（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。

22. 答：（1）由源主机发送的每个分组可能行走1 个跳段、2 个跳段或3 个跳段。走1 个跳段的概率为p ，走2 个跳段的概率为（1- p）p ，走3 个跳段的概率为（1- p）2 p。那么，一个分组平均通路长度的期望值为：

即每次发送一个分组的平均跳段数是 p2-3 p++3。

（2）一次发送成功（走完整个通路）的概率为( 1- p )2，令a = ( 1- p)2，两次发射成功的概率等于 ( 1- a) a，三次发射成功的概率等于 ( 1- a)2 a ，…，因此一个分组平均发送次数为：

即一个分组平均要发送1/(1- p )2次。

（3）最后，每一个接收到的分组行走的平均跳段数等于

23. First, the warning bit method explicitly sends a congestion notification to the source by setting a bit, whereas RED implicitly notifies the source by simply dropping one of its packets. Second, the warning bit method drops a packet only when there is no buffer space left, whereas RED drops packets before all the buffer are exhausted.

24. 答：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. 答：不可以发送任何大于1024 字节的分组。

26. 答：每5产生一个令牌，1 秒中可以发送200 000 个信元。每个信元含有48 个数据字节，即8×48= 384bit。

384×2×105 = × 106b/s

所以，最大的可持续的净数据速率为s。

27. 答：本题乍看起来，似乎以6Mb/s 速率发送用4/3 秒的时间可以发送完桶内8Mb 的数据，使漏桶变空。然而，这样回答是错误的，因为在这期间，已有更多的令牌到达。正确的答案应该使用公式S＝ C /(M-P )，这里的S表示以秒计量的突发时间长度，M 表示以每

秒字节计量的最大输出速率，C 表示以字节计的桶的容量，P 表示以每秒字节计量的令牌到达速率。则：

因此，计算机可以用完全速率6Mb/s 发送 s 的时间。

28. 答：令最大突发时间长度为 t 秒，在极端情况下，漏桶在突发期间的开始是充满的（1MB），在突发期间另有10 t MB 进入桶内。在传输突发期间的输出包含50 t MB。由等式1+10 t=50 t，得到 t=1/40s，即25ms。因此，以最大速率突发传送可维持25ms 的时间。

29. The bandwidths in MB/sec are as follows: A: 2, B: 0, C: 1, E: 3, H: 3, J: 3, K:2, and L: 1.

30. Here is 2 million and is million, so is , and from queueing theory, each packet experiences a delay four times what it would in an idle system. The time in an idle system is 500 nsec, here it is 2sec. With 10 routers along a path, the queueing plus service time is 20 sec.

31. There is no guarantee. If too many packets are expedited, their channel may have even worse performance than the regular channel.

32. 答：在这两种情况下都需要分割功能。即使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. 答：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. The initial IP datagram will be fragmented into two IP datagrams at I1. No other fragmentation will occur.

Link A-R1:

Length = 940; ID = x; DF = 0; MF = 0; Offset = 0 Link R1-R2:

(1) Length = 500; ID = x; DF = 0; MF = 1; Offset = 0 (2) Length = 460; ID = x; DF = 0; MF = 0; Offset = 60 Link 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. If the bit rate of the line is b, the number of packets/sec that the router can emit is b/8192, so the number of seconds it takes to emit a packet is 8192/b.

To put out 65,536 packets takes 229 /b sec. Equating this to the maximum packet lifetime, we get 229 /b = 10. Then, b is about 53,687,091 bps.

36. 答：因为为每一个分割的片段选择路由都需要该选项信息，因此该选项必须出现在每一个片段中。

37. 答：除去2 位作为前缀，将剩下18 位表示网络。概念上，网络数目可以有18 2 或262144 个。然而，全0 和全1 是特别地址，所以只有262142 个可供分配。

38. 答：The address is 答：对于一个B 类网络，高端16 位形成网络号，低端16 位是子网或主机域。在子网掩码的低端16 位中，最高有效4 位为1111，因此剩下12 位用于主机号。因此，存在4096 个主机地址。但由于全0 和全1 是特别地址，因此最大的主机数目为4094。

40. To start with, all the requests are rounded up to a power of two. The starting address, ending address, and mask are as follows: A: – written as – written as – written as – written as They can be aggregated to It is sufficient to add one new table entry: for the new block. If an incoming packet matches both and the longest one wins. This rule makes it possible to assign a large block to one outgoing line but make an exception for one or more small blocks within its range.

43. The packets are routed as follows: (a) Interface 1 (b) Interface 0 (c) Router 2 (d) Router 1 (e) Router 2

44. After NAT is installed, it is crucial that all the packets pertaining to a single connection pass in and out of the company via the same router, since that is where the mapping is kept. If each router has its own IP address and all traffic belonging to a given connection can be sent to the same router, the mapping can be done correctly and multihoming with NAT can be made to work.