Solutions Manual For Computer Networking Bible: [3 In 1] The Complete Crash Course To Effectively Design, Implement And Manage Networks. Including Sections On Security, Performance And Scalability 2024 By Rick C. Worley (Author) ALL Chapters

Solutions Manual For Computer Networking Bible: [3 In
1] The Complete Crash Course To Effectively Design,
Implement And Manage Networks. Including Sections
On Security, Performance And Scalability 2024
By Rick C. Worley (Author) ALL Chapters



.

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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 Tr Ip 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%

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 I S 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

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= (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.

0.16. 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
= (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 + 25 µs + 400 Ns + 800 Ns + 20 µs = 133.6 µs.

0.17. The Following Time Chart Outlines The Solution. Time Is Measured In Ms.
Star T Time Arrival Buffer Channel Arrival Buffer Host C

0 B1 00000
1 A1 00000 B1
2 B2 0000A1 B1
3 A2 0000B2 A1 B1
4 B3 000A2B2 A1 B1 00000
5 A3 000B3A2 B2 A1 B1
6 B4 00A3B3A2 B2 A1 00000
7 A4 000B4A3B3 A2 B2 A1
As The Time Chart Indicates, The Host C Receives The Packet B1 In 5 Ms.

0.18. As Shown In The Solution To Problem 11, The Packet A1 Reaches Host C In 7 Ms.

0.19. The Following Chart Outlines The Solution. Time Is Measured In Ms.