发布时间 : 星期四 文章计算机网络（第四版）答案更新完毕开始阅读c91dacbc1a37f111f1855b72
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 ?sec.
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.
第 5 章 网络层
1. 答：文件传送、远程登录和视频点播需要面向连接的服务。另一方面，信用卡验证和其他的销售点终端、电子资金转移，以及许多形式的远程数据库访问生来具有无连接的性质，在一个方向上传送查询，在另一个方向上返回应答。
2. 答：有。中断信号应该跳过在它前面的数据，进行不遵从顺序的投递。典型的例子是当一个终端用户键入退出（或kill）健时。由退出信号产生的分组应该立即发送，并且应该跳过当前队列中排在前面等待程序处理的任何数据（即已经键入但尚未被程序读取的数据）。
3. 答：不对。为了从任意源到任意目的地，为连接建立的分组选择路由，虚电路网络肯定需要这一能力。 4. 答：在连接建立的时候可能要协商窗口的大小、最大分组尺寸和超时值。
5. 答：虚电路实现需要在1000 秒内固定分配5*8=40 字节的存储器。数据报实现需要比虚电路实现多传送的头信息的容量等于(15-3 )? ×4×200＝9600字节-跳段。现在的问题就变成了40000 字节-秒的存储器对比9600 字节-跳段的电路容量。如果存储器的使用期为两年，即3600×8×5×52×2=1.7×107秒，一个字节-秒的代价为1/( 1.5×107)?= 6.7×10-8 分，那么40000 字节-秒的代价为2.7 毫分。另一方面，1 个字节-跳段代价是10-6 分，9600 个字节-跳段的代价为10-6 ×?9600=9.6×10-3分，即9.6 毫分，即在这1000 秒内的时间内便宜大约6.9 毫分。
6. 答：有可能。大的突发噪声可能破坏分组。使用k 位的检验和，差错仍然有2k的概率被漏检。如果分组的目的地段或虚电路号码被改变，分组将会被投递到错误的目的地，并可能被接收为正确的分组。换句话说，偶然的突发噪声可能把送往一个目的地的完全合法的分组改变成送往另一个目的地的也是完全合法的分组。
7. It will follow all of the following routes: ABCD, ABCF, ABEF, ABEG, AGHD, AGHF, AGEB, and AGEF. The number of hops used is 24.
8. 答：使用最短通路搜索算法选择一条路径，然后，删除刚找到的路径中的使用的所有的弧（对应各条链路）。接着，再运行一次最短通路搜索算法。这个第2 条路径在第1 条路径中有线路失效的情况下，可以作为替代路径启用；反之亦然。
9. 答：通过B 给出（11，6，14，18，12，8） 通过D 给出（19，15，9，3，12，13） 通过E 给出（12，11，8，14，5，9）
取到达每一目的地的最小值（C 除外）得到：（11，6，0，3，5，8） 输出线路是：（B，B，-，D，E，B）
10. 答：路由表的长度等于8*50=400bit。该表每秒钟在每条线路上发送2 次，因此400*2=800b/s，即在每条线路的每个方向上消耗的带宽都是800 bps。
11. 答：这个结论总是成立的。如果一个分组从某条线路上到达，必须确认包的到达。 如果线路上没有分组到达，它就是在发送确认。情况00 ( 没有分组到达并且不发送确认)和11 (到达和返回)逻辑上错误，因此不存在。
12. 所谓分级路由，就是将路由器按区（REGION）进行划分，每个路由器只须知道在自己的区内如何为分组选择路由到达目的地的细节，而不用知道其他区的内部结构。对于大的网络，也许两级结构是不够的，还可以把区组合成簇（CLUSTER），把簇再组合成域（ZONE），??对于等级式路由，在路由表中对应所有的本地路由器都有一个登录项，所有其他的区（本簇内）、簇（本域内）和域都缩减为单个路由器，因此减少了路由表的尺寸。
在本题中，4800=15*16*20。当选择15 个簇、16 个区，每个区20 个路由器时（或等效形式，例如20 个簇、16 个区，每个区15 个路由器），路由表尺寸最小，此时的路由表尺寸为15+16+20=51。
The minimum occurs at 15 clusters, each with 16 regions, each region having 20 routers, or one of the equivalent forms, e.g., 20 clusters of 16 regions of 15 routers. In all cases the table size is 15 + 16 + 20 = 51.
13. Conceivably it might go into promiscuous mode, reading all frames dropped onto the LAN, but this is very inefficient. Instead, what is normally done is that the home agent tricks the router into thinking it is the mobile host by responding to ARP requests. When the
router gets an IP packet destined for the mobile host, it broadcasts an ARP query asking for the 802.3 MAC-level address of the machine with that IP address. When the mobile host is not around, the home agent responds to the ARP, so the router associates the mobile user’s IP address with the home agent’s 802.3 MAC-level address.
14. 答：在一个子网中，从所有的源到一个指定的目的地的最佳路由的集合形成一棵以该目的地为根的树。这样的树就称作汇集树。汇集树不必是唯一的，其他具有相同通路长度的树可能存在。所有路由选择算法的目标都是要为所有的路由器寻找和使用汇集树。在广播形式的应用中，源主机需要向所有其他的主机发送报文。在称为反向通路转发的广播路由选择中，当广播分组到达路由器时，路由器对此分组进行检查，查看该分组是否来自于通常用于发送分组到广播源的线路，如果是，则此广播分组本身非常有可能是从源路由器来的第一个拷贝。
在这种情况下，路由器将此分组复制转发到进入线路以外的所有线路。然而，如果广播分组到来的线路不是到达源端的线路，那么分组就被当作副本而扔掉。
（1）反向通路转发算法，算法进行到5 个跳段后结束，总共产生28 个分组。 （2）使用汇集树算法，需要4 个跳段，总共产生14 个分组。
15. Node F currently has two descendants, A and D. It now acquires a third one, G, not circled because the packet that follows IFG is not on the sink tree. Node G acquires a second descendant, in addition to D, labeled F. This, too, is not circled as it does not come in on the sink tree.
16. Multiple spanning trees are possible. One of them is:
17. When H gets the packet, it broadcasts it. However, I knows how to get to I, so it does not broadcast. 18. Node H is three hops from B, so it takes three rounds to find the route.
19. It can do it approximately, but not exactly. Suppose that there are 1024 node identifiers. If node 300 is looking for node 800, it is probably better to go clockwise, but it could happen that there are 20 actual nodes between 300 and 800 going clockwise and only 16 actual nodes between them going counterclockwise.
The purpose of the cryptographic hashing function SHA-1 is to produce a very smooth distribution so that the node density is about the same all along the circle. But there will always be statistical fluctuations, so the straightforward choice may be wrong.
20. The node in entry 3 switches from 12 to 10.
21. 答：对时间以T 秒为单位分时隙。在时隙中，源路由器发送第一个分组。在时隙2 的开始，第2 个路由器收到了分组，但不能应答。在时隙3 的开始，第3 个路由器收到了分组，但也不能应答。这样，此后所有的路由器都不会应答。仅当目的地主机从目的地路由器取得分组时才会发送第1 个应答。现在确认应答开始往回传播。在源路由器可以发送第2 个分组之前，需要两次穿行该子网，需要花费的时间等于2(n-1)T 秒/分组，显然，这种协议的效率是很低的。