实例介绍
【实例简介】
MIT教材Data Networks全部的答案，不是一部分，1-6章都有。
@1993 Prentice-Hall Inc A Pearson Education Company Upper Saddle river, NJ 07458 All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher Printed in the United States of America ISBN-13-己0口24-己 Prentice-Hall International(UK)Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada Inc, Toronto Prentice-Hall Hispanoamericana, s.A., Mexico Prentice-Hall of india private limited, new delhi Prentice-Hall of Japan, Inc, Tokyo Pearson Education Asia Pte Ltd, Singapore Editora Prentice-Hall do Brasil, Ltda. Rio de janeiro Solutions Manual.Data Networks, 2/E by Dimitri Bertsekas and Robert Gallager TABLE OF CONTENTS CHAPTER 1 CHAPTER 2 chaPter 3 CHAPTER 4 ..…191 CHAPTER 5........ 114 ChaPTer 6............ 148 ACKNOWLEDGMENTS Several of our students have contributed to this solutions manual. We are particularly thankful to Rajesh Pankaj, Jane Simmons, John Spinelli, and Manos Varvarigos, CHAPTER 1 SOLUTIONS There are 250,000 pixels per square inch, and multiplying by the number of square inches and the number of bits per pixel gives 5.61 x 108bits a)There are 16x109 bits going into the network per hour. Thus there are 48 x 109 bits per hour traveling through the network, or 13.33 million bits per second. This requires 209 links of 64 kbit/sec. each b)since a telephone conversation requires two people, and 10% of the people are busy on the average, we have 50,000 simultaneous calls on the average, which requires 150,000 provide enough inks to avoid congestion(and to provide local access loops to eachto links on the average. Both the answer in a)and b)must be multiplied by some fact telephone), but the point of the problem is to illustrate how little data, both in absolute and comparative terms, is required for ordinary data transactions by people delayed or lost and can also get out of order in the network. In this interpretation, if a ily There are two possible interpretations of the problem. In the first, packets can be arbitra packet from a to B is sent at time t and not received by some later time t, there is no way to tell whether that packet will ever arrive later Thus if any data packet or protocol packet from a to B is lost, node b can never terminate with the assurance that it will never receive another packet In the second interpretation, packets can be arbitrarily delayed or lost, but cannot get out of order. Assume that each node is initially in a communication state, exchanging data packets. Then each node, perhaps at different times, goes into a state or set of states in which it sends protocol packets in an attempt to terminate. Assume that a node can enter the final termination state only on the receipt of one of these protocol packets(since timing information cannot help, since there is no side information, and since any data packet could be followed by another data packet). As in the three army problem, assume any particular ordering in which the two nodes receive protocol packets. The first node to receive a protocol packet cannot go to the final termination state since it has no assurance that any protocol packet will ever be received by the other node, and thus no assurance that the other The next protocol packet to be received then finds neither node n the final termination state Thus again the receiving node cannot terminate without the possibility that the other node will receive no more protocol packets and thus never terminate. The same situation occurs on each received protocol packet, and thus it is impossible to guarantee that both nodes can eventually terminate. This is essentially the same argument as used for the three army problem CHAPTER 2 SOLUTIONS Let x(t) be the output for the single pulse shown in Fig. 2.3(a)and let y(t) be the output for the sequence of pulses in Fig. 2.3(b). The input for 2. 3(b)is the sum of six input pulses of the type in 2. 3(a); the first such pulse is identical to that of 2.3(a), the second is delayed by t time units, the third is inverted and delayed by 2T time units, etc. From the ame invariance property, the response to the second pulse above is x(t-T)(i.e x(t)delayed by T); from the time invariance and linearity the response to the third pulse is x(t-2T. Using linearity to add the responses to the six pulses, the overall output is (t)=x(t)+x(t-T)-x(t-2T)+x(t-3Tx(t41)-x(t5T) To put the result in more explicit form, note that 0 <0 X 2y/T 0≤t<T (e-1)e .2t/T t≥T Thus the response from the ith pulse(1 sis 6)is zero up to time (i-1T. For t<O, then y(t)=0; fromOst<T y(t)=x(t)=1-e1;0≤t<T From t≤t<2T, y(t=x(t)+x(t-T) 2-1)e2+[1-e20T e-2VT Similarly, for 2Tst<3T y()=x()+x(tT)-x(t2T) (e2-1)e2T+(e2.1)e2TT-[1-e2(2/ -1+(2e4-1)e2 2T≤t<3T A similar analysis for each subsequent interval leads to y(t)=1-(2e6-2e4+1)e2 3T≤t<4T =-1+(2e8-2e6+2e4.1)e;4T≤t<6T 12-2e8+2e6-2e4+1)e;t≥6T The solution is continuous over t with slope discontinuities at0, 2T, 3T, 4T, and 6T; the value of y(t) at these points is y(0)=0; y(2T)=,982; y(3T)=-.732; y(4T)= 766; y(6T) to first find the response to a unit step and then view y(t)as the response to a sum or se x(t) 968. Another approach to the problem that gets the solution with less work is to displaced unit steps 2.2 From the convolution equation, Eq (2. 1), the output r(t)is T r(t= s(c)h(t-tdt= h(t-tdt Note that h(t-t)=oea(t-t)for t-t20, i.e. for t st), and h(t-t)=0 for t>t. Thus fort 0, h(t-t)=0 throughout the integration interval above. For ost<T, we then have (t)=ae a(t-t)dt+ odt =1-e ;0≤t<T Fort2T, h(t-c)=ae-a(t-t)over the entire integration interval and T -a(to)dt=e -a(t-t ae ;t≥T Thus the response increases towards l for 0stsT with the exponential decay factor a, and then, for t2T, decays toward o 2.3 From Eq(2.1) 2πfr rlt h(t-tdt Using t'=t-t as the variable of integration for any given t, r(t) 2πcf(t-τ 2元ft 2元rt h(t)dτ H(n where H(f) is as given in Eq ( 2.3) h(t)= H(ej2ritdf Since H(f is 1 from-fo to fo and 0 elsewhere we can integrate exp(j2ncft) from-fo to fo, obtaining h(t)= R2 exp(2fot)-exp(-12tfot]- sin(2fot t Note that this impulse response is unrealizable in the sense that the response starts before the impulse(and, even worse, starts an infinite time before the impulse). None the less, such ideal filters are useful abstractions in practice 2.5 The function s1(t)is compressed by a factor of B on the time axis as shown below (t) A A2 A2 f 2nft (t)e - di lIft ∫o=3( sOD f) Thus S(f) is attenuated by a factor of B in amplitude and expanded by a factor of p on the requency scale; compressing a function in time expands it in frequency and vice versa 2.6 a)We use the fact that cos(x)=[exp(jx)+exp(-jx)]/2. Thus the Fourier transform of s(t)cos(2πfo)is exp(2πfo)+exp(-j2πfo exp(-j2πft)dt s(t expl-Jzπ (f-fo)] dt exp[-j2(f+fo)t] dt s(f-fo) S(f+fo) b)Here we use the identity cos2(x)=[l+cos( 2x)1/2. Thus the Fourier transform of s(t)cos<(2pfot)is the Fourier transform of s(t)/2 plus the Fourier transform of s(t)cos[2p(2fo)t/2. USing the result in part a, this is S(f/2+ S(f-2fo)/4+S(f+2fo)/4 2.7 a)E(frame time on 9600 bps link)=1000 bits /9600 bps=0.104 sec E(frame time on 50,000bps link]=0.02 sec. b)E(time for 106 frames on 9600 bps link)=1.04:105 sec E(time for 106 frames on 50, 000 bps link)=2. 104 sec Since the frame lengths are statistically independent, the variance of the total number of bits in 106 frames is 106 times the variance for one frame. Thus the standard deviation of the total number of bits in 106 frames is 103 times the standard deviation of the bits in one frame or 5 100 bits. The standard deviation of the transmission time is then S D. time for 106 frames on 9600 bps link)=5. 105/9600=52 sec S D(time for 106 frames on 50,000 bps link)=5. 105/50,000=10 sec c)The point of all the above calculations is to see that, for a large number of frames, the expected time to transmit the frames is very much larger than the standard deviation of the transmission time; that is, the time per frame, averaged over a very long sequence of frames is close to the expected frame time with high probability. One's intuition would then suggest that the number of frames per unit dime, averaged over a very long time period, is close to the reciprocal of the expected frame time with high probability. This intuition is correct and follows either from renewal theory or from direct analysis. Thus the reciprocal of the expected frame time is the rate of frame transmissions in the usual sense of the word rate 2.8 Let xij be the bit in row i, column j. Then the ith horizontal parity check is h=习x where the summation is summation modulo 2. Summing both sides of this equation (modulo 2)over the rows i, we have ihi=习 This shows that the modulo 2 sum of all the horizontal parity checks is the same as the modulo 2 sum of all the data bits. The corresponding argument on columns shows that the modulo 2 sum of the vertical parity checks is the same 【实例截图】
【核心代码】
MIT教材Data Networks全部的答案，不是一部分，1-6章都有。
@1993 Prentice-Hall Inc A Pearson Education Company Upper Saddle river, NJ 07458 All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher Printed in the United States of America ISBN-13-己0口24-己 Prentice-Hall International(UK)Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall Canada Inc, Toronto Prentice-Hall Hispanoamericana, s.A., Mexico Prentice-Hall of india private limited, new delhi Prentice-Hall of Japan, Inc, Tokyo Pearson Education Asia Pte Ltd, Singapore Editora Prentice-Hall do Brasil, Ltda. Rio de janeiro Solutions Manual.Data Networks, 2/E by Dimitri Bertsekas and Robert Gallager TABLE OF CONTENTS CHAPTER 1 CHAPTER 2 chaPter 3 CHAPTER 4 ..…191 CHAPTER 5........ 114 ChaPTer 6............ 148 ACKNOWLEDGMENTS Several of our students have contributed to this solutions manual. We are particularly thankful to Rajesh Pankaj, Jane Simmons, John Spinelli, and Manos Varvarigos, CHAPTER 1 SOLUTIONS There are 250,000 pixels per square inch, and multiplying by the number of square inches and the number of bits per pixel gives 5.61 x 108bits a)There are 16x109 bits going into the network per hour. Thus there are 48 x 109 bits per hour traveling through the network, or 13.33 million bits per second. This requires 209 links of 64 kbit/sec. each b)since a telephone conversation requires two people, and 10% of the people are busy on the average, we have 50,000 simultaneous calls on the average, which requires 150,000 provide enough inks to avoid congestion(and to provide local access loops to eachto links on the average. Both the answer in a)and b)must be multiplied by some fact telephone), but the point of the problem is to illustrate how little data, both in absolute and comparative terms, is required for ordinary data transactions by people delayed or lost and can also get out of order in the network. In this interpretation, if a ily There are two possible interpretations of the problem. In the first, packets can be arbitra packet from a to B is sent at time t and not received by some later time t, there is no way to tell whether that packet will ever arrive later Thus if any data packet or protocol packet from a to B is lost, node b can never terminate with the assurance that it will never receive another packet In the second interpretation, packets can be arbitrarily delayed or lost, but cannot get out of order. Assume that each node is initially in a communication state, exchanging data packets. Then each node, perhaps at different times, goes into a state or set of states in which it sends protocol packets in an attempt to terminate. Assume that a node can enter the final termination state only on the receipt of one of these protocol packets(since timing information cannot help, since there is no side information, and since any data packet could be followed by another data packet). As in the three army problem, assume any particular ordering in which the two nodes receive protocol packets. The first node to receive a protocol packet cannot go to the final termination state since it has no assurance that any protocol packet will ever be received by the other node, and thus no assurance that the other The next protocol packet to be received then finds neither node n the final termination state Thus again the receiving node cannot terminate without the possibility that the other node will receive no more protocol packets and thus never terminate. The same situation occurs on each received protocol packet, and thus it is impossible to guarantee that both nodes can eventually terminate. This is essentially the same argument as used for the three army problem CHAPTER 2 SOLUTIONS Let x(t) be the output for the single pulse shown in Fig. 2.3(a)and let y(t) be the output for the sequence of pulses in Fig. 2.3(b). The input for 2. 3(b)is the sum of six input pulses of the type in 2. 3(a); the first such pulse is identical to that of 2.3(a), the second is delayed by t time units, the third is inverted and delayed by 2T time units, etc. From the ame invariance property, the response to the second pulse above is x(t-T)(i.e x(t)delayed by T); from the time invariance and linearity the response to the third pulse is x(t-2T. Using linearity to add the responses to the six pulses, the overall output is (t)=x(t)+x(t-T)-x(t-2T)+x(t-3Tx(t41)-x(t5T) To put the result in more explicit form, note that 0 <0 X 2y/T 0≤t<T (e-1)e .2t/T t≥T Thus the response from the ith pulse(1 sis 6)is zero up to time (i-1T. For t<O, then y(t)=0; fromOst<T y(t)=x(t)=1-e1;0≤t<T From t≤t<2T, y(t=x(t)+x(t-T) 2-1)e2+[1-e20T e-2VT Similarly, for 2Tst<3T y()=x()+x(tT)-x(t2T) (e2-1)e2T+(e2.1)e2TT-[1-e2(2/ -1+(2e4-1)e2 2T≤t<3T A similar analysis for each subsequent interval leads to y(t)=1-(2e6-2e4+1)e2 3T≤t<4T =-1+(2e8-2e6+2e4.1)e;4T≤t<6T 12-2e8+2e6-2e4+1)e;t≥6T The solution is continuous over t with slope discontinuities at0, 2T, 3T, 4T, and 6T; the value of y(t) at these points is y(0)=0; y(2T)=,982; y(3T)=-.732; y(4T)= 766; y(6T) to first find the response to a unit step and then view y(t)as the response to a sum or se x(t) 968. Another approach to the problem that gets the solution with less work is to displaced unit steps 2.2 From the convolution equation, Eq (2. 1), the output r(t)is T r(t= s(c)h(t-tdt= h(t-tdt Note that h(t-t)=oea(t-t)for t-t20, i.e. for t st), and h(t-t)=0 for t>t. Thus fort 0, h(t-t)=0 throughout the integration interval above. For ost<T, we then have (t)=ae a(t-t)dt+ odt =1-e ;0≤t<T Fort2T, h(t-c)=ae-a(t-t)over the entire integration interval and T -a(to)dt=e -a(t-t ae ;t≥T Thus the response increases towards l for 0stsT with the exponential decay factor a, and then, for t2T, decays toward o 2.3 From Eq(2.1) 2πfr rlt h(t-tdt Using t'=t-t as the variable of integration for any given t, r(t) 2πcf(t-τ 2元ft 2元rt h(t)dτ H(n where H(f) is as given in Eq ( 2.3) h(t)= H(ej2ritdf Since H(f is 1 from-fo to fo and 0 elsewhere we can integrate exp(j2ncft) from-fo to fo, obtaining h(t)= R2 exp(2fot)-exp(-12tfot]- sin(2fot t Note that this impulse response is unrealizable in the sense that the response starts before the impulse(and, even worse, starts an infinite time before the impulse). None the less, such ideal filters are useful abstractions in practice 2.5 The function s1(t)is compressed by a factor of B on the time axis as shown below (t) A A2 A2 f 2nft (t)e - di lIft ∫o=3( sOD f) Thus S(f) is attenuated by a factor of B in amplitude and expanded by a factor of p on the requency scale; compressing a function in time expands it in frequency and vice versa 2.6 a)We use the fact that cos(x)=[exp(jx)+exp(-jx)]/2. Thus the Fourier transform of s(t)cos(2πfo)is exp(2πfo)+exp(-j2πfo exp(-j2πft)dt s(t expl-Jzπ (f-fo)] dt exp[-j2(f+fo)t] dt s(f-fo) S(f+fo) b)Here we use the identity cos2(x)=[l+cos( 2x)1/2. Thus the Fourier transform of s(t)cos<(2pfot)is the Fourier transform of s(t)/2 plus the Fourier transform of s(t)cos[2p(2fo)t/2. USing the result in part a, this is S(f/2+ S(f-2fo)/4+S(f+2fo)/4 2.7 a)E(frame time on 9600 bps link)=1000 bits /9600 bps=0.104 sec E(frame time on 50,000bps link]=0.02 sec. b)E(time for 106 frames on 9600 bps link)=1.04:105 sec E(time for 106 frames on 50, 000 bps link)=2. 104 sec Since the frame lengths are statistically independent, the variance of the total number of bits in 106 frames is 106 times the variance for one frame. Thus the standard deviation of the total number of bits in 106 frames is 103 times the standard deviation of the bits in one frame or 5 100 bits. The standard deviation of the transmission time is then S D. time for 106 frames on 9600 bps link)=5. 105/9600=52 sec S D(time for 106 frames on 50,000 bps link)=5. 105/50,000=10 sec c)The point of all the above calculations is to see that, for a large number of frames, the expected time to transmit the frames is very much larger than the standard deviation of the transmission time; that is, the time per frame, averaged over a very long sequence of frames is close to the expected frame time with high probability. One's intuition would then suggest that the number of frames per unit dime, averaged over a very long time period, is close to the reciprocal of the expected frame time with high probability. This intuition is correct and follows either from renewal theory or from direct analysis. Thus the reciprocal of the expected frame time is the rate of frame transmissions in the usual sense of the word rate 2.8 Let xij be the bit in row i, column j. Then the ith horizontal parity check is h=习x where the summation is summation modulo 2. Summing both sides of this equation (modulo 2)over the rows i, we have ihi=习 This shows that the modulo 2 sum of all the horizontal parity checks is the same as the modulo 2 sum of all the data bits. The corresponding argument on columns shows that the modulo 2 sum of the vertical parity checks is the same 【实例截图】
【核心代码】
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