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得习题号

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 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 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 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得确认号不一定为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:应为当前拥塞窗口得一半,而不就是阈值得一半。

4.(P3)UDP and TCP use 1s plement for their checksums、Suppose you have the following

three 8-bit bytes: 01101010, 01001111, 、What is the 1s plement of the sum of these 8-bit byte? (Note that although UDP and TCP use 16-bit words in puting the checksum, for this problem you are being asked to consider 8-bit sums、) Show all work、、Why is it that UDP takes plement of the sum; that is, why not just use the sum? With the 1s plement 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。为了发现错误,接收端增加4个字组(3个原始得,1个取反后得),如果总数包含0,即有错误。所有得一位错误会发现,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得协议中,只有当接收到分组x+1时才能检测到分组x得丢失。如果传输x与传输x+1之间有很长得延时,那么在此协议中修复分组x需要很长得时间;如果要发送大量得数据,在仅有NAK得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?