计网第三章作业

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?
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、
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时刻发送得ACKs并进⼊窗⼝4,5,6;在t5时刻发送⽅接收接收⽅在t2时刻发送得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 在功能上就是⼀样得,窗⼝1会⾃动
排除有可能⽆序得包。

9.(P23)Consider transferring an enormous 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⼤⼩得头,共分段:232bytes/1460bytes=2941758段;
头⼤⼩与=2941758×66=194156028bytes;总共需传输194156028+232bytes=4489123324bytes=35912986592bits得数据;⽤10Mbps得速度传输则时间为3591s。

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慢启动得时间间隔就是[1,6]与[23,26];
b.运⾏TCP避免拥塞时得时间间隔就是[1,6]与[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,因此拥塞窗⼝得尺⼨被
设置为1。

e.Threshold得初始值设置为32,因为在这个窗⼝尺⼨就是慢启动停⽌,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得⼀半。

当在第16个周期检测
到丢失时,拥塞窗⼝得⼤⼩就是42,因此在第18个传输周期时threshold值为21。

g.当检测到报⽂段丢失时,threshold被设置为拥塞窗⼝值得⼀半。

当在第22个周期检测
到丢失时,拥塞窗⼝得⼤⼩就是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与threshold得值被设置为⽬前拥塞窗⼝长度8得⼀半。

因此
新得拥塞窗⼝与threshold得值为4、
11.(P38)Host A is sending an enormous Host B over a TCP connection、Over this connection
there is never any packet loss and the timers never expire、Denote the transmission rate of the link connecting Host A to the Internet by R bps、Suppose that the process in Host A is capable of sending data into its TCP socket at a rate S bps, where S = 10*R、Further suppose that the TCP receive buffer is large enough to hold the entire file, and the send buffer can hold only one percent of the file、What would prevent the process in Host A from continuously passing data to its TCP socket at rate S bps? TCP flow control? TCP congestion control? Or something else? Elaborate、
答:
在这个问题中,接收机没有溢出得危险,因为接收机得接收缓冲区可以承载整个⽂件。

同样得,因为在计时器计时完毕前,没有丢失与回复确认,TCP不会限制发送⽅。

但就是,在端系统A上得进程不会⼀直传输数据给套接字,因为发送缓冲区会很快被填满,⼀旦发送缓冲区满,那么进程就会以平均速率传输数据。