计算机网络英文习题
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How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 × 108 m/s, and transmission rate of 2Mbps? More generally, how long does it take packet of length L to propagate over a link of distance of d, propagation speed s, and transmission rate R bps? Does this delay depends on packet length? Does this delay depends on transmission rate? Answer: (a) The general expression of propagation delay is: dprop = (b) The propagation delay for the given data is: dprop = 2.5 × 106 = 0.01 s 2.5 × 108 d s
= 320 seconds
Page 3 of 16
Handout # 7
EE 471/ CS 471 /CS 573: Fall 2012-2013
Due on Sept 17, 2012
Problems From Chapter 1 of Textbook:
Ch-1, P-2 (Solution)
Ch-1, RQ-13 (Solution)
Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are R1 and R2 , respectively. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a packet of length L? (Ignore queuing, propagation delay and processing delay). Answer: First, realize that the store-and-forward packet switching means the packet switch completely receives the packet from the sending host before forwarding it to the receiving host. In this case, the end-to-end delay is the sum of transmission delays and propagation delays of the two links (i.e., one link between sender and the switch and other between the switch and the receiver). Let d1 denote the transmission delay of the link between sender and switch and d2 denote the transmission delay of the link between switch and receiver. Then, d1 = L R1 1 1 + R1 R2 R1 + R2 R1 R2 d2 = L R2
Ch-1, RQ-12 (Solution)
Why is it said that packet switching employs statistical multiplexing? Contrast statistical multiplexing with the multiplexing that takes place in TDM. Answer: In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.
Ch-1, RQ-19 (Solution)
Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 500 kbps, R2 = 2 Mbps and R3 = 1 Mbps. (a) Assuming no other traffic in the network, what is the throughput for the file transfer. (b) Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B? (c) Repeat (1) and (2), but now with R2 reduced to 100 kbps. Answer: (a) When there is no other network traffic, throughput will be min(R1 , R2 , R3 ) i.e., 500 kbps. (b) Transmission delay of 4 × 106 byte is calculated below. dtran = (32 × 106 ) ÷ (500 × 103 ) = (32 × 106 ) (500 × 103 )
Due on Sept 17, 2012
Review Questions From Chapter 1 of Textbook:
Ch-1, RQ-1 (Solution)
What is difference between a host and an end system? List the types of end system. Is a Web server an end system? Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internetconnected PDAs, WebTVs, etc.
(c) The propagation delay is independent of the packet length (see equation). (d) The propagation delay is independent of the transmission rate used to transmit data.
Consider an application which transmits data at a steady rate (for example, the sender generates a N -bit unit of data every k time units, where k is small and fixed). Also, when such an application starts, it will stay on for relatively long period of time. Answer the following questions, briefly justifying your answer: (a) Would a packet-switched network or a circuit-switched network be more appropriate for this application? Why? (b) Suppose that a packet-switched network is used and only traffic in this network comes from such application as described above. Furthermore, assume that the sum of the application data rates is less than the capacities of each and every link. Is some form of congestion control needed? Why? Answer: (a) A circuit-switched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and the traffic is not bursty, bandwidth can be reserved for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session. (b) Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications’ data rates, no congestion (very little queueing) will occur.
= 320 seconds
Page 3 of 16
Handout # 7
EE 471/ CS 471 /CS 573: Fall 2012-2013
Due on Sept 17, 2012
Problems From Chapter 1 of Textbook:
Ch-1, P-2 (Solution)
Ch-1, RQ-13 (Solution)
Suppose there is exactly one packet switch between a sending host and a receiving host. The transmission rates between the sending host and the switch and between the switch and the receiving host are R1 and R2 , respectively. Assuming that the switch uses store-and-forward packet switching, what is the total end-to-end delay to send a packet of length L? (Ignore queuing, propagation delay and processing delay). Answer: First, realize that the store-and-forward packet switching means the packet switch completely receives the packet from the sending host before forwarding it to the receiving host. In this case, the end-to-end delay is the sum of transmission delays and propagation delays of the two links (i.e., one link between sender and the switch and other between the switch and the receiver). Let d1 denote the transmission delay of the link between sender and switch and d2 denote the transmission delay of the link between switch and receiver. Then, d1 = L R1 1 1 + R1 R2 R1 + R2 R1 R2 d2 = L R2
Ch-1, RQ-12 (Solution)
Why is it said that packet switching employs statistical multiplexing? Contrast statistical multiplexing with the multiplexing that takes place in TDM. Answer: In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.
Ch-1, RQ-19 (Solution)
Suppose Host A wants to send a large file to Host B. The path from Host A to Host B has three links, of rates R1 = 500 kbps, R2 = 2 Mbps and R3 = 1 Mbps. (a) Assuming no other traffic in the network, what is the throughput for the file transfer. (b) Suppose the file is 4 million bytes. Dividing the file size by the throughput, roughly how long will it take to transfer the file to Host B? (c) Repeat (1) and (2), but now with R2 reduced to 100 kbps. Answer: (a) When there is no other network traffic, throughput will be min(R1 , R2 , R3 ) i.e., 500 kbps. (b) Transmission delay of 4 × 106 byte is calculated below. dtran = (32 × 106 ) ÷ (500 × 103 ) = (32 × 106 ) (500 × 103 )
Due on Sept 17, 2012
Review Questions From Chapter 1 of Textbook:
Ch-1, RQ-1 (Solution)
What is difference between a host and an end system? List the types of end system. Is a Web server an end system? Answer: There is no difference. Throughout this text, the words “host” and “end system” are used interchangeably. End systems include PCs, workstations, Web servers, mail servers, Internetconnected PDAs, WebTVs, etc.
(c) The propagation delay is independent of the packet length (see equation). (d) The propagation delay is independent of the transmission rate used to transmit data.
Consider an application which transmits data at a steady rate (for example, the sender generates a N -bit unit of data every k time units, where k is small and fixed). Also, when such an application starts, it will stay on for relatively long period of time. Answer the following questions, briefly justifying your answer: (a) Would a packet-switched network or a circuit-switched network be more appropriate for this application? Why? (b) Suppose that a packet-switched network is used and only traffic in this network comes from such application as described above. Furthermore, assume that the sum of the application data rates is less than the capacities of each and every link. Is some form of congestion control needed? Why? Answer: (a) A circuit-switched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and the traffic is not bursty, bandwidth can be reserved for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session. (b) Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications’ data rates, no congestion (very little queueing) will occur.
