BCS THE CHARTERED INSTITUTE FOR IT. BCS HIGHER EDUCATION QUALIFICATIONS BCS Level 5 Diploma in IT COMPUTER NETWORKS

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1 BCS THE CHARTERED INSTITUTE FOR IT BCS HIGHER EDUCATION QUALIFICATIONS BCS Level 5 Diploma in IT COMPUTER NETWORKS Friday 2 nd October 2015 Morning Answer any FOUR questions out of SIX. All questions carry equal marks Time: TWO hours Answer any Section A questions you attempt in Answer Book A Answer any Section B questions you attempt in Answer Book B The marks given in brackets are indicative of the weight given to each part of the question. Only non-programmable calculators are allowed in this examination. Section A Answer Section A questions in Answer Book A S O L U T I O N S A1. This question is about the features and operation of Asynchronous Transfer Mode (ATM) which has seen widespread adoption within the Universal Mobile Telecommunication System, 3 rd generation mobile phone networks. a) Show by means of a diagram, the cell format using within an ATM network. (5 marks)

2 1 octet Generic Flow Control Virtual Path identifier Virtual Path identifier Virtual Channel identifier Virtual Channel Identifier Virtual Channel identifier Payload Type CLP Header Error Control 5 octets Information Field (48 octets) (Marking scheme: 1 for a 5 octet header; 1 for 48 octet payload; 1 for Virtual Path identified; 1 for Virtual Channel identifier; 1 for Header Error Control) b) What is the difference between a Virtual Path and a Virtual Channel? (5 marks) A virtual channel defines a single point to point connection, identified by its virtual channel identifier (VCI). A virtual path however, is a bundle of virtual channels that share the same end-point. Hence, a virtual path can be considered as a container that contains several virtual channels. Each virtual path is identified by its unique virtual path identifier (VPI). (Marking scheme: 2 for the VC, 3 for the VP) c) When a connection is established over an ATM network a process called Connection Admission Control is used to ensure that the network is able to provide the required quality of service. Briefly explain how this process works and explain how a quality of service requirement is specified using a traffic contract. (6 marks) When an end-station wishes to establish an ATM connection it must issue a call request message that identifies the remote end of the connection and also the quality of service required from the connection. Each ATM switch on receiving this request will determine if it

3 has the resources to support such a connection. If it can then the request is passed onto the next switch along the route and so on until it reaches the intended destination. If any switch is unable to commit the necessary resources then the connection will fail. This process is called connection admission control. When requesting a connection the desired quality of service is defined within a traffic contract which forms part of the CAC process. This traffic contract will define a range of traffic parameters that define characteristics such as the peak cell rate, the cell delay variation tolerance, sustainable cell rate, maximum burst size etc. (Marking scheme: 1 for the fact that the connection request specifies the QoS required, 2 for the fact that switches check the request against the resources they have reserve capacity as required, 1 for the fact that the connection will fail if any switch cannot support the request, 2 marks for examples of typical traffic characteristic parameters) d) Explain the purpose and function of the ATM Adaptation Layer type 5 (AAL5) protocol. (9 marks) The lower layers of the ATM protocol suite are responsible for the transmission of the 53 octet ATM cells. At the higher layer we have the applications between transported over the network. Hence, there is a requirement in the middle to convert the application to and from an ATM cell stream. This is the function of the AAL layer. The AAL protocols are end to end protocols and hence, only present in the end-stations. The basic function of AAL5 is: to provide a way for variable bit rate, connectionless applications to send and receive data over an ATM network; to accept data from higher layer protocols (typically IP) and map this onto a stream of 53 octet cells; to Receive ATM cells and combine these into data structures that are acceptable to the higher layer protocol, i.e. convert an ATM cell stream back into an IP datagram; to ensure the delivery of each higher layer protocol message (IP datagram) by inserting an error check sequence (CRC) into a trailer that is added into the last ATM cell of the combined cell stream. This will allow lost cells to be detected and that in turn would result in deletion of the whole higher layer protocol message. (Marking scheme: 1 for AAL being between higher layer protocols and ATM cell layer; 1 for AAL being present in the end stations only, 1 for AAL5 being for variable bit rate, connectionless applications; 2 for mapping higher layer (IP datagram) onto a cell stream; 2 for reassembly of that cell stream, 2 for additional of a CRC in a trailer field to detect any transmission errors)

4 A2. This question is about physical layer transmission systems. a) A transmission system uses a data coding scheme that defines a symbol as a voltage that can have one of sixteen possible values. If the system operates at a transmission rate of 400 symbols per second, determine the data rate measured in: i) baud; (2 marks) Baud is defined as the number of symbols per second. Therefore if the system transmits at 400 symbols per second then the data rate is also 400 baud. (Marking scheme: 1 for Baud = 1 symbol per second, 1 for 400 baud) ii) bits per second. (4 marks) A symbol is a voltage level that can have one of 16 possible values. Sixteen levels can be represented by 4 bits. Therefore one symbol represents 4 data bits. If each symbol represents 4 bits then the transmission rate in bits per second will be 400 x 4 = 1600 bits per second. (Marking scheme: 2 for determining each symbol represents 4 bits and 2 for 1600 bits per second) b) Show by means of a diagram how a logic 1 and a logic 0 is represented by using Manchester encoding. (4 marks) (Marking scheme: 2 for the logic 1 encoding, 2 for the logic 0 encoding. The key feature is that there is a transition in the centre of the bit. Half marks if the transitions are the other way up.)

5 c) Show by means of a diagram how the bit sequence, would be transmitted using Manchester encoding. (4 marks) (Marking scheme: 0.5 for each bit. Important that a transition occurs within the centre of each bit with additional transitions where subsequent bits have the same value.) d) Wide Area Network (WAN) encapsulation protocols are used when connecting a router to an externally provided WAN service. These protocols are based on the High Level Data Link Control (HDLC) in which each message starts and ends with the unique flag sequence of In order to prevent this flag sequence from occurring at other parts of the message, a process known as zero bit insertion, or bit stuffing is used. By considering the transmission of the following 5 message data bytes show by means of a diagram how zero bit insertion is used when transmitting this message (9 marks) With zero bit insertion, when five consecutive 1s are detected then an additional zero bit is inserted in the data stream as shown below. Note that the transmitted bytes must be considered as a continuous data stream. (Marking scheme: 1 for knowing that zero bit insertion means adding logic 0s into the data stream; 2 for knowing that this must be done after 5 consecutive logic 1s; 2 for noting that a 0 needs to be inserted within byte 2; 2 for knowing that a 0 has to be inserted within byte 3 and 2 for byte 5 as shown above.)

6 e) When transmitting the data sequence shown in part (d), how many bits in total have to be sent? (2 marks) Note that an additional 3 logic 0s have had to be inserted. This means that (8x5) + 3 bits need to be sent = 43 bits. (59 when including the start and end flags) (Marking scheme: 1 for 43 and 1 mark for 59) A3. This question is about the Transmission Control Protocol (TCP) and User Datagram Protocol (UDP). a) Both the TCP and UDP protocols use port numbers. What are these port numbers used for and what is meant by the term well known port? (6 marks) Both TCP and UDP provide services to higher layer protocols however multiple higher layer protocols can be multiplexed onto a single UDP or TCP layer. Each of these higher layer protocols are then differentiated by means of UDP/TCP port numbers. Port numbers are 16 bits in length. Therefore the port number identifies the particular higher layer protocol to which a given data stream is destined. Some of these port numbers are pre-defined and are therefore referred to as "well known ports", for example port 80 refers to the higher layer http protocol. (Marking scheme: 2 for recognising that port numbers are 16 bits in length; 2 for knowing that they identify the higher layer protocol to which the data stream is destined, 2 for explaining what a well known port is) b) For each of the following applications determine whether you would use TCP or UDP and explain the reasons for your choice. i. File transfer This should be TCP The reason is that you want a file to be transmitted in its entirety without any errors, therefore the error detection and correction properties of TCP are needed. (Marking scheme: 1 mark for protocol selection; 2 marks for reason) ii. Watching a real time streamed video This should be UDP. The reason is that when watching a movie, delay is critical and therefore there simply isn't any time to seek the retransmission of any errors. The simplicity of UDP is therefore required.

7 (Marking scheme: 1 mark for protocol selection; 2 marks for reason) iii. Web browsing This should be TCP The reason is that web pages need to be delivered without error so that all content is properly formatted and presented. Therefore the error detection and correction properties of TCP are needed. (Marking scheme: 1 mark for protocol selection; 2 marks for reason) iv. A Voice over IP (VoIP) telephone conversation This should be UDP. The reason is that a telephone conversation has strict timing requirements for the transfer of data and seeking the retransmission of any errors would introduce too much delay. Therefore the simplicity of UDP is needed. (Marking scheme: 1 mark for protocol selection; 2 marks for reason) (4 x 3 = 12 marks) b) By considering the operation of the TCP protocol, briefly explain how it is able to overcome errors in the transmission and ensure that data is transferred reliably over a network. (7 marks) The following process is followed: Every octet transmitted through TCP is uniquely identified by a 32 bit sequence number which increases by one for each new octet. Data is transmitted and acknowledged by the receiving end station. Acknowledgements are identified by means of the ACK bit and acknowledgement number within the TCP header. A positive acknowledgement is indicated by virtue of the fact that the ACK bit is set and then the acknowledgement number will indicate the number of the first nonacknowledged octet. In other words all octets up to an including acknowledgement number -1 have been successfully received. If data is corrupted or lost in transit then this must be detected by the transmitter. If an acknowledgement has not been received within a given time determined by a timer then the transmitter simply sends the data again. It is the responsibility of the receiver to ignore any duplicates it receives. Hence, the transmitter will continue re-sending data until a positive acknowledgement is received. (Marking scheme: 1 for knowing that each octet is uniquely identified by a sequence number; 2 for the acknowledgement using an ACK bit and the operation of the acknowledgement number; 2 marks for ACK timeout at the transmitter and 2 marks for resending until a positive ACK is received)

8 Section B Answer Section B questions in Answer Book B S O L U T I O N S B4. This question is about IPv4 addressing. a) In classful addressing, the IP address space is divided into 5 classes. Indicate the classes of each of the following address expressed in binary. Indicate how the class was identified (8 marks) In the binary notation, the first few bits can immediately tell us the class of the address: Class A starts in 0 Class B starts in 10 Class C starts in 110 Class D starts in 1110 Class E starts in 1111 Therefore the address given are: Class A Class C Class B Class E (Marking scheme: 2 marks per address correctly identified, 8 marks in total) b) A host was given the /27 IP address, indicate: The network address to which the host belongs. The network broadcast address to which the host belongs. The total number of hosts available in the network (6 marks)

9 If we apply the AND function to the binary representation of the IP address and the network mask ( ) we will obtain the network address: Network address: The broadcast address is represented by putting as 1s all the bits belonging to the host portion: The total number of hosts is given by the following formula: 2 n 2, where n represents the number of bits available in the host portion. In this case n = 5, which means that we have hosts or 30. (Marking scheme: 2 marks per correct answer) c) Describe the concept of classless addressing, indicating the reason why it was proposed and providing an example of a classless IP address. (5 marks) With the growth of the Internet, it was clear that a larger address space was needed as a long-term solution. The larger address space, however, requires that the length of IP addresses to be increased, which means the format of the IP packets needs to be changed. Although the long-range solution has already been devised and is called IPv6, a short-term solution was also devised to use the same address space but to change the distribution of addresses to provide a fair share to each organization. The short-term solution still uses IPv4 addresses, but it is called classless addressing. In other words, the class privilege was removed from the distribution to compensate for the address depletion. (Marking scheme: 2 marks for describing the concept, 2 for the reason why it was proposed and 1 for the example) d) Considering classes addressing, an organization is granted a block of addresses with the beginning address /24. The organization needs to have 3 subblocks of addresses to use in its three subnets as shown below: One subblock of 120 addresses. One subblock of 60 addresses. One subblock of 10 addresses. Indicate the network IP address and the subnet mask for each of the subblocks. (6 marks) For 120 addresses we need 7 bits in the host portion, therefore the network address is: /25, leaving /25 available for other networks. For 60 addresses we need 6 bits in the host portion, therefore the network address is: /26, leaving /26 available for other networks. For 10 addresses we need 4 bits, therefore the network address is /28 leaving 3 more subblocks of 14 hosts each available.

10 (Marks: 2 marks per correct answer) B5. This question concerns wireless local area networks (WLAN) technology and IEEE standards. a) The data link layer in the IEEE standard is divided into two sublayers: LLC and MAC. Indicate the functions performed by each sublayer. (5 marks) Logical Link Control: handles framing, flow control, and error control. Media Access Control: defines the access method and the framing format specific to the corresponding LAN protocol. (Marking scheme: 1 mark per correct function in each sublayer) b) Draw the flow diagram of the Carrier Sense Multiple Access/Collision Avoidance mechanism used by (CSMA/CA). (6 marks) (Marking scheme: 2 marks for the back off mechanism, 2 marks for the persistent strategy, 2 marks for the collision detection mechanism) c) Indicate at least the reasons why CSMA/CD cannot be implemented by Wireless LANs. (6 marks) For collision detection a station must be able to send data and receive collision signals at the same time. This can mean costly stations and increased bandwidth requirements. Collision may not be detected because of the hidden station problem. We will discuss this problem later in the chapter. The distance between stations can be great. Signal fading could prevent a station at one end from hearing a collision at the other end.

11 (Marking scheme: 2 marks per reason given) d) The IEEE addressing mechanism specifies four cases, defined by the value of the two flags in the FC field, ToDs and FromDS. Explain the values those flags could take and the values the different addresses should take. Use the following table to provide your answer: (8 marks) ToDS FromDS Address 1 Address 2 Address 3 Address Destination Source BSS ID N/A 0 1 Destination Sending AP Source N/A 1 0 Receiving AP Source Destination N/A 1 1 Receiving AP Sending AP Destination Source (Marking scheme: 1 mark per correct cell value) B6. This question is about the concept of Quality of Service (QoS). a) Traditionally, four types of characteristics are attributed to a flow: reliability, delay, jitter, and bandwidth. Briefly explain how each concept is related to QoS. (8 marks) Reliability is a characteristic that a flow needs. Lack of reliability means losing a packet or acknowledgment, which entails retransmission. However, the sensitivity of application programs to reliability is not the same. For example, it is more important that electronic mail, file transfer, and Internet access have reliable transmissions than telephony or audio conferencing. Source-to-destination delay is another flow characteristic. Again applications can tolerate delay in different degrees. In this case, telephony, audio conferencing, video conferencing, and remote log-in need minimum delay, while delay in file transfer or e- mail is less important. Jitter is the variation in delay for packets belonging to the same flow. For example, if four packets depart at times 0, 1, 2, 3 and arrive at 20, 21, 22, 23, all have the same delay, 20 units of time. On the other hand, if the above four packets arrive at 21, 23, 21, and 28, they will have different delays: 21, 22, 19, and 24. For applications such as audio and video, the first case is completely acceptable; the second case is not. For these applications, it does not matter if the packets arrive with a short or long delay as long as the delay is the same for all packets. For this application, the second case is not acceptable. Jitter is defined as the variation in the packet delay. High jitter means the difference between delays is large; low jitter means the variation is small. Different applications need different bandwidths. In video conferencing we need to send millions of bits per second to refresh a colour screen while the total number of bits in an may not reach even a million.

12 (Marking scheme: 2 marks per explanation) b) Briefly explain the concept of RSVP and the three reservation styles defined by RSVP. (8 marks) In the Integrated Services model, an application program needs resource reservation. This means that if we want to use IntServ at the IP level, we need to create a flow, a kind of virtual-circuit network, out of the IP, which was originally designed as a datagram packet-switched network. A virtual-circuit network needs a signalling system to set up the virtual circuit before data traffic can start. The Resource Reservation Protocol (RSVP) is a signalling protocol to help IP create a flow and consequently make a resource reservation. When there is more than one flow, the router needs to make a reservation to accommodate all of them. RSVP defines three types of reservation styles: Wild Card Filter Style In this style, the router creates a single reservation for all senders. The reservation is based on the largest request. This type of style is used when the flows from different senders do not occur at the same time. Fixed Filter Style In this style, the router creates a distinct reservation for each flow. This means that if there are n flows, n different reservations are made. This type of style is used when there is a high probability that flows from different senders will occur at the same time. Shared Explicit Style In this style, the router creates a single reservation that can be shared by a set of flows. (Marking scheme: 2 marks per description of RSVP, 2 marks per style) c) Describe two problems with Integrated Services (4 marks) Scalability. The Integrated Services model requires that each router keep information for each flow. As the Internet is growing every day, this is a serious problem. Service-Type Limitation. The Integrated Services model provides only two types of services, guaranteed and control-load. Those opposing this model argue that applications may need more than these two types of services. (Marking scheme: 2 marks per problem) d) Briefly explain the concept of Differentiated Services and one of the per-hop behaviours specified by it. (5 marks) Differentiated Services (DS or Diffserv) was introduced by the IETF (Internet Engineering Task Force) to handle the shortcomings of Integrated Services. Two fundamental changes were made: 1. The main processing was moved from the core of the network to the edge of the network. This solves the scalability problem. The routers do not have to store information about flows. The applications, or hosts, define the type of service they need each time they send a packet.

13 2. The per-flow service is changed to per-class service. The router routes the packet based on the class of service defined in the packet, not the flow. This solves the service-type limitation problem. We can define different types of classes based on the needs of applications. The Diffser model defines three per-hop behaviours (PHBs) for each node that receives a packet. DE PHB. The DE PHB (default PHB) is the same as best-effort delivery, which is compatible with TOS. EF PHB. The EF PHB (expedited forwarding PHB) provides the following services: Low loss Low latency Ensured bandwidth This is the same as having a virtual connection between the source and destination. AF PHB. The AF PHB (assured forwarding PHB) delivers the packet with a high assurance as long as the class traffic does not exceed the traffic profile of the node. The users of the network need to be aware that some packets may be discarded. (Marking scheme: 3 marks for explaining DiffServ and 2 marks for explaining one of the PHBs.

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