24. Some books quote the maximum size of an Ethernet frame as 1518 bytesinstead of 1500 bytes. Are they wrong? Explain your answer.（E）
以太网一帧中数据占用是1500 bytes，但是把目的地地址，源地址，类型/长度域以及校验和域也算上，帧总长就为1518 bytes
26. How many frames per second can gigabit Ethernet handle? Think carefullyand take into account all the relevant cases. Hint: the fact that it is gigabit Ethernetmatters.(E)
最小的以太网帧是64bytes = 512 bits，所以依题1 Gbps 的带宽可得1,953,125 =2*106 frames/sec，然而，这只是在充满最小的帧时是这样，如果没有充满帧，填充短帧至4096 bits，这时每秒处理的帧的最大数量为244,140 bytes，对于最大的帧长12,144 bits，每秒处理的帧的最大数量为82,345 frames/sec.
28. In Fig. 4-27, four stations, A, B, C, and D, are shown. Which of the last twostations do you think is closest to A and why?(E)
站C最接近 A。因为C最先听到A发出的RTS并且通过插入一个NAV信号作为回应。D 对其没有回应，说明它不在A的频率范围内。
29. Suppose that an 11-Mbps 802.11b LAN is transmitting 64-byte frames
back-to-back over a radio channel with a bit error rate of 10-7. How many frames per second will be damaged on average?（E）
一帧是64bytes=512 bits，位出错率为p=10-7 ，所有512 位正确到达的概率为(1- p)512 = 0.9999488，所以帧被破坏的概率约为5*10-5，每秒钟发送的帧数为11*106/512 = 21,484frames/sec ，将上两个数乘一下，大约每秒钟有一帧被破坏。
30. An 802.16 network has a channel width of 20 MHz. How many bits/sec can besent to a subscriber station?（E）
这取决于离子站有多远。如果子站就在附近，那么使用QAM-64 可得带宽120Mbps；中等距离时，使用QAM-16 可得带宽80 Mbps；远程距离，QPSK 可得带宽40 Mbps.
（原题给出的是20mhz 的带宽，要求的是数据率，按照前面的Nyquist定理，最大数据率应该是：2HlogN,但是答案没有乘以2。）
31. IEEE 802.16 supports four service classes.Which service class is the bestchoice for sending uncompressed video?（E）
未压缩的视频有一个固定的位速率。每帧都有与前一帧相同的点数量，因此，可能要准确计
算需要的带宽。最后，最好选用固定位速率服务。
Chapter 5
1. Give two example computer applications for which connection-oriented service
is appropriate. Now give two examples for which connectionless service is best.(E)
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。
5. Consider the following design problem concerning implementation ofvirtual-circuit service. If virtual circuits are used internal to the subnet, each datapacket must have a 3-byte header and each router must tie up 8 bytes of storage forcircuit identification. If datagrams are used internally, 15-byte headers are neededbut no router table space is
required. Transmission capacity costs 1 cent per 106bytes, per hop. Very fast router memory can be purchased for 1 cent per byte and isdepreciated over two years, assuming a 40-hour business week. The statisticallyaverage session runs for 1000 sec, in which time 200 packets are transmitted. Themean packet requires four hops. Which implementation is cheaper, and by howmuch?（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-秒的存储器对比9600 字节-跳段的电路容量的开销。如果存储器的使用期为两年，即3600×8×5×52×2=1.5×107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。
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）
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
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 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。
17. In Fig. 5-20, do nodes H or I ever broadcast on the lookup shown starting at A?（E）
在d中，E，H，I接收到了广播信息之后阴影节点是新的接收节点；箭头显示了可能的逆向路由路径。H 收到分组A后，它广播A；然而，I知道了如何到达I，所以I不广播收到的分组。
18. Suppose that node B in Fig. 5-20 has just rebooted and has no routing
information in its tables. It suddenly needs a route to H. It sends out broadcasts with TTL set to 1, 2, 3, and so on. How many rounds does it take to find a route?（E） 从结点B到H 需要3跳，因此要花3圈来找到路由线路。
24. Give an argument why the leaky bucket algorithm should allow just onepacket per tick, independent of how large the packet is.（M）
通常计算机能够以很高的速率产生数据，网络也可以用同样的速率运行。然而，路由器却只能在短时间内以同样高的速率处理数据。对于排在队列中的一个分组，不管它有多大，路由器必须做大约相同分量的工作。显然，处理10 个100 字节长的分组所作的工作比处理1 个1000 字节长的分组要做的工作多得多。
25. The byte-counting variant of the leaky bucket algorithm is used in a particular system. The rule is that one 1024-byte packet, or two 512-byte packets, etc., may be sent on each tick. Give a serious restriction of this system that was not mentioned in the text.（E）
不可以发送任何大于1024 字节的分组。
28. Imagine a flow specification that has a maximum packet size of 1000 bytes, a token bucket rate of 10 million bytes/sec, a token bucket size of 1 million bytes, and a maximum transmission rate of 50 million bytes/sec. How long can a burst at
maximum speed last?（E）
令最大突发时间长度为Δ t 秒。在极端情况下，漏桶在突发期间的开始是充满的（1MB），这期间数据流入桶内10Δ t MB，流出包含50Δ t MB，由等式1+10Δ t=50Δt，得到Δ t=1/40s，即25ms。因此，以最大速率突发传送可维持25ms 的时间。
32. Is fragmentation needed in concatenated virtual-circuit internets or only indatagram systems?（E）
都需要分割功能。即使是在一个串接的虚电路网络中，沿通路的某些网络可能接受1024 字节分组，而另一些网络可能仅接受48字节分组，分割功能仍然是需要的。
33. Tunneling through a concatenated virtual-circuit subnet is straightforward: the multiprotocol router at one end just sets up a virtual circuit to the other end and passes packets through it. Can tunneling also be used in datagram subnets? If so, how?（E）
可以。只需把分组封装在属于所经过的子网的数据报的载荷段中，并进行发送。
34. Suppose that host A is connected to a router R 1, R 1 is connected to anotherrouter, R 2, and R 2 is connected to host B. Suppose that a TCP message thatcontains 900 bytes of data and 20 bytes of TCP header is passed to the IP code athost A for delivery to B. Show the Total length, Identification, DF, MF, andFragment offset fields of the IP header in each packet transmitted over the threelinks. Assume that link A-R1 can support a maximum frame size of 1024 bytesincluding a 14-byte frame header, link R1-R2 can support a maximum frame size of512 bytes, including an 8-byte frame header, and link R2-B can support a maximumframe size of 512 bytes including a 12-byte frame header.（M） 在I1 最初的IP数据报会被分割成两个IP数据报，以后不会再分割了。 链路A-R1：Length = 940; ID = x; DF = 0; MF = 0; Offset = 0 链路R1-R2：
(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0
(2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60 链路R2-B:
(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60
36．An IP datagram using the strict source routing option has to be fragmented. Do you think the option iscopied into each fragment, or is it sufficient to just put it in the first fragment? Explain your answer.
Since the information is needed to route every fragment, the option mustappear in every fragment.
38．Convert IP address whose hexadecimal notation is C22F1582 to dotted decimal notation? The address is 194.47.21.130.
39．A network on the Internet has a subnet mask of 255.255.240.0. What is the maximum number ofhosts it can handle?
The mask is 20 bits long, so the network part is 20 bits. The remaining 12bits are for the host, so 4096 host addresses exist.
40. A large number of consecutive IP address are available starting at 198.16.0.0.Suppose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000addresses, respectively, and in that order. For each of these, give the first IP addressassigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.（M） A：4000à212 ；B：2000à211 ；C：4000à212 ；D：8000à213 ； 始地址，尾地址，和子网掩码如下：
A：198.16.0.0 –198.16.15.255子网写作198.16.0.0/20
B：198.16.16.0 – 198.16.23.255 子网写作198.16.16.0/21 C：198.16.32.0 – 198.16.47.255 子网写作198.16.32.0/20
43．A router has the following (CIDR) entries in its routing table: Address/maskNext hop 135.46.56.0/22Interface 0 135.46.60.0/22Interface 1 192.53.40.0/23 Router 1
DefaultRouter 2
For each of the following IP addresses, what does the router do if a packet with that address arrives? (a) 135.46.63.10 (b) 135.46.57.14 (c) 135.46.52.2 (d) 192.53.40.7
(e) 192.53.56.7
The packets are routed as follows: (a) Interface 1