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端口 B1 到 LAN 3 需要被重新标记为 GW. 可以工作。进入核心域的帧都会是合法帧，因此帧会在第一个核心交换机里被加 上标记，可以用 MAC 或 IP 地址。 类似的，在出口处，交换机去掉这些标记后再 输出帧。 42. Briefly describe the difference between store-and-forward and cut-through Chapter 5 The Network Layer Problems switches.（E） 1. Give two example computer applications for which connection-oriented service 一个存储转发交换机在它的表项里存储进来的每一帧，然后检查并且转发它；直 is appropriate. Now give two examples for which connectionless service is best.(E)
通型交换机帧一进来就完全转发掉。只要目的地地址是可用的，转发就可以开始。 43. Store-and-forward switches have an advantage over cut-through switches with 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 respect to damaged frames. Explain what it is.（E） 存储转发型交换机存储整个帧然后转发它们。一个帧到来后，可以验证校验和， 如果帧被破坏了，马上丢掉坏帧。用直通型交换机，坏帧不能被交换机丢掉，因为 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. Are there any circumstances when connection-oriented service will (or at least should) deliver packets out of order? Explain.(M) 那时检测错误的同时帧就已经转掉了。 有。中断信号不遵从顺序的投递，它会跳过在它前面的数据。例如是当一个终端 44. To make VLANs work, configuration tables are needed in the switches and 用户键入退出（或 kill）健时，由退出信号产生的分组被立即发送，并且跳过了当 bridges. What if the VLANs of Fig. 4-49(a) use hubs rather than multidrop cables? 前队列中等待程序处理的排在前面任何数据（即已经键入但没被程序读取的数据）。 Do the hubs need configuration tables, too? Why or why not?（E） 3. Datagram subnets route each packet as a separate unit, independent of all 不做路由，进入到集线器的每一帧分发到所有其它的线路上。 others. Virtual-circuit subnets do not have to do this, since each data packet follows 45. In Fig. 4-50 the switch in the legacy end domain on the right is a VLAN-aware a predetermined route. Does this observation mean that virtual-circuit subnets do
not need the capability to route isolated packets from an arbitrary source to an switch. Would it be possible to use a legacy switch there? If so, how would that work? arbitrary destination? Explain your answer.（E） If not, why not?（E）
不对。为了使分组能从任意源到达任意目的地，连接建立时要选择路由，虚电路 网络也需要这一能力。 不需要，集线器只是将所有的输入线收集在一起，并没有进行配置。在集线器中 4. Give three examples of protocol parameters that might be negotiated when a connection is set up.（E）
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。
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5. Consider the following design problem concerning implementation of
virtual-circuit service. If virtual circuits are used internal to the subnet, each data packet must have a 3-byte header and each router must tie up 8 bytes of storage for circuit identification. If datagrams are used internally, 15-byte headers are needed but no router table space is required. Transmission capacity costs 1 cent per 106 bytes, per hop. Very fast router memory can be purchased for 1 cent per byte and is depreciated over two years, assuming a 40-hour business week. The statistically average session runs for 1000 sec, in which time 200 packets are transmitted. The mean packet requires four hops. Which implementation is cheaper, and by how much?（H） 4 跳意味着引入了 5 个路由器。实现虚电路需要在 1000 秒内固定分配 5*8=40 字 节的存储器。实现数据报需要比实现虚电路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-跳段的电路容量的 7
所有的路由选择如下： ABCD, ABCF, ABEF, ABEG, AGHD, AGHF, AGEB, and AGEF，所以总跳数为 24 8. Give a simple heuristic for finding two paths through a network from a given source to a given destination that can survive the loss of any communication line (assuming two such paths exist). The routers are considered reliable enough, so it is not necessary to worry about the possibility of router crashes.（E）
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。
字节-秒的存储器对比 9600开销。如果存储器的使用期为两年，即 3600×8×5×52×2=1.5×10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， -61 个字节-跳段代价是 10分，9600 个字节-跳段的代价为 10-6×\慇2X9600=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。 9. Consider the subnet of Fig. 5-13(a). Distance vector routing is used, and the following vectors have just come in to router C: from B: (5, 0, 8, 12, 6, 2); from D:
6. Assuming that all routers and hosts are working properly and that all software (16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9, 0, 4). The measured delays to B, D, and E, in both is free of all errors, is there any chance, however small, that a packet will be are 6, 3, and 5, respectively. What is C's new routing table? Give both the outgoing
line to use and the expected delay.（M） delivered to the wrong destination?（E） 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 2-k 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. Consider the network of Fig. 5-7, but ignore the weights on the lines. Suppose that it uses flooding as the routing algorithm. If a packet sent by A to D has a maximum hop count of 3, list all the routes it will take. Also tell how many hops worth of bandwidth it consumes.（E）
A， B，C，D， E，F
通过 B 给出（11，6，14，18，12，8）
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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. If delays are recorded as 8-bit numbers in a 50-router network, and delay vectors are exchanged twice a second, how much bandwidth per (full-duplex) line is chewed up by the distributed routing algorithm? Assume that each router has three lines to other routers.（E） 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 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. In Fig. 5-14 the Boolean OR of the two sets of ACF bits are 111 in every row. Is this just an accident here, or does it hold for all subnets under all circumstances? （M） 它区的路由，14 个表项用于远程的群，这时路由表尺寸最小为 20+15+14。
次，因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. For hierarchical routing with 4800 routers, what region and cluster sizes should be chosen to minimize the size of the routing table for a three-layer hierarchy? A good starting place is the hypothesis that a solution with k clusters of k regions of k routers is close to optimal, which means that k is about the cube root of 4800 (around 16). Use trial and error to check out combinations where all three parameters are in the general vicinity of 16.
依题可选择 15 个群、16 个区，每个区 20 个路由器时，即使得
4800=15*16*20，
这时每个路由器需要 20 个表项记录本地路由器，15 个表项记录用于到同一群内其