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Solution Manual For Computer Networks, 4th Edition

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    COMPUTER NETWORKS FOURTH EDITION PROBLEM SOLUTIONS ANDREW S. TANENBAUM Vrije Universiteit Amsterdam, The Netherlands

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    PROBLEM SOLUTIONS 1 SOLUTIONS TO CHAPTER 1 PROBLEMS 1. The dog can carry 21 gigabytes, or 168 gigabits. A speed of 18 km/hour equals 0.005 km/sec. The time to travel distance x km is x / 0.005 = 200 x sec, yielding a data rate of 168 / 200 x Gbps or 840 /x Mbps. For x < 5.6 km, the dog has a higher rate than the communication line. 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 is probably cheaper. It provides more computing power and better interactive interfaces. 3. A transcontinental fiber link might have many gigabits/sec of bandwidth, but the latency will also be high due to the speed of light propagation over thousands of kilometers. In contrast, a 56-kbps modem calling a computer in the same building has low bandwidth and low latency. 4. A uniform delivery time is needed for voice, so the amount of jitter in the net- work is important. This could be expressed as the standard deviation of the delivery time. Having short delay but large variability is actually worse than a somewhat longer delay and low variability. 5. No. The speed of propagation is 200,000 km/sec or 200 meters/ μ sec. In 10 μ sec 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. 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, and social systems from being turned upside down every time a different party comes to power. Also, many people hold strong opinions on controversial social issues, without really knowing the facts of the matter. Allowing poorly reasoned opinions be to written into law may be undesirable. The potential effects of advertising campaigns by special interest groups of one kind or another also have to be considered. Another major issue is security. A lot of people might worry about some 14-year kid hacking the system and falsifying the results.

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    2 PROBLEM SOLUTIONS FOR CHAPTER 1 8. Call the routers A , B , C , D , and E. There are ten potential lines: AB , AC , AD , AE , BC , BD , BE , CD , CE , and DE . 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. The mean router-router path is twice the mean router-root path. Number the levels of the tree with the root as 1 and the deepest level as n. The path from the root to level n requires n − 1 hops, and 0.50 of the routers are at this level. The path from the root to level n − 1 has 0.25 of the routers and a length of n − 2 hops. Hence, the mean path length, l , is given by l = 0.5 × ( n − 1) + 0.25 × ( n − 2) + 0.125 × ( n − 3) + . . . or l = i = 1 Σ ∞ n (0.5) i − i = 1 Σ ∞ i (0.5) i This expression reduces to l = n − 2 . The mean router-router path is thus 2 n − 4 . 10. Distinguish n + 2 events. Events 1 through n consist of the corresponding host successfully attempting to use the channel, i.e., without a collision. The probability of each of these events is p (1 − p ) n − 1 . Event n + 1 is an idle channel, with probability (1 − p ) n . Event n + 2 is a collision. Since these n + 2 events are exhaustive, their probabilities must sum to unity. The proba- bility of a collision, which is equal to the fraction of slots wasted, is then just 1 − np (1 − p ) n − 1 − (1 − p ) n . 11. Among other reasons for using layered protocols, using them leads to break- ing up the design problem into smaller, more manageable pieces, and layering means that protocols can be changed without affecting higher or lower ones, 12. No. In the ISO protocol model, physical communication takes place only in the lowest layer, not in every layer. 13. Connection-oriented communication has three phases. In the establishment phase a request is made to set up a connection. Only after this phase has been successfully completed can the data transfer phase be started and data trans- ported. Then comes the release phase. Connectionless communication does not have these phases. It just sends the data. 14. Message and byte streams are different. In a message stream, the network keeps track of message boundaries. In a byte stream, it does not. For exam- ple, suppose a process writes 1024 bytes to a connection and then a little later 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 bytes

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    PROBLEM SOLUTIONS FOR CHAPTER 1 3 each. With a byte stream, the message boundaries do not count and the receiver will get the full 2048 bytes as a single unit. The fact that there were originally two distinct messages is lost. 15. 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 example, but there are many others. 16. The service shown is the service offered by layer k to layer k + 1. Another service that must be present is below layer k , namely, the service offered to layer k by the underlying layer k − 1. 17. The probability, P k , of a frame requiring exactly k transmissions is the proba- bility of the first k − 1 attempts failing, p k − 1 , times the probability of the k -th transmission succeeding, (1 − p ) . The mean number of transmission is then just k = 1 Σ ∞ kP k = k = 1 Σ ∞ k (1 − p ) p k − 1 = 1 − p 1 33333 18. (a) Data link layer. (b) Network layer. 19. Frames encapsulate packets. When a packet arrives at the data link layer, the entire thing, header, data, and all, is used as the data field of a frame. The entire packet is put in an envelope (the frame), so to speak (assuming it fits). 20. With n layers and h bytes added per layer, the total number of header bytes per message is hn , so the space wasted on headers is hn . The total message size is M + nh , so the fraction of bandwidth wasted on headers is hn / ( M + hn ). 21. Both models are based on layered protocols. Both have a network, transport, and application layer. In both models, the transport service can provide a reliable end-to-end byte stream. On the other hand, they differ in several ways. The number of layers is different, the TCP/IP does not have session or presentation layers, OSI does not support internetworking, and OSI has both connection-oriented and connectionless service in the network layer. 22. TCP is connection oriented, whereas UDP is a connectionless service. 23. The two nodes in the upper-right corner can be disconnected from the rest by three bombs knocking out the three nodes to which they are connected. The system can withstand the loss of any two nodes. 24. 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 6.4 billion hosts. My intuition says that is much too conservative, since by then probably every television in the world and possibly billions of other appliances will be on home LANs connected to

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    4 PROBLEM SOLUTIONS FOR CHAPTER 1 the Internet. The average person in the developed world may have dozens of Internet hosts by then. 25. If the network tends to lose packets, it is better to acknowledge each one separately, so the lost packets can be retransmitted. On the other hand, if the network is highly reliable, sending one acknowledgement at the end of the entire transfer saves bandwidth in the normal case (but requires the entire file to be retransmitted if even a single packet is lost). 26. Small, fixed-length cells can be routed through switches quickly, and com- pletely in hardware. Small, fixed-size cells also make it easier to build hardware that handles many cells in parallel. Also, they do not block transmission lines for very long, making it easier to provide quality-of-service guarantees. 27. The speed of light in coax is about 200,000 km/sec, which is 200 meters/ μ sec. At 10 Mbps, it takes 0.1 μ sec to transmit a bit. Thus, the bit lasts 0.1 μ sec in time, during which it propagates 20 meters. Thus, a bit is 20 meters long here. 28. The image is 1024 × 768 × 3 bytes or 2,359,296 bytes. This is 18,874,368 bits. At 56,000 bits/sec, it takes about 337.042 sec. At 1,000,000 bits/sec, it takes about 18.874 sec. At 10,000,000 bits/sec, it takes about 1.887 sec. At 100,000,000 bits/sec, it takes about 0.189 sec. 29. Think about the hidden terminal problem. Imagine a wireless network of five stations, A through E , such that each one is in range of only its immediate neighbors. Then A can talk to B at the same time D is talking to E . Wireless networks have potential parallelism, and in this way differ from Ethernet. 30. One disadvantage is security. Every random delivery man who happens to be in 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. 31. One advantage is that if everyone uses the standard, everyone can talk to everyone. Another advantage is that widespread use of any standard will give it economies of scale, as with VLSI chips. A disadvantage is that the political compromises necessary to achieve standardization frequently lead to poor standards. Another disadvantage is that once a standard has been widely adopted, it is difficult to change,, even if new and better techniques or methods are discovered. Also, by the time it has been accepted, it may be obsolete. 32. There are many examples, of course. Some systems for which there is inter- national standardization include compact disc players and their discs, Walk- man tape players and audio cassettes, cameras and 35mm film, and automated

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    PROBLEM SOLUTIONS FOR CHAPTER 1 5 teller machines and bank cards. Areas where such international standardiza- tion is lacking include VCRs and videotapes (NTSC VHS in the U.S., PAL VHS in parts of Europe, SECAM VHS in other countries), portable tele- phones, lamps and lightbulbs (different voltages in different countries), electr- ical sockets and appliance plugs (every country does it differently), photo- copiers and paper (8.5 x 11 inches in the U.S., A4 everywhere else), nuts and bolts (English versus metric pitch), etc. SOLUTIONS TO CHAPTER 2 PROBLEMS 1. a n = π n − 1 333 , b n = 0, c = 1 . 2. A noiseless channel can carry an arbitrarily large amount of information, no matter how often it is sampled. Just send a lot of data per sample. For the 4 kHz channel, make 8000 samples/sec. If each sample is 16 bits, the channel can send 128 kbps. If each sample is 1024 bits, the channel can send 8.2 Mbps. The key word here is ‘‘noiseless.’’ With a normal 4 kHz channel, the Shannon limit would not allow this. 3. Using the Nyquist theorem, we can sample 12 million times/sec. Four-level signals provide 2 bits per sample, for a total data rate of 24 Mbps. 4. A signal-to-noise ratio of 20 dB means S/N = 100. Since log 2 101 is about 6.658, the Shannon limit is about 19.975 kbps. The Nyquist limit is 6 kbps. The bottleneck is therefore the Nyquist limit, giving a maximum channel capacity of 6 kbps. 5. To send a T1 signal we need H log 2 (1 + S /N ) = 1.544 × 10 6 with H = 50, 000 . This yields S /N = 2 30 − 1, which is about 93 dB. 6. A passive star has no electronics. The light from one fiber illuminates a number of others. An active repeater converts the optical signal to an electri- cal one for further processing. 7. Use ∆ f = c ∆λ / λ 2 with ∆λ = 10 − 7 meters and λ = 10 − 6 meters. This gives a bandwidth ( ∆ f ) of 30,000 GHz. 8. The data rate is 480 × 640 × 24 × 60 bps, which is 442 Mbps. For simplicity, let us assume 1 bps per Hz. From Eq. (2-3) we get ∆λ = λ 2 ∆ f /c . We have ∆ f = 4.42 × 10 8 , so ∆λ = 2.5 × 10 − 6 microns. The range of wavelengths used is very short. 9. The Nyquist theorem is a property of mathematics and has nothing to do with technology. It says that if you have a function whose Fourier spectrum does not contain any sines or cosines above f , then by sampling the function at a

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    6 PROBLEM SOLUTIONS FOR CHAPTER 2 frequency of 2 f you capture all the information there is. Thus, the Nyquist theorem is true for all media. 10. In the text it was stated that the bandwidths (i.e., the frequency ranges) of the three bands were approximately equal. From the formula ∆ f = c ∆λ / λ 2 , it is clear that to get a constant ∆ f , the higher the frequency, the larger ∆λ has to be. The x-axis in the figure is λ , so the higher the frequency, the more ∆λ you need. In fact, ∆λ is quadratic in λ . The fact that the bands are approxi- mately equal is an accidental property of the kind of silicon used. 11. Start with λ f = c . We know that c is 3 × 10 8 m/s. For λ = 1 cm, we get 30 GHz. For λ = 5 m, we get 60 MHz. Thus, the band covered is 60 MHz to 30 GHz. 12. At 1 GHz, the waves are 30 cm long. If one wave travels 15 cm more than the other, they will arrive out of phase. The fact that the link is 50 km long is irrelevant. 13. If the beam is off by 1 mm at the end, it misses the detector. This amounts to a triangle with base 100 m and height 0.001 m. The angle is one whose tangent is thus 0.00001. This angle is about 0.00057 degrees. 14. With 66/6 or 11 satellites per necklace, every 90 minutes 11 satellites pass overhead. This means there is a transit every 491 seconds. Thus, there will be a handoff about every 8 minutes and 11 seconds. 15. The satellite moves from being directly overhead toward the southern hor- izon, with a maximum excursion from the vertical of 2 φ . It takes 24 hours to go from directly overhead to maximum excursion and then back. 16. The number of area codes was 8 × 2 × 10, which is 160. The number of prefixes was 8 × 8 × 10, or 640. Thus, the number of end offices was limited to 102,400. This limit is not a problem. 17. With a 10-digit telephone number, there could be 10 10 numbers, although many of the area codes are illegal, such as 000. However, a much tighter limit is given by the number of end offices. There are 22,000 end offices, each with a maximum of 10,000 lines. This gives a maximum of 220 million telephones. There is simply no place to connect more of them. This could never be achieved in practice because some end offices are not full. An end office in a small town in Wyoming may not have 10,000 customers near it, so those lines are wasted. 18. Each telephone makes 0.5 calls/hour at 6 minutes each. Thus, a telephone occupies a circuit for 3 minutes/hour. Twenty telephones can share a circuit, although having the load be close to 100% ( ρ = 1 in queueing terms) implies very long wait times). Since 10% of the calls are long distance, it takes 200 telephones to occupy a long-distance circuit full time. The interoffice trunk

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    PROBLEM SOLUTIONS FOR CHAPTER 2 7 has 1,000,000/4000 = 250 circuits multiplexed onto it. With 200 telephones per circuit, an end office can support 200 × 250 = 50,000 telephones. 19. The cross-section of each strand of a twisted pair is π / 4 square mm. A 10-km length of this material, with two strands per pair has a volume of 2 π / 4 × 10 − 2 m 3 . This volume is about 15,708 cm 3 . With a specific gravity of 9.0, each local loop has a mass of 141 kg. The phone company thus owns 1.4 × 10 9 kg of copper. At 3 dollars each, the copper is worth about 4.2 bil- lion dollars. 20. Like a single railroad track, it is half duplex. Oil can flow in either direction, but not both ways at once. 21. Traditionally, bits have been sent over the line without any error correcting scheme in the physical layer. The presence of a CPU in each modem makes it possible to include an error correcting code in layer 1 to greatly reduce the effective error rate seen by layer 2. The error handling by the modems can be done totally transparently to layer 2. Many modems now have built in error correction. 22. There are four legal values per baud, so the bit rate is twice the baud rate. At 1200 baud, the data rate is 2400 bps. 23. The phase shift is always 0, but two amplitudes are used, so this is straight amplitude modulation. 24. If all the points are equidistant from the origin, they all have the same ampli- tude, so amplitude modulation is not being used. Frequency modulation is never used in constellation diagrams, so the encoding is pure phase shift key- ing. 25. Two, one for upstream and one for downstream. The modulation scheme itself just uses amplitude and phase. The frequency is not modulated. 26. There are 256 channels in all, minus 6 for POTS and 2 for control, leaving 248 for data. If 3/4 of these are for downstream, that gives 186 channels for downstream. ADSL modulation is at 4000 baud, so with QAM-64 (6 bits/baud) we have 24,000 bps in each of the 186 channels. The total bandwidth is then 4.464 Mbps downstream. 27. A 5-KB Web page has 40,000 bits. The download time over a 36 Mbps chan- nel is 1.1 msec. If the queueing delay is also 1.1 msec, the total time is 2.2 msec. Over ADSL there is no queueing delay, so the download time at 1 Mbps is 40 msec. At 56 kbps it is 714 msec. 28. There are ten 4000 Hz signals. We need nine guard bands to avoid any interference. The minimum bandwidth required is 4000 × 10 + 400 × 9 = 43,600 Hz.

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    8 PROBLEM SOLUTIONS FOR CHAPTER 2 29. A sampling time of 125 μ sec corresponds to 8000 samples per second. According to the Nyquist theorem, this is the sampling frequency needed to capture all the information in a 4 kHz channel, such as a telephone channel. (Actually the nominal bandwidth is somewhat less, but the cutoff is not sharp.) 30. The end users get 7 × 24 = 168 of the 193 bits in a frame. The overhead is therefore 25/193 = 13%. 31. In both cases 8000 samples/sec are possible. With dibit encoding, two bits are sent per sample. With T1, 7 bits are sent per period. The respective data rates are 16 kbps and 56 kbps. 32. Ten frames. The probability of some random pattern being 0101010101 (on a digital channel) is 1/1024. 33. A coder accepts an arbitrary analog signal and generates a digital signal from it. A demodulator accepts a modulated sine wave only and generates a digital signal. 34. (a) 64 kbps. (b) 32 kbps. (c) 8 kbps. 35. The signal must go from 0 to A in one quarter of a wave—that is, in a time T / 4 . In order to track the signal, 8 steps must fit into the quarter wave, or 32 samples per full wave. The time per sample is 1 /x so the full period must be long enough to contain 32 samples—that is, T > 32 /x or f max = x / 32 . 36. A drift rate of 10 − 9 means 1 second in 10 9 seconds or 1 nsec per second. At OC-1 speed, say, 50 Mbps, for simplicity, a bit lasts for 20 nsec. This means it takes only 20 seconds for the clock to drift off by one bit. Consequently, the clocks must be continuously synchronized to keep them from getting too far apart. Certainly every 10 sec, preferably much more often. 37. Of the 90 columns, 86 are available for user data in OC-1. Thus, the user capacity is 86 × 9 = 774 bytes/frame. With 8 bits/byte, 8000 frames/sec, and 3 OC-1 carriers multiplexed together, the total user capacity is 3 × 774 × 8 × 8000, or 148.608 Mbps. 38. 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 accommodate 8000 frames/sec × 4 columns × 9 rows × 8 bits = 2.304 Mbps. It can be used to accommodate European CEPT-1 service. VT6 can accommodate 8000 frames/sec × 12 columns × 9 rows × 8 bits = 6.912 Mbps. It can be used to accommodate DS-2 service. 39. Message switching sends data units that can be arbitrarily long. Packet switching has a maximum packet size. Any message longer than that is split up into multiple packets.

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    PROBLEM SOLUTIONS FOR CHAPTER 2 9 40. 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 an SPE of 1044 columns. One SPE column is taken up by path overhead, leaving 1043 columns for user data. Since each column holds 9 bytes of 8 bits, an OC-12c frame holds 75,096 user data bits. With 8000 frames/sec, the user data rate is 600.768 Mbps. 41. 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 42. With circuit switching, at t = s the circuit is set up; at t = s + x /b the last bit is sent; at t = s + x /b + kd the message arrives. With packet switching, the last bit is sent at t = x /b. To get to the final destination, the last packet must be retransmitted k − 1 times by intermediate routers, each retransmission tak- ing p /b sec, so the total delay is x /b + ( k − 1) p /b + kd. Packet switching is faster if s > ( k − 1) p /b. 43. The total number of packets needed is x /p , so the total data + header traffic is ( p + h ) x /p bits. The source requires ( p + h ) x /pb sec to transmit these bits. The retransmissions of the last packet by the intermediate routers take up a total of ( k − 1)( p + h ) /b sec. Adding up the time for the source to send all the bits, plus the time for the routers to carry the last packet to the destination, thus clearing the pipeline, we get a total time of ( p + h ) x /pb + ( p + h )( k − 1) /b sec. Minimizing this quantity with respect to p , we find p = √ 7 7777777 hx / ( k − 1) . 44. Each cell has six neighbors. If the central cell uses frequency group A , its six neighbors can use B , C , B , C , B , and C respectively. In other words, only 3 unique cells are needed. Consequently, each cell can have 280 frequencies. 45. First, initial deployment simply placed cells in regions where there was high density of human or vehicle population. Once they were there, the operator often did not want to go to the trouble of moving them. Second, antennas are typically placed on tall buildings or mountains. Depending on the exact loca- tion of such structures, the area covered by a cell may be irregular due to obs- tacles near the transmitter. Third, some communities or property owners do not allow building a tower at a location where the center of a cell falls. In such cases, directional antennas are placed at a location not at the cell center. 46. If we assume that each microcell is a circle 100 m in diameter, then each cell has an area of 2500 π . If we take the area of San Francisco, 1.2 × 10 8 m 2 and divide it by the area of 1 microcell, we get 15,279 microcells. Of course, it is impossible to tile the plane with circles (and San Francisco is decidedly three-dimensional), but with 20,000 microcells we could probably do the job.

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    10 PROBLEM SOLUTIONS FOR CHAPTER 2 47. Frequencies cannot be reused in adjacent cells, so when a user moves from one cell to another, a new frequency must be allocated for the call. If a user moves into a cell, all of whose frequencies are currently in use, the user’s call must be terminated. 48. It is not caused directly by the need for backward compatibility. The 30 kHz channel was indeed a requirement, but the designers of D-AMPS did not have to stuff three users into it. They could have put two users in each channel, increasing the payload before error correction from 260 × 50 = 13 kbps to 260 × 75 = 19.5 kbps. Thus, the quality loss was an intentional trade-off to put more users per cell and thus get away with bigger cells. 49. D-AMPS uses 832 channels (in each direction) with three users sharing a sin- gle channel. This allows D-AMPS to support up to 2496 users simultane- ously per cell. GSM uses 124 channels with eight users sharing a single channel. This allows GSM to support up to 992 users simultaneously. Both systems use about the same amount of spectrum (25 MHz in each direction). D-AMPS uses 30 KHz × 892 = 26.76 MHz. GSM uses 200 KHz × 124 = 24.80 MHz. The difference can be mainly attributed to the better speech quality provided by GSM (13 Kbps per user) over D-AMPS (8 Kbps per user). 50. The result is obtained by negating each of A , B , and C and then adding the three chip sequences. Alternatively the three can be added and then negated. The result is (+3 +1 +1 − 1 − 3 − 1 − 1 +1). 51. By definition S d T ≡ m 1 3 33 i = 1 Σ m S i T i If T sends a 0 bit instead of 1 bit, its chip sequence is negated, with the i -th element becoming − T i . Thus, S d T ≡ m 1 3 33 i = 1 Σ m S i ( − T i ) = − m 1 3 33 i = 1 Σ m S i T i = 0 52. When two elements match, their product is +1. When they do not match, their product is − 1. To make the sum 0, there must be as many matches as mismatches. Thus, two chip sequences are orthogonal if exactly half of the corresponding elements match and exactly half do not match. 53. Just compute the four normalized inner products: ( − 1 +1 − 3 +1 − 1 − 3 +1 +1) d ( − 1 − 1 − 1 +1 +1 − 1 +1 +1)/8 = 1 ( − 1 +1 − 3 +1 − 1 − 3 +1 +1) d ( − 1 − 1 +1 − 1 +1 +1 +1 − 1)/8 = − 1

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    PROBLEM SOLUTIONS FOR CHAPTER 2 11 ( − 1 +1 − 3 +1 − 1 − 3 +1 +1) d ( − 1 +1 − 1 +1 +1 +1 − 1 − 1)/8 = 0 ( − 1 +1 − 3 +1 − 1 − 3 +1 +1) d ( − 1 +1 − 1 − 1 − 1 − 1 +1 − 1)/8 = 1 The result is that A and D sent 1 bits, B sent a 0 bit, and C was silent. 54. Ignoring speech compression, a digital PCM telephone needs 64 kbps. If we divide 10 Gbps by 64 kbps we get 156,250 houses per cable. Current systems have hundreds of houses per cable. 55. It is both. Each of the 100 channels is assigned its own frequency band (FDM), and on each channel the two logical streams are intermixed by TDM. This example is the same as the AM radio example given in the text, but nei- ther is a fantastic example of TDM because the alternation is irregular. 56. A 2-Mbps downstream bandwidth guarantee to each house implies at most 50 houses per coaxial cable. Thus, the cable company will need to split up the existing cable into 100 coaxial cables and connect each of them directly to a fiber node. 57. The upstream bandwidth is 37 MHz. Using QPSK with 2 bits/Hz, we get 74 Mbps upstream. Downstream we have 200 MHz. Using QAM-64, this is 1200 Mbps. Using QAM-256, this is 1600 Mbps. 58. Even if the downstream channel works at 27 Mbps, the user interface is nearly always 10-Mbps Ethernet. There is no way to get bits to the computer any faster than 10-Mbps under these circumstances. If the connection between the PC and cable modem is fast Ethernet, then the full 27 Mbps may be available. Usually, cable operators specify 10 Mbps Ethernet because they do not want one user sucking up the entire bandwidth. SOLUTIONS TO CHAPTER 3 PROBLEMS 1. Since each frame has a chance of 0.8 of getting through, the chance of the whole message getting through is 0.8 10 , which is about 0.107. Call this value p . The expected number of transmissions for an entire message is then E = i = 1 Σ ∞ ip (1 − p ) i − 1 = p i = 1 Σ ∞ i (1 − p ) i − 1 To reduce this, use the well-known formula for the sum of an infinite geometric series, S = 1 = 1 Σ ∞ α i = 1 − α 1 33333 Differentiate both sides with respect to α to get S ′ = i = 1 Σ ∞ i α i − 1 = (1 − α ) 2 1 3 3333333

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    12 PROBLEM SOLUTIONS FOR CHAPTER 3 Now use α = 1 − p to get E = 1 /p . Thus, it takes an average of 1/0.107, or about 9.3 transmissions. 2. The solution is (a) 00000100 01000111 11100011 11100000 01111110 (b) 01111110 01000111 11100011 11100000 11100000 11100000 01111110 01111110 (c) 01111110 01000111 110100011 111000000 011111010 01111110 3. After stuffing, we get A B ESC ESC C ESC ESC ESC FLAG ESC FLAG D. 4. If you could always count on an endless stream of frames, one flag byte might be enough. But what if a frame ends (with a flag byte) and there are no new frames for 15 minutes. How will the receiver know that the next byte is actu- ally the start of a new frame and not just noise on the line? The protocol is much simpler with starting and ending flag bytes. 5. The output is 011110111110011111010. 6. It is possible. Suppose that the original text contains the bit sequence 01111110 as data. After bit stuffing, this sequence will be rendered as 011111010. If the second 0 is lost due to a transmission error, what is received is 01111110, which the receiver sees as end of frame. It then looks just before the end of the frame for the checksum and verifies it. If the check- sum is 16 bits, there is 1 chance in 2 16 that it will accidentally be correct, leading to an incorrect frame being accepted. The longer the checksum, the lower the probability of an error getting through undetected, but the probabil- ity is never zero. 7. If the propagation delay is very long, as in the case of a space probe on or near Mars or Venus, forward error correction is indicated. It is also appropri- ate, in a military situation in which the receiver does not want to disclose his location by transmitting. If the error rate is low enough that an error- correcting code is good enough, it may also be simpler. Finally, real-time systems cannot tolerate 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 two changes 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, so messages that do not extend beyond bit 31 (including the parity bits) fit. Thus, five parity bits are sufficient. The bit pattern transmitted is 011010110011001110101 10. The encoded value is 101001001111.

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    PROBLEM SOLUTIONS FOR CHAPTER 3 13 11. If we number the bits from left to right starting at bit 1, in this example, bit 2 (a parity bit) is incorrect. The 12-bit value transmitted (after Hamming encoding) was 0xA4F. The original 8-bit data value was 0xAF. 12. A single error will cause both the horizontal and vertical parity checks to be wrong. 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 parity will catch them. Three errors might slip by undetected, for example, if some bit is inverted along with its row and column parity bits. Even the corner bit will not catch this. 13. Describe an error pattern as a matrix of n rows by k columns. Each of the correct bits is a 0, and each of the incorrect bits is a 1. With four errors per block, each block will have exactly four 1s. How many such blocks are there? There are nk ways to choose where to put the first 1 bit, nk − 1 ways to choose the second, and so on, so the number of blocks is nk ( nk − 1)( nk − 2)( nk − 3). Undetected errors only occur when the four 1 bits are at the vertices of a rectangle. Using Cartesian coordinates, every 1 bit is at a coordinate ( x , y ), where 0 ≤ x < k and 0 ≤ y < n . Suppose that the bit closest to the origin (the lower-left vertex) is at ( p , q ). The number of legal rectangles is ( k − p − 1)( n − q − 1). Then the total number of rectangles can be found by summing this formula for all possible p and q . The probability of an undetected error is then the number of such rectangles divided by the number of ways to distribute the four bits: nk ( nk − 1)( nk − 2)( nk − 3) p = 0 Σ k − 2 q = 0 Σ n − 2 ( k − p − 1)( n − q − 1) 3 33333333333333333333333 14. The remainder is x 2 + x + 1 . 15. The frame is 10011101. The generator is 1001. The message after appending three zeros is 10011101000. The remainder on dividing 10011101000 by 1001 is 100. So, the actual bit string transmitted is 10011101100. The received bit stream with an error in the third bit from the left is 10111101100. Dividing this by 1001 produces a remainder 100, which is different from zero. Thus, the receiver detects the error and can ask for a retransmission. 16. The CRC is computed during transmission and appended to the output stream as soon as the last bit goes out onto the wire. If the CRC were in the header, it would be necessary to make a pass over the frame to compute the CRC before transmitting. This would require each byte to be handled twice—once for checksumming and once for transmitting. Using the trailer cuts the work in half.

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