计算机网络第四章作业参考答案.doc

1. For this problem, use a formula from this chapter, but first state the formula. Frames arrive randomly at a 100-Mbps channel for transmission. If the channel is busy when a frame arrives, it waits its turn in a queue. Frame length is exponentially distributed with a mean of 10,000 bits/frame. For each of the following frame arrival rates, give the delay experienced by the average frame, including both queueing time and transmission time.(a) 90 frames/sec.(b) 900 frames/sec.(c) 9000 frames/sec.Solution：The formula is TMean time delayCA channel of capacity C bpsThe average length of frames is 1/ bitsAn average arrival rate framds/sFrom the problem we can get:C=100M1/=10,000= (1) (2) (3) 4. Measurements of a slotted ALOHA channel with an infinite number of users show that 10% of the slots are idle.(a) What is the channel load, G?(b) What is the throughput?(c) Is the channel under loaded or overloaded?Solution：(1) And according to the Poisson distribution , the probability of no other traffic during the same slot as our test frame is then .(2) In the slotted ALOHA channel, the relation between S(the throughput per frame time) and G(attempts per packet time) is . (3) G=2.3 frames per frame time 1 So the channel is overloaded.6. What is the length of a contention slot in CSMA/CD for(a) a 2-km twin-lead cable(signal propagation speed is 82% of the signal propagation speed in vacuum)?(b) a 40-km multimode fiber optic cable (signal propagation speed is 65% of the signal propagation speed in vacuum)?Solution：The length of a contention slot should be twice as the time for a signal to propagate between the two farthest stations.Suppose the signal propagation speed in vacuum is = m/s.(1) (2) 13.What is the baud rate of classic 10-Mbps Ethernet?Solution：The classic Ethernet use Manchester encoding which two symbols carry a bit. So the baud rate of classic 10M-bps Ethernet should be 20M symbols/s. 14. Sketch the Manchester encoding on a classic Ethernet for the bit stream 0001110101.Solution：（the figure come from 谭棋）We can know that：in the middle of clock cycle，signal which changes from high to low denotes 1，and signal which changes from low to high denotes 0。So：15. A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3) has a propagation speed of 200 m/sec. Repeaters are not allowed in this system. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to send a 32-bit acknowledgement frame. What is the effective data rate, excluding overhead, assuming that there are no collisions?Solution：The length of a contention slot Time to seize the channel: Time to transmit data: Time of data propagate in the channel: Time for the receiver to seize the channel: Time to transmit acknowledgement: Time of acknowledgement propagates in the channel: Total effective data=(256-32)bit=224bitsthe effective data rate= 16.Consider building a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters. The signal speed in the cable is 200,000 km/sec. What is the minimum frame size?Solution：The length of a contention slot The minimum frame size=1Gbps t=1Gbps 0.01ms=10000b=1250B18.Ethernet frames must be at least 64 bytes long to ensure that the transmitter is still going in the event of a collision at the far end of the cable. Fast Ethernet has the same 64-byte minimum frame size but can get the bits out ten times faster. How is it possible to maintain the same minimum frame size?Solution：EthernetFast EthernetThe length of contention slottThe bandwidth of the channel According to the problem, And So：If we make , it can maintain the same minimum frame size.(s is the maximum length of the wire)Or：（come from 谭棋）We can know that the minimum frame size 2DC/S.D for the distance of channel, S for the propagation speed of signal and C for the capacity of channel. If frame size is the same, but bit rate is ten times faster ,maybe the channel length is shorter, or the propagation speed is faster.The fast Ethernet is the former. Maximum length is 1/10 of the Ethernet.36.A switch designed for use with fast Ethernet has a backplane that can move 10 Gbps. How many frames/sec can it handle in the worst case?Solution：（from 谭棋）In the worst case, every frame will be transmitted using the shortest length, thats 64byte.Suppose it can handle x frames/s. 64B8bit/framex frame/s=10Gbps x=10109/512=19531250(frame)38. Consider the extended LAN connected using bridges B1 and B2 in Fig. 4-41(b). Suppose the hash tables in the two bridges are empty. List all ports on which a packet will be forwarded for the following sequence of data transmissions:(a) A sends a packet to C.(b) E sends a packet to F.(c) F sends a packet to E.(d) G sends a packet to E.(e) D sends a packet to A.(f) B sends a packet to F.Solution:Bridges use backward learning algorithm to maintain the table.(a)ACB1table Receive frame from Port1 and flooding frame to Port2,3,4. A1B2table Receive frame from Port4. and flooding frame to Port1,2,3A4(b)EFB2 receive frame from Port2 and flooding it to Port1,3,4A4E2B1 receive flooding frame from Port4 and flooding it to Port1,2,3A1E4 (c)FEB2 receive frame from Port2 and discard it.and B2table is:A4E2F2B1 can not receive the frame, and its tableA1E4 (d) GEB2 Receive frame from Port3 and forward it to Port2. and B2s table is:A4E2F2G3B1 receive no frame this timeA1E4 (e)DAB2 Receive frame and forward it to Port 4 according to its tableA4E2F2G3D1B1 Receive the frame from Port4 and forward to port 1.A1E4D4 (f)BFB1 Receive frame and flooding it to Port1,3,4A1E4D4B2B2 Receive frame from Port4 and forward it to port2 according to its table.A4E2F2G3D1B439. Store-and-forward switches have an advantage over cut-through switches with respect to damaged frames. Explain what it is.Answer: (merge 谭棋s answer)Because bridges only look at the MAC addresses to decide how to forward frames, it is possible to start forwarding as soon as the destination header field has come in, before the rest of the frame has arrived. Cut-through switches start to transmit frames once the head of frame arrives. No check work will be done before transmitting .So damaged frame cannot be discards.But Store-and-forward switches will wait for the whole frame arrive the detect it .If error occurs, switch will ask for retransmitting, or discard it.In conclusion, cut-through switches will make more cost if frame is damaged. and Store-and-forward has a higher reliability rather than cut-through switch.44. Please describe the working principle of a switch.(请描述交换机的工作原理) Answer：4 Keywords: Discard, Forward, Flooding, Learning. The routing procedure for an incoming frame depends on the port it arrives on (the source port) and the address to which it is destined (the destination address). The procedure is as follows.1. DiscardIf the port for the destination address is the same as the source port, discard the frame.2. ForwardIf the port for the destination address and the source port are different ,forward the frame on to the destination port.3. FloodingIf the destination port is unknown, use flooding and send the frame on all ports except the source port. Learningthe bridges operate in promiscuous mode, so they see e