SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
Solutions Manual For Introduction to Computer Networks
and Cybersecurity 1st Edition By Chwan-Hwa (John) Wu;
J. David Irwin 9781138071896 ALL Chapters
pg. 1
, SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
Introduction to Computer Networks and Cybersecurity Chwan-Hwa
(John) Wu Auburn University J. David Irwin
Chapter 0
0.1. SM may not be capable of meeting an unusual peak demand for
bandwidth since SM is targeted at normal, average bandwidth
operation. Therefore, an ISP may set different classes of services
and let only the high-paying customers have Internet access when
there is a traffic jam.
0.2. Transmission delay results from finite bandwidth in a
transmission line. Propagation delay is due to the fact that
electromagnetic waves are limited by the speed of light. Any bit
of information will encounter both delays on a path from
transmitter to receiver.
0.3.The Overhead = (100 + 4)/ (1000 + 100 +4) = 9.42%
0.4.The Overhead = (80 + 4)/ (100 + 80 + 4) = 45.65%
0.5.The Overhead = (60 + 4)/ (100 + 60 + 4) = 39.02%
0.6. The total delays = round trip delay for establishing a TCP connection +
sending the file =
2* 1 ms + 1000 * 8/ (1536000) + 1 ms = 2 + 5.2
ms + 1 ms = 8.2 ms The Overhead = 2* 1 ms/
23.83 ms = 24.39%
pg. 2
,SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
0.7. The processing delay in each router is 0.001 seconds. The
queuing delay is 0. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 0.007 seconds, and the propagation delay is d/ s = (2 x
105)/ (2 x 108) = 0.001 seconds. Therefore, the total delay is 7 ms
+1 ms. + 0 + 7 ms. + 1 ms. + 1 ms + 0 + 7 ms = 24 ms.
0.8. The processing delay in each router is 1 ms. The queuing delay
is specified as 5 ms. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 7 ms, and the propagation delay is d/ s = (2 x 105)/ (2
x 108) = 1 ms. Therefore, the total delay is 7 ms +1 ms + 5 ms +7
ms + 1 ms + 1 ms + 5 ms + 7 ms = 34 ms.
0.9. The processing delay in each router is 0.002 seconds. The
queuing delay is 0. 002 seconds. The transmission delay is L/ R =
(10Kbits)/ (1Mbps) = 0.010 seconds, and the propagation delay is
d/ s = (2 x 106)/ (2 x 108) = 0.01 seconds. Therefore, the total
delay is 10 ms + 2 ms + 2 ms + 10 ms. + 10 ms. + 2 ms + 2 ms +
10 ms = 48 ms.
0.10. The processing delay in each router is 100 µs. The queuing
delay is specified as 0.5 ms. The transmission delay is L/ R =
(5Kbits)/ (2Mbps) = 2.5 ms, and the propagation delay is d/ s =
(5 x 104)/ (2 x 108) = 0.25 ms. Therefore, the total delay is 2.5
ms +0.1ms + 0.5 ms +2.5 ms + 0.25 ms + 0.1ms + 0.5 ms +2.5
ms + 0.25 ms = 9.2 ms.
0.11. The processing delay in each router is 400 ns. The queuing delay is
specified as 800 ns. The transmission delay is L/ R = (3.1Kbits)/
(155Mbps) = 20 µs, and the propagation delay is d/ s
pg. 3
, SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
= (5 x 103)/ (2 x 108) = 25 µs. Therefore, the total delay is 20 µs + 400
ns + 800 ns +20 µs + 25 µs + 400 ns + 800 ns + 20 µs = 87.4 µs.
0.12. The processing delay in each router is 0.001 seconds. The queuing
delay is 0. The transmission delay is L/ R = (7Kbits)/ (1Mbps) =
0.007 seconds, and the propagation delay is d/ s = (2 x 105)/ (2 x 108)
= 0.001 seconds. Therefore, the total delay is 7 ms +1 ms. + 0 + 7
ms. + 1 ms. + 1 ms +0 +7 ms + 1ms + 1 ms + 0 + 7 ms = 33 ms.
0.13. The processing delay in each router is 1 ms. The queuing delay is
specified as 5 ms. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 7 ms, and the propagation delay is d/ s = (2 x 105)/ (2 x
108) = 1 ms. Therefore, the total delay is 7 ms +1 ms + 5 ms +7 ms +
1 ms + 1 ms + 5 ms + 7 ms + 1 ms + 1 ms + 5 ms + 7 ms = 48 ms.
0.14. The processing delay in each router is 0.002 seconds. The queuing
delay is 0. 002 seconds. The transmission delay is L/ R = (10Kbits)/
(1Mbps) = 0.010 seconds, and the propagation delay is d/ s = (2 x
106)/ (2 x 108) = 0.01 seconds. Therefore, the total delay is 10 ms +2
ms + 2 ms + 10 ms. + 10 ms. + 2 ms. + 2 ms + 10 ms + 10 ms + 2 ms
+ 2 ms + 10 ms = 72 ms.
0.15. The processing delay in each router is 100 us. The queuing delay is
specified as 0.5 ms. The transmission delay is L/ R = (5Kbits)/
(2Mbps) = 2.5 ms, and the propagation delay is d/ s = (5 x 104)/ (2 x
108) = 0.25 ms. Therefore, the total delay is 2.5 ms +0.1ms + 0.5 ms
+2.5 ms + 0.25 ms + 0.1 ms + 0.5 ms + 2.5 ms + 0.25 ms + 0.1 ms +
0.5 ms + 2.5 ms = 12.3 ms.
pg. 4
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
Solutions Manual For Introduction to Computer Networks
and Cybersecurity 1st Edition By Chwan-Hwa (John) Wu;
J. David Irwin 9781138071896 ALL Chapters
pg. 1
, SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
Introduction to Computer Networks and Cybersecurity Chwan-Hwa
(John) Wu Auburn University J. David Irwin
Chapter 0
0.1. SM may not be capable of meeting an unusual peak demand for
bandwidth since SM is targeted at normal, average bandwidth
operation. Therefore, an ISP may set different classes of services
and let only the high-paying customers have Internet access when
there is a traffic jam.
0.2. Transmission delay results from finite bandwidth in a
transmission line. Propagation delay is due to the fact that
electromagnetic waves are limited by the speed of light. Any bit
of information will encounter both delays on a path from
transmitter to receiver.
0.3.The Overhead = (100 + 4)/ (1000 + 100 +4) = 9.42%
0.4.The Overhead = (80 + 4)/ (100 + 80 + 4) = 45.65%
0.5.The Overhead = (60 + 4)/ (100 + 60 + 4) = 39.02%
0.6. The total delays = round trip delay for establishing a TCP connection +
sending the file =
2* 1 ms + 1000 * 8/ (1536000) + 1 ms = 2 + 5.2
ms + 1 ms = 8.2 ms The Overhead = 2* 1 ms/
23.83 ms = 24.39%
pg. 2
,SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
0.7. The processing delay in each router is 0.001 seconds. The
queuing delay is 0. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 0.007 seconds, and the propagation delay is d/ s = (2 x
105)/ (2 x 108) = 0.001 seconds. Therefore, the total delay is 7 ms
+1 ms. + 0 + 7 ms. + 1 ms. + 1 ms + 0 + 7 ms = 24 ms.
0.8. The processing delay in each router is 1 ms. The queuing delay
is specified as 5 ms. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 7 ms, and the propagation delay is d/ s = (2 x 105)/ (2
x 108) = 1 ms. Therefore, the total delay is 7 ms +1 ms + 5 ms +7
ms + 1 ms + 1 ms + 5 ms + 7 ms = 34 ms.
0.9. The processing delay in each router is 0.002 seconds. The
queuing delay is 0. 002 seconds. The transmission delay is L/ R =
(10Kbits)/ (1Mbps) = 0.010 seconds, and the propagation delay is
d/ s = (2 x 106)/ (2 x 108) = 0.01 seconds. Therefore, the total
delay is 10 ms + 2 ms + 2 ms + 10 ms. + 10 ms. + 2 ms + 2 ms +
10 ms = 48 ms.
0.10. The processing delay in each router is 100 µs. The queuing
delay is specified as 0.5 ms. The transmission delay is L/ R =
(5Kbits)/ (2Mbps) = 2.5 ms, and the propagation delay is d/ s =
(5 x 104)/ (2 x 108) = 0.25 ms. Therefore, the total delay is 2.5
ms +0.1ms + 0.5 ms +2.5 ms + 0.25 ms + 0.1ms + 0.5 ms +2.5
ms + 0.25 ms = 9.2 ms.
0.11. The processing delay in each router is 400 ns. The queuing delay is
specified as 800 ns. The transmission delay is L/ R = (3.1Kbits)/
(155Mbps) = 20 µs, and the propagation delay is d/ s
pg. 3
, SOLUTIONS MANUAL FOR INTRODUCTION TO COMPUTER NETWORKS AND CYBERSECURITY 1ST
EDITION BY CHWAN-HWA (JOHN) WU; J. DAVID IRWIN 9781138071896 ALL CHAPTERS
= (5 x 103)/ (2 x 108) = 25 µs. Therefore, the total delay is 20 µs + 400
ns + 800 ns +20 µs + 25 µs + 400 ns + 800 ns + 20 µs = 87.4 µs.
0.12. The processing delay in each router is 0.001 seconds. The queuing
delay is 0. The transmission delay is L/ R = (7Kbits)/ (1Mbps) =
0.007 seconds, and the propagation delay is d/ s = (2 x 105)/ (2 x 108)
= 0.001 seconds. Therefore, the total delay is 7 ms +1 ms. + 0 + 7
ms. + 1 ms. + 1 ms +0 +7 ms + 1ms + 1 ms + 0 + 7 ms = 33 ms.
0.13. The processing delay in each router is 1 ms. The queuing delay is
specified as 5 ms. The transmission delay is L/ R = (7Kbits)/
(1Mbps) = 7 ms, and the propagation delay is d/ s = (2 x 105)/ (2 x
108) = 1 ms. Therefore, the total delay is 7 ms +1 ms + 5 ms +7 ms +
1 ms + 1 ms + 5 ms + 7 ms + 1 ms + 1 ms + 5 ms + 7 ms = 48 ms.
0.14. The processing delay in each router is 0.002 seconds. The queuing
delay is 0. 002 seconds. The transmission delay is L/ R = (10Kbits)/
(1Mbps) = 0.010 seconds, and the propagation delay is d/ s = (2 x
106)/ (2 x 108) = 0.01 seconds. Therefore, the total delay is 10 ms +2
ms + 2 ms + 10 ms. + 10 ms. + 2 ms. + 2 ms + 10 ms + 10 ms + 2 ms
+ 2 ms + 10 ms = 72 ms.
0.15. The processing delay in each router is 100 us. The queuing delay is
specified as 0.5 ms. The transmission delay is L/ R = (5Kbits)/
(2Mbps) = 2.5 ms, and the propagation delay is d/ s = (5 x 104)/ (2 x
108) = 0.25 ms. Therefore, the total delay is 2.5 ms +0.1ms + 0.5 ms
+2.5 ms + 0.25 ms + 0.1 ms + 0.5 ms + 2.5 ms + 0.25 ms + 0.1 ms +
0.5 ms + 2.5 ms = 12.3 ms.
pg. 4
