总共2 n -1个父节点，所以，
因为 2n >> 1 所以p1≈2- n
在共享父节点的条件下遍历树，从第二级开始每一级访问两个节点，这样遍历树所走过的节点总数n1 = 1++2+…+2+2=1=2n，接下来，我们考察两个发送站共享祖父节点的概率p2和遍历树所走过的节点总数n2。此时在每个父节点下面仅可能有一个站发送。两个发送站共享一个指定的祖父节点的概率是1/ C 22n-1。
共有2 n -2个祖父节点
遍历树比1 n 减少两个节点，即
通过类似的分析和计算，可以得到，两个发送站共享曾祖父节点（属n-3 级祖先节点）的概率是 p3= 2-n++2
遍历树所经过的节点总数比n2又少两个节点，
因此，最坏的情形是2n+1 个时隙（共享父节点），对应于i=0；
最好的情形是3 个时隙，对应于i=n-1 （两个发送站分别位于左半树和右半树），所以平均时隙数等于
该表达式可以简化为
12. 答：如果所有站的发射有效范围都很大，以至于任一站都可以收到所有其他站发送的信号，那么任一站都可以与其他站以广播方式通信。在这样的条件下，CSMA/CD 可以工作的很好。
13. 答：WDMA（wave length division multiple access）是一个波分多路访问协议。每个站点分配2 个信道；其中窄信道是控制信道，接收其他站发给该站的控制信号；宽信道用作该站点输出数据帧的信道。每个信道被划分成许多个时隙组。时隙0 用某种特殊的方式标记，以便于后继时隙的识别。所有的信道均用同一个全局时钟来同步。每个站点都有2 个发送端和2个接收端，它们分别是：
（1）一个波长固定不变的接收端，它用来侦听本站点的控制信道。 （2）一个波长可调的发送端，它用于向其他站点的控制信道发送帧。 （3）一个波长固定不变的发送端，它用于输出数据帧。 （4）一个波长可调的接收端，它用来选择要侦听的数据发送端。
也就是说，每个站点都侦听自己的控制信道，看是否有请求产生，并将接收端的波长调为发送端的波长，从而得到数据。
GSM（Global system for mobile communication）是一种数字蜂窝无线电系统信道分配方案。
系统中每个蜂窝最多可拥有200 多个全双工信道，每个信道包括下行链路频率（从基站到可移动站）和上行链路频率（从可移动站到基站），每个频段宽200kHz。每一个信道均可采用时分复用技术，支持多个独立的连接。
两种协议都使用FDM 和TDM 结合的方法，它们都可以提供专用的频道（波长），并且都划分时隙，实现TDM。
14. Yes. Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F.
15. (a) Number the floors 1-7. In the star configuration, the router is in the middle of floor 4. Cables are needed to each of the 7 × 15 . 1 = 104 sites. The total length of these cables is
The total length is about 1832 meters.
(b) For , 7 horizontal cables 56 m long are needed, plus one vertical cable 24 m long, for a total of 416 m.
16. 答：以太网使用曼彻斯特编码，这就意味着发送的每一位都有两个信号周期。标准以太网的数据率为10Mb/s，因此波特率是数据率的两倍，即20MBaud。
17. The signal is a square wave with two values, high (H) and low (L). The pattern is LHLHLHHLHLHLLHHLLHHL.
18. The pattern this time is HLHLHLLHHLLHLHHLHLLH.
19. The round-trip propagation time of the cable is 10 ìsec. A complete transmission has six phases:
transmitter seizes cable (10 ìsec) transmit data ìsec)
Delay for last bit to get to the end ìsec) receiver seizes cable (10 ìsec) acknowledgement sent ìsec)
Delay for last bit to get to the end ìsec)
The sum of these is ìsec. In this period, 224 data bits are sent, for a rate of about Mbps.
20. 答：把获得通道的尝试从1 开始编号。第i 次尝试分布在2 i-1 个时隙中。因此，i 次尝试碰撞的概率是2-(i-1)，开头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 . 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 = . The probability of all 512 of them surviving correctly is (1 . p)512, which is about .
The fraction damaged is thus about 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
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