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computer networks fifth edition problem solutions andrew s. tanenbaum vrije computer networks manual pdf by tanenbaum 5th edition solutions free download. COMPUTER NETWORKS FIFTH EDITION PROBLEM SOLUTIONS ANDREW S. TANENBAUM Vrije Universiteit Amsterdam, The Netherlands and DAVID WETHERALL University of Washington Seattle, WA PRENTICE Save this PDF as. Instructor Solutions Manual for Computer Networks, 5th Edition. Andrew S. Tanenbaum, Vrije University, Amsterdam, The Netherlands. David J. Wetherall.

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COMPUTER NETWORKS. FIFTH EDITION. PROBLEM SOLUTIONS. ANDREW S . TANENBAUM. Vrije Universiteit. Amsterdam, The Netherlands and. Computer Networks Tanenbaum 5th Edition Solution - [PDF] [EPUB] Computer Computer Networks 5th Edition Textbook Solutions - Chegg. Tanenbaum 5th Edition Solution [PDF] [EPUB] Mohammad Monzurul. Download with Computer Networks 5th Edition Andrew S. Tanenbaum.

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 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 For an OC line: It can be used to accommodate DS It can be used to accommodate DS-2 service.

This leaves an SPE of columns. One SPE column is taken up by path overhead,. Since each column holds 9 bytes of 8 bits, an OCc frame holds 75, user data bits. The three networks have the following properties: Full interconnect: 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.

This makes frequency. 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.

Just compute the four normalized inner products: 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. Usually, cable operators specify Mbps Ethernet because they do not want one user sucking up the entire bandwidth.

Since each frame has a chance of 0. The solution is a b. 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 actually 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 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 appropriate in a military situation in which the receiver does not want to disclose its location by transmitting.

If the error rate is low enough that an error-correcting code is good enough, it may also be simpler. Finally, realtime systems cannot tolerate waiting for retransmissions. Making one change to any valid character cannot generate another valid character 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 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, 5 parity bits are sufficient. The bit pattern transmitted 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 will also be detected.

If they are in the same row or column, that row s or column s parity will catch them. If two errors are in the same row, the column parity of at least one of them will catch the error.

Instructor Solutions Manual for Computer Networks, 5th Edition

If two errors are in the same column, the row parity of at least one of them will catch the error. A 4-bit error in which the four error bits lie on the four corners of a rectangle cannot be caught. Total bits transmitted per block is bits. In case of error detection mechanism, one parity bit is transmitted per block.

However, a block may encounter a bit error x times. Every time an error is encountered, bits have to be retransmitted. So, total bits transmitted per block is x bits. 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. 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. 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 4 bits: When the first 1 goes in, 11 comes out and S 1 stores the 1.

Then 0 goes in and 01 comes out, with S 2 now storing a 1 and S 1 storing the 0. The complete output sequence, including these initial values is To obtain the checksum, we need to calculate the ones complement sum of words, which is same as sum modulo 2 4 and adding any overflow of high order bits back into low-order bits: 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 of , which is different from 0.

Thus, the receiver detects the error and can ask for a retransmission.

If the. A trivial example is if all ones in the bit stream are inverted to zeros. The CRC checksum polynomial is or degree 32, so a Yes. CRC catches all single-bit errors. CRC catches all double-bit errors for any reasonably long message. CRC may not be able catch all even number of isolated bit errors.

CRC catches all odd number of isolated bit errors. CRC catches all burst errors with burst lengths less than or equal to CRC may not be able to catch a burst error with burst length greater than Yes, it is possible. The reason is that an acknowledgement frame may arrive correctly, but after the sender s timer has expired. This can happen if the receiver gets delayed in sending the acknowledgement frame, because its CPU is overloaded processing other jobs in the system.

For frame sizes above bits, stop-and-wait is reasonably efficient. 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.

To operate efficiently, the sequence space actually, the sender s 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 Mbps excluding the 1 header bit , a byte frame takes msec. 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 A frame takes 0. Seven-bit sequence numbers are needed. Let the sender s window be S l, S u and the receiver s be R l, R u. Let the window size be W.

The relations that must hold are: 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 was accepted. 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 then send a NAK. If this packet 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. A sends 0 to B. A times out and repeats 0, but now B expects 1, so it sends a NAK. If A merely resent r. 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 it would be 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. The receiver s window now contains 4 and 0. If 4 is lost and 0 arrives, it will be put in buffer 0 and arrived[0] will be set to true.

The loop in the code for FrameArrival will be executed once, and an out-of-order message will be delivered to the host. This protocol requires MaxSeq to be odd to work properly. However, other implementations 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 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. 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 2 flag bytes, 1 protocol byte, and 2 checksum bytes, for a total of 5 overhead bytes per frame. For maximum overhead, 2 flag bytes, 1 byte each for address and control, 2 bytes for protocol and 4 bytes for checksum.

This totals to 10 overhead bytes. To make this frame size a multiple of 48, the number of padding bytes will be This will result in an AAL5 frame of size bytes.

This can fit in three ATM cells. For the three arrival rates, we get a 0. At low load, no collisions are expected so the transmission is likely to be successful. This introduces half a slot time of delay.

Tanenbaum & Wetherall, Instructor Solutions Manual for Computer Networks | Pearson

The E events are separated by E 1 intervals of four slots each, so the delay is 4 e G 1. 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. Signal propagation time for 2 km is 8. The worst case is where all stations want to send and s is the lowest-numbered station.

If a higher-numbered station and a lower-numbered station have packets to send at the same time, the higher-numbered station will always win the bid. Thus, a lower-numbered station will be starved from sending its packets if there is a continuous stream of higher-numbered stations ready to send their packets.

Stations 2, 3, 5, 7, 11, and 13 want to send. Eleven slots are needed, with the contents of each slot being as follows: Slot 1: So, no other communication is possible in this case. 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 middle of floor 4. Classic Ethernet uses Manchester encoding, which means it has two signal periods per bit sent. The data rate 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. A complete transmission has six phases: Transmit data Delay for last bit to get to the end 5. Acknowledgement sent 3. In this period, data bits are sent, for a rate of about 3. Number the acquisition attempts starting at 1. Attempt i is distributed among 2 i 1 slots.

Thus, the probability of a collision on attempt i is 2 i 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. 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.

D did not respond, so it must be outside A s radio range. So, given the scenario in Figure b , MACA protocol will allow simultaneous communication, B to A and C to D, but will allow only one of these communications to take place at a time. So, the 6 Mbps stations will get 0. 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, QAM is used for Mbps. For medium distances, QAM is used for 80 Mbps. One reason is the need for real-time quality of service.

If an error is discovered, there is no time for 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. However, unlike , WiMAX base stations are much more powerful than access points.

Also, transmissions in WiMAX are carefully scheduled by the base station for each subscriber. Allowing this would create 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. Second, the access code at the start of the frame is derived from the master s identity. This is how slaves tell which message belongs to which piconet. 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. So, a maximum of bits can be transmitted in a 3-slot frame.

Out of this, to bits are overhead bits, leaving a maximum of to bits for the data field. Bluetooth uses FHSS, just as does. Out of this, a maximum of bits are for data. In case of repetition encoding, data is replicated thrice, so the actual data transmitted is about bits. They do not. The dwell time in is not standardized, so it has to be announced to new stations that arrive.

There is no need to announce this. All Bluetooth devices have this hardwired into the chip. Bluetooth was designed to be cheap, and fixing the hop rate and dwell time leads to a simpler chip.

We want to maximize the probability that one and only one tag responds in a given slot. This occurs when the reader sets Q equal to 10 slots. Consulting Fig. One key security concern is unauthorized tracking of RFID tags. This becomes quite serious if the item is sensitive in nature, for example, a passport, and the tag can be used to retrieve further information, for example, the nationality and other personal information of the person holding the passport.

Another security concern is the ability of a reader to change tag information. This can be used by an adversary to, for example, change the price of a tagged item he plans to buy. Stallings, Data and Computer Communications, 6 th ed. Tannenbaum, Computer Networks, 4 th ed. Given the following. Ungraded A noiseless 2-kHz channel is sampled every 5 ms.

What is the maximum data rate? In this problem, the channel being sampled gives us the. The Physical Layer: Data Transmission Basics Encode data as energy at the data information source and transmit the encoded energy using transmitter hardware: Possible Energy Forms: Electrical, light,.

Wireless Physical Layer Q1. Is it possible to transmit a digital signal, e. It is not possible to transmit. Determine the minimum transmission rate. Data Communications Prof. Analog vs.

Digital Transmission Compare at two levels: Data continuous audio vs. Signaling continuously varying electromagnetic wave vs. Also Transmission.

A Date: Chapter 3: What is an unacknowledged connectionless service? What is an acknowledged. Lecture 3: Data Link Layer 1 Principal service: Transferring data from the network layer of the source machine to the one of the destination machine Virtual communication versus actual communication: Specific functions. Interference is avoided under frequency.

Define the term Computer Networks. A Computer network is a number if computers interconnected by one or more transmission paths. The transmission path often is the telephone. Lecture 4 Communications Channels Definition: Classification criteria: Basic technology Next class: Green chapter 4, 6, 7, Data Transmission Concepts and terminology Transmission terminology Transmission from transmitter to receiver goes over some transmission medium using electromagnetic waves Guided media.

Waves are guided. Broadband Networks Prof. Explain the operation of Cellular. A typical. Network Performance: Networks must be fast What are the essential network performance metrics: Channel Rates and Shared Media You are entrusted with the design of a network to interconnect a set of geographically. Introduction Computer Network. A transmission system, usually private owned, very speedy and secure, covering a geographical area in the range of kilometres, comprising a shared transmission medium and a set.

EECS Chapter 1 Review Questions R1. What is the difference between a host and an end system? List several different types of end systems. Is a Web server an end system? There is no difference. Computer Networks Homework 1 Assigned: January 29, Due: February 7, 1. Suppose a Mbps point-to-point link is being set up between Earth and a new lunar colony.

The distance from the. Transport Layer Protocols Version. Transport layer performs two main tasks for the application layer by using the network layer. It provides end to end communication between two applications, and implements. Part I: The problem set consists of two parts: The problem specifications pages. In the succeeding years, the underlying technology has gone through three phases, known as generations.

Network administrators must design a network to operate within an acceptable delay budget. This topic. Org http: The T-Carrier specification defines. This document serves to explain the sources of latency on a switched Ethernet network and describe how to calculate cumulative latency. The work is protected by local and international copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning.

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