CS 455. Week 9. Reading: 1. Data Communications and Networking Behrouz A Forouzan Edition: 4.
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1 CS 455 Week 9 Reading: 1. Data Communications and Networking Behrouz A Forouzan Edition: 4. Objectives: 1. To test the students knowledge on the concepts covered from Week 1 through Week 7. Concepts: 1. Midterm Examination 2. Midterm Solution Outline: References: CS 455 Week 9 Page 1
2 Midterm Examination Duration : 45 mins Max Marks : 50 Mark the following as True/False ( Each carries 1 mark ) 1.DOCSIS - Data Over Cable Service Interface Standard True / False 2. A DNS uses a UDP on port number to serve requests. True / False 3. Classless Inter-Domain Routing is based on Variable-Length True / False Supernet Masking (VLSM) 4. The IPv6 provides an address space of addresses. True / False 5. RIPv1 lacks support for variable length subnet masks. True / False 6. Cell relay uses variable length packets. True / False 7. Gateways are referred to as a translation mechanism True / False from one router to another. 8. A limitation of the PCM is the Nyquist frequency. True / False 9. QPSK encodes 4 bits per symbol. True / False 10. When a pre amplifier is used prior to A-D conversion, True / False it affects the SNR ratio 11. ARP and RARP have the same packet format. True / False 12. UDP provides reliable service. True / False 13. A star topology performs better than a ring topology True / False under heavy network loads. 14. The TCP/IP suite uses encapsulation to provide abstraction. True / False of protocols and services 15. Virtual circuit switching is a packet switching technology True / False that may emulate packet switching. CS 455 Week 9 Page 2
3 Short Answer Questions 1. In relation to a connectionless packet-mode network, explain the meaning of the following terms: (Each carries 3 marks) (I) best-effort service, (II) store-and-forward delay, (III) mean packet transfer delay, (IV) jitter. 2. How is a switch different from a bridge? What are the advantages of a cut-through switch? (carries 3 marks) Solve the following. Each carries 10 marks. 1. A series of information frames with a mean length of 100 bits is to be transmitted across the following data links using an idle RQ protocol. If the velocity of propagation of the links is 2 x 10 8 m/s, determine the link efficiency (utilization) for each type of link: T ix = (i) (ii) N(bitsInFrame) R(bitrateBPS) T p = S(meters) V(velocity) 1 U= 2*Tp Tix a 10km link with a data transmission rate of 9600bps, a 500m link with a data transmission rate of 10Mbps. 2. Assuming a bit rate of 56kbps, derive the line signaling rate in baud for the following modulation schemes: (i) (ii) two bits per signal 4-PSK, 4-QAM four bits per signal 16-QAM CS 455 Week 9 Page 3
4 Midterm Solution Mark the following as True/False 1.DOCSIS - Data Over Cable Service Interface Standard True / False 2. A DNS uses a UDP on port number to serve requests. True / False 3. Classless Inter-Domain Routing is based on Variable-Length True / False Supernet Masking (VLSM) 4. The IPv6 provides an address space of addresses. True / False 5. RIPv1 lacks support for variable length subnet masks. True / False 6. Cell relay uses variable length packets. True / False 7. Gateways are reffered to as a translation mechanism True / False from one router to another. 8. A limitation of the PCM is the Nyquist frequency. True / False 9. QPSK encodes 4 bits per symbol. True / False 10. When a pre amplifier is used prior to A-D conversion, True /False it affects the SNR ratio 11. ARP and RARP have the same packet format. True / False 12. UDP provides reliable service. True / False 13. A star topology performs better than a ring topology True / False under heavy network loads. 14. The TCP/IP suite uses encapsulation to provide abstraction. True /False of protocols and services 15. Virtual circuit switching is a packet switching technology True / False that may emulate packet switching. CS 455 Week 9 Page 4
5 Short Answer Questions. 3. In relation to a connectionless packet-mode network, explain the meaning of the following terms: (I) best-effort service, (II) store-and-forward delay, (III) mean packet transfer delay, (IV) jitter. ANS. (I)The Transmission mechanism used by the TCP/IP protocols is the Internetworking Protocol. It is unreliable and connectionless protocol said as best effort delivery service. The term effort means that IP provides no error checking or tracking. IP assumes the underlying layers and does its best to get the transmission through to its destination but it there will be no guarantee. As known there are two kinds of networks: Connectionless network and Connection oriented network. The connectionless network does not require a connection and the exchange of data between two computers is said to done whenever they wish. Internally, the term Router is implemented in the connectionless network so as to transmit the date to the destination required. On both type of the networks each packet is received by the router on an incoming link and stored in a memory buffer. A check is performed to determine if any transmissions or bit errors are present in the packet header. If the signal that is used to represent a binary 0 is corrupted, then it is interpreted by the receiver as a binary 1 and vice versa. And if an error is detected then the packet is discarded. Hence, this service by a connection less network is said to be best-effort service. (II) If there are no errors detected the addresses carried in the packet header are read to know the outgoing link that should be used and packet is placed in a queue which is ready for formatting on the selected outgoing link. Total packets present are transmitted at a maximum bit rate. The transmission phenomena is performed in this manner forming a sequence of packets received on number of incoming links of which requires forwarding on the same outgoing link. Due to this, a packet may experience the delay when it is in the sequence of output. This particular operation is known Store and forward delay (i.e. a process of storing for certain time and then after the forwarding of the packets is handled due to the occurrence of delay) (III) Relating to the Store and Forward delay in each PSE/router contributes to the whole transfer delay of the packet across the network and this average of delay is said to be Mean Packet Transfer Delay. CS 455 Week 9 Page 5
6 (IV) The variation about the mean is said to be the Delay variation or JITTER. For example: There are three packets which should arrive to the destination, the first packet arrives at 00:00:02(2 seconds delay), the second packet arrives at 00:00:17(7 seconds delay) and the third arrives at 00:00:28(8 seconds delay). If the receiver starts playing the first packet at 00:00:02, it will finish at 00:00:12. But, the second packet has not yet arrived and it arrives 5 seconds later. This strategy is called Jitter. The time stamp is used as one solution to Jitter. 2. How is a switch different from a bridge? What are the advantages of a cutthrough switch? Switches and bridges are similar in many ways with slight differences: a) A bridge discards frames that have the source and destination on the same segment; a switch switches all frames it receives b) Switches use duplex operation on all ports Advantage of cut through switch: The store-and-forward mode of operation is not implemented by some switches; when the required output is available the switch starts to forward the frame as soon as the destination MAC address has been received which is possible through the cut-through switch. This is used to achieve the forward delays which are low and even there is chance that the transmission is done with errors. Solve the following. 1. A series of information frames with a mean length of 100 bits is to be transmitted across the following data links using an idle RQ protocol. If the velocity of propagation of the links is 2 x 10 8 m/s, determine the link efficiency (utilization) for each type of link: T ix = N(bitsInFrame) R(bitrateBPS) U= 1 2*T Tix p T p = S(meters) V(velocity) CS 455 Week 9 Page 6
7 (i)a 10km link with a data transmission rate of 9600bps, 100 bits T ix = = 0.01seconds 9600 bps T p = 10* 10 3 meters = 5*10-5 seconds= seconds 2*10 8 m / sec U= = = = * (ii)a 500m link with a data transmission rate of 10Mbps. 100 bits T ix = 10*10 bps T p = 500 meters = 25*10-7 seconds = seconds 2*10 m / sec U = = = = * Assuming a bit rate of 56kbps, derive the line signaling rate in baud for the following modulation schemes: (i)two bits per signal 4-PSK, 4-QAM Given, Data bit rate=56 kbps No of bits per signal, m=2 Calculate Line signaling rate, R s R= R s *m 56= R s *2 R s = 56 / 2 = 28 baud (ii)four bits per signal 16-QAM Given, Data bit rate=56 kbps No of bits per signal, m=4 Calculate Line signaling rate, R s R=R s * m 56= R s * 4 R s =56 / 4 = 14 baud CS 455 Week 9 Page 7
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