1:
1. 第一章课后题第11题 (第四版13题 )
What is the principal difference betweenconnectionless communication and connection-oriented communication?
Connection-orientedcommunication has three phases. In the establishment phase a request is made toset up a connection. Only after this phase has been successfully completed canthe data transfer phase be started and data transported. Then comes the releasephase. Connectionless communication does not have these phases. It just sendsthe data.
知识点:
面向连接与无连接,常识
2. 第一章课后题第18题 (第四版 20题)
A system has an n-layerprotocol hierarchy. Applications generate messages of length M bytes. At each of the layers, an h-byte header is added. What fraction of the networkbandwidth is filled with headers?
Withn layers and h bytes added per layer, thetotal number of header bytes per message is hn, so the space wasted on headers is hn. The total message sizeis M + nh, so the fraction of bandwidth wasted on headers is hn /(M + hn).
知识点:
分层模型,网络传输消息的基本方式
3. 第二章课后题第3题。(第四版 4题)
If a binary signal is sent over a 3-kHzchannel whose signal-to-noise ratio is 20 dB, what is the maximum achievabledata rate?
A signal-to-noise ratio of 20 dB means S/N= 100. Since log2101 is about 6.658, the Shannonlimit is about 19.975 kbps. The Nyquist limit is 6 kbps. The bottleneck istherefore the Nyquist limit, giving a maximum channel capacity of 6 kbps.
知识点: 数据最大传输速率.Shannon定理和Nyquist定理。
4. 第二章课后题第36题。(第四版 42题)
Compare the delay in sending an x-bit message over a k-hoppath in a circuit-switched network and in a (lightly loaded) packet-switchednetwork. The circuit setup time is s sec, thepropagation delay is d sec per hop, the packetsize is p bits, and the data rate is b bps. Under what conditions does the packet networkhave a lower delay?
With circuit switching, at t = s thecircuit is set up; at t = s + x /b the last bit is sent;at t = s + x /b + kd the message arrives.
With packet switching, the last bit is sentat t = x /b. To get to the final destination, the last packetmust be retransmitted k − 1 times by intermediate routers, eachretransmission taking
p /b sec, sothe total delay is x /b + (k − 1)p /b + kd. Packetswitching is faster if s > (k − 1)p /b.
知识点：电路交换和包交换
5. 第三章课后题第7题。 (第四版 10题)
An 8-bit byte with binary value 10101111 isto be encoded using an even-parity Hamming code. What is the binary value afterencoding?
Theencoded value is 101001001111.
知识点：纠错码
2:任选3题完成（至少有1题必须是第三章的）
1. 第三章课后题第15题。 (第四版 15题)
A bit stream 10011101 is transmitted usingthe standard CRC method described in the text. The generator polynomial is x3 + 1. Show the actual bit stringtransmitted. Suppose the third bit from the left is inverted duringtransmission. Show that this error is detected at the receiver's end.
The frame is 10011101. The generator is1001. The message after appending three zeros is 10011101000. The remainder ondividing 10011101000 by 1001 is 100. So, the actual bit string transmitted is10011101100.
The received bit stream with an error inthe third bit from the left is 10111101100. Dividing this by 1001 produces aremainder 100, which is different from zero. Thus, the receiver detects theerror and can ask for a retransmission.
知识点：检错码,易集成
2. 第三章课后题第26题。 (第四版 25题)
In protocol 6 the code for frame_arrival has asection used for NAKs. This section is invoked if the incoming frame is a NAKand another condition is met. Give a scenario where the presence of this othercondition is essential.
Consider the following scenario. A sends0 to B. B gets it and sends an ACK, but the ACK gets lost. A timesout and repeats 0, but now B expects 1, so it sends a NAK. If A merelyre-sent r.ack+1, it would be sending frame 1, which it has not got yet.
知识点：可靠传输原理
3. 第四章课后题第15题。（第四版19题）
A 1-km-long, 10-Mbps CSMA/CD LAN (not802.3) has a propagation speed of 200 m/µsec. Repeaters arenot allowed in this system. Data frames are 256 bits long, including 32 bits ofheader, checksum, and other overhead. The first bit slot after a successfultransmission is reserved for the receiver to capture the channel in order tosend a 32-bit acknowledgement frame. What is the effective data rate, excludingoverhead, assuming that there are no collisions?
The round-trip propagation time of thecable is 10 μsec. A complete transmission has six phases:
transmitter seizes cable (10 μsec)
transmit data (25.6 μsec)
Delay for last bit to get to the end (5.0μsec)
receiver seizes cable (10 μsec)
acknowledgement sent (3.2 μsec)
Delay for last bit to get to the end (5.0μsec)
The sum of these is 58.8 μsec. Inthis period, 224 data bits are sent, for a rate of about 3.8 Mbps.
知识点：CDMA/CD原理
4. 第四章课后题第22题。（第四版28题）
In Fig.4-27, four stations, A, B, C, and D, are shown. Which of the last two stations do youthink is closest to A and why?
Station C is the closest to A sinceit heard the RTS and responded to it by asserting its NAV signal. D didnot respond so it must be outside A’s radio range.
知识点：802.11原理及实际协议
5. 第四章课后题第42题。（第四版45题）
In Fig.4-50 the switch in the legacy end domain on the right is a VLAN-awareswitch. Would it be possible to use a legacy switch there? If so, how wouldthat work? If not, why not?
It would work. Frames entering the coredomain would all be legacy frames, so it would be up to the first core switchto tag them. It could do this by using MAC addresses or IP addresses. Similarly,on the way out, that switch would have to untag outgoing frames.
知识点：VLAN
建议课后题，五版38题9题，4版37题，10题（不交这四题）
4版37题：
Consider the interconnected LANs showns in Fig.4-44. Assume that hosts a and b are on LAN 1, c is on LAN 2, and d is onLAN 8. Initially, hash tables in all bridges are empty and the spanning treeshown in Fig4-44(b) is used. Show how the hash tables of different bridges change aftereach of the following events happen in sequence, first (a) then (b) and so on.
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- (a) a sends to d.
- (b) c sends to a.
- (c) d sends to c.
- (d) d moves to LAN 6.
- (e) d sends to a
The first frame will be forwarded by everybridge. After this transmission, each bridge will have an entry for destinationa with appropriate port in its hash table. For example, D’s hashtable will now have an entry to forward frames destined to a on LAN 2.The second message will be seen by bridges B, D, and A. Thesebridges will append a new entry in their hash table for frames destined for c.For example bridge D’s hash table will now have another entry to forwardframes destined to c on LAN 2. The third message will be seen by bridgesH, D, A, and B. These bridges will append a new entryin their hash table for frames destined for d. The fifth message will beseen by bridges E, C, B, D, and A. Bridges Eand C will append a new entry in their hash table for framesdestined for d, while bridges D, B, and A will updatetheir hash table entry for destination d.
知识点：网桥生成树
10题：
Sixteen stations, numbered 1 through 16, arecontending for the use of a shared channel by using the adaptive tree walkprotocol. If all the stations whose addresses are prime numbers suddenly becomeready at once, how many bit slots are needed to resolve the contention?
Stations2, 3, 5, 7, 11, and 13 want to send. Eleven slots are needed, with the
contentsof each slot being as follows:
slot1: 2, 3, 5, 7, 11, 13
slot2: 2, 3, 5, 7
slot3: 2, 3
slot4: 2
slot5: 3
slot6: 5, 7
slot7: 5
slot8: 7
slot9: 11, 13
slot10: 11
slot11: 13
知识点：自适应树MAC协议
3：任选3题完成
1. 第五版第5题或者第4版第9题
Consider the subnet of Fig.5-13(a). Distance vector routing is used, and the following vectors havejust come in to router C: from B: (5, 0, 8, 12, 6, 2); from D:(16, 12, 6, 0, 9, 10); and from E: (7, 6, 3, 9,0, 4). The measured delays to B, D, and E, are 6, 3, and5, respectively. What is C's new routing table?Give both the outgoing line to use and the expected delay？
Goingvia B gives (11, 6, 14, 18,12, 8).
Goingvia D gives (19, 15, 9, 3, 9,10).
Goingvia E gives (12, 11, 8, 14, 5,9).
Takingthe minimum for each destination except C gives (11, 6, 0, 3, 5, 8).
Theoutgoing lines are (B, B, –, D, E, B).
知识点：RIPv2
2. 第五版第9题或者第4版第14题
Looking at the subnet of Fig.5-6, how many packets are generated by a broadcast from B, using
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- (a) reverse path forwarding?
- (b) the sink tree?
(a)The reverse path forwarding algorithm takes five rounds to finish. The packetrecipients on these rounds are AC, DFIJ, DEGHIJKN, GHKN, and LMO, respectively. A totalof 21 packets are generated.
(b)The sink tree needs four rounds and 14 packets.
知识点：广播路由
3. 第五版第20题或者第4版第34题
Suppose that host A is connected to arouter R1, R1 is connected to another router, R2, and R2 is connected to hostB. Suppose that a TCP message that contains 900 bytes of data and 20 bytes ofTCP header is passed to the IP code at host A for delivery to B. Show the Totallength, Identification, DF, MF, and Fragment offset fields of the IP header ineach packet transmitted over the three links. Assume that link A-R1 can supporta maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2can support a maximum frame size of 512 bytes, including an 8-byte frameheader, and link R2-B can support a maximum frame size of 512 bytes including a12-byte frame header.
Theinitial IP datagram will be fragmented into two IP datagrams at I1. No otherfragmentation will occur.
LinkA-R1:
Length = 940; ID= x; DF = 0; MF = 0; Offset = 0
LinkR1-R2:
(1)Length = 500; ID = x; DF = 0; MF = 1; Offset = 0
(2)Length = 460; ID = x; DF = 0; MF = 0; Offset = 60
LinkR2-B:
(1)Length = 500; ID = x; DF = 0; MF = 1; Offset = 0
(2)Length = 460; ID = x; DF = 0; MF = 0; Offset = 60
知识点：IP分段
4. 第五版第22题或者第4版第36题
An IP datagram using the Strict source routing option has to be fragmented. Doyou think the option is copied into each fragment, or is it sufficient to justput it in the first fragment? Explain your answer.
Sincethe information is needed to route every fragment, the option must
appearin every fragment.
知识点：IP选项字段
5. 第五版第25题或者第4版第39题
A network on the Internet has a subnet maskof 255.255.240.0. What is the maximum number of hosts it can handle?
Themask is 20 bits long, so the network part is 20 bits. The remaining 12 bits arefor the host, so 4096 host addresses exist.
知识点：IP子网
6. 第五版第27题或者第4版第40题
A large number of consecutive IP address areavailable starting at 198.16.0.0. Suppose that four organizations, A, B, C, and D, request 4000,2000, 4000, and 8000 addresses, respectively, and in that order. For each ofthese, give the first IP address assigned, the last IP address assigned, andthe mask in the w.x.y.z/snotation.
To start with, all the requests are roundedup to a power of two. The starting
address, ending address, and mask are asfollows:
4000: 198.16.0.0/20 198.16.0.0 –198.16.15.255
2000: 198.16.16.0/21 198.16.16.0 –198.16.23.255
4000: 198.16.32.0/20 198.16.32.0—198.16.47.255
8000: 198.16.64.0/19 198.16.64.0 –198.16.95.255
知识点：IP无类域划分
7. 第五版第28题或者第4版第41题
A router has just received the followingnew IP addresses: 57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21, and57.6.120.0/21. If all of them use the same outgoing line, can they beaggregated? If so, to what? If not, why not?
Theycan be aggregated to 57.6.96/19.
知识点：路由聚合
8. 第五版第30题或者第4版第43题
A router has thefollowing (CIDR) entries in its routing table:
| Address/mask | Next hop |
| 135.46.56.0/22 | Interface 0 |
| 135.46.60.0/22 | Interface 1 |
| 192.53.40.0/23 | Router 1 |
| default | Router 2 |
For each of thefollowing IP addresses, what does the router do if a packet with that addressarrives?
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- (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
Thepackets are routed as follows:
(a)Interface 1
(b)Interface 0
(c)Router 2
(d)Router 1
(e)Router 2
知识点：路由表
9. 第五版第32题或者第4版第45题
You have just explained the ARP protocol toa friend. When you are all done, he says: ''I've got it. ARP provides a serviceto the network layer, so it is part of the data link layer.'' What do you sayto him?
You say that ARP does not provide a serviceto the network layer, it is part of the network layer and helps provide a serviceto the transport layer. The issue of IP addressing does not occur in the datalink layer. Data link layer protocols are like protocols 1 through 6 in Chap. 3, HDLC, PPP, etc. They move bitsfrom one end of a line to the other.
知识点：分层概念及IP操作流程
1. 面向连接与非连接，电路交换与包交换的原理，理解传播延迟和传输延迟
2. 无线局域网MAC层原理及实际协议
3. 以太网MAC层原理及实际协议
4. 帧分界，纠错码、检错码，单工有噪信道协议的可靠传输
5. IP操作规范以及使用ARP协议后IP传输数据包的详细流程
6. IP地址分配，路由聚合，最长前缀匹配转发
7. RIP协议，包括计数到无穷，水平分割，毒性逆转，触发更新
8. OSPF协议，图抽象过程，树建立，形成转发表。
9. BGP协议原理，路径属性的作用。
10. TCP三次握手建立连接，拆除连接过程。
11. TCP头部各个字段的含义及其在数据传输过程中的变化。
12. TCP拥塞控制原理，慢启动及冲突避免算法。
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