Solution Manual For Computer Networks, 5th Edition

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COMPUTER NETWORKSFIFTH EDITIONPROBLEM SOLUTIONSANDREW S. TANENBAUMVrije UniversiteitAmsterdam, The NetherlandsandDAVID J. WETHERALLUniversity of WashingtonSeattle, WA

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PROBLEM SOLUTIONS1SOLUTIONS TO CHAPTER 1 PROBLEMS1.The dog can carry 21 gigabytes, or 168 gigabits.A speed of 18 km/hourequals 0.005 km/sec. The time to travel distancexkm isx/0. 005=200xsec,yielding a data rate of 168/200xGbps or 840/xMbps.Forx< 5. 6 km, thedog has a higher rate than the communication line.(i)If dog’s speed is doubled, maximum value ofxis also doubled.(ii)If tape capacity is doubled, value ofxis also doubled.(iii)If data rate of the transmission line is doubled, value ofxis halved.2.The LAN model can be grown incrementally. If the LAN is just a long cable.it cannot be brought down by a single failure (if the servers are replicated) It isprobably cheaper. It provides more computing power and better interactive in-terfaces.3.A transcontinental fiber link might have many gigabits/sec of bandwidth, butthe latency will also be high due to the speed of light propagation over thou-sands of kilometers. Similarly, a satellite link may run at megabits/sec but havea high latency to send a signal into orbit and back.In contrast, a 56-kbpsmodem calling a computer in the same building has low bandwidth and lowlatency. So do low-end local and personal area wireless techologies such asZigbee.4.A uniform delivery time is needed for voice as well as video, so the amount ofjitter in the network is important.This could be expressed as the standarddeviation of the delivery time. Having short delay but large variability is ac-tually worse than a somewhat longer delay and low variability. For financialtransaction traffic, reliability and security are very important.5.No. The speed of propagation is 200,000 km/sec or 200 meters/μsec. In 10μsec the signal travels 2 km. Thus, each switch adds the equivalent of 2 km ofextra cable. If the client and server are separated by 5000 km, traversing even50 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 airandvacuumis300,000km/sec,sothepropagationdelayaloneis160,000/300,000 sec or about 533 msec.7.There is obviously no single correct answer here, but the following pointsseem relevant. The present system has a great deal of inertia (checks and bal-ances) built into it.This inertia may serve to keep the legal, economic, andsocial systems from being turned upside down every time a different partycomes to power.Also, many people hold strong opinions on controversial

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2PROBLEM SOLUTIONS FOR CHAPTER 1social issues, without really knowing the facts of the matter. Allowing poorlyreasoned opinions be to written into law may be undesirable. The potential ef-fects of advertising campaigns by special interest groups of one kind or anoth-er also have to be considered. Another major issue is security. A lot of peoplemight 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 (threespeeds or no line), so the total number of topologies is 410=1, 048, 576. At100 ms each, it takes 104,857.6 sec, or slightly more than 29 hours to inspectthem all.9.Distinguishn+2 events. Events 1 throughnconsist of the corresponding hostsuccessfully attempting to use the channel, i.e., without a collision. The prob-ability of each of these events isp(1−p)n−1. Eventn+1 is an idle channel,with probability (1−p)n. Eventn+2 is a collision. Since thesen+2 eventsare exhaustive, their probabilities must sum to unity. The probability of a col-lision,whichisequaltothefractionofslotswasted,isthenjust1−np(1−p)n−1−(1−p)n.10.Among other reasons for using layered protocols, using them leads to breakingup the design problem into smaller, more manageable pieces, and layeringmeans that protocols can be changed without affecting higher or lower ones.One possible disadvantage is the performance of a layered system is likely tobeworsethantheperformanceofamonolithicsystem,althoughitisextremely difficult to implement and manage a monolithic system.11.In the OSI protocol model, physical communication between peers takes placeonly in the lowest layer, not in every layer.12.Message and byte streams are different.In a message stream, the networkkeeps track of message boundaries.In a byte stream, it does not.For ex-ample, suppose a process writes 1024 bytes to a connection and then a littlelater writes another 1024 bytes. The receiver then does a read for 2048 bytes.With a message stream, the receiver will get two messages, of 1024 byteseach.With a byte stream, the message boundaries do not count and the re-ceiver will get the full 2048 bytes as a single unit.The fact that there wereoriginally two distinct messages is lost.13.Negotiation has to do with getting both sides to agree on some parameters orvalues to be used during the communication. Maximum packet size is one ex-ample, but there are many others.14.The service shown is the service offered by layerkto layerk+1.Anotherservice that must be present is below layerk, namely, the service offered tolayerkby the underlying layerk−1.

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PROBLEM SOLUTIONS FOR CHAPTER 1315.The probability,Pk, of a frame requiring exactlyktransmissions is theprobability of the firstk−1 attempts failing,pk−1, times the probability of thek-th transmission succeeding, (1−p).The mean number of transmission isthen justinfinityk=1ΣkPk=infinityk=1Σk(1−p)pk−1=11−pOr, more directly, if the probability of a message getting through is 1−pthenthe expected number of transmissions per successful message is 1 / ( 1−p).16.Withnlayers andhbytes added per layer, the total number of header bytes permessage ishn, so the space wasted on headers ishn. The total message size isM+nh, so the fraction of bandwidth wasted on headers ishn/(M+hn). Thisestimate does not take into account fragmentation (one higher layer message issent as multiple lower layer messages) or aggregation (multiple higher layermessages 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 willlower the overhead.17.TCP is connection oriented, whereas UDP is a connectionless service. Alterna-tively, TCP provides a reliable service, whereas UDP provides an unreliableservice.18.Observe that many nodes are connected to three other nodes; the others areconnected 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, thetwo nodes in the upper-right corner can be disconnected from the rest by threebombs knocking out the three nodes to which they are connected. The systemcan withstand the loss of any two nodes.19.Doubling every 18 months means a factor of four gain in 3 years. In 9 years,the gain is then 43or 64, leading to 38.4 billion hosts. That sounds like a lot,but if every television, cellphone, camera, car, and appliance in the world isonline, maybe it is plausible.It would require the average person to havedozens of hosts by then given that the estimate is much greater than theexpected world population.20.If the network tends to lose packets, it is better to acknowledge each one sepa-rately, so the lost packets can be retransmitted. On the other hand, if the net-work is highly reliable, sending one acknowledgement at the end of the entiretransfer saves bandwidth in the normal case (but requires the entire file to beretransmitted if even a single packet is lost).

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4PROBLEM SOLUTIONS FOR CHAPTER 121.Having mobile phone operators know the location of users lets the operatorslearn much personal information about users, such as where they sleep, work,travel and shop. This information might be sold to others or stolen; it could letthe government monitor citizens. On the other hand, knowing the location ofthe user lets the operator send help to the right place in an emergency. It mightalso be used to deter fraud, since a person who claims to be you will usuallybe near your mobile phone.22.The speed of light in coax is about 200,000 km/sec, which is 200 meters/μsec.At 10 Mbps, it takes 0.1μsec to transmit a bit. Thus, the bit lasts 0.1μsec intime, during which it propagates 20 meters. Thus, a bit is 20 meters long here.23.The image is 1600×1200×3 bytes or 5,760,000 bytes. This is 46,080,000bits. At 56,000 bits/sec, it takes about 822.857 sec. At 1,000,000 bits/sec, ittakes 46.080 sec. At 10,000,000 bits/sec, it takes 4.608 sec. At 100,000,000bits/sec, it takes about 0.461 sec. At 1,000,000,000 bits/sec it takes about 46msec.24.Think about the hidden terminal problem. Imagine a wireless network of fivestations,AthroughE, such that each one is in range of only its immediateneighbors. ThenAcan talk toBat the same timeDis talking toE. Wirelessnetworks have potential parallelism, and in this way differ from Ethernet.25.One advantage is that if everyone uses the standard, everyone can talk toeveryone. Another advantage is that widespread use of any standard will giveit economies of scale, as with VLSI chips. A disadvantage is that the politicalcompromises necessary to achieve standardization frequently lead to poorstandards. Another disadvantage is that once a standard has been widely adop-ted, it is difficult to change,, even if new and better techniques or methods arediscovered. Also, by the time it has been accepted, it may be obsolete.26.There are many examples, of course. Some systems for which there is interna-tional standardization include DVD players and their discs, digital camerasand their storage cards, and automated teller machines and bank cards. Areaswhere such international standardization is lacking include broadcast televi-sion (NTSC in the U.S., PAL in parts of Europe, SECAM in other countries),lamps and lightbulbs (different voltages in different countries), electrical sock-ets and appliance plugs (every country does it differently), photocopiers andpaper (8.5 x 11 inches in the U.S., A4 everywhere else), nuts and bolts(English versus metric pitch), etc.27.This has no impact on the operations at layers k-1 or k+1.28.There is no impact at layer k-1, but operations in k+1 have to be reimple-mented.

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PROBLEM SOLUTIONS FOR CHAPTER 1529.One reason is request or response messages may get corrupted or lost duringtransmission, which will necessitate a retransmission delay. Another reason isthe processing unit in the client may get overloaded processing several re-quests at once. Yet another reason is that the request or response messagesmay be queued in the network along with other messages.30.Small-sized cells result in large header-to-payload overhead. Fixed-size cellsresult in wastage of unused bytes in the payload.

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6PROBLEM SOLUTIONS FOR CHAPTER 2SOLUTIONS TO CHAPTER 2 PROBLEMS1.an= −1πn,bn=0,c=1.2.A noiseless channel can carry an arbitrarily large amount of information, nomatter how often it is sampled.Just send a lot of data per sample.For the4-kHz channel, make 8000 samples/sec. If each sample is 16 bits, the channelcan send 128 kbps.If each sample is 1024 bits, the channel can send 8.2Mbps. The key word here is ‘‘noiseless.’’ With a normal 4 kHz channel, theShannon limit would not allow this.A signal-to-noise ratio of 30 dB meansS/N = 1000. Using Eq. (2-3) withB=4000 we get a maximum data rate ofabout 39.86 kbps.3.Using the Nyquist theorem, we can sample 12 million times/sec.Four-levelsignals provide 2 bits per sample, for a total data rate of 24 Mbps.4.A signal-to-noise ratio of 20 dB meansS/N=100.Since log2101 is about6.658, the Shannon limit is about 19.975 kbps. The Nyquist limit is 6 kbps fora binary signal (with 1 bit per symbol).The bottleneck is therefore theNyquist limit, giving a maximum channel capacity of 6 kbps.5.TosendaT1signalweneedHlog2(1+S/N)=1. 544×106withH=50, 000. This yieldsS/N=230.88−1, which is about 93 dB.6.Fiber has many advantages over copper. It can handle much higher bandwidththan copper. It is not affected by power surges, electromagnetic interference,power failures, or corrosive chemicals in the air. It does not leak light and isquite difficult to tap. Finally, it is thin and lightweight, resulting in much lowerinstallation costs. There are some downsides of using fiber over copper. It canbe damaged easily by being bent too much.And fiber interfaces cost morethan electrical interfaces.7.Use Eq. (2-4) to convert wavelengths of 1 micron plus/minus 0.05 microns tofrequency. We haveflow=3×108/ (1. 05×10−6)=3/1. 05×1014. Similarlyfhigh=3/0. 95×1014. Thus∆f=(3/0. 95−3/1. 05)×1014=3×1013. This isa bandwidth of 30,000 GHz.8.The data rate is 2560×1600×24×60 bps, which is 5898 Mbps. For simpli-city, let us assume 1 bps per Hz, so that we require roughly 6×109MHz ofbandwidth. From Eq. (2-4) we can convert wavelengths to bandwidth. Let thewavelengths range from 1.3 microns to 1. 3+λmicrons. Then with Eq. (2-4)we have: 3×108/1. 3×10−6−3×108/ [ (1. 3+λ)×10−6]=6×109. Solv-ing,λ=3. 3×10−5microns. The range of wavelengths used is very short,even with slightly different assumptions.

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PROBLEM SOLUTIONS FOR CHAPTER 279.The Nyquist theorem is a property of mathematics and has nothing to do withtechnology. It says that if you have a function whose Fourier spectrum doesnot contain any sines or cosines abovef, by sampling the function at a frequen-cy of 2fyou capture all the information there is. Thus, the Nyquist theorem istrue for all media.10.Start withλf=c. We know thatcis 3×108m/s.Forλ=1 cm, we get 30GHz. Forλ=5 m, we get 60 MHz. Thus, the band covered is 60 MHz to 30GHz.11.If the beam is off by 1 mm at the end, it misses the detector. This amounts to atriangle with base 100 m and height 0.001 m. The angle is one whose tangentis thus 0.00001. This angle is about 0.00057 degrees.12.With 66/6 or 11 satellites per necklace, every 90 minutes 11 satellites passoverhead. This means there is a transit every 491 seconds. Thus, there will bea handoff about every 8 minutes and 11 seconds.13.Transit time = 2×(Altitude/Speed of light). The speed of light in air or vac-uum is 300,000 km/sec. This evaluates to 239 msec for GEO, 120 msec forMEO, and 5 msec for LEO satellites.14.The call travels from the North Pole to the satellite directly overhead, and thentransits through four other satellites to reach the satellite directly above theSouth Pole. Down it goes down to earth to the South Pole. The total distancetraveled is 2×750+0. 5×circumferenceataltitude750 km. Circumfer-ence at altitude 750 km is 2×π×(6371+750)=44, 720 km. So, the totaldistancetraveledis23,860km.Timetotravelthisdistance=23860/300000=79. 5 msec. In addition, switching occurs at six satellites.So, the total switching time is 60usec. So, the total latency is about 79.56msec.15.In NRZ, the signal completes a cycle at most every 2 bits (alternating 1s and0s). So, the minimum bandwidth need to achieveBbits/sec data rate is B/2Hz. In MLT-3, the signal completes a cycle at most every 4 bits (a sequence of1s), thus requiring at leastB/4 Hz to achieve B bits/sec data rate. Finally, inManchester encoding, the signal completes a cycle in every bit, thus requiringat leastBHz to achieveBbits/sec data rate.16.Since 4B/5B encoding uses NRZI, there is a signal transition every time a 1 issent. Furthermore, the 4B/5B mapping (see Figure 2-21) ensures that a se-quence of consecutive 0s cannot be longer than 3. To see this, note that: i) nocodeword has more than 2 consecutive zeros; ii) the longest leading sequenceis 1 zero; and iii) the longest trailing sequence is 2 zeros. Combining ii and iiiallows 3 consecutive 0s across two symbols. In this worst case, the transmittedbits will have a sequence 10001, resulting in a signal transition in 4 bits.

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8PROBLEM SOLUTIONS FOR CHAPTER 217.The number of area codes was 8×2×10, which is 160. The number of pre-fixes was 8×8×10, or 640. Thus, the number of end offices was limited to102,400. This limit is not a problem.18.Each telephone makes 0.5 calls/hour at 6 minutes each.Thus, a telephoneoccupies a circuit for 3 minutes/hour. Twenty telephones can share a circuit,although having the load be close to 100% (ρ=1 in queuing terms) impliesvery long wait times.Since 10% of the calls are long distance, it takes 200telephones to occupy a long-distance circuit full time.The interoffice trunkhas 1,000,000/4000 = 250 circuits multiplexed onto it.With 200 telephonesper circuit, an end office can support 200×250=50, 000 telephones. Sup-porting such a large number of telephones may result in significantly long waittimes. For example, if 5,000 (10% of 50,000) users decide to make a long-dis-tance telephone call at the same time and each call lasts 3 minutes, the worst-case wait time will be 57 minutes. This will clearly result in unhappy custom-ers.19.The cross-section of each strand of a twisted pair isπ/4 square mm. A 10-kmlengthofthismaterial,withtwostrandsperpairhasavolumeof2π/4×10−2m3. This volume is about 15,708 cm3. With a specific gravity of9.0, each local loop has a mass of 141 kg.The phone company thus owns1. 4×109kg of copper. At $6 each, the copper is worth about 8.5 billion dol-lars.20.Like a single railroad track, it is half duplex. Oil can flow in either direction,but not both ways at once.A river is an example of a simplex connectionwhile a walkie-talkie is another example of a half-duplex connection.21.Traditionally, bits have been sent over the line without any error- correctingscheme in the physical layer. The presence of a CPU in each modem makes itpossible to include an error-correcting code in layer 1 to greatly reduce the ef-fective error rate seen by layer 2. The error handling by the modems can bedone totally transparently to layer 2.Many modems now have built-in errorcorrection.While this significantly reduces the effective error rate seen atlayer 2, errors at layer 2 are still possible. This can happen, for example, be-cause of loss of data as it is transferred from layer 1 to layer 2 due lack of buff-er space.22.There are four legal values per baud, so the bit rate is twice the baud rate. At1200 baud, the data rate is 2400 bps.23.Since there are 32 symbols, 5 bits can be encoded. At 1200 baud, this provides5×1200=6000 bps.24.Two, one for upstream and one for downstream. The modulation scheme itselfjust uses amplitude and phase. The frequency is not modulated.

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PROBLEM SOLUTIONS FOR CHAPTER 2925.There are 10 4000 Hz signals. We need nine guard bands to avoid any inter-ference. The minimum bandwidth required is 4000×10+400×9 = 43, 600Hz.26.A sampling time of 125μsec corresponds to 8000 samples per second.According to the Nyquist theorem, this is the sampling frequency needed tocapture all the information in a 4-kHz channel, such as a telephone channel.(Actually the nominal bandwidth is less, but the cutoff is not sharp.)27.With a modern T1 line, the end users get 8×24=192 of the 193 bits in aframe. (With older T1s only 7 of the 8 bits per channel could be used for data.)The overhead is therefore 1 / 193 = 0.5%. From Figure 2-41, percent overheadin OC-1 is (51.84−49.536)/51.84 = 3.63%. In OC-768, percent overhead is(39813.12−38043.648)/39813.12 = 4.44%.28.In both cases 8000 samples/sec are possible. With QPSK encoding, 2 bits aresent per sample. With modern T1, 8 bits are sent per period. (With older T1sonly 7 of the 8 bits could be used for data.) The respective data rates are 16kbps and 64 kbps.29.Ten frames.The probability of some random 10-bit pattern being 10 bits ofthe framing pattern (of 0101010101) on a digital channel) is 1/210or 1/1024.This is the shortest number of frames for which the probability is less than1/1000.30.A coder accepts an arbitrary analog signal and generates a digital signal fromit. A demodulator accepts a modulated sine wave only and generates a digitalsignal.31.A drift rate of 10−9means 1 second in 109seconds or 1 nsec per second. AtOC-1 speed, say, 50 Mbps, for simplicity, a bit lasts for 20 nsec. This means ittakes only 20 seconds for the clock to drift off by 1 bit, and less for fasterspeeds. Consequently, the clocks must be continuously synchronized to keepthem from getting too far apart. Certainly every 10 sec, preferably much moreoften.32.The lowest bandwidth link (1 Mbps) is the bottleneck. The transmission timeto send 1 GB at 1 Mbps is 1024×8 seconds. The one-way latency =4×(35800 / 300000)=480msec.Thusthetotaltimeis1. 2+1024×8+0. 48=8193. 68 seconds.33.The lowest bandwidth link (1 Mbps) is the bottleneck. With 64 KB packetsthere are 16 packets in a 1 GB file. This adds 16×32=512 bytes of headersto transmit. The transmission time is thus 1024. 5×8 seconds. The one-waylatency is the propagation delay of 4×(35800 / 300000)=480 msec plusthreeswitchingtimesof0.01msec.Thusthetotaltimeis1024. 5×8+0. 48+3×0. 00001=8196. 48003 seconds.

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10PROBLEM SOLUTIONS FOR CHAPTER 234.Of the 90 columns, 86 are available for user data in OC-1. Thus, the user ca-pacity is 86×9=774 bytes/frame. With 8 bits/byte, 8000 frames/sec, and 3OC-1carriersmultiplexedtogether,thetotalusercapacityis3×774×8×8000, or 148.608 Mbps. For an OC-3072 line:Gross data rate=51. 84×3072=159252. 48 Mbps.SPE data rate=50. 112×3072=153944. 064 Mbps.User data rate=49. 536×3072=152174. 592 Mbps.35.VT1.5 can accommodate 8000 frames/sec×3 columns×9 rows×8 bits =1.728 Mbps. It can be used to accommodate DS-1. VT2 can accommodate8000 frames/sec×4 columns×9 rows×8 bits = 2.304 Mbps. It can be used toaccommodateEuropeanCEPT-1service.VT6canaccommodate8000frames/sec×12 columns×9 rows×8 bits = 6.912 Mbps. It can be used toaccommodate DS-2 service.36.TheOC-12cframesare12×90=1080columnsof9rows.Ofthese,12×3=36 columns are taken up by line and section overhead. This leaves anSPE of 1044 columns. One SPE column is taken up by path overhead, leaving1043 columns for user data.Since each column holds 9 bytes of 8 bits, anOC-12c frame holds 75,096 user data bits.With 8000 frames/sec, the userdata rate is 600.768 Mbps.37.The three networks have the following properties:Star: best case = 2, average case = 2, worst case = 2.Ring: best case = 1, average case =n/4, worst case =n/2.Full interconnect: best case = 1, average case = 1, worst case = 1.38.With circuit switching, att=sthe circuit is set up, att=s+x/bthe last bit issent, att=s+x/b+kdthe message arrives. With packet switching, the lastbit is sent att=x/b.To get to the final destination, the last packet must beretransmittedk−1 times by intermediate routers, with each retransmissiontakingp/bsec, so the total delay isx/b+(k−1)p/b+kd. Packet switchingis faster ifs> (k−1)p/b.In addition to the faster transmission under theseconditions, packet switching is preferable when fault-tolerant transmission inthe presence of switch failures is desired.39.The total number of packets needed isx/p, so the total data + header traffic is(p+h)x/pbits.The source requires (p+h)x/pbsec to transmit these bits.The retransmissions of the last packet by the intermediate routers take up atotal of (k−1)(p+h)/bsec. Adding up the time for the source to send all thebits, plus the time for the routers to carry the last packet to the destination, thusclearing the pipeline, we get a total time of (p+h)x/pb+ (p+h)(k−1)/bsec. Minimizing this quantity with respect top, we findp=√hx/(k−1).

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PROBLEM SOLUTIONS FOR CHAPTER 21140.Each cell has six neighbors. If the central cell uses frequency groupA, its sixneighbors can useB,C,B,C,B, andC, respectively.In other words, onlythree unique cells are needed. Consequently, each cell can have 280 frequen-cies.41.First, initial deployment simply placed cells in regions where there was a highdensity of human or vehicle population. Once they were there, the operatorsoften did not want to go to the trouble of moving them. Second, antennas aretypically placed on tall buildings or mountains. Depending on the exact loca-tion of such a structure, the area covered by a cell may be irregular due toobstacles near the transmitter.Third, some communities or property ownersdo not allow building a tower at a location where the center of a cell falls. Insuch cases, directional antennas are placed at a location not at the cell center.In the case of regular shapes, there is typically a buffer two cells wide where afrequency assigned to a cell is not reused in order to provide good separationand low interference. When the shapes of cells are irregular, the number ofcells separating two cells that are using the same frequency is variable, de-pending on the width of the intermediate cells. This makes frequency assign-ment much more complicated.42.If we assume that each microcell is a circle 100 m in diameter, each cell has anarea of 2500π. If we take the area of San Francisco, 1. 2×108m2, and divideit by the area of 1 microcell, we get 15,279 microcells. Of course, it is impos-sible to tile the plane with circles (and San Francisco is decidedly three-dimen-sional), but with 20,000 microcells we could probably do the job.43.Frequencies cannot be reused in adjacent cells, so when a user moves from onecell to another, a new frequency must be allocated for the call. If a user movesinto a cell, all of whose frequencies are currently in use, the user’s call must beterminated.44.The result is obtained by negating each ofA,B, andCand then adding thethree chip sequences. Alternatively, the three can be added and then negated.The result is (+3 +1 +1−1−3−1−1 +1).45.When two elements match, their product is +1. When they do not match, theirproduct is−1.To make the sum 0, there must be as many matches as mis-matches. Thus, two chip sequences are orthogonal if exactly half of the cor-responding elements match and exactly half do not match.46.Just compute the four normalized inner products:(−1 +1−3 +1−1−3 +1 +1) • (−1−1−1 +1 +1−1 +1 +1)/8 = 1(−1 +1−3 +1−1−3 +1 +1) • (−1−1 +1−1 +1 +1 +1−1)/8 =−1(−1 +1−3 +1−1−3 +1 +1) • (−1 +1−1 +1 +1 +1−1−1)/8 = 0(−1 +1−3 +1−1−3 +1 +1) • (−1 +1−1−1−1−1 +1−1)/8 = 1

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12PROBLEM SOLUTIONS FOR CHAPTER 2The result is thatAandDsent 1 bits,Bsent a 0 bit, andCwas silent.47.Here are the chip sequences:(+1 +1 +1 +1 +1 +1 +1 +1)(+1−1 +1−1 +1−1 +1−1)(+1 +1−1−1 +1, +1−1−1)(+1−1−1 +1 +1−1−1 +1)48.Ignoring speech compression, a digital PCM telephone needs 64 kbps. If wedivide 10 Gbps by 64 kbps we get 156,250 houses per cable. Current systemshave hundreds of houses per cable.49.A 2-Mbps downstream bandwidth guarantee to each house implies at most 50houses per coaxial cable.Thus, the cable company will need to split up theexisting cable into 100 coaxial cables and connect each of them directly to afiber node.50.The upstream bandwidth is 37 MHz. Using QPSK with 2 bits/Hz, we get 74Mbps upstream.Using QAM-128, this is 259 Mbps.Downstream we have200 MHz. Using QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600Mbps.51.The downstream data rate of a cable user is the smaller of the downstreamcable bandwidth and the bandwidth of the communication medium betweenthe cable modem and the user’s PC. If the downstream cable channel works at27 Mbps, the downstream data rate of the cable user will be(a) 10 Mbps.(b) 27 Mbps.(c) 27 Mbps.This is assuming that the communication medium between cable modem andthe user’s PC is not shared with any other user.

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PROBLEM SOLUTIONS FOR CHAPTER 313SOLUTIONS TO CHAPTER 3 PROBLEMS1.Since each frame has a chance of 0.8 of getting through, the chance of thewhole message getting through is 0. 810, which is about 0.107. Call this valuep. The expected number of transmission attempts before successful receptionis thenE=1/p. Thus, it takes an average of 1/0.107, or about 9.3 transmis-sions.2.The solution is(a) 00000101 01000111 11100011 11100000 01111110(b) 01111110 01000111 11100011 11100000 11100000 11100000 01111110 01111110(c) 01111110 01000111 110100011 111000000 011111010 011111103.After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D.4.The maximum overhead occurs when the payload consists of only ESC andFLAG bytes. In this case, there will be 100% overhead.5.If you could always count on an endless stream of frames, one flag byte mightbe enough. But what if a frame ends (with a flag byte) and there are no newframes for 15 minutes? How will the receiver know that the next byte is ac-tually the start of a new frame and not just noise on the line? The protocol ismuch simpler with starting and ending flag bytes.6.The output is 011110111110011111010.7.If the propagation delay is very long, as in the case of a space probe on or nearMars or Venus, forward error correction is indicated. It is also appropriate in amilitary situation in which the receiver does not want to disclose its locationby transmitting. If the error rate is low enough that an error-correcting code isgood enough, it may also be simpler. Finally, real-time systems cannot toler-ate waiting for retransmissions.8.Making one change to any valid character cannot generate another valid char-acter due to the nature of parity bits. Making two changes to even bits or twochanges to odd bits will give another valid character, so the distance is 2.9.Parity bits are needed at positions 1, 2, 4, 8, and 16 for messages that do notextend beyond 31 bits (including the parity bits). Our message has 16 data bitsand5paritybits,foratotalof21bits.Writethismessageab1c101d0011001e10101, from position 1 through 21 whereathroughearethe even parity bits.ais the sum of all positions with a 20term except itself, orodd positions except 1. Thusais 1+1+1+0+1+0+1+1+1+1=0.bisthe sum of positions with a 21term except 2, or positions 3, 6, 7, 10, 11, 14,15, 18, and 19.bis 1+0+1+0+1+0+1+0+1=1.cis the sum of allother positions with a 22term, or positions 5 to 7, 12 to 15, 20, and 21.cis1+0+1+1+0+0+1+0+1=1.dis the sum of all other positions with a

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Solution Manual For Computer Networks, 5th Edition - Page 16 preview image

14PROBLEM SOLUTIONS FOR CHAPTER 323term, or positions 9 to 15.dis 0+0+1+1+0+0+1=1.eis the sum ofallotherpositionswitha24term,orpositions17to21.eis1+0+1+0+1=1. The bit pattern transmitted is 011110110011001110101.10.If we number the bits from left to right starting at bit 1, the received messagewas 111001001111. (Note that different bit orderings will give differentanswers.) The parity bits are at positions 1, 2, 4 and 8. Recomputing the paritysums, we find the syndrome to be 0010. (The 23term is on the left, the 20termis on the right.) This means that the bit in position 2, a parity bit, is incorrect.Flipping this bit, we get the 12-bit value transmitted (after Hamming de-coding) of 0xA4F. Removing the parity bits we get the original 8-bit datavalue of 0xAF.11.A single error will cause both the horizontal and vertical parity checks to bewrong. Two errors will also be easily detected. If they are in different rows,the row parity will catch them. If they are in the same row, the column paritywill catch them. Three errors will also be detected. If they are in the same rowor column, that row’s or column’s parity will catch them. If two errors are inthe same row, the column parity of at least one of them will catch the error. Iftwo errors are in the same column, the row parity of at least one of them willcatch the error. A 4-bit error in which the four error bits lie on the four cornersof a rectangle cannot be caught since there are two errors in each row and col-umn.12.From Eq. (3-1), we know that 10 check bits are needed for each block in caseof using Hamming code. Total bits transmitted per block is 1010 bits. In caseof error detection mechanism, one parity bit is transmitted per block. Supposeerror rate isxper bit. However, a block may encounter a bit error 1000xtimes.Every time an error is encountered, 1001 bits have to be retransmitted. So,total bits transmitted per block is 1001+1000x×1001 bits. For error detec-tion and retransmission to be better, 1001+1000x×1001 < 1010. So, theerror rate must be less than 9×10−6.13.A block consists ofnrows bykcolumns, ornkdifferent positions. The num-ber of 4-bit errors is the number of ways we can choose 4 of these positions,‘‘nk choose 4’’ ornk(nk−1)(nk−2)(nk−3) / 4! Undetected errors occurwhen the 4 errors are at the vertices of a rectangle. To compute how manyundetected errors there are we want to know the number of ways we canchoose two vertices that are neither in the same row nor column; the remainingtwo vertices of the rectangle are then fully determined. The first vertex can beany of thenkpositions. The second vertex can be at any position in a differentrow and column, or (n−1)(k−1) positions. There arenk(n−1)(k−1) / 2positions. The probability of an undetected 4-bit error is:12(n−1)(k−1)(nk−1)(nk−2)(nk−3)

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