17. The probability, Pk , of a frame requiring exactly k transmissions is the probability of the first k-1 attempts failing, pk-1 , times the probability of the k-th transmission succeeding, (1-p) . The mean number of transmission is then
just
18. OSI 的哪一层分别处理以下问题？
把传输的比特流划分为帧——数据链路层
决定使用哪条路径通过子网——网络层.
24. Doubling every 18 months means a factor of four gain in 3 years. In 9 years, the gain is then 43 or 64, leading to 6.4 billion hosts. My intuition says that is much too conservative, since by then probably every television in the world and possibly billions of other appliances will be on home LANs connected to the Internet. The average person in the developed world may have dozens of Internet hosts by then.
26. Small, fixed-length cells can be routed through switches quickly, and completely in hardware. Small, fixed-size cells also make it easier to build hardware that handles many cells in parallel. Also, they do not block transmission lines for very long, making it easier to provide quality-of-service guarantees.
27. The speed of light in coax is about 200,000 km/sec, which is 200 meters/sec. At 10 Mbps, it takes 0.1 sec to transmit a bit. Thus, the bit lasts 0.1 sec in time, during which it propagates 20 meters. Thus, a bit is 20 meters long here.
28. The image is 1024 768 3 bytes or 2,359,296 bytes. This is 18,874,368 bits. At 56,000 bits/sec, it takes about 337.042 sec. At 1,000,000 bits/sec, it takes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887 sec. At 100,000,000 bits/sec, it takes about 0.189 sec.
29. Think about the hidden terminal problem. Imagine a wireless network of five stations, A through E, such that each one is in range of only its immediate neighbors. Then A can talk to B at the same time D is talking to E. Wireless networks have potential parallelism, and in this way differ from Ethernet.
30. One disadvantage is security. Every random delivery man who happens to be in the building can listen in on the network. Another disadvantage is reliability. Wireless networks make lots of errors. A third potential problem is battery life, since most wireless devices tend to be mobile.
If the beam is off by 1 mm at the end, it misses the detector. This amounts to a triangle with base 100 m and height 0.001 m. The angle is one whose tangent is thus 0.00001. This angle is about 0.00057 degrees.
14. With 66/6 or 11 satellites per necklace, every 90 minutes 11 satellites pass overhead. This means there is a transit every 491 seconds. Thus, there will be a handoff about every 8 minutes and 11 seconds.
15. The satellite moves from being directly overhead toward the southern horizon, with a maximum excursion from the vertical of 2. It takes 24 hours to go from directly overhead to maximum excursion and then back.
16. The number of area codes was 8×2×10, which is 160. The number of prefixes was 8×8 ×10, or 640. Thus, the number of end offices was limited to 102,400. This limit is not a problem.
17. With a 10-digit telephone number, there could be 1010 numbers, although many of the area codes are illegal, such as 000. However, a much tighter limit is given by the number of end offices. There are 22,000 end offices, each with a maximum of 10,000 lines. This gives a maximum of 220 million telephones. There is simply no place to connect more of them. This could never be achieved in practice because some end offices are not full. An end office in a small town in Wyoming may not have 10,000 customers near it, so those lines are wasted.
24. If all the points are equidistant from the origin, they all have the same amplitude, so amplitude modulation is not being used. Frequency modulation is never used in constellation diagrams, so the encoding is pure phase shift keying.
25. Two, one for upstream and one for downstream. The modulation scheme itself just uses amplitude and phase. The frequency is not modulated.
26. There are 256 channels in all, minus 6 for POTS and 2 for control, leaving 248 for data. If 3/4 of these are for downstream, that gives 186 channels for downstream. ADSL modulation is at 4000 baud, so with QAM-64 (6 bits/baud) we have 24,000 bps in each of the 186 channels. The total bandwidth is then 4.464 Mbps downstream.
27. A 5-KB Web page has 40,000 bits. The download time over a 36 Mbps channel is 1.1 msec. If the queueing delay is also 1.1 msec, the total time is 2.2 msec. Over ADSL there is no queueing delay, so the download time at 1 Mbps is 40 msec. At 56 kbps it is 714 msec.
28. There are ten 4000 Hz signals. We need nine guard bands to avoid any interference. The minimum bandwidth required is 4000×10 400×9 =43,600 Hz.
29. 答：125 的采样时间对应于每秒8000 次采样。一个典型的电话通道为4kHz。根据奈奎斯特定理，为获取一个4kHz 的通道中的全部信息需要每秒8000 次的采样频率。
(Actually the nominal bandwidth is somewhat less, but the cutoff is not sharp.)
31. 答：比较使用如下方案的无噪声4kHz 信道的最大数据传输率:
(a) 每次采样2 比特的模拟编码 ——16kbps
(b) T1 PCM 系统——56kbps
In both cases 8000 samples/sec are possible. With dibit encoding, two bits are sent per sample. With T1, 7 bits are sent per period. The respective data rates are 16 kbps and 56 kbps.
38. VT1.5 can accommodate 8000 frames/sec ×3 columns×9 rows×8 bits =1.728 Mbps. It can be used to accommodate DS-1. VT2 can accommodate 8000 frames/sec ×4 columns×9 rows ×8 bits = 2.304 Mbps. It can be used to accommodate European CEPT-1 service. VT6 can accommodate 8000 frames/sec×12 columns×9 rows×8 bits = 6.912 Mbps. It can be used to accommodate DS-2 service.
39. Message switching sends data units that can be arbitrarily long. Packet switching has a maximum packet size. Any message longer than that is split up into multiple packets.
41. 答：The three networks have the following properties:
星型：最好为2，最差为2，平均为2；
环型：最好为1，最差为n/2，平均为n/4
如果考虑n 为奇偶数，
则n 为奇数时，最坏为（n-1）/2，平均为（n+1）/4
n 为偶数时，最坏为 n/2，平均为n2/4(n1)
全连接：最好为1，最差为1，平均为1。
42. 对于电路交换， t= s时电路建立起来；t＝s+ x /d 时报文的最后一位发送完毕；t= s+x/b+kd时报文到达目的地。而对于分组交换，最后一位在t=x/b时发送完毕。
为到达最终目的地，最后一个分组必须被中间的路由器重发k1次，每次重发花时间p/ b，所以总的延迟为
为了使分组交换比电路交换快，必须：
所以：
43. 答：所需要的分组总数是x /p ，因此总的数据加上头信息交通量为(p+h)x/p位。
源端发送这些位需要时间为(p+h)x/pb
中间的路由器重传最后一个分组所花的总时间为(k-1)(p+h)/ b
因此我们得到的总的延迟为
对该函数求p 的导数，得到
令
得到
因为p＞0，所以
故
时能使总的延迟最小。
44. Each cell has six neighbors. If the central cell uses frequency group A, its six neighbors can use B, C, B, C, B, and C respectively. In other words, only 3 unique cells are needed. Consequently, each cell can have 280 frequencies.
45. First, initial deployment simply placed cells in regions where there was high density of human or vehicle population. Once they were there, the operator often did not want to go to the trouble of moving them. Second, antennas are typically placed on tall buildings or mountains. Depending on the exact location of such structures, the area covered by a cell may be irregular due to obstacles near the transmitter. Third, some communities or property owners do not allow building a tower at a location where the center of a cell falls. In such cases, directional antennas are placed at a location not at the cell center.
46. If we assume that each microcell is a circle 100 m in diameter, then each cell has an area of 2500. If we take the area of San Francisco, 1.2 108 m2 and divide it by the area of 1 microcell, we get 15,279 microcells. Of course, it is impossible to tile the plane with circles (and San Francisco is decidedly three-dimensional), but with 20,000 microcells we could probably do the job.
47. Frequencies cannot be reused in adjacent cells, so when a user moves from one cell to another, a new frequency must be allocated for the call. If a user moves into a cell, all of whose frequencies are currently in use, the user’s call must be terminated.
48. It is not caused directly by the need for backward compatibility. The 30 kHz channel was indeed a requirement, but the designers of D-AMPS did not have to stuff three users into it. They could have put two users in each channel, increasing the payload before error correction from 260 ×50=13 kbps to 260×75 =19.5 kbps. Thus, the quality loss was an intentional trade-off to put more users per cell and thus get away with bigger cells.
49. D-AMPS uses 832 channels (in each direction) with three users sharing a single channel. This allows D-AMPS to support up to 2496 users simultaneously per cell. GSM uses 124 channels with eight users sharing a single channel. This allows GSM to support up to 992 users simultaneously. Both systems use about the same amount of spectrum (25 MHz in each direction).
D-AMPS uses 30 KHz×892 = 26.76 MHz. GSM uses 200 KHz ×124 =24.80 MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over D-AMPS (8 Kbps per user).
50. The result is obtained by negating each of A, B, and C and then adding the three chip sequences. Alternatively the three can be added and then negated.
The result is (+3 +1 +1 1 3 1 1 +1).
51. By definition
If T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the i-th element becoming Ti . Thus,
52. When two elements match, their product is +1. When they do not match, their product is 1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match.
53. Just compute the four normalized inner products:
(1 +1 3 +1 1 3 +1 +1) d (1 1 1 +1 +1 1 +1 +1)/8 = 1
(1 +1 3 +1 1 3 +1 +1) d (1 1 +1 1 +1 +1 +1 1)/8 = 1
(1 +1 3 +1 1 3 +1 +1) d (1 +1 1 +1 +1 +1 1 1)/8 = 0
(1 +1 3 +1 1 3 +1 +1) d (1 +1 1 1 1 1 +1 1)/8 = 1
The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent.
55. 答：它既像TDM，也像FDM。100 个频道中的每一个都分配有自己的频带（FDM），在每个频道上又都有两个逻辑流通过TDM 交织播放（节目和广告交替使用频道）。
This example is the same as the AM radio example given in the text, but neither is a fantastic example of TDM because the alternation is irregular.
56. A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a fiber node.
57. The upstream bandwidth is 37 MHz. Using QPSK with 2 bits/Hz, we get 74 Mbps upstream. Downstream we have 200 MHz. Using QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600 Mbps.
58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10-Mbps Ethernet. There is no way to get bits to the computer any faster than 10-Mbps under these circumstances. If the connection between the PC and cable modem is fast Ethernet, then the full 27 Mbps may be available. Usually, cable operators specify 10 Mbps Ethernet because they do not want one user sucking up the entire bandwidth.
第 3 章 数据链路层
1. 答：由于每一帧有0.8 的概率正确到达，整个信息正确到达的概率为 p=0.810=0.107。
为使信息完整的到达接收方，发送一次成功的概率是p ，二次成功的概率是(1-p)p，三次成功的概率为(1-p)2 p，i 次成功的概率为(1-p)i-1 p，因此平均的发送次数等于：
3. After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.
4. If you could always count on an endless stream of frames, one flag byte might be enough. But what if a frame ends (with a flag byte) and there are no new frames for 15 minutes. How will the receiver know that the next byte is actually the start of a new frame and not just noise on the line? The protocol is much simpler with starting and ending flag bytes.
