计算机网络（第4版） 清华大学出版社 习题答案（中文版）(6)

13. 答：WDMA（wave length division multiple access）是一个波分多路访问协议。每个站点分配2 个信道；其中窄信道是控制信道，接收其他站发给该站的控制信号；宽信道用作该站点输出数据帧的信道。每个信道被划分成许多个时隙组。时隙0 用某种特殊的方式标记，以便于后继时隙的识别。所有的信道均用同一个全局时钟来同步。每个站点都有2 个发送端和2个接收端，它们分别是：

（1）一个波长固定不变的接收端，它用来侦听本站点的控制信道。 （2）一个波长可调的发送端，它用于向其他站点的控制信道发送帧。 （3）一个波长固定不变的发送端，它用于输出数据帧。

（4）一个波长可调的接收端，它用来选择要侦听的数据发送端。

也就是说，每个站点都侦听自己的控制信道，看是否有请求产生，并将接收端的波长调为发送端的波长，从而得到数据。

GSM（Global system for mobile communication）是一种数字蜂窝无线电系统信道分配方案。

系统中每个蜂窝最多可拥有200 多个全双工信道，每个信道包括下行链路频率（从基站到可移动站）和上行链路频率（从可移动站到基站），每个频段宽200kHz。每一个信道均可采用时分复用技术，支持多个独立的连接。

两种协议都使用FDM 和TDM 结合的方法，它们都可以提供专用的频道（波长），并且都划分时隙，实现TDM。

14. Yes. Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F.

15. (a) Number the floors 1-7. In the star configuration, the router is in the middle of floor 4. Cables are needed to each of the 7 × 15 . 1 = 104 sites. The total length of these cables is

The total length is about 1832 meters.

(b) For 802.3, 7 horizontal cables 56 m long are needed, plus one vertical cable 24 m long, for a total of 416 m.

16. 答：以太网使用曼彻斯特编码，这就意味着发送的每一位都有两个信号周期。标准以太网的数据率为10Mb/s，因此波特率是数据率的两倍，即20MBaud。

17. The signal is a square wave with two values, high (H) and low (L). The pattern is LHLHLHHLHLHLLHHLLHHL.

18. The pattern this time is HLHLHLLHHLLHLHHLHLLH.

19. The round-trip propagation time of the cable is 10 ìsec. A complete transmission has six phases:

transmitter seizes cable (10 ìsec) transmit data (25.6 ìsec)

Delay for last bit to get to the end (5.0 ìsec) receiver seizes cable (10 ìsec) acknowledgement sent (3.2 ìsec)

Delay for last bit to get to the end (5.0 ìsec)

The sum of these is 58.8 ìsec. In this period, 224 data bits are sent, for a rate of about 3.8 Mbps.

i-1 20. 答：把获得通道的尝试从1 开始编号。第i 次尝试分布在2 个时隙中。因此，i 次尝试碰撞的概率是2-(i-1)，开头k-1 次尝试失败，紧接着第k 次尝试成功的概率是：

即：

上式可简化为：

所以每个竞争周期的平均竞争次数是∑ kpk 21. 答：对于1km 电缆，单程传播时间为1/200000 =5×10-6 s，即5，来回路程传播时间为2t =10。为了能够按照CSMA/CD 工作，最小帧的发射时间不能小于10。以1Gb/s 速率工作，10可以发送的比特数等于：

因此，最小帧是10 000 bit 或1250 字节长。

22. The minimum Ethernet frame is 64 bytes, including both addresses in the Ethernet frame header, the type/length field, and the checksum. Since the header fields occupy 18 bytes and the packet is 60 bytes, the total frame size is 78 bytes, which exceeds the 64-byte minimum. Therefore, no padding is used.

23. The maximum wire length in fast Ethernet is 1/10 as long as in Ethernet.

24. The payload is 1500 bytes, but when the destination address, source address, type/length, and checksum fields are counted too, the total is indeed 1518.

25. The encoding is only 80% efficient. It takes 10 bits of transmitted data to represent 8 bits of actual data. In one second, 1250 megabits are transmitted, which means 125 million codewords. Each codeword represents 8 data bits, so the true data rate is indeed 1000 megabits/sec.

26. The smallest Ethernet frame is 512 bits, so at 1 Gbps we get 1,953,125 or almost 2 million frames/sec. However, this only works when frame bursting is operating. Without frame bursting, short frames are padded to 4096 bits, in which case the maximum number is 244,140. For the largest frame (12,144 bits), there can be as many as 82,345 frames/sec.

27. Gigabit Ethernet has it and so does 802.16. It is useful for bandwidth efficiency (one preamble, etc.) but also when there is a lower limit on frame size.

28. Station C is the closest to A since it heard the RTS and responded to it by asserting its NAV signal. D did not respond so it must be outside A’s radio range.

29. A frame contains 512 bits. The bit error rate is p = 10.7. The probability of all 512 of them surviving correctly is (1 . p)512, which is about 0.9999488.

The fraction damaged is thus about 5 × 10.5. The number of frames/sec is 11 × 106 /512 or about 21,484. Multiplying these two numbers together, we get about 1 damaged frame per second.

30. It depends how far away the subscriber is. If the subscriber is close in, QAM-64 is used for 120 Mbps. For medium distances, QAM-16 is used for 80 Mbps. For distant stations, QPSK is used for 40 Mbps.

31. Uncompressed video has a constant bit rate. Each frame has the same number of pixels as the previous frame. Thus, it is possible to compute very accurately how much bandwidth will be needed and when. Consequently, constant bit rate service is the best choice.

32. One reason is the need for real-time quality of service. If an error is discovered, there is no time to get a retransmission. The show must go on. Forward error correction can be used here. Another reason is that on very low quality lines (e.g., wireless channels), the error rate can be so high that practically all frames would have to be retransmitted, and the retransmission would

probably damaged as well. To avoid this, forward error correction is used to increase the fraction of frames that arrive correctly.

33. It is impossible for a device to be master in two piconets at the same time. There are two problems. First, only 3 address bits are available in the header while as many as seven slaves

could be in each piconet. Thus, there would be no way to uniquely address each slave. Second, the access code at the start of the frame is derived from the master’s identity. This is how slaves tell which message belongs to which piconet. If two overlapping piconets used the same access code, there would be no way to tell which frame belonged to which piconet. In effect, the two piconets would be merged into one big piconet instead of two separate ones.

34. Bluetooth uses FHSS, just as 802.11 does. The biggest difference is that Bluetooth hops at a rate of 1600 hops/sec, far faster than 802.11.

35. An ACL channel is asynchronous, with frames arriving irregularly as data are produced. An SCO channel is synchronous, with frames arriving periodically at a well-defined rate.

36. They do not. The dwell time in 802.11 is not standardized, so it has to be announced to new stations that arrive. In Bluetooth this is always 625 ìsec. There is no need to announce this. All Bluetooth devices have this hardwired into the chip. Bluetooth was designed to be cheap, and fixing the hop rate and dwell time leads to a simpler chip.

37. The first frame will be forwarded by every bridge. After this transmission, each bridge will have an entry for destination a with appropriate port in its hash table. For example, D’s hash table will now have an entry to forward frames destined to a on LAN 2. The second message will be seen by bridges B, D, and A. These bridges will append a new entry in their hash table for frames destined for c. For example bridge D’s hash table will now have another entry to forward frames destined to c on LAN 2. The third message will be seen by bridges H, D, A, and B. These bridges will append a new entry in their hash table for frames destined for d. The fifth message will be seen by bridges E, C, B, D, and A. Bridges E and C will append a new entry in their hash table for frames destined for d, while bridges D, B, and A will update their hash table entry for destination d.

38. Bridges G, I and J are not used for forwarding any frames. The main reason for having loops in an extended LAN is to increase reliability. If any bridge in the current spanning tree fails, the (dynamic) spanning tree algorithm reconfigures the spanning tree into a new one that may include one or more of these bridges that were not a part of the previous spanning tree.

39. The simplest choice is to do nothing special. Every incoming frame is put onto the

backplane and sent to the destination card, which might be the source card. In this case, intracard traffic goes over the switch backplane. The other choice is to recognize this case and treat it specially, sending the frame out directly and not going over the backplane.

40. The worst case is an endless stream of 64-byte (512-bit) frames. If the backplane can handle 109 bps, the number of frames it can handle is 109 /512. This is 1,953,125 frames/sec.

41. The port on B1 to LAN 3 would need to be relabeled as GW.

42. A store-and-forward switch stores each incoming frame in its entirety, then examines it and forwards it. A cut-through switch starts to forward incoming frames before they have arrived completely. As soon as the destination address is in, the forwarding can begin.

43. Store-and-forward switches store entire frames before forwarding them. After a frame comes in, the checksum can be verified. If the frame is damaged, it is discarded immediately. With cut=through, damaged frames cannot be discarded by the switch because by the time the error is detected, the frame is already gone. Trying to deal with the problem is like locking the barn door after the horse has escaped.

44. No. Hubs just connect all the incoming lines together electrically. There is nothing to configure. No routing is done in a hub. Every frame coming into the hub goes out on all the other lines.

45. It would work. Frames entering the core domain would all be legacy frames, so it would be up to the first core switch to tag them. It could do this by using MAC addresses or IP addresses. Similarly, on the way out, that switch would have to untag outgoing frames.

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