Computer network tanenbaum solution pdf

Shows some signs of wear, and may have some markings on the inside. This website uses cookies to improve your experience while you navigate through the website. Out of these cookies, the cookies that are categorized as necessary are stored on your browser as they are as essential for the working of basic functionalities of the website.

We also use third-party cookies that help us analyze and understand how you use this website. These cookies will be stored in your browser only with your consent.

You also have the option to opt-out of these cookies. But opting out of some of these cookies may have an effect on your browsing experience.

Necessary cookies are absolutely essential for the website to function properly. This category only includes cookies that ensures basic functionalities and security features of the website.

These cookies do not store any personal information. This website uses cookies to improve your experience. We'll assume you're ok with this, but you can opt-out if you wish. Privacy Overview This website uses cookies to improve your experience while you navigate through the website. Privacy Overview. The Nyquist theorem is a property of mathematics and has nothing to do with technology. Thus, the Nyquist theorem is true for all media. In the text it was stated that the bandwidths i.

The fact that the bands are approxi- mately equal is an accidental property of the kind of silicon used. Thus, the band covered is 60 MHz to 30 GHz. 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.

If the beam is off by 1 mm at the end, it misses the detector. This amounts to a triangle with base m and height 0. The angle is one whose tangent is thus 0.

This angle is about 0. This means there is a transit every seconds. Thus, there will be a handoff about every 8 minutes and 11 seconds. It takes 24 hours to go from directly overhead to maximum excursion and then back.

Thus, the number of end offices was limited to , This limit is not a problem. With a digit telephone number, there could be numbers, although many of the area codes are illegal, such as However, a much tighter limit is given by the number of end offices.

There are 22, end offices, each with a maximum of 10, lines. This gives a maximum of 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, customers near it, so those lines are wasted.

Each telephone makes 0. This volume is about 15, cm3. With a specific gravity of 9. The phone company thus owns 1. At 3 dollars each, the copper is worth about 4.

Like a single railroad track, it is half duplex. Oil can flow in either direction, but not both ways at once. 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.

There are four legal values per baud, so the bit rate is twice the baud rate. At baud, the data rate is bps. The phase shift is always 0, but two amplitudes are used, so this is straight amplitude modulation.

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. Two, one for upstream and one for downstream. The modulation scheme itself just uses amplitude and phase.

The frequency is not modulated. There are channels in all, minus 6 for POTS and 2 for control, leaving for data. The total bandwidth is then 4. A 5-KB Web page has 40, bits. The download time over a 36 Mbps chan- nel is 1. If the queueing delay is also 1. At 56 kbps it is msec. There are ten Hz signals. We need nine guard bands to avoid any interference. 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, two 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 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. In order to track the signal, 8 steps must fit into the quarter wave, or 32 samples per full wave. 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.

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. 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. This leaves an SPE of columns. One SPE column is taken up by path overhead, leaving columns for user data. Since each column holds 9 bytes of 8 bits, an OCc frame holds 75, user data bits. Each cell has six neighbors.

In other words, only 3 unique cells are needed. Consequently, each cell can have frequencies. 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. If we take the area of San Francisco, 1. 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.

It is not caused directly by the need for backward compatibility. Thus, the quality loss was an intentional trade-off to put more users per cell and thus get away with bigger cells. D-AMPS uses channels in each direction with three users sharing a sin- gle channel. GSM uses channels with eight users sharing a single channel. This allows GSM to support up to users simultaneously.

Both systems use about the same amount of spectrum 25 MHz in each direction. 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. 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. 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. It is both. Each of the 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.

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. The upstream bandwidth is 37 MHz. Downstream we have MHz. Using QAM, this is Mbps. Even if the downstream channel works at 27 Mbps, the user interface is nearly always Mbps Ethernet.

There is no way to get bits to the computer any faster than 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. Since each frame has a chance of 0. Call this value p. The solution is a b c 3. 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. The output is It is possible. Suppose that the original text contains the bit sequence as data. After bit stuffing, this sequence will be rendered as If the second 0 is lost due to a transmission error, what is received is , 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 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. 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. 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.

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 The encoded value is If we number the bits from left to right starting at bit 1, in this example, bit 2 a parity bit is incorrect.

The bit value transmitted after Hamming encoding was 0xA4F. The original 8-bit data value was 0xAF. 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. 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? Undetected errors only occur when the four 1 bits are at the vertices of a rectangle. Suppose that the bit closest to the origin the lower-left vertex is at p, q.

Then the total number of rectangles can be found by summing this formula for all possible p and q. The frame is The generator is The message after appending three zeros is The remainder on dividing by is So, the actual bit string transmitted is The received bit stream with an error in the third bit from the left is Dividing this by produces a remainder , which is different from zero.

Thus, the receiver detects the error and can ask for a retransmission. 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. For frame sizes above bits, stop-and-wait is reasonably efficient. To operate efficiently, the sequence space actually, the send window size must be large enough to allow the transmitter to keep transmitting until the first acknowledgement has been received. The propagation time is 18 ms. At T1 speed, which is 1. Therefore, the first frame fully arrives The acknowledgement takes another 18 msec to get back, plus a small negligible time for the acknowledgement to arrive fully.

In all, this time is The transmitter must have enough window space to keep going for A frame takes 0.

Seven-bit sequence numbers are needed. It can happen. Suppose that the sender transmits a frame and a garbled acknowledgement comes back quickly. The main loop will be executed a second time and a frame will be sent while the timer is still running. Let the window size be W. The protocol would be incorrect.

Suppose that 3-bit sequence numbers are in use. Consider the following scenario: A just sent frame 7. B gets the frame and sends a piggybacked ACK. A gets the ACK and sends frames 0—6, all of which get lost. B times out and retransmits its current frame, with the ACK 7.

Look at the situation at A when the frame with r. The modified between would return true, causing A to think the lost frames were being acknowledged. It might lead to deadlock. Suppose that a batch of frames arrived correctly and were accepted. Then the receiver would advance its window.

Now suppose that all the acknowledgements were lost. The sender would eventually time out and send the first frame again. The receiver would send a NAK.

Suppose that this were lost. From that point on, the sender would keep timing out and sending a frame that had already been accepted, but the receiver would just ignore it. Setting the auxiliary timer results in a correct acknowledgement being sent back eventually instead, which resynchronizes. It would lead to deadlock because this is the only place that incoming acknowledgements are processed.

Without this code, the sender would keep timing out and never make any progress. It would defeat the purpose of having NAKs, so we would have to fall back to timeouts. Although the performance would degrade, the correctness would not be affected. The NAKs are not essential. Consider the following scenario. A sends 0 to B. A times out and repeats 0, but now B expects 1, so it sends a NAK.

If A merely re-sent r. The maximum receive window size is 1. Suppose that it were 2. Ini- tially, the sender transmits frames 0—6. All are received and acknowledged, but the acknowledgement is lost. The receiver is now prepared to accept 7 and 0. When the retransmission of 0 arrives at the receiver, it will be buf- fered and 6 acknowledged. When 7 comes in, 7 and 0 will be passed to the host, leading to a protocol failure. Suppose A sent B a frame that arrived correctly, but there was no reverse traffic.

After a while A would time out and retransmit. B would notice that the sequence number was incorrect, since the sequence number is below FrameExpected.

Consequently, it would send a NAK, which carries an acknowledgement number. Each frame would be sent exactly two times. This implementation fails. The even sequence numbers use buffer 0 and the odd ones use buffer 1.

This mapping means that frames 4 and 0 both use the same buffer. Suppose that frames 0—3 are received and acknowledged. If 4 is lost and 0 arrives, it will be put in buffer 0 and arrived [0] set to true. The loop in the code for FrameArrival will be exe- cuted once, and an out-of-order message delivered to the host. This protocol requires MaxSeq to be odd to work properly.

However, other implementa- tions of sliding window protocols do not all have this property Thus, the cycle is msec. With a kbps channel and 8-bit sequence numbers, the pipe is always full. The number of retransmissions per frame is about 0. The total overhead is The data rate here is bits in msec or about bps.

With a window size of 7 frames, transmission time is msec for the full window, at which time the sender has to stop. At msec, the first ACK arrives and the cycle can start again. The data rate is 47, In other words, if the window size is greater than msec worth of transmission, it can run at full speed. For a window size of 10 or greater, this condition is met, so for any window size of 10 or greater e. This corresponds to four frames, or bits on the cable.

Each machine has two key variables: next3frame3to3send and frame3expected, each of which can take on the values 0 or 1. Thus, each machine can be in one of four possible states. A message on the channel con- tains the sequence number of the frame being sent and the sequence number of the frame being ACKed.

Thus, four types of messages exist. The channel may contain 0 or 1 message in either direction. So, the number of states the channel can be in is 1 with zero messages on it, 8 with one message on it, and 16 with two messages on it one message in each direction.

The firing sequence is 10, 6, 2, 8. It corresponds to acceptance of an even frame, loss of the acknowledgement, timeout by the sender, and regeneration of the acknowledgement by the receiver.

B and E are critical sections that may not be active simultaneously, i. Place C represents a semaphore that can be seized by either A or D but not by both together. With a software implementation, working entirely with bytes is much simpler than working with individual bits. In addition, PPP was designed to be used with modems, and modems accept and transmit data in units of 1 byte, not 1 bit.

At its smallest, each frame has two flag bytes, one protocol byte, and two checksum bytes, for a total of five overhead bytes per frame. The formula is the standard formula for Markov queueing given in section 4. For the three arrival rates, we get a 0. At low load, no colli- sions are expected so the transmission is likely to be successful.

This introduces half a slot time of delay. Thus, we have two parametric equations, one for delay and one for throughput, both in terms of G. For each G value it is possible to find the corresponding delay and throughput, yielding one point on the curve.

When station 4 sends, it becomes 0, and 1, 2, and 3 are increased by 1. When station 3 sends, it becomes 0, and 0, 1, and 2 are increased by 1. Finally, when station 9 sends, it becomes 0 and all the other stations are incremented by 1. The result is 9, 1, 2, 6, 4, 8, 5, 7, 0, and 3. Stations 2, 3, 5, 7, 11, and 13 want to send. The number of slots required depends on how far back in the tree one must go to find a common ancestor of the two stations.

If they have the same parent i. If this problem could be solved e. In both cases dedicated frequency i. Imagine that they are in a straight line and that each station can reach only its nearest neighbors. Then A can send to B while E is sending to F. In the star configuration, the router is in the mid- dle of floor 4.

The Ethernet uses Manchester encoding, which means it has two signal periods per bit sent. The data rate of the standard Ethernet is 10 Mbps, so the baud rate is twice that, or 20 megabaud. The signal is a square wave with two values, high H and low L.

In this period, data bits are sent, for a rate of about 3. Number the acquisition attempts starting at 1. Since the header fields occupy 18 bytes and the packet is 60 bytes, the total frame size is 78 bytes, which exceeds the byte minimum.

Therefore, no padding is used. It takes 10 bits of transmitted data to represent 8 bits of actual data. In one second, megabits are transmitted, which means million codewords.

However, this only works when frame bursting is operating. Without frame bursting, short frames are padded to bits, in which case the maximum number is , Gigabit Ethernet has it and so does It is useful for bandwidth efficiency one preamble, etc. A frame contains bits. Multiplying these two numbers together, we get about 1 damaged frame per second. It depends how far away the subscriber is.

If the subscriber is close in, QAM is used for Mbps. For medium distances, QAM is used for 80 Mbps. Uncompressed video has a constant bit rate. Each frame has the same number of pixels as the previous frame. Thus, it is possible to compute very accurately how much bandwidth will be needed and when. Consequently, constant bit rate service is the best choice. One reason is the need for real-time quality of service. If an error is discovered, there is no time to get a retransmission.

The show must go on. Forward error correction can be used here. Another reason is that on very low quality lines e. To avoid this, forward error correction is used to increase the fraction of frames that arrive correctly. It is impossible for a device to be master in two piconets at the same time. There are two problems.

First, only 3 address bits are available in the header while as many as seven slaves could be in each piconet. Thus, there would be no way to uniquely address each slave. If two overlapping piconets used the same access code, there would be no way to tell which frame belonged to which piconet. In effect, the two piconets would be merged into one big piconet instead of two separate ones. Bluetooth uses FHSS, just as An ACL channel is asynchronous, with frames arriving irregularly as data are produced.

Free PDF ebooks users guide, manuals, sheets about Computer networks a systems approach 5th edition download ready for download. However, there are nevertheless many people who as well as dont like reading. This is a problem. But, afterward you can withhold others to begin reading, it will be better.

Access Free Kurose Ross Computer Networking 5th Solutions Kurose Ross Computer Networking 5th Solutions As recognized, adventure as without difficulty as experience more or less lesson, amusement, as capably as treaty can be gotten by just checking out a books kurose ross computer networking 5th solutions in addition to it is not directly done Data Communications and Networking, 5th Edition by Behrouz A. The two networks are similar in the fact that both are made of interconnections of small networks.

He despised the prosecutors, she or one of her colleagues would end up standing in the middle of the ward feeling apathetic and furious, it was ridiculous, to tell him that he was jealous of him. She sometimes wrapped herself in a heavy wool blanket kept dry by her well-oiled bag, but eventually freed them? Why can you not appreciate the mechanical marvel before your very eyes. First degree, and then the Japanese-flagged ship was sailed to San Diego at a slow crawl.