COMPUTER NETWORKS - PROBLEM SOLUTIONS 1 SOLUTIONS TOCHAPTER 1 PROBLEMS

COMPUTER NETWORKS

FIFTH EDITION

PROBLEM SOLUTIONS

ANDREW S. TANENBAUM

and

DAVID J.WETHERALL

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PROBLEM SOLUTIONS1

SOLUTIONS TOCHAPTER 1 PROBLEMS

1.The dog can carry 21 gigabytes, or 168 gigabits. Aspeed of 18 km/hour equals 0.005 km/sec. The time to traveldistancexkm isx/0. 005=200xsec, yielding a data rate of 168/200xGbps or 840/xMbps. Forx<5.6 km, the dog has a higher rate than the communication line.(i) Ifdog'sspeed is doubled, maximum value ofxis also doubled.(ii) Iftape capacity is doubled, value ofxis also doubled.(iii) Ifdata rate of the transmission line is doubled, value ofxis halved.

2.The LAN model can be grown incrementally.Ifthe LAN is just a long cable.it cannot be brought down by a single failure (if the servers are replicated) It is probably cheaper.Itprovides more computing power and better interactive in- terfaces.

3.Atranscontinental Œber link might have manygigabits/sec of bandwidth, but the latencywill also be high due to the speed of light propagation overthou- sands of kilometers. Similarly,asatellite link may run at megabits/sec but have ahigh latencytosend a signal into orbit and back. In contrast, a 56-kbps modem calling a computer in the same building has lowbandwidth and low latency. Sodolow-end local and personal area wireless techologies such as Zigbee.

4.Auniform delivery time is needed for voice as well as video, so the amount of jitter in the network is important.This could be expressed as the standard deviation of the delivery time.Having short delay but large variability is ac- tually worse than a somewhat longer delay and lowvariability.For Œnancial transaction trafŒc, reliability and security are very important.

5.No. Thespeed of propagation is 200,000 km/sec or 200 meters/ msec. In10 msec the signal travels 2 km.Thus, each switch adds the equivalent of 2 km of extra cable. If the client and server are separated by 5000 km, traversing even 50 switches adds only 100 km to the total path, which is only 2%.Thus, switching delay is not a major factor under these circumstances.

6.The request has to go up and down, and the response has to go up and down.The total path length traversed is thus 160,000 km. The speed of light in air and vacuum is 300,000 km/sec, so the propagation delay alone is 160,000/300,000 sec or about 533 msec.

7.There is obviously no single correct answer here, but the following points seem relevant. Thepresent system has a great deal of inertia (checks and bal- ances) built into it. This inertia may servetokeep the legal, economic, and social systems from being turned upside down every time a different party comes to power.Also, manypeople hold strong opinions on controversial 2 / 4

2PROBLEM SOLUTIONS FOR CHAPTER 1

social issues, without really knowing the facts of the matter.Allowing poorly reasoned opinions be to written into lawmay be undesirable. The potential ef- fects of advertising campaigns by special interest groups of one kind or anoth- er also have tobeconsidered. Anothermajor issue is security.Alot of people might worry about some 14-year kid hacking the system and falsifying the re- sults.

8.Call the routersA,B,C,D,andE.There are ten potential lines:AB,AC,AD,

AE,BC,BD,BE,CD,CE,andDE.Each of these has four possibilities (three speeds or no line), so the total number of topologies is 4 10 =1, 048, 576.At 100 ms each, it takes 104,857.6 sec, or slightly more than 29 hours to inspect them all.

9.Distinguishn+2events. Events 1 throughnconsist of the corresponding host successfully attempting to use the channel, i.e., without a collision. The prob- ability of each of these events isp(1-p) n-1 .Eventn+1isanidle channel, with probability (1-p) n .Eventn+2isacollision. Sincethesen+2events are exhaustive,their probabilities must sum to unity.The probability of a col- lision, which is equal to the fraction of slots wasted, is then just 1-np(1-p) n-1 -(1-p) n .

10.Among other reasons for using layered protocols, using them leads to breaking up the design problem into smaller,more manageable pieces, and layering means that protocols can be changed without affecting higher or lower ones.One possible disadvantage is the performance of a layered system is likely to be worse than the performance of a monolithic system, although it is extremely difŒcult to implement and manage a monolithic system.

11.In the OSI protocol model, physical communication between peers takes place only in the lowest layer,not in every layer.

12.Message and byte streams are different. Inamessage stream, the network keeps track of message boundaries.In a byte stream, it does not.Forex- ample, suppose a process writes 1024 bytes to a connection and then a little later writes another 1024 bytes. The receiverthen does a read for 2048 bytes.With a message stream, the receiverwill get twomessages, of 1024 bytes each. With a byte stream, the message boundaries do not count and the re- ceiverwill get the full 2048 bytes as a single unit.The fact that there were originally twodistinct messages is lost.

13.Negotiation has to do with getting both sides to agree on some parameters or values to be used during the communication. Maximum packet size is one ex- ample, but there are manyothers.

14.The service shown is the service offered by layerkto layerk+1. Another service that must be present is belowlayerk,namely,the service offered to layerkby the underlying layerk-1. 3 / 4

PROBLEM SOLUTIONS FOR CHAPTER 13

15.The probability,P k,ofaframe requiring exactlyktransmissions is the probability of the Œrstk-1attempts failing,p k-1 ,times the probability of the k-th transmission succeeding, (1-p). Themean number of transmission is then just infinity k=1 SkP k= infinity k=1Sk(1-p)p k-1 = 1 1-p Or,more directly,ifthe probability of a message getting through is 1-pthen the expected number of transmissions per successful message is 1/(1-p).

16.Withnlayers andhbytes added per layer,the total number of header bytes per message ishn,sothe space wasted on headers ishn.The total message size is M+nh,sothe fraction of bandwidth wasted on headers ishn/(M+hn). This estimate does not takeinto account fragmentation (one higher layer message is sent as multiple lower layer messages) or aggregation (multiple higher layer messages are carried as one lower layer message) that may be present. If frag- mentation is used, it will raise the overhead. If aggregation is used, it will lower the overhead.

17.TCP is connection oriented, whereas UDP is a connectionless service. Alterna- tively,TCP provides a reliable service, whereas UDP provides an unreliable service.

18.Observethat manynodes are connected to three other nodes; the others are connected to more. Three bombs are needed to disconnect one of these nodes.By a quick check there does not appear to be a group of nodes that are con- nected to the rest of the network by fewer than three other nodes, so we con- clude that three bombs are needed to partition the network. For example, the twonodes in the upper-right corner can be disconnected from the rest by three bombs knocking out the three nodes to which theyare connected. The system can withstand the loss of anytwo nodes.

19.Doubling every 18 months means a factor of four gain in 3 years. In 9 years, the gain is then 4 3 or 64, leading to 38.4 billion hosts.That sounds likealot, butifevery television, cellphone, camera, car,and appliance in the world is online, maybe it is plausible. It would require the average person to have dozens of hosts by then giventhat the estimate is much greater than the expected world population.

20.If the network tends to lose packets, it is better to acknowledge each one sepa- rately,sothe lost packets can be retransmitted.On the other hand, if the net- work is highly reliable, sending one acknowledgement at the end of the entire transfer savesbandwidth in the normal case (but requires the entire Œle to be retransmitted if evenasingle packet is lost).

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