bereave计网-第三章作业

2021年1月27日发(作者：alonzo)
Chapter 3

注：括弧中标题号为第四版教材中对应的习题号


1.

(R14)Suppose HostA sends two TCP segments back to back to Host B over a TCPconnection.
The first segment has sequence number 90; the second hassequence number 110.
a. How much data is in the first segment?
e
that
the
first
segment
is
lost
but
the
second
segment
arrives
at

the
acknowledgment that Host B sends to Host A, what will be theacknowledgment 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 durationof the connection.
b.
suppose
Host
A
is
sending
Host
B
a
large
file
over
a
TCP
connection.
Thenumber
of
unacknowledged bytes that A sends cannot exceed the size ofthe receive buffer.
c. HostA is sending Host B a large file over a TCP connection. Assume HostB has no data to
send
Host
A.
Host
B
will
not
send
acknowledgments
toHost
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.
Ifthe
sequence
number
for
a
segment
of
this
connection
is
m,
then
thesequence
number
for
the
subsequent
segment will necessarily be m + 1.
f. Suppose that the last
SampleRTT
in a TCP connection is equal to 1 current value of
TimeoutInterval
for the connection will necessarilybe>=1 sec.
e Host A sends one segment with sequence number 38 and 4 bytesof data over a TCP
connection to Host B. In this same segment theacknowledgment 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 atthe 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
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< br>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< br>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?


答：

U=n
×
L/R/(RTT+L/R)
≥
80%
n
≥
3001

8.

(P19)Answer true or false to the following questions and briefly justify youranswer:
a. With the SR protocol, it is possible for the sender to receive an ACK for apacket that falls
outside of its current window
b. With GBN, it is possible for the sender to receive an ACK for a packet thatfalls outside of
its current window.
c.
The
alternating-bit
protocol
is
the
same
as
the
SR
protocol
with
a
senderand
receiver
window size of 1.
d.
The
alternating-bit
protocol
is
the
same
as
the
GBN
protocol
with
a
senderand
receiver
window size of 1.