to retrieve (phc hi) files from a remote computer for use inthe local computer, and to manage or control files in a remote computer locally.

- Mail services. This application provides the basis for e-mail forwarding andstorage.

- Directory services. This application provides distributed database sources andaccess for global information about various objects and services.

11. Question (1.0 mark):What are the differences between propagation delay andtransmission delay?

Propagation delaymeasures the time required for a bit to travel from the source to the

destination. The propagation delay is calculated by dividing the distance by the

propagation speed:

distancepropagation delay

propagation speed

Transmission delayis the time required for transmission of a message symbol depends

on the size of the message and the bandwidth of the channel:

message sizetransmission delay

bandwidth


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12. Question (1.0 mark):What are multiplexing and its major goal?Multiplexing is the set of techniques that allows the simultaneous transmission of

multiple signal across a single data link.

The major goal of Multiplexing is to share an expensive resource.

13. Question (1.0 mark):Why doesstatistical TDM have higher bandwidth efficiencythan synchronous TDM?.

Because in statistical TDM, slots are dramatically allocated. Only when an input line

has a slots worth of data to send is it given a slot in the output frame. Therefore, thenumber of slots in each frame is less than the number of input lines. In synchronous

TDM, however, if a source does not have data to send, the corresponding slot in the

output frame is empty. Thus, the number of slots in each frame is always equal to thenumber of input lines.

14. Question (1.0 mark):Discuss the efficiency of circuit switching and packetswitching?

Circuit switching:

The circuit-switching networks are not as efficient as the packet-switching network

because resources are allocated during the entire duration of the connection. Therefore,these resources are unavailable to other connections.

Moreover, since the route from one point to another is decided before communication is

started, the communication remains on the same route from the beginning to the end of

process even if that route is not the most efficient one.

Packet switching:

The efficiency of the packet switching network is better than that of the circuit

switching network because resources are allocated only when there are packets to be

transferred. Therefore, these resources can be shared to other connections.

Moreover, each packet are allowed to find its way to the destination by the most

efficient route possible.

15. Question (1.0 mark): Why is ATM so called virtual-circuit switching?Because ATM uses two types of connection: Permanent virtual-circuit connection

(PVC), and Switch virtual-circuit connection (SVC).

16. Question (1.0 mark):What are the roles of MAC sublayer?


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- Frame delimiting and recognition- To perform the control of access to media.- It performs the unique addressing to stations directly connected to LAN.- Detection of error

17. Question (1.0 mark):Why are random access protocols also known as contentionprotocols?

In random access method, no rules specify (quy nh) which station should send next.Stations competes with one another to access the medium. This is why this method is

also known as contention method(phng php u tranh).

18. Question (1.0 mark):Why does slotted-ALOHA provide better performance thanpure-ALOHA ?Because

In pure ALOHA, the stations transmit frames whenever they have data to send. When

two or more stations transmit simultaneously, there is a collision and the frames are

distroyed.

In slotted ALOHA, however, the time of the shared channel is divided into discrete

intervals called slots. If any station is not able to place the frame onto the channel at the

beginning of the slot, the station has to wait until the beginning of the next time slot. By

this method, slotted ALOHA can reduce the chances of collision so it provide better

performance than the pure ALOHA.

http://ecomputernotes.com/computernetworkingnotes/communication-networks/what-is-aloha

19. Question (1.0 mark):Distinguish the major differences of CSMA/CD andCSMA/CA.

The major differences between CSMA/CD and CSMA/CA are:

- CSMA/CD is typically used in wired networks while CSMA/CA is used inwireless networks.

- CSMA/CD is standardized in IEEE 802.3 and CSMA/CA is standardized inIEEE 802.11.

- CSMA/CD takes effect after a collision while CSMA/CA acts to preventcollisions before they happen.

- CSMA/CA reduces the possibility of a collision while CSMA/CD onlyminimizes the recovery time.
http://ecomputernotes.com/computernetworkingnotes/communication-networks/what-is-alohahttp://ecomputernotes.com/computernetworkingnotes/communication-networks/what-is-alohahttp://ecomputernotes.com/computernetworkingnotes/communication-networks/what-is-aloha


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http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-

csma-cd/

http://www.differencebetween.com/difference-between-csma-cd-and-vs-csma-ca/#ixzz32E2EAe55

20. Question (1.0 mark):What are Clear-to-Send (CTS) and Request-to-Send (RTS)of wireless LAN designed for ?

CTS and RTS of wireless LAN are designed for reducing frame collision

http://en.wikipedia.org/wiki/IEEE_802.11_RTS/CTS

21. Question (1.0 mark):Why is ATM called a asynchronous network ?ATM uses statistical (asynchronous) time-division multiplexing,that is why it iscalled

Asynchronous Transfer Mode

22. Question (1.0 mark):Explain why connection at ATM layer is considered asconnection-oriented mode.

Because in ATM, a virtual circuit must be established between two endpoints before the

actual data exchange begins.

23. Question (1.0 mark):Why is an IPv4 address considered unique and universal?Ipv4 address is uniquebecause each address is defined one, and only one connectiontwo the Internet.

Ipv4 address is universalbecause each addressing system must be accepted by any host

that wants to be connected to the Internet.

24. Question (1.0 mark):How to distinguish IPv4address classes in the classfuladdressing method? How many classes are there ?

In the classful addressing method, there are 5 classes: A, B, C, D and E.

The distinction of these five classes bases on their first bits (if the address is given in

binary notation) as follow:

- If the first bit is 0, it defines class A.- Class B always starts with two bits 10.- The first three bits of class C is always 110.- The first four bits of class D is always 1110.- And 1111 is the first four bits of class E.
http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-csma-cd/http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-csma-cd/http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-csma-cd/http://www.differencebetween.com/difference-between-csma-cd-and-vs-csma-ca/#ixzz32E2EAe55http://www.differencebetween.com/difference-between-csma-cd-and-vs-csma-ca/#ixzz32E2EAe55http://en.wikipedia.org/wiki/IEEE_802.11_RTS/CTShttp://en.wikipedia.org/wiki/IEEE_802.11_RTS/CTShttp://en.wikipedia.org/wiki/IEEE_802.11_RTS/CTShttp://www.differencebetween.com/difference-between-csma-cd-and-vs-csma-ca/#ixzz32E2EAe55http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-csma-cd/http://www.differencebetween.net/technology/protocols-formats/difference-between-csma-ca-and-csma-cd/


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25. Question (1.0 mark):What are the problems of classful addressing ?The problem of classful addressing is that each class is devided into a fixed number of

blocks with each block having a fixed size as shown in the table below:

Since the addresses were not distributed properly, resulting in no more addresses

available for organizations and individuals that needed to be connected to the Internet.

To understand the problem, let us think about class A. This class can be assigned to

only 128 very large organizations in the world but each needs to have a single network

with 16,777,216 nodes.A block in class A address is too large for almost any

organizations. This means most of the address in class A were wasted (unused). Class B

addresses were designed for midsize organizations, but many of the addresses in this

class also remained unused. A block in class C is probably too small for many

organizations. Class E were almost never used, wasting the whole class.

26. Question (1.0 mark):Describe briefly the classless addressing method.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 varies based on

the nature and size of the entity.

Restrictions:

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 (1, 2, 4, 8,...)3. The first address must be evenly divisible by the number of addresses.

27. Question (1.0 mark):List address sets for private networks. Are they unique?


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They are unique inside the organization, but they are not unique globally.

28. Question (1.0 mark):What is NAT?Network Address Translation or NAT is technology that allows a private network to use

a set of private addresses for internal communication and a set of global Internetaddresses for externalcommunication.

29. Question (1.0 mark):By using a figure, describe briefly how NAT operates.

The figure above shows a NAT implementation. As the figure shows, the private

network uses private addresses.The router that connects the network to the global

address uses one private address and one global address. The private network is

invisible to the rest of the Internet; the rest of the Internet sees only the NAT router with

the address 200.24.5.8.

30. Question (1.0 mark):Explain why communication at the network layer in Internetis connectionless.

Because in connectionless service, the network layer protocol treats each packet

independently, with each packet having no relationship to any other packet. The packets

in a message may or may not travel the same path to their destination. This type service

is used in the datagram approach to packet switching. The Internet has chosen this of

service the network layer.

The reason for this decision is that the Internet is made of so many heterogeneous

networks that it is almost impossible to create a connection from the source to

destination without knowing the nature of the networks in advance.

31. Question (1.0 mark):Explain why IPv4 is considered as an unreliable andconnectionless datagram protocol (a best-effort delivery service)?

Ipv4 is considered as an unreliable and connectionless datagram protocol because it

provides no error control or flow control (except for error detection on the header).

IPv4 is also a connectionless protocol for a packet-switching network, each datagram ishandled independently, and each datagram can follow a different route to the


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destination. The datagrams sent by the same source to the same destination could arrive

out of order.

32. Question (1.0 mark):What is the role of the Address Resolution Protocol.The Address Resolution Protocol (ARP) is used toglue the network and data-link layersin mapping network-layer addresses to link-layeraddresses. (19.1)

33. Question (1.0 mark):What are RIP, OSPF, BGP?- The Routing Information Protocol (RIP) is an intradomain routing protocol used

inside an autonomous system. It is a very simple based on distance vector

routing.

- The Open Shortest Path First or OSPF protocol is an intradomain routingprotocol based on link state routing. Its domain is also an autonomous system.

- Border Gateway Protocol (BGP) is an interdomain routing protocol using pathvector routing.

34. Question (1.0 mark):What is distance vector routing?Distance vector routing is routing method in which each router sends its neighbors a list

ofnetworks it can reach and the distance to each network.

35. Question (1.0 mark):What is link state routing?Link state routing is a routing method in which each router shares its knowledge of

charges in its neighborhood with all other routers.

36. Question (1.0 mark):Inshortest path slectionalgorithm, what does shortest pathmean?

There are many ways for a packet to go to node B from node A. There may be some

intermediate node from A to B such as node C, D, ... The distance between two adjacent

nodes is called cost, or weight.The shortest path from A to B is the one whose

cumulative cost, or weight is smallest.

37. Question (1.0 mark):Explain why UDP is considered as connectionless transportlayer protocol.

Because the packets (user datagrams) are sent from one party to another with no need

for connection establishment or connection release. This means that each user datagramcan travel on a different path. The packets are not numbered, they may be delayed or

lost or may arrive out of sequence.


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38. Question (1.0 mark):Explain why UDP is not a reliable transport protocol.Because, there is no flow control and error control mechanism in UDP.

39. Question (1.0 mark):Explain why TCP is considered as connection-orientedtransport layer protocol.

Because if two TCPs want to send data to each other, they have to establish the

connection between them before sending data.

40. Question (1.0 mark):List some reasons causing congestion in networks.Congestion results from applications sending more data than the network devices (e.g.,

routers and switches) can accommodate (chac), thus causing the buffers on suchdevices to fill up and possibly overflow. A buffer is a portion of a device's memory that

is set aside as a temporary holding place for data that is being sent to or received from

another device. This can result in delayed or lost packets, thus causing applications to

retransmit the data, thereby adding more traffic and further increasing the congestion.

41. Question (1.0 mark):Briefly describe how Slow Start algorithm operates for TCPcongestion control?

The slow-start algorithmis based on the idea that the size of the congestion

window(cwnd) starts with one maximum segment size(MSS), but it increases one MSS

eachtime an acknowledgment arrives. This algorithm starts slowly but grows

exponentially.


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(RTT: Round-time trip)

Start cwnd = 1 = 20After 1 RTT cwnd = cwnd + 1 = 1 + 1 = 2 = 21After 2 RTT

cwnd = cwnd + 2 = 2 + 2 = 4 = 2

2

After 3 RTT cwnd = cwnd+ 4 = 4 + 4 = 8 = 23

42. Question (1.0 mark):Briefly describe how Congestion Avoidance algorithmoperates for TCP congestion control?

Congrestion Avoidance algorithm increases the cwnd (congestion window) additively

until congestion is detected. In this algorithm, each time the whole window of

segments is acknowledged, the size of the congesstion window is increased by 1.

(RTT: Round-trip time)

Start cwnd = iAfter 1 RTT cwnd = i + 1After 2 RTT cwnd = i + 2After 3 RTT cwnd = i + 3


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43. Question (1.0 mark):Distinguish traffic parameters such as average data rate, peakdata rate and maximum burst size.

- Average data rate is the number of bits sent during a period of time divided by thenumber of second in that period.

amount of dataAverage data ratetime

- Peak data rate: the maximum data rate of the traffic.- Maximum burst size: refers to the maximum length of the time the traffic is

generated at the peak rate.

44. Question (1.0 mark):For best-effort traffic, which performance metric (delay,throughput) is more important. Why ?

Throughput. Because for best-effort traffic, we want to transmit data as much aspossible but not care much about whether the data are delay or not.

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