陕西师范大学 计算机网络课后习题指导 Solutions for Problem Sets of Computer Networking 王涛秦石醉张阳阳张敏江等编著 6/25/2015 本书提供教材：Computer Networking A Top-down Approach6 th edition的配套习题指导， 习题及相关资源的原始版权归原作者所有
陕西师范大学 计算机网络课后习题指导 Solutions for Problem Sets of Computer Networking 王涛 秦石醉 张阳阳 张敏江等 编著 6/25/2015 本书提供教材：Computer Networking A Top-down Approach 6th edition 的配套习题指导， 习题及相关资源的原始版权归原作者所有
说明 本书系陕西师范大学内部教学参考使用，编者对教程所包含内容没有权利要求， 一切权利归Computer Networking A Top--down Approach6 th edition原作者及出版 社所有。 本书由王涛、秦石醉、张阳阳、张敏江等人编著，部分内容参考自网络公开资源。 编写过程难免有错误，请谅解。 本书习题基于英文经典计算机网络教材Computer Networking A Top-dowm Approach”第六版
说明 本书系陕西师范大学内部教学参考使用，编者对教程所包含内容没有权利要求， 一切权利归 Computer Networking A Top-down Approach 6th edition 原作者及出版 社所有。 本书由王涛、秦石醉、张阳阳、张敏江等人编著，部分内容参考自网络公开资源。 编写过程难免有错误，请谅解。 本书习题基于英文经典计算机网络教材“Computer Networking A Top-down Approach”第六版
Table of Contents 说明 0 Chapter 1 Computer Networks and the Internet. .1 Chapter 2 Application Layer .19 Chapter 3 Transport Layer..... 36 Chapter 4 The Network Layer........ 69 Chapter 5 The Link Layer:Links,Access Networks,and LANs..............93 Chapter 6 Wireless and Mobile Networks... 110 Chapter 7 Multimedia Networking..... 117 Chapter 8 Security in Computer Networks.... .133 Chapter 9 Network Management....... 145
Table of Contents 说明.................................................................................................................... 0 Chapter 1 Computer Networks and the Internet ......................................... 1 Chapter 2 Application Layer ........................................................................19 Chapter 3 Transport Layer...........................................................................36 Chapter 4 The Network Layer......................................................................69 Chapter 5 The Link Layer: Links, Access Networks, and LANs..............93 Chapter 6 Wireless and Mobile Networks.................................................110 Chapter 7 Multimedia Networking ............................................................117 Chapter 8 Security in Computer Networks...............................................133 Chapter 9 Network Management...............................................................145
计算机网络课后习题指导 陕西师范大学 Chapter 1 Computer Networks and the Internet PI.Design and describe an application-level protocol to be used between an automatic teller machine and a bank's centralized computer.Your protocol should allow a user's card and password to be anaccount withdrawal to be made (that is,money disbursed to the user).Your protocol entities should be able to handle the all-too-common case in which there is not enough money in the account to cover the withdrawal.Specify your protocol by listing the messages exchanged and the action taken by theautomatic teller machine or the bank's centralized computer on transmission and messaes Sketch the operation of your simple withdrawal with diagram similar to that in Figure 12.Explicitly state the assumptions made by your protocol about the underlying end-to-end transport service. 答：本题不止一个答案，很多协议都能解决这个问趣，下面是一个简单的例 子： HELO 2 now that there。card in the user all done ATM(p) Me purpose opn.) ERR
计算机网络课后习题指导 陕西师范大学 1 Chapter 1 Computer Networks and the Internet P1. Design and describe an application-level protocol to be used between an automatic teller machine and a bank’s centralized computer. Your protocol should allow a user’s card and password to be verified, the account balance (which is maintained at the centralized computer) to be queried, and anaccount withdrawal to be made (that is, money disbursed to the user). Your protocol entities should be able to handle the all-too-common case in which there is not enough money in the account to cover the withdrawal. Specify your protocol by listing the messages exchanged and the action taken by theautomatic teller machine or the bank’s centralized computer on transmission and receipt of messages. Sketch the operation of your protocol for the case of a simple withdrawal with no errors, using a diagram similar to that in Figure 1.2. Explicitly state the assumptions made by your protocol about the underlying end-to-end transport service. 答：本题不止一个答案，很多协议都能解决这个问题，下面是一个简单的例 子：
计算机网络课后习题指导 陕西师范大学 Correct operation client server HE10(u银Rxid) 华 BALANCE Insitiwhen theret eh moy Ho世aerid) k it valid uaerid) PASSKD〈gawd BALANCE ERR (not enough funds) P2.Equation 1.1 gives a formula for the end-to-end delay of sending one packet of length L over N links of transmission rateR.Generalize this formula for sending Psuch packets back-to-back over the N links 答：由一个分组端到端时延公式：dn-o-end=N可得，当连续发送P个分 组时，得到时延为：demd-to-end=N后×P。 P3.Consider an application that transmits data at a steady rate (for example,the sender generates an N-bit unit of data every k time units,where k is small and fixed).Also,when such an application starts,it will continue running for a relatively long period of time.Answer the following questions, briefly justifying your answer: a Would a packet-switched network or a circuit-switched network be more appropriate for this application?Why? b.Suppose that a packet-switched network is used and the only traffic in this network comes from such applications as described above.Furthermore.assume that the sum of the application data rates is less than the capacities of each and every link.Is some form of congestion control needed? Why? 答：a.电路交换网更适合所描述的应用，因为这个应用要求在可预测的平滑带宽上进行长期 的会话。由于传输速率是已知，且波动不大 因此可 以给 应用会话话路预留带宽而不会有 太多的浪费。另外，我们不需要太过担心由长时间典型会话应用积累起来的，建立和拆除电 路时耗费的开销时间。 b.由于所给的带宽足够大，因此该网络中不需要拥塞控制机制。最坏的情况下（几乎可能拥 2
计算机网络课后习题指导 陕西师范大学 2 P2. Equation 1.1 gives a formula for the end-to-end delay of sending one packet of length L over N links of transmission rate R. Generalize this formula for sending P such packets back-to-back over the N links. 答：由一个分组端到端时延公式：d = 可得，当连续发送 P 个分 组时，得到时延为：d = × 。 P3. Consider an application that transmits data at a steady rate (for example, the sender generates an N-bit unit of data every k time units, where k is small and fixed). Also, when such an application starts, it will continue running for a relatively long period of time. Answer the following questions, briefly justifying your answer: a. Would a packet-switched network or a circuit-switched network be more appropriate for this application? Why? b. Suppose that a packet-switched network is used and the only traffic in this network comes from such applications as described above. Furthermore, assume that the sum of the application data rates is less than the capacities of each and every link. Is some form of congestion control needed? Why? 答：a. 电路交换网更适合所描述的应用，因为这个应用要求在可预测的平滑带宽上进行长期 的会话。由于传输速率是已知，且波动不大，因此可以给各应用会话话路预留带宽而不会有 太多的浪费。另外，我们不需要太过担心由长时间典型会话应用积累起来的，建立和拆除电 路时耗费的开销时间。 b. 由于所给的带宽足够大，因此该网络中不需要拥塞控制机制。最坏的情况下（几乎可能拥
计算机网络课后习翘指导 陕西师范大学 塞)，所有的应用分别从一条或名条特定的网络链路传输。而由于每条链路的带宽足够处理 所有的应用数据，因此不会发生拥塞现象（只会有非常小的队列）。 P4.Consider the ciruit-switched network in Figure 1.13.Recall that there are 4 circuits on each link Label the four switches A.B.C and D,going in the clockwise direction. a.What is the maximum number of simultaneous connections that can be in progress at any one time in this network? b.Suppose that all between switchesAand C.What is the maximum number of simultaneous connections that can be in progress? e.Suppose we want to make four connections between switches A and C,and another four connections between switches B and D.Can we route these calls through the four links to accommodate all eight connections? 答：a因为这4对相邻交换机，每对之间可以建立4条连接，因此最多可以建立16条连 接。 b.我们可以通过右上角交换机有四条连接，通过左下角交换机有四条连接。 一共有8个连接 c是的。对于A和C之间的链接，我们可以选择两条路径通过B两条路径通过D。对于B 和D之间的链接，我们可以选择两条路径通过A两条路径通过C。在这种情况下，任意一条 链路至少有四个连接。 P5.Review the car-caravan analogy in Section 1.4.Assume a propagation speed km/hour. a.Suppose the caravan travels 150 km,beginning in front of one tollbooth,passing through a second tollbooth.and finishing iust after a third tollbooth.What is the end-to-end delav? b.Repeat (a),now assuming that there are eight cars in the caravan instead of ten. 答：由于收费站间隔150km,车速100kmh,收费站以每12s通过一辆汽车的速度提供服 务。 a10辆车，第一个收费站要花费1205，即2分钟来处理。每一辆车要达到第 个收费站都 会有75100-0.75小时，即45分钟的传播延时，因此所有的车在通过第一个收费站时花费47 分钟，在通过第二个收费站和第二条链路时同样花费47分钟，通过第三个收费站花费2分 钟。因此，（端到端）总延时为96分钟。 要站之间的延时为8x2秒+5分=6分36称，端到强总延时 倍再加上第 (46min+36sec)×2+8×12sec=94min+38seg P6.This elementary problem begins to explore propagation delay and transmission delay.two central concepts in data networking Consider two hosts, Aand B.conected by a single link of rate Rbps Suppose that the wo hostsare separated bymmeters,and suppose the propagation speed long th link is s meters/sec.Host A is to send a packet of size L bits to Host B a.Express the propasation delav.dpop in terms of m and s. b.Determine the transmission time of the packet,dirans,in terms of L and R aeanmdqueuinegdelhysobtainaneypresionforteendoenddely d.Suppose ost A begins to transmit the packet at time0.At time where is the last bit of the packet e.Suppose dorop is greater than dirars.At time /drans.where is the first bit of the packet? f.Supposedpp is less than At time where is the first bit of the packet? 3
计算机网络课后习题指导 陕西师范大学 3 塞），所有的应用分别从一条或多条特定的网络链路传输。而由于每条链路的带宽足够处理 所有的应用数据，因此不会发生拥塞现象（只会有非常小的队列）。 P4. Consider the circuit-switched network in Figure 1.13. Recall that there are 4 circuits on each link. Label the four switches A, B, C and D, going in the clockwise direction. a. What is the maximum number of simultaneous connections that can be in progress at any one time in this network? b. Suppose that all connections are between switches A and C. What is the maximum number of simultaneous connections that can be in progress? c. Suppose we want to make four connections between switches A and C, and another four connections between switches B and D. Can we route these calls through the four links to accommodate all eight connections? 答：a. 因为这 4 对相邻交换机，每对之间可以建立 4 条连接，因此最多可以建立 16 条连 接。 b. 我们可以通过右上角交换机有四条连接，通过左下角交换机有四条连接，一共有 8 个连接。 c. 是的。对于 A 和 C 之间的链接，我们可以选择两条路径通过 B 两条路径通过 D。对于 B 和 D 之间的链接，我们可以选择两条路径通过 A 两条路径通过 C。在这种情况下，任意一条 链路至少有四个连接。 P5. Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 100 km/hour. a. Suppose the caravan travels 150 km, beginning in front of one tollbooth, passing through a second tollbooth, and finishing just after a third tollbooth. What is the end-to-end delay? b. Repeat (a), now assuming that there are eight cars in the caravan instead of ten. 答：由于收费站间隔 150km，车速 100km/h，收费站以每 12s 通过一辆汽车的速度提供服 务。 a. 10 辆车，第一个收费站要花费 120s，即 2 分钟来处理。每一辆车要达到第二个收费站都 会有 75/100=0.75 小时，即 45 分钟的传播延时，因此所有的车在通过第一个收费站时花费 47 分钟，在通过第二个收费站和第二条链路时同样花费 47 分钟，通过第三个收费站花费 2 分 钟。因此，（端到端）总延时为 96 分钟。 b. 每两个收费站之间的延时为 8×12 秒＋45 分＝46 分 36 秒， 端到端总延时是该时延两 倍再加上第三个收费站传输时延，即： (46 + 36) × 2+8 × 12 = 94 + 38 P6. This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host A is to send a packet of size L bits to Host B. a. Express the propagation delay, d prop, in terms of m and s. b. Determine the transmission time of the packet, dtrans, in terms of L and R. c. Ignoring processing and queuing delays, obtain an expression for the endto-end delay. d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans, where is the last bit of the packet? e. Suppose dprop is greater than dtrans. At time t = dtrans, where is the first bit of the packet? f. Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of the packet?
计算机网络课后习题指导 陕西师范大学 g.Suppose s-2.5 10s,L-120 bits,and R=56 kbps.Find the distance m so that dpopequals 答：a传播时延：drop=m/s秒 b.传输时延dtrans=L/R秒： c.端到端时延demd-ta-end=m/s+L/R秒： d.该分组的最后一个bit刚刚离开主机A: c第一个比特距离主机A为dras×s米，还没有到达B: 第一个比特已经到达B: g由g=得，m=片×s=0×2.5×10m≈536km。 P7.In this problem.we consider sending real-time voice from Host a to Host b over a packet-switched network (VolP)Host analog voice toa digital 6 kbps bit stre on the fly.Hostthen groups the bits into 56-byte packets.There is one link between Hosts A and B:its transmission rate is 2 Mbps and its propagation delay is 10 msec.As soon as Host A gathers a packet.it sends it to Host B.As soon as Host B receives an entire packet,it converts the packet's bits to an analog signal How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)? 答：考虑分组第一个比特bi。在该b传输之前，需要将分组所有的b都产生，这个分组产 生时延为： 64x10s=7×10-3s=7ms 然后是分组传输时延为： =6s=0.224ms 最后由题意可得分组在链路上的传播时延为：10ms 综上，节点总时延为上面三个时延之和： =7+0.224+10=17.224ms P8.Suppose users share a Mbps link.Also suppose each user requires 150 kbps when transmitting. but each user transmits only 10 percent of the time.(See the discussion of packet switching versus circuit switching in Section 1 3 a When circuit switching is used,how many users can be su orted? b.For f this packet switching is used.Find the given user is transmitting. c.Suppose there are 120 users.Find the probability that at any given time.exactly n users are transmitting simultaneously.(Hint:Use the binomial distribution.) d.Find the probability that there are 21 or more users transmitting simultaneously 答：a当使用电路交换时，由于每一个用户共享带宽，所以用户数量为： 0=20个： b.由题知每个用户仅有10%的时间在传输，所以任意时刻某用户传输概率为10%： c.当有120个用户，其中恰好有n个用户同时传输的概率为：
计算机网络课后习题指导 陕西师范大学 4 g. Suppose s = 2.5 · 108, L = 120 bits, and R = 56 kbps. Find the distance m so that dprop equals dtrans. 答：a.传播时延: = /秒； b.传输时延 = /秒； c.端到端时延 = m/s +L/R秒； d.该分组的最后一个 bit 刚刚离开主机 A； e.第一个比特距离主机 A 为d × 米，还没有到达 B； f.第一个比特已经到达 B； g.由 = 得，m = × = × × 2.5 × 10 ≈ 536。 P7. In this problem, we consider sending real-time voice from Host A to Host B over a packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-byte packets. There is one link between Hosts A and B; its transmission rate is 2 Mbps and its propagation delay is 10 msec. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at Host A) until the bit is decoded (as part of the analog signal at Host B)? 答：考虑分组第一个比特 bit。在该 bit 传输之前，需要将分组所有的 bit 都产生，这个分组产 生时延为： × × = 7 × 10 = 7 然后是分组传输时延为： = × × = 0.224 最后由题意可得分组在链路上的传播时延为：10ms 综上，节点总时延为上面三个时延之和： d = 7 + 0.224 + 10 = 17.224 P8. Suppose users share a 3 Mbps link. Also suppose each user requires 150 kbps when transmitting, but each user transmits only 10 percent of the time. (See the discussion of packet switching versus circuit switching in Section 1.3.) a. When circuit switching is used, how many users can be supported? b. For the remainder of this problem, suppose packet switching is used. Find the probability that a given user is transmitting. c. Suppose there are 120 users. Find the probability that at any given time, exactly n users are transmitting simultaneously. (Hint: Use the binomial distribution.) d. Find the probability that there are 21 or more users transmitting simultaneously. 答：a. 当使用电路交换时，由于每一个用户共享带宽，所以用户数量为： = 20个； b. 由题知每个用户仅有 10%的时间在传输，所以任意时刻某用户传输概率为 10%； c. 当有 120 个用户，其中恰好有 n 个用户同时传输的概率为：
计算机网络课后习邈指导 陕西师范大学 p=(9)×p”×(1-pm)120- d.有21个或者更多用户在传输的概率为： P=1-∑（C20)p(1-p120- 我们用中心极限定理来求此概率，令X为独立随机变量，则P(X=1)=p。 P(21个或更多的用户")=1- 信-（品 =P≤3)=P≤270=097 当Z为标准正态随机变量时，所求概率P(21个或更多的用户")≈0.003。 P9.Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an isprovided witha1 Mbps link.Us swhen bus but are busy generating data only with probabilityp.1.Suppose that the 1 Mbps link is replaced by a I Gbps linl a.What is N.the maximum number of users that can be supported simultaneously under circuit switching? b Now onsider packet switchingand auser popuation formua (n terms ofp an Nu are sending dat 含：a链客布变为1Gps.用户传输速率为10OKPs,则可支持提大的用户数量 1Gps 10bps N=100Kbps=10X10bps=10000台： b.令传输用户数量为，则多于N用户发送数据的概率为： 或者 1-20ra-r P10.Consider a packet of length L which begins at end system A and travels over three links to a destination end system.These three links are acket switches.Let ds and dente the lnsth propon pd.and the rin1.2.3.The pcke switch delays each packet by dAssuming no queuing delays,in terms of d.s.R.(i=1.2.3).and L.what is the total end-to-end delay for the packet?Suppose now the packet is 1.500 bytes,the propagation 5
计算机网络课后习题指导 陕西师范大学 5 p = × × (1 − ) d. 有 21 个或者更多用户在传输的概率为： P=1 − (1 − ) 我们用中心极限定理来求此概率，令X为独立随机变量，则P = 1 = 。 P"21 个或更多的用户" = 1 − P ≤ 21 P ≤ 21 = ∑ −12 √120 × 0.1 × 0.9 ≤ 9 √120 × 0.1 × 0.9 ≈ P ≤ 9 3.286 = ( ≤ 2.74) = 0.997 当 Z 为标准正态随机变量时，所求概率P"21 个或更多的用户" ≈ 0.003。 P9. Consider the discussion in Section 1.3 of packet switching versus circuit switching in which an example is provided with a 1 Mbps link. Users are generating data at a rate of 100 kbps when busy, but are busy generating data only with probability p = 0.1. Suppose that the 1 Mbps link is replaced by a 1 Gbps link. a. What is N, the maximum number of users that can be supported simultaneously under circuit switching? b. Now consider packet switching and a user population of M users. Give a formula (in terms of p, M, N) for the probability that more than N users are sending data. 答：a. 链路带宽为 1Gbps，用户传输速率为 100Kbps，则可支持最大的用户数量为： N = 1 100 = 10 100 × 10 = 10000 台； b. 令传输用户数量为 n，则多于 N 用户发送数据的概率为： (1 − ) 或者 1− (1 − ) P10. Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let di, si, and Ri denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2, 3. The packet switch delays each packet by dproc. Assuming no queuing delays, in terms of di, si, Ri,(i = 1,2,3), and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation
计算机网络课后习遐指导 陕西师范大学 speed on all three links is 2.510'8 m/s,the transmission rates of all three links are 2 Mbps.the end-to-end delay 答：由题意，设两个端系统为A和B,分组交换器为S,和S2,链路传输速率为R,由题意总 时延由A,S,和S2上的传输时延，在交换器S,和S的处理时延以及在三条链路上的传播时延 构成，则传输时延为： +L1500×8×3b =18ms 传搭时延为： 4ra=4+d+4_500+40+100)×10n =40ms 2.5×108m/s 则总时延为 dcotal drram +dprop +dproc=18+0+3+3=64ms P11.In the above problem,suppose R=Ra=Rs=R and dpoe=0.Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire nacket to a ive.Wha is the nd-ten delay? 答：由题意可知 分组链路 速率相 7点处理 延为0，路由 不再存储转发而是在 接收到后立即发送每一个比特b,所以在两个路由器上就不存在传输时延，类似的，传输 时延为： 1500×8bi drm=片-2x10ps6ms 传播时延不变 4rp-4+4+色-50o0+40ot10o0x102m=40nmg 2.5×108m/s 端到端的总时延为：6+40=46ms P1.Apacket switch receivesa packet and determnes the oubound link to which the packet should be forwarded.When the packet arrives,one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted.Packets are transmitted in bps.What is the queuing delay for the packet?More generally,what is the queuing delay when all packets have length L,the transmission rate isR.x bits of the currently-being-transmitted packet have been transmitted,and n packets are already in the queue? 答：由题意，新到达的分组必须等待它之前到达的分组传输完成才能传输，之前分组包括 等待传输的分组和正在传输的分组两 部分，则： a.新的分组需要等待1500×4+1500÷2=6750 bytesf的传输，即其排队时延为 1500×4+1500÷2)×86i=27ms 2×106bps b.更一般的情况，当前分组已传输bits,队列有n个分组等待传输，则新到达的分组排队
计算机网络课后习题指导 陕西师范大学 6 speed on all three links is 2.5 ·10*8 m/s, the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay? 答：由题意，设两个端系统为 A 和 B，分组交换器为和，链路传输速率为 R，由题意总 时延由 A，和上的传输时延，在交换器和的处理时延以及在三条链路上的传播时延 构成，则传输时延为： = + + = 1500 × 8 × 3 2 × 10 = 18ms 传播时延为： = + + = (5000 +4000 +1000) × 10 2.5 × 10/ = 40 则总时延为： d = + + d = 18 + 40 +3 + 3 = 64 P11. In the above problem, suppose R1 = R2 = R3 = R and dproc = 0. Further suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting for the entire packet to arrive. What is the end-to-end delay? 答：由题意可知，分组链路传输速率相同，节点处理时延为 0，路由器不再存储转发而是在 接收到后立即发送每一个比特 bit，所以在两个路由器上就不存在传输时延，类似的，传输 时延为： = = 1500 × 8 2 × 10 = 6ms 传播时延不变： = + + = (5000 +4000 +1000) × 10 2.5 × 10/ = 40 端到端的总时延为：6 +40 = 46ms P12. A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue? 答：由题意，新到达的分组必须等待它之前到达的分组传输完成才能传输，之前分组包括 等待传输的分组和正在传输的分组两部分，则： a. 新的分组需要等待1500 × 4 + 1500 ÷ 2 = 6750bytes的传输，即其排队时延为： (1500 × 4 +1500 ÷2) × 8 2 × 10 = 27 b. 更一般的情况，当前分组已传输 x bits，队列有 n 个分组等待传输，则新到达的分组排队
计算机网络课后习邈指导 陕西师范大学 时延为： d=nxL+- P13.(a)Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued.Each packet is of lengthLand the link has transmission rate R.What is the average queuing delay for the Npackets? (b)Now se that Nsuch packets arrive to the link every LNR seconds.What is the average queuing delay ofa packet? 答：a由题意，第一个分组没有排队时延，第一个分组排队时延为0，第二个分组排队时 延为5第N个分组排队时延为-：，则平均排队时延为： R 2R b.由题意，N个分组的传输时间为秒，第1,2,3，，N个分组排队时延分别为： 0,告，业，则其平均排队时延为： N-1 R 2 P14.Consider the queuing delay in a router buffer.LetI denote traffic intensity,that/R b.Plot the total delay as a function of L/R. 答：a传输时延为LR,总时延为： daa=due+dramR-n+员FR-刀 b.令LRx,则 drotal=1 ax 当x=0时，总时延为0：当x逐渐增大是，总时廷doa也随之增大，当x接近1a时，趋近 于无穷大。 P15.Leta denote the rate of packets arriving at a link in packets/sec,and letdenote the link's transmission rate in packets/sec.Based on the formula for the total delay(ie.,the queuing delay plus the transmission delay)derived in the previous problem,derive a formula for the total delay 答：由题，a表示链路到达速率，单位为packets/sec:表示传输速率，单位为packets/ec, 共倒数μ即传输时延：由14题可知，总时延为：dot=祭流通强度I=片带入a和 以得到总时延为：
计算机网络课后习题指导 陕西师范大学 7 时延为： d = × + ( − ) P13. (a) Suppose N packets arrive simultaneously to a link at which no packets are currently being transmitted or queued. Each packet is of length L and the link has transmission rate R. What is the average queuing delay for the N packets? (b) Now suppose that N such packets arrive to the link every LN/R seconds. What is the average queuing delay of a packet? 答：a. 由题意，第一个分组没有排队时延，第一个分组排队时延为 0，第二个分组排队时 延为 ，第 N 个分组排队时延为()× ，则平均排队时延为： 0 + L + 2L + ⋯ + ( − 1) × × 1 = ( − 1) 2 b. 由题意， N 个分组的传输时间为 秒，第 1，2，3，···，N 个分组排队时延分别为： 0, , ···, () ，则其平均排队时延为： 1 ( − 1) × = 1 × × = × − 1 2 P14. Consider the queuing delay in a router buffer. Let I denote traffic intensity; that is, I = La/R. Suppose that the queuing delay takes the form IL/R (1 – I) for I < 1. a. Provide a formula for the total delay, that is, the queuing delay plus the transmission delay. b. Plot the total delay as a function of L/R. 答：a. 传输时延为 L/R，总时延为： = + = (1 −) + = L (1 − ) b. 令 L/R=x，则 = 1− 当 x=0 时，总时延为 0；当 x 逐渐增大是，总时延也随之增大，当 x 接近 1/a 时，趋近 于无穷大。 P15. Let a denote the rate of packets arriving at a link in packets/sec, and let μ denote the link’s transmission rate in packets/sec. Based on the formula for the total delay (i.e., the queuing delay plus the transmission delay) derived in the previous problem, derive a formula for the total delay in terms of a and μ. 答：由题，a 表示链路到达速率，单位为 packets/sec；μ 表示传输速率，单位为 packets/sec， 其倒数 1/μ 即传输时延；由 14 题可知，总时延为： = / ，流通强度I = ，带入 a 和 μ 得到总时延为：