Solution Manual for Computer Networks, 6th Edition

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COMPUTER NETWORKSSIXTH EDITIONPROBLEM SOLUTIONSANDREW S. TANENBAUMVrije UniversiteitAmsterdam, The NetherlandsNICK FEAMSTERUniversity of ChicagoChicago, IllinoisDAVID WETHERELLGoogle

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PROBLEM SOLUTIONS1SOLUTIONS TO CHAPTER 1 PROBLEMS1.Because the raven flies at an average speed of 40 km/h, it needs 160/40 = 4hours for each one-way trip.(i) The raven makes only one trip because 1.8 terabytes exactly fits on onescroll.1800GB4h×3600=18GB/s=1Gbps(ii) To communicate 3.6 TB of data, the raven has to fly back to pick up asecond scroll. This means that it needs to fly a total of 3×4=12 hours.3600GB12h×3600=112GB/s=23Gbps(iii) The receiving castle receives 1.8 terabytes of data every 8 hours.1800GB8h×3600=116GB/s=12Gbps2.Therearemultiplecorrectanswers.Asignificantdisadvantageistheincreased risk of invading people’s privacy. The increase in the number of net-worked devices means a larger attack surface for malicious parties trying toobtain personal information.If the information is not stolen, companies thatprocess and store data from IoT devices could sell it to third parties such asadvertising companies.3.Secondly, wireless networks allow people to move around, instead of tyingthem to a wall.Secondly, although wireless networks provide lower band-width than wired networks, their bandwidth has become large enough to sup-portapplicationsthatpeoplefindmeaningful.Examplesincludemediastreaming and video conferencing. Finally, installing wires in (old) buildingscan be expensive.4.An advantage for the company is that they do not have to pay the up-front costwhen buying expensive hardware. They lease machines from the data center,paying only for what they use.A disadvantage for the company is that theymay not know the underlaying infrastructure used by the data center, making itmore difficult to obtain high performance from their applications.The largeamount of resources available in data centers makes it is easier for the com-pany to scale with user demand, which is an advantage for both. A disadvan-tage for the users is that it becomes more difficult to track their own data, andwhat it is used for.

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2PROBLEM SOLUTIONS FOR CHAPTER 15.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). Itis probably cheaper. It provides more computing power and better interactiveinterfaces.6.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 technologies such asZigbee.7.No. The speed of propagation is 200,000 km/sec or 400 meters/μsec. In 20μsec, the signal travels 4 km. Thus, each switch adds the equivalent of 4 kmof extra cable.If the client and server are separated by 5000 km, traversingeven 50 switches adds only 200 km to the total path, which is only 4%. Thus,switching delay is not a major factor under these circumstances.8.The delay is 1% of the total time, which means100μs×n29, 700km300, 000km/s+100μs×n=0. 01, wherenis the number of satellites.29, 700km300, 000km/s+100μs×n=100×100μs×n29, 700km300, 000km/s=99×100μs×n29, 700km300, 000km/s×99×100μs=10=nThis means the signal must pass 10 satellites for the switching delay to be 1%of the total delay.9.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.10.Traveling at 2/3 the speed of light means 200,000 km/sec. The signal travelsfor 100 milliseconds, or 0.1 seconds.This means the signal traversed a dis-tance of 200, 000×0. 1=4000 km.

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PROBLEM SOLUTIONS FOR CHAPTER 1311.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 controversialsocial 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.12.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. At 50ms each, it takes 52,428.8 sec, or about 14.6 hours to inspect them all.13.The mean router-router path is twice the mean router-root path. Number thelevels of the tree with the root as 1 and the deepest level asn. The path fromthe root to levelnrequiresn<1 hops and 0.50 of the routers are at this level.The path from the root to leveln<1 has 0.25 of the routers and a length ofn<2 hops. Hence, the mean path length,l, is given byl=0. 5×(n<1)+0. 25×(n<2)+0. 125×(n<3)+. . .orl=infinityi=1Yn(0. 5)i<infinityi=1Yi(0. 5)iThis expression reduces tol=n<2.The mean router-router path is thus2n<4.14.Distinguishn+2 events. Events 1 throughnconsist of the corresponding hostsuccessfully attempting to use the channel, i.e., without a collision.Theprobability of each of these events isp(1<p)n<1. Eventn+1 is an idle chan-nel, with probability (1<p)n.Eventn+2 is a collision.Since thesen+2events are exhaustive, their probabilities must sum to unity. The probability ofa collision, which is equal to the fraction of slots wasted, is then just1<np(1<p)n<1<(1<p)n.15.Instead of trying to foresee bad things and avoid them from happening, suc-cessful networks are fault-tolerant. They allow bad things to happen but iso-late or hide them from the rest of the system. Examples include error correc-tion, error detection, and network routing.

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4PROBLEM SOLUTIONS FOR CHAPTER 116.Because the responsibility of networking is distributed over multiple layers,each layer only has partial knowledge of where the data needs to go. The linklayer only knows to which machine the data should be sent next. The networklayer knows which machine on the entire network is the correct destination.The transport layer knows to which process on the destination machine todeliver the data.17.Guarantee LayerBest effort delivery NetworkReliable Delivery TransportIn-order Delivery TransportByte-stream abstraction TransportPoint-to-point link abstraction Link18.FunctionInterfacesend bits over link(bits)Physical layersend bytes to process(dst, src, bytes)Transport layersend bytes over link(dst, src, bytes)Link layersend bytes to machine(dst, src, bytes)Network layer19.5×1500 = 7,500 bytes per 100 milliseconds. So, the rate is 75,000 bytes persecond.20.In the OSI protocol model, physical communication between peers takes placeonly in the lowest layer, not in every layer.21.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.22.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.

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PROBLEM SOLUTIONS FOR CHAPTER 1523.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<1.24.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 just'k=1YkPk='k=1Yk(1<p)pk<1=11<pOr, more directly, if the probability of a message getting through is 1<p, thenthe expected number of transmissions per successful message is 1 / ( 1<p).25.OSI model: (a) Data link layer. (b) Network layer.TCP/IP model: (a) Link layer. (b) Internet layer.26.Frames encapsulate packets. When a packet arrives at the data link layer, theentire thing—header, data, and all—is used as the data field of a frame. Theentire packet is put in an envelope (the frame), so to speak (assuming it fits).27.Each layer considers the data from the layer above it as its payload, and addstheir header and/or trailer on the outside of this payload. The resulting orderof headers and trailers is: [1][3][5][M][6][4][2].28.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.29.Many answers are possible. Most have to do with connecting a otherwise iso-lated devices to the Internet. (a) A wireless modem is the access point for aWiFi network and connects to the cable provider’s network. (b) A phone con-nected to 3G or WiMAX and functioning as the access point for a WiFi net-work. (c) A phone that is connected to a WiFi network and at the same timeacts as the master in a Bluetooth network. (d) A router at an IXP that connectsan ISP’s network to that from other ISPs. (e) A phone that is connected to twobase stations at the same time during a soft handover. This last answer doesnot connect another device to the Internet, but rather creates the illusion of acontinuous connection to the user.

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6PROBLEM SOLUTIONS FOR CHAPTER 130.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.31.Doubling every 18 months means a factor of four gain in 3 years. In 9 years,the gain is then 43or 64, leading to 64 billion hosts. That sounds like a lot, butif every television, cellphone, camera, car, and appliance in the world is online,maybe it is plausible. It would require the average person to have 10 hosts bythen given that the estimate is much greater than the expected world popula-tion. But if half the world is connected and they average 20 devices per per-son, it is at least plausible.32.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).33.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.34.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.35.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.36.The image is 3840×2160×3 bytes or 8,294,400 bytes. This is 66,355,200bits. At 56,000 bits/sec, it takes about 1184.91 sec. At 1,000,000 bits/sec, ittakes about 66.36 sec.At 10,000,000 bits/sec, it takes about 6.64 sec.At100,000,000 bits/sec, it takes about 0.663 sec.At 1,000,000,000 bits/sec ittakes about 64.5 msec.

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PROBLEM SOLUTIONS FOR CHAPTER 1737.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.38.One disadvantage is security. Every random delivery man who happens to bein the building can listen in on the network. Another disadvantage is reliabil-ity. Wireless networks make lots of errors. A third potential problem is bat-tery life, since most wireless devices tend to be mobile.39.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 widelyadopted, it is difficult to change,, even if new and better techniques or methodsare discovered. Also, by the time it has been accepted, it may be obsolete.40.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 televis-ion (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 (Eng-lish versus metric pitch), etc.41.Networks are used in a large number of different environments, and each envi-ronment imposes different requirements on its protocols. For example, chan-nels with high error rates may need to use error correction codes to achievereasonable effective throughput, public networks may required additionalsecurity not needed in home networks, and commercial networks may need tooffer Quality of Service guarantees to paying customers.42.This has no impact on the operations at layers k-1 or k+1.43.There is no impact at layer k-1, but operations in k+1 have to be reimple-mented.44.Navigating to a webpage will likely trigger multiple GET requests.Each ofthese requests asks for a certain resource, such as the HTML document, theCSS document, and external resources such as images. Doing these requestsseparately can improve performance by, for instance, obtaining and renderingthe page HTML while downloading an external image.

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8PROBLEM SOLUTIONS FOR CHAPTER 145.Below are ten possible answers, although many more exist.1.Buying consumer goods through an online store.2.Watching video-on-demand.3.Listening to music using a streaming service.4.Communicating with others through email or instant messaging.5.Storing documents and photos on cloud-based storage.6.Reading the news on the Web.7.Handing in homework on your school’s online learning system8.Doing your taxes on the government’s website.9.Transferring money via a banking application on your phone.10.Playing video games with friends.

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PROBLEM SOLUTIONS FOR CHAPTER 29SOLUTIONS TO CHAPTER 2 PROBLEMS1.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.2.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.3.Use Eq. (2-1) 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. Thus6f=(3/0. 95<3/1. 05)×1014=3×1013. This isa bandwidth of 30,000 GHz.4.The data rate is 3840×2160×24×60 bps, which is about 11,944 Mbps or11.944 Gbps.5.In the text it was stated that the bandwidths (i.e., the frequency ranges) of thethree bands were approximately equal. From the formula in the text we havef=c/h.For a range of wavelengthshtoh+xwe haveflower=c/(h+x)andfupper=c/h. For a fixed value ofx, the range will be larger ifhis smaller.This is why the range of wavelengths on the left is smaller even though therange of frequencies is approximately equal in all bands.6.This would be a major scientific breakthrough. Compared to electrons, whichare subject to electromagnetic forces, it is difficult to make photons interact.Assuming there are computers that work this way, the maximum data ratesobtained from information theory would not change. However, it is no longernecessary to translate between optical and electrical signals when using opticalfiber communication. This makes it possible to obtain much higher data ratesusing the same optical fibers.7.Start withhf=c. We know thatcis 3×108m/s. Forh=1 cm, we get 30GHz. Forh=1 m, we get 300 MHz. Thus, the band covered is 300 MHz to30 GHz.8.At 1 GHz, the waves are 30 cm long. If one wave travels 15 cm more than theother, they will arrive out of phase.The fact that the link is 100 km long isirrelevant.

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10PROBLEM SOLUTIONS FOR CHAPTER 29.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.10.an=<1/n,bn=0,c=1.11.Shannon’s theorem gives 5×log2(1+10000)566. 44 Gbps. Nyquist theo-rem gives 2×5log2(2)=10 Gbps. 10 Gbps is the lowest upper bound. TheNyquist theorem yields a lower upper-bound because the signal is binary. On achannel with a 40 dB signal-to-noise ratio, a higher data rate can be achievedby using more signal levels.12.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 the3-kHz channel, make 6000 samples/sec. If each sample is 16 bits, the channelcan send 96 kbps. If each sample is 1024 bits, the channel can send 8.2 Mbps.The key word here is ‘‘noiseless.’’ With a normal 3 kHz channel, the Shannonlimit would not allow this.A signal-to-noise ratio of 30 dB means S/N =1000. WithB=3000 we get a maximum data rate of about 29.895 kbps.13.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.14.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.15.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.16.For every four data bits, 5 bits are sent. The channel sends 80 million bits persecond, which can be done using a 40 MHz channel.17.Only the distance from the origin differs, thus Amplitude Shift Keying is used.18.Yes. QAM-16 uses 16 distinct symbols. Each of these symbols can be assign-ed to a bit sequence, in which case it can send 4 bits per symbol.Alterna-tively, some of the symbols can be used as control signals, leaving fewer sig-nals to send bit sequences.19.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, in

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PROBLEM SOLUTIONS FOR CHAPTER 211Manchester encoding, the signal completes a cycle in every bit, thus requiringat leastBHz to achieveBbits/sec data rate.20.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.21.There are four legal values per baud, so the bit rate is twice the baud rate. At1200 baud, the data rate is 2400 bps.22.Two, one for upstream and one for downstream. The modulation scheme itselfjust uses amplitude and phase. The frequency is not modulated.23.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.24.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).25.By definitionS•T>1mmi=1YSiTiIfTsends a 0 bit instead of 1 bit, its chip sequence is negated, with thei-thelement becoming<Ti. Thus,S•T>1mmi=1YSi(<Ti)=<1mmi=1YSiTi=026.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.27.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 = 1The result is thatAandDsent 1 bits,Bsent a 0 bit, andCwas silent.

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12PROBLEM SOLUTIONS FOR CHAPTER 228.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)29.A normalized inner product of 1/4 means that the inner product is 128/4=32.Each pair of chips is either equal or unequal. The number of equal pairs (E)minus the number of unequal pairs (U) must be equal to 32:E<U=32. Weknow thatE+U=128AU=128<E.Substituting 128<EforUgivesE<(128<E)=32AE=80. Probability ofE=80 equals£¤12880¥¦212850. 0013Probability of a normalized inner product of 14or higherequals128i=80Y£¤128i¥¦212850. 003The notation£¤xy¥¦indicates a binomial coefficient.30.Yes, they can. A network that includes local star-shaped topologies does notneed to be fully hierarchical. The backbone of the network can follow the de-sign of Baran’s distributed switching system.31.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.32.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% (l=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.

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PROBLEM SOLUTIONS FOR CHAPTER 21333.The cross-section of each strand of a twisted pair is//4 square mm. A 10-kmlengthofthismaterial,withtwostrandsperpairhasavolumeof2//4×104cm3.This volume for 15 million pairs is about 2. 356x1011cm3.With a specific gravity of 9.0, this many cubic centimeters of copper weighsabout 2. 12×1012grams.The phone company thus owns 2. 12×109kg ofcopper. At $6 each, the copper is worth about 12.7 billion dollars.34.Since there are 32 symbols, 5 bits can be encoded. At 4800 baud, this provides5×4800=24, 000 bps.35.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.36.There are 256 channels in all, minus 6 for POTS and 2 for control, leaving 248for data. If 3/4 of these are for downstream, that gives 186 channels for down-stream. ADSL modulation is at 4000 baud, so with QAM-64 (6 bits/baud) wehave 24,000 bps in each of the 186 channels.The total bandwidth is then4.464 Mbps downstream.37.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.)38.TosendaT1signalweneedHlog2(1+S/N)=1. 544×106withH=1, 000, 000. This yieldsS/N=21.544<1, which is about 2.82 dB.39.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.40.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 than1/1000.41.With a T1 line, the end users get 8×24=192 of the 193 bits in a frame. (Witholder T1s only 7 of the 8 bits per channel could be used for data.) The over-head is therefore 1 / 193 = 0.5%. From Figure 2-41, percent overhead in OC-1

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14PROBLEM SOLUTIONS FOR CHAPTER 2is (51.84<49.536)/51.84 = 3.63%. In OC-768, percent overhead is (39813.12<38043.648)/39813.12 = 4.44%.42.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.43.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.44.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.45.The OC-12c frames are 12×90=1080 columns of 9 rows.Of these,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.46.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.47.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 = 1.48.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 retransmission

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