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=50Δ
t，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- 23 - 间。
29. The network of Fig. 5-37 uses RSVP with multicast trees for hosts 1 and 2 as shown. Suppose that host 3 requests a channel of bandwidth 2 MB/sec for a flow from host 1 and another channel of bandwidth 1 MB/sec for a flow from host 2. At the same time, host 4 requests a channel of bandwidth 2 MB/sec for a flow from host
1 and host 5 requests a channel of bandwidth 1 MB/sec for a flow from host 2. How 34. Suppose that host A is connected to a router R 1, R 1 is connected to another much total bandwidth will be reserved for these requests at routers A, B, C, E, H, J, router, R 2, and R 2 is connected to host B. Suppose that a TCP message that K, and L?(E) contains 900 bytes of data and 20 bytes of TCP header is passed to the IP code at host A for delivery to B. Show the Total length, Identification, DF, MF, and
Fragment offset fields of the IP header in each packet transmitted over the three
links. Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of
512 bytes, including an 8-byte frame header, and link R2-B can support a maximum
frame 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
35. A router is blasting out IP packets whose total length (data plus header) is 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。
带宽流量如下：A?2，B?0，C?1，E?3，H?3，J?3，K?2 和 L?1 30. The CPU in a router can process 2 million packets/sec. The load offered to it is 1.5 million packets/sec. If a route from source to destination contains 10 routers, how much time is spent being queued and serviced by the CPUs?(E)
依题知 =2 million， =1.5 million，可知 =0.75，从排队理论可知，每个分 组经历的延迟是空系统中的四倍。空系统中的时延是 500 nsec，这里为 2 sec. 经过 10 个路由器，排队总时间为 20 sec.
31. Consider the user of differentiated services with expedited forwarding. Is there a guarantee that expedited packets experience a shorter delay than regular packets? Why or why not?（E）
无法保证，如果快速的分组太多，它们分配的带宽可能会比常规的分组的性能更
差。
32. Is fragmentation needed in concatenated virtual-circuit internets or only in datagram 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）
1024 bytes. Assuming that packets live for 10 sec, what is the maximum line speed the router can operate at without danger of cycling through the IP datagram ID number space?（E）
addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, and the mask in the w.x.y.z/s notation.（M） A：4000?212；B：2000?211；C：4000?212；D：8000?213；
如果线路的比特率为 b，则每秒钟分组的数量为 b/8192，那么发送一个分组所需 始地址，尾地址，和子网掩码如下： 的时间为 8192/b；输出 65,536 个分组要花费 229 /b sec，依题分组的生存期为 A：198.16.0.0 –198.16.15.255 子网写作 198.16.0.0/20 10s，
B：198.16.16.0 – 198.16.23.255 子网写作 198.16.16.0/21
将 229 /b=10，可得 b 为 53,687,091 bps
C：198.16.32.0 – 198.16.47.255 子网写作 198.16.32.0/20
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 8000
- 24 -
D：198.16.64.0 – 198..16.95.255 子网写作 198.16.64.0/19
46. ARP and RARP both map addresses from one space to another. In this respect, they are similar. However, their implementations are fundamentally different. In what major way do they differ?（E） 在 RARP 的实现中有一个 RARP 服务器负责回答查询请求。 在 ARP 的实现中没有这样的服务器，主机自己回答 ARP 查询。 52. The Protocol field used in the IPv4 header is not present in the fixed IPv6 header. Why not?（E） 设置协议段的目的是要告诉目的地主机把 IP 分组交给那一个协议处理程序。中 途的路由器并不需要这一信息，因此不必把它放在主头中。实际上，这个信息存在 于头中，但被伪装了。最后一个（扩展）头的下一个字段就用于这一目的。 Chapter 6 The Transport Layer Problems 1. In our example transport primitives of Fig. 6-2, LISTEN is a blocking call. Is this strictly necessary? If not, explain how a nonblocking primitive could be used. What advantage would this have over the scheme described in the text?(E) 从―被动连接建立在进行中‖到―已建立‖的虚线不再依确认的传输情况而定。该变 迁可立即发生。实质上，―被动连接建立在进行中‖状态已经消失，因为它们什么时 候都不可见。 3. In both parts of Fig. 6-6, there is a comment that the value of SERVER_PORT must be the same in both client and server. Why is this so important?（E） 如果客户机发送一个分组给 SERVER_PORT 并且服务器当时并没有侦听这一端 口，那么这个分组将不会被投递给服务器。 4. Suppose that the clock-driven scheme for generating initial sequence numbers is used with a 15-bit wide clock counter. The clock ticks once every 100 msec, and the maximum packet lifetime is 60 sec. How often need resynchronization take place a. (a) in the worst case? b. (b) when the data consumes 240 sequence numbers/min?（M） 时钟驱动方案的基本思想是同一时间不会有两个活动的 TPDUs 使用相同的序列 号。序列号空间应该足够大，使得当编号循环一周时，具有相同号码的旧的 TPDU 已 经不复存在。
（a） 时钟大循环周期是 215，即 32768 100ms，即 0.1
秒，所以
滴答，每滴答
不是。事实上，LISTEN 调用可以表明建立新连接的意愿，但不封锁。当有了建 立连接的尝试时，调用程序可以被提供一个信号。然后，它执行，比如说，OK 或 REJECT 来接受或拒绝连接。然而，在原先的封锁性方案中，就缺乏这种灵活性。 2. In the model underlying Fig. 6-4, it is assumed that packets may be lost by the network layer and thus must be individually acknowledged. Suppose that the network layer is 100 percent reliable and never loses packets. What changes, if any,
are needed to Fig. 6-4?(E)