Chapter 1 :answer of the problems

Problem 2

a) 电路交换网络更适合这种应用。



应用的持续时间很长，并且其对带宽的需求比较
平稳，突发很小，因此可以利用资源预留预先为
应用在传输路径的每一条链路上预留带宽，而不
会招致多大的浪费。
另外，由于应用持续时间较长，建立和拆除连接
的开销相对变小。
b) 在如题目所属的条件下，分组交换网络中
也没有必要进行拥塞控制。

在最坏情况下，即，所有应用同时传输数据时，
链路也有足够的带宽予以容纳，因此，不会产生
拥塞。
Chapter 1 : answer of the problems

Problem 3


a) We can n connections between
each of the four pairs of adjacent
switches. This gives a maximum
of 4n connections.
b) We can n connections passing
through the switch in the upperright-hand corner and another n
connections passing through the
switch in the lower-left-hand
corner, giving a total of 2n
connections.
Chapter 1 : answer of the problems

P63.4

Tollbooths are 100 km apart, and the cars propagate at
100km/hr. A tollbooth services a car at a rate of one car
every 12 seconds. There are ten cars.


a) It takes 120 seconds, or two minutes, for the first
tollbooth to service the 10 cars. Each of these cars have a
propagation delay of 60 minutes before arriving at the
second tollbooth. Thus, all the cars are lined up before the
second tollbooth after 62 minutes. The whole process
repeats itself for traveling between the second and third
tollbooths. Thus the total delay is 124 minutes.
b) Delay between tollbooths is 7*12 seconds plus 60
minutes, i.e., 61 minutes and 24 seconds. The total delay is
twice this amount, i.e., 122 minutes and 48 seconds.
Chapter 1 : answer of the problems

P63. 5

a) Packet Switched Virtual-Circuit Networks
( F  h) / R

The time to transmit one packet onto a link is .

The time to deliver the packet over Q links is.
t s  Q ( F  h) / R
Thus the total latency is .


b)Q(Packet
Switched Networks
F  2h) / R

c) Circuit-Switched Networks

Q ( F  h) / R
Because there is no store-and-forward
delays at the
t s  (h  F ) / R
links, the total delay is:
Chapter 1 : answer of the problems

P64: 7




Consider the first bit in a packet. Before this bit can be
transmitted, all of the bits in the packet must be generated.
This requires
48 * 8
sec=6msec.
64  103
The time required to transmit the packet is
48 * 8sec=
384 sec
1 10 6
Propagation delay = 2 msec.
The delay until decoding is
6msec +
+ 2msec = 8.384msec
384 sec
Chapter 1 : answer of the problems
P65: 14




bandwidth-delay product=R*tprop=40,000 bits
maximum bits=min(R*tprop, L)=40,000 bits
the bandwidth-delay product of a link is the
maximum number of bits that can be in the link



Bandwidth*Delay=number of bits in flight to fill link
1-bit width=propgation speed* ttransmits a bit
=10-6*2.5*108=250 m
s/R
Chapter 1 : answer of the problems

P66. 15



1-bit width=s/R
s/R=m, so R=s/m=2.5*108/107=25bps
P66. 16




bandwidth-delay product=
R*tprop=109*107/(2.5*108)=40,000,000 bits
maximum bits=min(R*tprop, L)=min(40,000,000,
400,000)=400,000 bits
the bandwidth-delay product of a link is the maximum
number of bits that can be in the link
1-bit width=propgation speed* ttransmits a bit
=2.5*108/109=0.25m