COMPUTER NETWORKS. FIFTH EDITION. PROBLEM SOLUTIONS. ANDREW S . TANENBAUM. Vrije Universiteit. Amsterdam, The Netherlands and. Computer Networks Book by Andrew S. Tanenbaum pdf download. solution manual computer networks 5th edition tanenbaum. table of contents. chapter. Computer Networks Tanenbaum 5th Edition Solution - [PDF] [EPUB] Computer Computer Networks 5th Edition Textbook Solutions - Chegg.

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acm/Computer Networks - A Tanenbaum - 5th solution manual computer networks 5th edition tanenbaum. table of contents. chapter. Computer networks / Andrew S. Tanenbaum, David J. Wetherall. -- 5th ed. p. cm. .. Now, in the fifth edition, networks are about content dis- . Solutions manual. Instructor Solutions Manual for Computer Networks, 5th Edition. Andrew S. Tanenbaum, Vrije University, Amsterdam, The Netherlands. David J. Wetherall.

The dog can carry 21 gigabytes, or gigabits. The LAN model can be grown incrementally. If the LAN is just a long cable. It provides more computing power and better interactive interfaces. In contrast, a kbps modem calling a computer in the same building has low bandwidth and low latency.

A uniform delivery time is needed for voice as well as video, so the amount of jitter in the network is important.

This could be expressed as the standard deviation of the delivery time. Having short delay but large variability is actually worse than a somewhat longer delay and low variability. For financial transaction traffic, reliability and security are very important. Thus, each switch adds the equivalent of 2 km of extra cable.

Thus, switching delay is not a major factor under these circumstances. The request has to go up and down, and the response has to go up and down.

The total path length traversed is thus , km. 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 balances 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.

Another major issue is security. A lot of people might worry about some year kid hacking the system and falsifying the results. At ms each, it takes , Events 1 through n consist of the corresponding host successfully attempting to use the channel, i. The probability of each of these events is p 1 p n 1. The probability of a collision, which is equal to the fraction of slots wasted, is then just 1 np 1 p n 1 1 p n.

Among other reasons for using layered protocols, using them leads to breaking up the design problem into smaller, more manageable pieces, and layering means that protocols can be changed without affecting higher or lower ones. One possible disadvantage is the performance of a layered system is likely to be worse than the performance of a monolithic system, although it is extremely difficult to implement and manage a monolithic system. In the ISO protocol model, physical communication takes place only in the lowest layer, not in every layer.

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 example, suppose a process writes bytes to a connection and then a little later writes another bytes. The receiver then does a read for bytes.

With a message stream, the receiver will get two messages, of bytes each. With a byte stream, the message boundaries do not count and the receiver will get the full bytes as a single unit. The fact that there were originally two distinct messages is lost. 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. Another service that must be present is below layer k, namely, the service offered to layer k by the underlying layer k 1. 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 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. 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 That sounds like a lot, but if every television, cellphone, camera, car, and appliance in the world is online, maybe it is plausible.

The average person may have dozens of hosts by then. 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. Having mobile phone operators know the location of users lets the operators learn 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 let the government monitor citizens. On the other hand, knowing the location of the user lets the operator send help to the right place in an emergency. It might also be used to deter fraud, since a person who claims to be you will usually be near your mobile phone.

Instructor Solutions Manual for Computer Networks, 5th Edition

At 10 Mbps, it takes 0. Thus, the bit lasts 0. Thus, a bit is 20 meters long here.

The image is bytes or 5,, bytes. This is 46,, bits. 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.

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.

There are many examples, of course. Some systems for which there is international standardization include compact disc players and their discs, digital cameras and their storage cards, and automated teller machines and bank cards.

One reason is request or response messages may get corrupted or lost during transmission. Another reason is the processing unit in the satellite may get overloaded processing several requests from different clients.

Small-sized cells result in large header-to-payload overhead. Fixed-size cells result in wastage of unused bytes in the payload. Just send a lot of data per sample. If each sample is 16 bits, the channel can send kbps. If each sample is bits, the channel can send 8.

The key word here is noiseless. With a normal 4 khz channel, the Shannon limit would not allow this. Four-level signals provide 2 bits per sample, for a total data rate of 24 Mbps. Since log is about 6. The Nyquist limit is 6 kbps. The bottleneck is therefore the Nyquist limit, giving a maximum channel capacity of 6 kbps.

Fiber has many advantages over copper.

It can handle much higher bandwidth than copper. It is not affected by power surges, electromagnetic interference, power failures, or corrosive chemicals in the air.

Instructor Solutions Manual for Computer Networks, 5th Edition

It does not leak light and is quite difficult to tap. Finally, it is thin and lightweight, resulting in much lower installation costs. There are some downsides of using fiber over copper. First, it can be damaged easily by being bent too much.

Second, optical communication is unidirectional, thus requiring either two fibers or two frequency bands on one fiber for two-way communication. Finally, fiber interfaces cost more than electrical interfaces. The data rate is bps, which is Mbps.

For simplicity, let us assume 1 bps per Hz. From Eq.

The range of wavelengths used is very short. 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, by sampling the function at a frequency of 2f you capture all the information there is. Thus, the Nyquist theorem is true for all media.

Thus, the band covered is 60 MHz to 30 GHz. This amounts to a triangle with base m and height m. The angle is one whose tangent is thus This angle is about degrees. This means there is a transit every seconds.

Thus, there will be a handoff about every 8 minutes and 11 seconds. The call travels from the North Pole to the satellite directly overhead, and then transits through four other satellites to reach the satellite directly above the South Pole. Down it goes down to earth to the South Pole. The total distance traveled is circumference at altitude km. So, the total distance traveled is 23, km. In addition, switching occurs at six satellites.

So, the total switching time is 60 usec. So, the total latency is about msec. In NRZ, the signal completes a cycle at most every 2 bits alternating 1s and 0s. Thus, in the worst case, the transmitted bits will have a sequence , resulting in a signal transition in 4 bits.

The number of area codes was , which is The number of prefixes was , or Thus, the number of end offices was limited to , This limit is not a problem. Each telephone makes 0. This will clearly result in unhappy customers. This volume is about 15, cm 3. With a specific gravity of 9. The phone company thus owns kg of copper. 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 connection while a walkie-talkie is another example of a half-duplex connection.

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. While this significantly reduces the effective error rate seen at layer 2, errors at layer 2 are still possible. This can happen, for example, because of loss of data as it is transferred from layer 1 to layer 2 due lack of buffer space. There are four legal values per baud, so the bit rate is twice the baud rate. At baud, the data rate is bps.

Since there are 32 symbols, 5 bits can be encoded. Two, one for upstream and one for downstream. The modulation scheme itself just uses amplitude and phase. The frequency is not modulated. There are Hz signals. We need nine guard bands to avoid any interference.

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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. With dibit encoding, 2 bits are sent per sample. With T1, 7 bits are sent per period. The respective data rates are 16 kbps and 56 kbps. Ten frames. A demodulator accepts a modulated sine wave only and generates a digital signal.

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 1 bit. Consequently, the clocks must be continuously synchronized to keep them from getting too far apart. Certainly every 10 sec, preferably much more often. The lowest bandwidth link 1 Mbps is the bottleneck. Again, the lowest-bandwidth link is the bottleneck. Of the 90 columns, 86 are available for user data in OC It can be used to accommodate DS It can be used to accommodate DS-2 service.

This leaves an SPE of columns. Since each column holds 9 bytes of 8 bits, an OCc frame holds 75, user data bits. In addition to the faster transmission under these conditions, packet switching is preferable when fault-tolerant transmission in the presence of switch failures is desired. Each cell has six neighbors. In other words, only three unique cells are needed. Consequently, each cell can have frequencies.

First, initial deployment simply placed cells in regions where there was a high density of human or vehicle population. Once they were there, the operators 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 location of such a structure, the area covered by a cell may be irregular due to obstacles 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. In the case of regular shapes, there is typically a buffer two cells wide where a frequency assigned to a cell is not reused in order to provide good separation and low interference.

When the shapes of cells are irregular, the number of cells separating two cells that are using the same frequency is variable, depending on the width of the intermediate cells. If we take the area of San Francisco, m 2, and divide it by the area of 1 microcell, we get 15, microcells. Of course, it is impossible to tile the plane with circles and San Francisco is decidedly three-dimensional , but with 20, microcells we could probably do the job.

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. 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. 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. Here are the chip sequences: , Ignoring speech compression, a digital PCM telephone needs 64 kbps.

If we divide 10 Gbps by 64 kbps we get , houses per cable. Current systems have hundreds of houses per cable. 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 coaxial cables and connect each of them directly to a fiber node. Downstream we have MHz. Using QAM, this is Mbps. The downstream data rate of a cable user is the smaller of the downstream cable bandwidth and the bandwidth of the communication medium between the cable modem and the user s PC.

If the downstream cable channel works at 27 Mbps, the downstream data rate of the cable user will be a 10 Mbps. This is assuming that the communication medium between cable modem and the user s PC is not shared with any other user. If you're interested in creating a cost-saving package for your students, contact your Pearson rep.

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You have successfully signed out and will be required to sign back in should you need to download more resources. Andrew S. Wetherall, University of Washington. If You're an Educator Download instructor resources Additional order info. Overview Order Downloadable Resources Overview.Each cell has six neighbors. An Introduction to Computer Networks, Release 1. This information might be sold to others or stolen; it could let the government monitor citizens. Besides the basic computer, some special devices and software may be required especially for computer graphics.

The worst case is where all stations want to send and s is the lowest-numbered station.

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