计网-第三章作业

Chapter 3 注 括弧中标题号为第四版教材中对应的习题号注 括弧中标题号为第四版教材中对应的习题号 1 R14 Suppose Host A sends two TCP segments back to back to Host B over a TCP connection The first segment has sequence number 90 the second has sequence number 110 a How much data is in the first segment b Suppose that the first segment is lost but the second segment arrives at B In the acknowledgment that Host B sends to Host A what will be the acknowledgment number 答 a 90 109 20bytes b ack number 90 对第一个报文段确认 2 R15 True or false a The size of the TCP RcvWindow never changes throughout the duration of the connection b suppose Host A is sending Host B a large file over a TCP connection The number of unacknowledged bytes that A sends cannot exceed the size of the receive buffer c Host A is sending Host B a large file over a TCP connection Assume Host B has no data to send Host A Host B will not send acknowledgments to Host A because Host B cannot piggyback the acknowledgment on data d The TCP segment has a field in its header for RcvWindow e Suppose Host A is sending a large file to Host B over a TCP connection If the sequence number for a segment of this connection is m then the sequence number for the subsequent segment will necessarily be m 1 f Suppose that the last SampleRTT in a TCP connection is equal to 1 sec The current value of TimeoutInterval for the connection will necessarily be 1 sec g Suppose Host A sends one segment with sequence number 38 and 4 bytes of data over a TCP connection to Host B In this same segment the acknowledgment number is necessarily 42 答 a F b T c F 即使没有数据传送 也会进行单独确认 d T e F 按字节编号 不按报文段编号 f F g F B A 的确认号不一定为 38 4 42 3 R17 True or false Consider congestion control in TCP When the timer expires at the sender the threshold is set to one half of its previous value 答 F 应为当前拥塞窗口的一半 而不是阈值的一半 4 P3 UDP and TCP use 1s complement for their checksums Suppose you have the following three 8 bit bytes 01101010 01001111 01110011 What is the 1s complement of the sum of 2 these 8 bit byte Note that although UDP and TCP use 16 bit words in computing the checksum for this problem you are being asked to consider 8 bit sums Show all work Why is it that UDP takes complement of the sum that is why not just use the sum With the 1s complement scheme how does the receiver detect errors Is it possible that a 1 bit error will go undetected How about a 2 bit error 答 01101010 01001111 11000101 11000101 01110011 00010001 取反为 11101110 为了发现错误 接收端增加 4 个字组 3 个原始的 1 个取反后的 如果总数包含 0 即有错误 所有的一位错误会发现 但两位错误有可能不会被发现 5 P7 Draw the FSM for the receiver side of protocol rdt3 0 答 6 P13 Consider a reliable data transfer protocol that uses only negative acknowledgements Suppose the sender sends data only infrequently Would a NAK only protocol be preferable to a protocol to that uses ACKs Why Now suppose the sender has a lot of data to send and the end to end connection experiences few losses In this second case would a NAK only protocol be preferable to a protocol that uses ACKs Why 答 在仅使用 NAK 的协议中 只有当接收到分组 x 1 时才能检测到分组 x 的丢失 如果传 输 x 和传输 x 1 之间有很长的延时 那么在此协议中修复分组 x 需要很长的时间 如 果要发送大量的数据 在仅有 NAK 的协议中修复速度很快 如果错误很少 那么 NAK 只偶尔发送 ACK 则会明显减少反馈时间 7 P14 Consider the cross country example shown in Figure 3 17 How big would the window size have to be for the channel utilization to be greater than 80 percent 3 答 U n L R RTT L R 80 n 3001 8 P19 Answer true or false to the following questions and briefly justify your answer a With the SR protocol it is possible for the sender to receive an ACK for a packet that falls outside of its current window b With GBN it is possible for the sender to receive an ACK for a packet that falls outside of its current window c The alternating bit protocol is the same as the SR protocol with a sender and receiver window size of 1 d The alternating bit protocol is the same as the GBN protocol with a sender and receiver window size of 1 4 答 a T 在 t0时刻发送方窗口 3 发送包 1 2 3 在 t1时刻接收方接收 ACKs1 2 3 在 t2时 刻发送方延时并重新发送 1 2 3 在 t3时刻接收方接收包并重新发送确认 1 2 3 在 t4时刻发送方接收接收方在 t1时刻发送的 ACKs 并进入窗口 4 5 6 在 t5时刻发送方 接收接收方在 t2时刻发送的 ACKs1 2 3 这些 ACKs 在窗口之外 b T 见 a c T d T 在窗口 1 时 SR GBN the alternating bit protocol 在功能上是一样的 窗口 1 会 自动排除有可能无序的包 9 P23 Consider transferring an enormous file of L bytes from Host A to Host B Assume an MSS of 1 460 bytes a What is the maximum value of L such that TCP sequence numbers are not exhausted Recall that the TCP sequence number fields has 4 bytes b For the L you obtain in a find how long it takes to transmit the file Assume that a total of 66 bytes of transport network and data link header are added to each segment before the resulting packet is sent our over a 100 Mbps link Ignore flow control and congestion control so A can pump out the segments back to back and continuously 答 a TCP 序号范围为 4bytes LMAX 232bytes b 传输速度 155Mbps 每段加 66bytes 大小的头 共分段 232bytes 1460bytes 2941758 段 头大小和 2941758 66 194156028bytes 总共需 传输 194156028 232bytes 4489123324bytes 35912986592bits 的数据 用 10Mbps 的 速度传输则时间为 3591s 5 10 P34 Consider the following plot of TCP window size as a function of time Assuming TCP Reno is the protocol experiencing the behavior shown above answer the following questions In all cases you should provide a short discussion justifying your answer a Identify the intervals of time when TCP slow start is operating b Identify the intervals of time when TCP congestion avoidance is operating c After the 16th transmission round is segment loss detected by a triple duplicate ACK or by a timeout d After the 22nd transmission round is segment loss detected by a triple duplicate ACK or by a timeout e What is the initial value of Threshold at the first transmission round f What is the value of Threshold at the 18th transmission round g What is the value of Threshold at the 24th transmission round h During what transmission round is the 70th segment sent i Assuming a packet loss is detected after the 26th round by the receipt of a triple duplicate ACK what will be the values of the congestion window size and of Threshold 答 a 运行 TCP 慢启动的时间间隔是 1 6 和 23 26 b 运行 TCP 避免拥塞时的时间间隔是 1 6 和 17 22 c 在第 16 个传输周期后 通过 3 个冗余 ACK 能够检测到一个报文段丢失 如果有一 个超时 拥塞窗口尺寸将减小为 1 d 在第 22 个传输周期后 因为超时能够检测到一个报文段丢失 因此拥塞窗口的尺 寸被设置为 1 e Threshold 的初始值设置为 32 因为在这个窗口尺寸是慢启动停止 避免拥塞开始 f 当检测到报文段丢失时 threshold 被设置为拥塞窗口值的一半 当在第 16 个周期 检测到丢失时 拥塞窗口的大小是 42 因此在第 18 个传输周期时 threshold 值为 21 6 g 当检测到报文段丢失时 threshold 被设置为拥塞窗口值的一半 当在第 22 个周期 检测到丢失时 拥塞窗口的大小是 26 因此在第 24 个传输周期时 threshold 值为 13 h 在第 1 个传输周期内 报文段 1 被传送 在第 2 个传输周期发送报文段 2 3 在第 3 个传输周期发送报文段 4 7 在第 4 个传输周期发送 8 15 在第 5 个传输周期发 送 16 31 在第 6 个传输周期发送 32 63 在第 7 个传输周期发送 64 96 因此 报 文段 70 在第 7 个传送周期内发送 i 当丢失出现时拥塞窗口和 threshold 的值被设置为目前拥塞窗口长度 8 的一半 因此 新的拥塞窗口和 threshold 的值为 4 11 P38 Host A is sending an enormous file to Host B over a TCP c