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计算机网络6版教师专用复习题与问题解答
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《计算机网络:自顶向下方法》第六版的复习问题和问题解答文档是一个专门为教育工作者设计的教学辅助工具。这份文档包含了第五版Jim Kurose和Keith Ross所著的经典教材《计算机网络:自顶向下方法》的详细答案,旨在帮助教师在教学过程中为学生提供参考。它强调了这些解决方案仅供教师使用,禁止私自复制或分发给其他人员,包括其他教师,并且不得将内容公开发布到互联网上。 文档发布日期为2012年5月,反映了当时版本的解决方案。作者对多年来参与准备和提供反馈的学生、同事表示感谢,特别提到了Hong Gang Zhang、Rakesh Kumar、Prithula Dhungel和Vijay Annapureddy等人的贡献。同时,也感谢所有读者的建议和纠错,他们共同提升了文档的质量。 在第一章的复习问题中,"主机"和"端系统"这两个术语被互换使用,这表明在该书的讨论中,它们具有同等含义。端系统包括个人电脑(PC)、工作站、Web服务器以及邮件服务器等各类设备,这些是构成计算机网络的基本元素。书中采用自顶向下的方法来阐述网络体系结构,从全局视角出发,逐步分解各个层次的功能和协议,让读者更好地理解网络通信的原理。 通过查阅这份解决方案,教师可以确保学生正确理解和掌握课程中的关键概念,如网络层模型、数据传输、路由选择、网络安全等。对于那些希望深入学习或准备考试的学生来说,这份资料也是一个宝贵的补充资源。然而,为了尊重版权和维护教学秩序,仅限于授权人员获取并用于教学目的。
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b) In this example of traceroutes from one city in France and from another city in
Germany to the same host in United States, three links are in common including the
transatlantic link.
Traceroutes to two different cities in China from same host in United States
c) Five links are common in the two traceroutes. The two traceroutes diverge before
reaching China
Problem 20
Throughput = min{R
s
, R
c
, R/M}
Problem 21
If only use one path, the max throughput is given by:
}},,,min{,},,,,min{},,,,max{min{
21
22
2
2
1
11
2
1
1
M
N
MM
NN
RRRRRRRRR
.
If use all paths, the max throughput is given by
M
k
k
N
kk
RRR
1
21
},,,min{
.
Problem 22
Probability of successfully receiving a packet is: p
s
= (1-p)
N
.
The number of transmissions needed to be performed until the packet is successfully
received by the client is a geometric random variable with success probability p
s
. Thus,
the average number of transmissions needed is given by: 1/p
s
. Then, the average number
of re-transmissions needed is given by: 1/p
s
-1.
Problem 23
Let’s call the first packet A and call the second packet B.
a) If the bottleneck link is the first link, then packet B is queued at the first link waiting
for the transmission of packet A. So the packet inter-arrival time at the destination is
simply L/R
s
.
b) If the second link is the bottleneck link and both packets are sent back to back, it must
be true that the second packet arrives at the input queue of the second link before the
second link finishes the transmission of the first packet. That is,
L/R
s
+ L/R
s
+ d
prop
< L/R
s
+ d
prop
+ L/R
c
The left hand side of the above inequality represents the time needed by the second
packet to arrive at the input queue of the second link (the second link has not started
transmitting the second packet yet). The right hand side represents the time needed by
the first packet to finish its transmission onto the second link.
If we send the second packet T seconds later, we will ensure that there is no queuing
delay for the second packet at the second link if we have:
L/R
s
+ L/R
s
+ d
prop
+ T >= L/R
s
+ d
prop
+ L/R
c
Thus, the minimum value of T is L/R
c
L/R
s
.
Problem 24
40 terabytes = 40 * 10
12
* 8 bits. So, if using the dedicated link, it will take 40 * 10
12
* 8 /
(100 *10
6
) =3200000 seconds = 37 days. But with FedEx overnight delivery, you can
guarantee the data arrives in one day, and it should cost less than $100.
Problem 25
a) 160,000 bits
b) 160,000 bits
c) The bandwidth-delay product of a link is the maximum number of bits that can be in
the link.
d) the width of a bit = length of link / bandwidth-delay product, so 1 bit is 125 meters
long, which is longer than a football field
e) s/R
Problem 26
s/R=20000km, then R=s/20000km= 2.5*10
8
/(2*10
7
)= 12.5 bps
Problem 27
a) 80,000,000 bits
b) 800,000 bits, this is because that the maximum number of bits that will be in the link
at any given time = min(bandwidth delay product, packet size) = 800,000 bits.
c) .25 meters
Problem 28
a) t
trans
+ t
prop
= 400 msec + 80 msec = 480 msec.
b) 20 * (t
trans
+ 2 t
prop
) = 20*(20 msec + 80 msec) = 2 sec.
c) Breaking up a file takes longer to transmit because each data packet and its
corresponding acknowledgement packet add their own propagation delays.
Problem 29
Recall geostationary satellite is 36,000 kilometers away from earth surface.
a) 150 msec
b) 1,500,000 bits
c) 600,000,000 bits
Problem 30
Let’s suppose the passenger and his/her bags correspond to the data unit arriving to the
top of the protocol stack. When the passenger checks in, his/her bags are checked, and a
tag is attached to the bags and ticket. This is additional information added in the
Baggage layer if Figure 1.20 that allows the Baggage layer to implement the service or
separating the passengers and baggage on the sending side, and then reuniting them
(hopefully!) on the destination side. When a passenger then passes through security and
additional stamp is often added to his/her ticket, indicating that the passenger has passed
through a security check. This information is used to ensure (e.g., by later checks for the
security information) secure transfer of people.
Problem 31
a) Time to send message from source host to first packet switch =
sec4sec
102
108
6
6
With store-and-forward switching, the total time to move message from source host
to destination host =
sec123sec4 hops
b) Time to send 1
st
packet from source host to first packet switch = .
sec5sec
102
101
6
4
m
. Time at which 2
nd
packet is received at the first switch = time
at which 1
st
packet is received at the second switch =
sec10sec52 mm
c) Time at which 1
st
packet is received at the destination host =
sec153sec5 mhopsm
. After this, every 5msec one packet will be received; thus
time at which last (800
th
) packet is received =
sec01.4sec5*799sec15 mm
. It
can be seen that delay in using message segmentation is significantly less (almost
1/3
rd
).
d)
i. Without message segmentation, if bit errors are not tolerated, if there is a
single bit error, the whole message has to be retransmitted (rather than a single
packet).
ii. Without message segmentation, huge packets (containing HD videos, for
example) are sent into the network. Routers have to accommodate these huge
packets. Smaller packets have to queue behind enormous packets and suffer
unfair delays.
e)
i. Packets have to be put in sequence at the destination.
ii. Message segmentation results in many smaller packets. Since header size is
usually the same for all packets regardless of their size, with message
segmentation the total amount of header bytes is more.
Problem 32
Yes, the delays in the applet correspond to the delays in the Problem 31.The propagation
delays affect the overall end-to-end delays both for packet switching and message
switching equally.
Problem 33
There are F/S packets. Each packet is S=80 bits. Time at which the last packet is received
at the first router is
S
F
R
S
80
sec. At this time, the first F/S-2 packets are at the
destination, and the F/S-1 packet is at the second router. The last packet must then be
transmitted by the first router and the second router, with each transmission taking
R
S 80
sec. Thus delay in sending the whole file is
)2(
80
S
F
R
S
delay
To calculate the value of S which leads to the minimum delay,
FSdelay
dS
d
400
Problem 34
The circuit-switched telephone networks and the Internet are connected together at
"gateways". When a Skype user (connected to the Internet) calls an ordinary telephone, a
circuit is established between a gateway and the telephone user over the circuit switched
network. The skype user's voice is sent in packets over the Internet to the gateway. At the
gateway, the voice signal is reconstructed and then sent over the circuit. In the other
direction, the voice signal is sent over the circuit switched network to the gateway. The
gateway packetizes the voice signal and sends the voice packets to the Skype user.
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