第一章
1.(Q1) What is the difference between a host and an end system? List the types ofend
systems. Is a Web rver an end system?C0000218
Answer: There is no difference. Throughout this text, the words “host” and “end system” are ud interchangeably. End systems include PCs, workstations, Web rvers, mail rvers, Internet-connected PDAs, WebTVs, etc.
2.(Q2) The word protocol is often ud to describe diplomatic relations. Give an example of a
diplomatic protocol.
Answer: Suppo Alice, an ambassador of country A wants to invite Bob, an ambassador of country B, over for dinner. Alice doesn’t simply just call Bob on the phone and say, come to our dinner table now”. Instead, she calls Bob and sugges ts a date and time. Bob may respond by saying he’s not available that particular date, but he is available another date. Alice and Bob continue to nd “messages” back and forth until they agree on a date and time. Bob then shows up at the embassy on the agreed date, hopefully not more than 15 minutes before or after the agreed time. Diplomatic protocols also allow for
either Alice or Bob to politely cancel the engagement if they have reasonable excus.
3.(Q3) What is a client program? What is a rver program? Does a rver programrequest
and receive rvices from a client program?
Answer: A networking program usually has two programs, each running on a different host, communicating with each other. The program that initiates the communication is the client.
Typically, the client program requests and receives rvices from the rver program.
4.(Q4) List six access technologies. Classify each one as residential access, company access, or
mobile access.
Answer:1. Dial-up modem over telephone line: residential; 2. DSL over telephone line: residential or small office; 3. Cable to HFC: residential; 4. 100 Mbps switched Etherent: company; 5. Wireless LAN: mobile; 6. Cellular mobile access (for example, 3G/4G): mobile
5.(Q5) List the available residential access technologies in your city. For each type of access,
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provide the advertid downstream rate, upstream rate, and monthly price.
Answer: Current possibilities include: dial-up (up to 56kbps); DSL (up to 1 Mbps upstream, up to 8 Mbps downstream); cable modem (up to 30Mbps downstream, 2 Mbps upstream.
6.(Q7) What are some of the physical media that Ethernet can run over?
Answer: Ethernet most commonly runs over twisted-pair copper wire and “thin” coaxial cable.It also can run over fibers optic links and thick coaxial cable.
7.(Q8)Dial-up modems, HFC, and DSL are all ud for residential access. For each of the
access technologies, provide a range of transmission rates and comment on whether the transmission rate is shared or dedicated.
Answer:Dial up modems: up to 56 Kbps, bandwidth is dedicated; ISDN: up to 128 kbps, bandwidth is dedicated; ADSL: downstream channel is .5-8 Mbps, upstream channel is up to
1 Mbps, bandwidth is dedicated; HFC, downstream channel is 10-30 Mbps and upstream
channel is usually less than a few Mbps, bandwidth is shared.
8.(Q13)Why is it said that packet switching employs statistical multiplexing? Contrast
statistical multiplexing with the multiplexing that takes place in TDM.
Answer:In a packet switched network, the packets from different sources flowing on a link do not follow any fixed, pre-defined pattern. In TDM circuit switching, each host gets the same slot in a revolving TDM frame.
9.(Q14) Suppo urs share a 2Mbps link. Also suppo each ur requires 1Mbps when
transmitting, but each ur transmits only 20 percent of the time. (See the discussion of statistical multiplexing in Section 1.3.)
a.When circuit switching is ud, how many urs can be supported?
b.For the remainder of this problem, suppo packet switching is ud. Why will there be
esntially no queuing delay before the link if two or fewer urs transmit at the same time? Why will there be a queuing delay if three urs transmit at the same time?
c.Find the probability that a given ur is transmitting.
d.Suppo now there are three urs. Find the probability that at any given time, all
three urs are transmitting simultaneously. Find the fraction of time during which the queue grows.
Answer:
a. 2 urs can be supported becau each ur requires half of the link bandwidth.
b.Since each ur requires 1Mbps when transmitting, if two or fewer urs transmit
simultaneously, a maximum of 2Mbps will be required. Since the available bandwidth of the shared link is 2Mbps, there will be no queuing delay before the link. Whereas, if three urs transmit simultaneously, the bandwidth required will be 3Mbps which is more than the available bandwidth of the shared link. In this ca, there will be queuing delay before the link.
c.Probability that a given ur is transmitting = 0.2
d.Probability that all three urs are transmitting simultaneously=33p3(1−p)0=
0.23=0.008. Since the queue grows when all the urs are transmitting, the fraction of
time during which the queue grows (which is equal to the probability that all three urs are transmitting simultaneously) is 0.008.
10.(Q16)Consider nding a packet from a source host to a destination host over a fixed route.
List the delay components in the end-to-end delay. Which of the delays are constant and which are variable?
Answer:The delay components are processing delays, transmission delays, propagation delays, and queuing delays. All of the delays are fixed, except for the queuing delays, which are variable.
11.(Q19) Suppo Host A wants to nd a large file to Host B. The path from Host A to Host B
has three links, of rates R1 = 250 kbps, R2 = 500 kbps, and R3 = 1 Mbps.
a.Assuming no other traffic in the network, what is the throughput for the file transfer.
b.Suppo the file is 2 million bytes. Roughly, how long will it take to transfer the file to
Host B?
c.Repeat (a) and (b), but now with R2 reduced to 200 kbps.
Answer:
a.250 kbps
b.64 conds
c.200 kbps; 80 conds
12.(P2)Consider the circuit-switched network in Figure 1.8. Recall that there are n circuits on
车的组词each link.
a.What is the maximum number of simultaneous connections that can be in progress at
any one time in this network?
b.Suppo that all connections are between the switch in the upper-left-hand corner
and the switch in the lower-right-hand corner. What is the maximum number of
simultaneous connections that can be in progress?
Answer:
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 upper-right-hand corner
andanother n connections passing through the switch in the lower-left-hand corner, giving a total of 2n connections.
13.(P4) Review the car-caravan analogy in Section 1.4. Assume a propagation speed of 50
km/hour.
a.Suppo the caravan travels 150 km, beginning in front of one tollbooth, passing
through a cond tollbooth, and finishing just before a third tollbooth. What is the
end-to-end delay?
b.Repeat (a), now assuming that there are five cars in the caravan instead of ten.
Answer: Tollbooths are 150 km apart, and the cars propagate at 50 km/hr, A tollbooth rvices a car at a rate of one car every 12 conds.
a.There are ten cars. It takes 120 conds, or two minutes, for the first tollbooth to rvice
the 10 cars. Each of the cars has a propagation delay of 180 minutes before arriving at the cond tollbooth. Thus, all the cars are lined up before the cond tollbooth after 182 minutes. The whole process repeats itlf for traveling between the cond and third tollbooths. Thus the total delay is 364 minutes.
b.Delay between tollbooths is 5*12 conds plus 180 minutes, i.e., 181minutes. The total
delay is twice this amount, i.e., 362 minutes.
14.(P5) This elementary problem begins to explore propagation delay and transmission delay,
two central concepts in data networking. Consider two hosts, A and B, connected by a single link of rate R bps. Suppo that the two hosts are parated by m meters, and suppo the propagation speed along the link is s meters/c. Host A is to nd a packet of size L bits to Host B.
a.Express the propagation delay, d prop , in terms of m and s.
b.Determine the transmission time of the packet, d trans, in terms of L and R.
c.Ignoring processing and queuing delays, obtain an expression for the end-to-end
delay.
d.Suppo Host A begins to transmit the packet at time t = 0. At time t = d trans, where is
the last bit of the packet?
e.Suppo d prop is greater than d trans. At time t = d trans, where is the first bit of the
packet?
f.Suppo d prop is less than d trans.At time t = d trans,where is the first bit of the packet?
晋江博物馆g.Suppo s = 2.5*108, L = 100bits, and R = 28kbps. Find the distance m so that d
prop equals d trans .
Answer:
a. d prop = m/s conds.
b. d trans = L/R conds.
c. d end-to-end = (m/s + L/R) conds.
d.The bit is just leaving Host A.
e.The first bit is in the link and has not reached Host B.
f.The first bit has reached Host B.
g.Want
m=L
R
S=
节约用水英语作文
100
28∗103
2.5∗108=893 km.
15.(P6) In this problem we consider nding real-time voice from Host A to Host B over a
packet-switched network (VoIP). Host A converts analog voice to a digital 64 kbps bit stream on the fly. Host A then groups the bits into 56-Byte packets. There is one link between Host A and B; its transmission rate is 500 kbps and its propagation delay is 2 mc.
As soon as Host A gathers a packet, it nds it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elaps from the time a bit is created (from the original analog signal at Host A) until the bit is decoded(as part of the analog signal at Host B)?
Answer: Consider the first bit in a packet. Before this bit can be transmitted, all of the bits in thepacket must be generated. This requires
56∗8
c=7 mc
64∗103
The time required to transmit the packet is
56∗8
c=896 μc
500∗103
Propagation delay = 2 mc.
The delay until decoding is
7mc +896μc + 2mc = 9.896mc
A similar analysis shows that all bits experience a delay of 9.896 mc.
16.(P9) Consider a packet of length L which begins at end system A, travels over one link to a
packet switch, and travels from the packet switch over a cond link to a destination end system. Let d i, s i, and R i denote the length, propagation speed, and the transmission rate of link i, for i= 1, 2. The packet switch delays each packet by d proc. Assuming no queuing delays, in terms of d i, s i, R i, (i= 1, 2), and L, what is the total end-to-end delay for the packet? Suppo now the packet Length is 1,000 bytes, the propagation speed on both links is 2.5 * 108m/s, the transmission rates of both links is 1 Mbps, the packet switch processing delay is 2 mc, the length of the first link is 6,000 km, and the length of the last link is 3,000 km. For the values, what is the end-to-end delay?
Answer: The first end system requires L/R1to transmit the packet onto the first link; the packet propagates over the first link in d1/s1; the packet switch adds a processing delay of
西兰花的吃法d proc; after receiving th
e entire packet, the packet switch requires L/R2to transmit the packet
onto the cond link; the packet propagates over the cond link in d2/s2. Adding the five delays gives
d end-end = L/R1 + L/R2 + d1/s1 + d2/s2 + d proc
ps网格快捷键
To answer the cond question, we simply plug the values into the equation to get 8 + 8 +
骷髅山24 + 12 + 2= 54 mc.
17.(P10) In the above problem, suppo R1 = R2 = R and d proc= 0. Further suppo the packet
switch does not store-and-forward packets but instead immediately transmits each bit it receivers before waiting for the packet to arrive. What is the end-to-end delay?
Answer: Becau bits are immediately transmitted, the packet switch does not introduce any delay;in particular, it does not introduce a transmission delay. Thus,
d end-end = L/R + d1/s1 + d2/s2
