北交计算机网络原理第三章作业

Chapter 3注：括弧中标题号为第四版教材中对应的习题号1.2.(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 theacknowledgment that Host B sends to Host A, what will be the acknowledgment number?a) 20 bytesb) ack number = 903.(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 ofunacknowledged 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 tosend 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 sequencenumber 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 valueof 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 aTCP connection to Host B. In this same segment the acknowledgment number is necessarily42.a)F b)T c)F d) T e)F f)F g)F4.(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.false5.(P3)UDP and TCP use 1s complement for their checksums. Suppose you have the followingthree 8-bit bytes: 01101010, 01001111, 01110011. What is the 1s complement of the sum of 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?0110101001001111---------------------1011100101110011---------------------1001011001----------------------00101101为了检测错误，接收方将四个字相加（三个原始字和一个检测字）。

如果结果包含0，那么接收方就知道分组中存在错误。

所有的1bit 错误都将被检测出来，但是2个比特的错误有可能被忽略。

6.(P7)Draw the FSM for the receiver side of protocol rdt3.0.7.(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 toa protocol to that uses ACKs? Why? Now suppose the sender has a lot of data to send and theend-to-end connection experiences few losses. In this second case, would a NAK-only protocol be preferable to a protocol that uses ACKs? Why?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。

也就是说接收方接收到x-1 然后接收到x+1,只有当接收方接收到x+1 时才发现x 的丢失。

如果在传输x 和传输x+1 之间有很长时间的延时，那么在只有NAK 的协议中，x 的修复要花费很长的时间。

如果要发送大量的数据，那么在只有NAK 的协议中修复的速度将很快。

并且，如果错误很少，那么NAK 只是偶尔发送，并且从不发送ACK。

与只有ACK 的情况相比，只有NAK 的情况将明显减少反馈时间。

8.(P14)Consider the cross-country example shown in Figure 3.17. How big would the windowsize have to be for the channel utilization to be greater than 80 percent?U=n*L/R/(RTT+L/R)>=80%n>=30019.(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 fallsoutside of its current windowb. With GBN, it is possible for the sender to receive an ACK for a packet that falls outside ofits current window.c. The alternating-bit protocol is the same as the SR protocol with a sender and receiverwindow size of 1.d. The alternating-bit protocol is the same as the GBN protocol with a sender and receiverwindow size of 1.a)正确,假设发送方窗口大小为3,在t0 时刻发送分组1,2,3.在t1(t1>t0)时刻接收方确认1,2,3.在t2(t2>t1)时刻发送方计时器超时,重发1,2,3.在t3 时刻接收到重复的分组并重新确认1,2,3.在t4 时刻发送方接收到接收方在t1 时刻发送的ACK,并将其窗口前移到4,5,6.在t5 时刻发送方接收到接收方在t2 发送的ACK1,2,3.这些ACK 是在当前窗口之外的报文的ACK.b) Truec) Trued) True当窗口尺寸为1 时,SR,GBN,的比特交替协议在功能上相同.窗口尺寸1 排除了失序分组的可能性.在这种情况下,一个累积的ACK 就是一个普通的ACK.因为在窗口内它只能与一个分组有关.10.(P23)Consider transferring an enormous file of L bytes from Host A to Host B. Assume anMSS 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 totalof 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.a) 因为TCP序号范围有4bytes，所以L最大为2^32 bytesb) 传输速度为155 Mbps，每段加66 bytes大小的头，首先计算一共分多少段：2^32 bytes / 1460 bytes = 2941758 段。