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1. A channel has a bit rate of 4 kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50%?answer: 发送一帧的时间等于信道的传播延迟的 2 倍，信道的利用率为 50%，所以，在帧长满足发送时间大于延迟时间的 2 倍是，效率将会高于 50%。由于4kbps=4000bps 故4000*20*0.001*2=160bit只有在帧长不小于 160bit 时，停止等待协议的效率才会至少是 50%。解此题可供参考的公式有两个，下面这两种情况下都可以得到答案：一个是效率= 其中 P 是传输一帧所需要的时间，t 是端到端传送时延。所以可以由 +2 +2=50%，解出帧 N=160bit；二是 中第 42 张中讲到的公式，公式综合考虑了多种因素，信道丢失率，帧头的大小 ，以及 ACK 的发送时间 。在不考虑数据帧的处理时间 tproc 和 nn n/ACK 发送时间的情况下，我们可以推出帧的最小大小为 160bit。答案详情参看上次文档。作业中得到错误答案有两个（1）N=80bit 得出此答案的同学没有弄清楚停等协议，在停等协议公式 1 分母下面的 2t 是往返时间，而不是 t。（2 ） 160kbit，单位换算错误。我可以负责任的告诉你没有哪个网络里面帧能够有十几万比特大小的，以太网里最大帧不过 1526 字节。2. Imagine a sliding window protocol using so many bits for sequence numbers that wraparound never occurs. What relations must hold among the four window edges and the window size, which is constant and the same for both the sender and the receiver?answer: 假设发送者的窗口为（S1 ，Sn）,接受者的窗口为（ R1，Rn ） ，窗口大小为 W，则需满足：0=Sl 但一定有 Rl1 时，信道总是过载的，因此在这里信道是过载的。4What is the baud rate of the standard 10 Mbps Ethernet?答：以太网使用曼彻斯特编码，意味着发送每一位都有两个信号周期，标准以太网的数据率为10MB/S，一次波特率是数据率的两倍，为20MBaud。5A 1-km-long, 10-Mbps CSMA/CD LAN (not 802.3)has a propagation speed of 200 m/?sec. Repeaters are not allowed in this system. Data frames are 256 bits long, including 32 bits of header, checksum, and other overhead. The first bit slot after a successful transmission is reserved for the receiver to capture the channel in order to Send a 32-bit acknowledgement frame. What is the effective data rate, excluding overhead, assuming that there are no collisions?答：依题意知道一公里的铜电缆中单程的传播时间为1、200000=5 usec，往返的时间为2t=10 usec，我们知道，一次完整的传输分为六步，发送者侦听铜电缆的时间为10 usec，若线路可用发送数据帧传输时间为256 bits、10MPS=25.6usec，数据帧最后一位到达时传播的延迟为5.0usec，接听者侦听铜电缆的时间为10 usec，若线路可用接听者发送确认帧所用的时间为3.2 usec，确认帧最后一位到达时的传播延迟为5.0 usec，总共58.8 sec,在这期间发送了224 bits的数据，所以数据率为3.8MPS 。6Two CSMA/CD stations are each trying to transmit long (multiframe) files. After each frame is sent, they contend for the channel, using the binaryexponential backoff algorithm. What is the probability that the contention ends on round k, and what is the mean number of rounds per contention period?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：Pk=（1-2-（k-1） ）2-0*2*-1*2-(k-2)=（1-2-（k-1） ）2-(k-1)(k-2)/2 所以每个竞争周期的平均竞争次数是kpk(k=1,2,3)8An IP packet to be transmitted by Ethernet is 60 bytes long, including all its headers. If LLC is not in use, is padding needed in the Ethernet frame, andif so, how many bytes?答：最小的以太帧是 64bytes，包括了以太帧头部的二者地址、类型/长度域、校验和。因为头部域占用 18 bytes 报文是 60 bytes，总的帧长度是 78 bytes, 已经超过了 64-byte 的最小限制。因此，不需要填补。1. Describe distance vector (DV) algorithm. Discuss the feature of the DV routing algorithm.Solution: (1) The basic idea of DV algorithmEach node x begins with Dx(y), an estimate of thecost of the least-cost path from itself to node y, for all nodes in N. Let Dx= Dx(y): yin N be node xs distance vector, which is the vector of cost estimates from x to allother nodes, y, in N. With the DV algorithm, each node x maintains the followingrouting information: For each neighbor v, the cost c(x,v) from x to directly attached neighborv Node xs distance vector, that is, Dx= Dx(y): y in N, containing xs estimate ofits cost to all destinations, y, in N The distance vectors of each of its neighbors, that is, Dv= Dv(y): y in N for eachneighbor v of xIn the distributed, asynchronous algorithm, from time to time, each node sendsa copy of its distance vector to each of its neighbors. When a node x receives anew distance vector from any of its neighbors v, it saves vs distance vector, andthen uses the Bellman-Ford equation to update its own distance vector as follows:Dx(y) _ minvc(x,v) + Dv(y) for each node y in NIf node xs distance vector has changed as a result of this update step, node x willthen send its updated distance vector to each of its neighbors, which can in turnupdate their own distance vectors. Miraculously enough, as long as all the nodescontinue to exchange their distance vectors in an asynchronous fashion, each costestimate Dx(y) converges to dx(y), the actual cost of the least-cost path from node xto node y(2) The feature of the DV routing algorithmThe distancevector(DV) algorithm is iterative, asynchronous, and distributed. It is distributedin that each node receives some information from one or more of its directlyattached neighbors, performs a calculation, and then distributes the results of itscalculation back to its neighbors.It is iterative in that this process continueson until no more information is exchanged between neighbors. (Interestingly, thealgorithm is also self-terminatingthere is no signal that the computation shouldstop; it just stops.) The algorithm is asynchronous in that it does not require all ofthe nodes to operate in lockstep with each other.2. Consider a configuration in which packets are sent from computers on a LAN to systems on other networks. All of these packets must pass through a router that connects the LAN to a widearea network and hence to the outside world.Let us look at the traffic from the LAN through the router. Packets arrive with a mean arrivalrate of 5 per second. The average packet length is 144 bytes, and it is assumed that packet length is exponentially distributed. Line speed from the router to the wide-area network is9600 bps. The following questions are asked:(a) What is the utilization of the link of the router?(b) What is the mean residence time in the router?(c) How many packets are in the router, including those waiting for transmission and the one currently being transmitted (if any), on the average?Solution:(a) Mean arrival rate(throughput): X=5 packets/secAverage service time: S=(144bytes/packet)*(8bits/byte)/9600bps=0.12sec/packetUtilization(time the router is busy): U=X*S=(5 packets/sec)*(0.12 sec/packet)=0.6(b) The mean residence time is T=S/(1-U)=(0.12 sec/packet)/(1-0.6)=0.3 sec/packet(c) Number of packets in the router is En=U/(1-U)=1.5 packets2. Consider the arrival traffic characterized by a token bucket with parameters (average rate) = 1 Mbps, M (maximum output rate) = 2 Mbps, and C (token capacity) = 200Kb. What is the minimum rate r that needs to be allocated by a router in order to guarantee a delay no larger than 50ms?Solution:We build the equation according to the rule: the bits flowed in the router are equal to the bits flowed out the router. Let S be burst length, the maximum accumulative amount of arrival traffic to the router is C+S=MS.We get S=C/(M-). When the router deal the arrival traffic at the minimum rate r with a delay no larger than 50ms, let D=50ms and the equation is MS=r*(S+D). So,/()2*0/(21)0.41.60.5/ .52MSCMbpsKbpsMbMpsDs s3. Describe the border gateway protocol (BGP) and discuss how a packet would be transmitted among different autonomous system (AS).Solution:Border GatewayProtocol version 4 (BGP4)We just learned how ISPs use RIP and OSPF to determine optimal paths for sourcedestinationpairs that are internal to the same AS. Lets now examine how paths aredetermined for source-destination pairs that span multiple ASs. BGPprovides each AS a means to1. Obtain subnet reachability information from neighboring ASs.2. Propagate the reachability information to all routers internal to the AS.3. Determine “good” routes to subnets based on the reachability information andon AS policy.4. Suppose that frames are 1250 bytes long including 25 bytes of head. Also assume that ACK frame are 25 bytes long. Calculate the efficiency of stop-and-wait ARQ in a system that transmits at R=1Mbps and with reaction time of 1ms for channels with a bit error of 10-6, 10-5, 10-4.Solution:From the above figure and condition, we know the total time to transmit 1 frame is t0=2(tprop+tproc)+tf+ta.Here, 2(tprop+tproc) is the reaction time of 1ms, tf is the time to transmit the fames nf=1250 bytes with the head nh=25 bytes and ta is the time to transmit the ACK frame na=25 bytes. And the useful size of the frame is (nf-nh). Moreover, the probability of transmitting a frame without errors is (1-Pe). So the transmission efficiency is as follows.0()/(1)(1) (1)(2) ()(2/()(1250*8/=(1)0.*ef fh fhe e eprocprofafh fhe eprocprofaprocrpfafhereactionfaRntnPPPRRtttttntRnbysitbyPRsM (1)(25)/7(1)8 ee PeitsWhen Pe=10-6,=87.50%, when Pe=10-5,=87.50%, when Pe=10-4,=87.49%.5. Describe the TCP congestion control scheme. Derive the delay modeling for TCP traffic in fixed congestion window when WS/R 10 active less than .0004Q: how did we get value 0.0004?4.Caravan analogy Cars “propagate” at 100 km/hr Toll booth takes 12 sec to service a car (transmission time) carbit; caravan packet Q: How long until caravan is lined up before 2nd toll booth? Time to “push” entire caravan through toll booth onto highway = 12*10 = 120 sec Time for last car to propagate from 1st to 2nd toll both: 100km/(100km/hr)= 1 hr A: 62 minutes Cars now “propagate” at 1000 km/hr Toll booth now takes 1 min to service a car Q: Will cars arrive to 2nd booth before all cars serviced at 1st booth? Yes! After 7 min, 1st car at 2nd booth and 3 cars still at 1st booth. 1st bit of packet can arrive at 2nd router before packet is fully transmitted at 1st router! See Ethernet applet at AWL Web site5.6.Could the congestion problem be solved with a large buffer space?Too little memory:too much traffic will lead to buffer overflow and packet loss Too much memory:the queues and the delays can get so long that by the time the packets come out of the switch, most of them have already timed out and have been retransmitted by higher layers packets (or their retransmissions) have to be dropped after they have consumed precious network re-sources. Too much memory in the intermediate nodes is as harmful as too little memory.7.Could the congestion problem be solved with high-speed links? Introducing high-speed links without proper congestion control can lead to reduced performance the same speed(a) The time to transfer a particular file was five minutes.(b) The link between the first two nodes was replace by a fast 1 Mbits link, the transfer time increased to 20 Minutes! With the high-speed link, the arrival rate to the first router became much higher than the departure rate, leading to long queues, buffer overflows, and packet losses that caused the transfer time to increase. The protocols have to be designed specifically to ensure that this increasing range of link speeds does not degrade the performance.8.Could the congestion problem be solved with high-speed processors? Similar to that for links. Introduction of a high-speed processor in an existing network may increase the mismatch of speeds and the chances of congestion. Introducing high-speed links without proper schemes Congestion occurs even if all links and processors are of the same speed. An example of the balanced configuration - Assume all processors and links have a throughput capacity of 1 Gbits. A simultaneous transfer of data from nodes A and B to node C can lead to a total input rate of 2 Gbits per second at the router R while the output rate is only 1 Gbits per second, thereby, causing congestion.6. solutions? Congestion in networks is a dynamic problem. It cannot be solved with static solutions alone. We need protocol designs that protect networks in the event of congestion. The explosion of high-speed networks has led to more unbalanced networks that are causing congestion. In particular, packet loss due to buffer shortage is a symptom not a cause of congestion. Solution: proper protocols and mechanisms design, e.g. Admission Control, Scheduling, et. al7. “end to end” and “point to point”a) end to end communications:i. Data communications on a path between the source node and the destination node. The path possibly comprises multiple linksb) point to point communications: i. Data communications on a link connecting the adjacent nodes8. Performance Issue Stop-and-wait Without error t0 =2tprop+2tproc+tf+tack=2tprop+2tproc+nf/R+na/R Let Pf be the probability that a frame transmission has errors and needs to be retransmitted. tsw=t0/(1-Pf)Efficiency is: sw=(nf-no)/tsw)/R Where nf, nano are the total number of bits in the frame,number of bits in the ack, and number of bits in the overhead. R is the bandwidth.9. Constrains on Windows Size(a sliding window protocol)n bits sequence numberThe size of sending window: WsThe size of receiving window: WrWs+ Wr2n10. Binary exponential backoff When a collision occureach station waits 0 or 1 slots before try again In general, after i collision, a random number between 0 and 2i -1 is chosen. Maximum is 1023 After 16 collisions, the controller gives up and reports failure. 11. Three types of switching fabrics12. Input Port Queuingr Fabric slower than input ports combined -queueing may occur at input queues r Head-of-the-Line (HOL) blocking: queued datagram at front of queue prevents others in queue from moving forwardr queueing delay and loss due to input buffer overflow!16IP Fragmentation and Reassembly
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