Solutions Manual For
Computer Networks: A Systems Approach
Author: Larry L. Peterson,Bruce S. Davie
6th Edition
,Chapter 1 1
Solutions For Chapter 1
3. Success Here Depends Largely On The Ability Of Ones Search Tool To Separate Out
The Chaff. I Thought A Naive Search For Ethernet Would Be Hardest, But I Now Think
It’s Mpeg.
Mbone Www.Mbone.Com
Atm
Www.Atmforum.Co
m
Mpeg Try Searching For “Mpeg Format”, Or (1999)
Drogo.Cselt.Stet.It/Mpeg Ipv6 Playground.Sun.Com/Ipng,
Www.Ipv6.Com
Ethernet Good Luck.
5. We Will Count The Transfer As Completed When The Last Data Bit Arrives At Its
Destination. An Alternative Interpretation Would Be To Count Until The Last Ack Arrives
Back At The Sender, In Which Case The Time Would Be Half An Rtt (50 Ms) Longer.
(a) 2 Initial Rtt’s (200ms) + 1000kb/1.5mbps (Transmit) + Rtt/2 (Propagation)
≈ 0.25 + 8mbit/1.5mbps = 0.25 + 5.33 Sec = 5.58 Sec. If We Pay More Careful
Attention To When A Mega Is 106 Versus 220, We Get 8,192,000 Bits/1,500,000
Bits/Sec = 5.46 Sec, For A Total Delay Of 5.71 Sec.
(b) To The Above We Add The Time For 999 Rtts (The Number Of Rtts Between When
Packet 1 Arrives And Packet 1000 Arrives), For A Total Of 5.71 + 99.9 = 105.61.
(c) This Is 49.5 Rtts, Plus The Initial 2, For 5.15 Seconds.
(d) Right After The Handshaking Is Done We Send One Packet. One Rtt After The
Handshaking We Send Two Packets. At N Rtts Past The Initial Handshaking We
Have Sent 1 + 2 + 4 +
+2· ·n· = 2n+1 1 Packets.
− At N = 9 We Have Thus Been Able To Send All 1,000 Packets;
The Last Batch Arrives 0.5 Rtt Later. Total Time Is 2+9.5 Rtts, Or 1.15 Sec.
6. The Answer Is In The Book.
7. Propagation Delay Is 2× 103 M/(2× 108 M/Sec) = × 1 10−5 Sec = 10 µs. 100 Bytes/10 µs
Is 10 Bytes/µs, Or 10 Mb/Sec, Or 80 Mbit/Sec. For 512-Byte Packets, This Rises To 409.6
Mbit/Sec.
8. The Answer Is In The Book.
9. Postal Addresses Are Strongly Hierarchical (With A Geographical Hierarchy, Which
Network Ad- Dressing May Or May Not Use). Addresses Also Provide Embedded “Routing
Information”. Un- Like Typical Network Addresses, Postal Addresses Are Long And Of
Variable Length And Contain A Certain Amount Of Redundant Information. This Last
Attribute Makes Them More Tolerant Of Minor Errors And Inconsistencies. Telephone
Numbers Are More Similar To Network Addresses (Although Phone Numbers Are
Nowadays Apparently More Like Network Host Names Than Ad- Dresses): They Are
(Geographically) Hierarchical, Fixed-Length, Administratively Assigned, And In More-Or-
Less One-To-One Correspondence With Nodes.
,2 Chapter 1
10. One Might Want Addresses To Serve As Locators, Providing Hints As To How Data
Should Be Routed. One Approach For This Is To Make Addresses Hierarchical.
Another Property Might Be Administratively Assigned, Versus, Say, The Factory-Assigned
Ad- Dresses Used By Ethernet. Other Address Attributes That Might Be Relevant Are
Fixed-Length V. Variable-Length, And Absolute V. Relative (Like File Names).
, Chapter 1 3
If You Phone A Toll-Free Number For A Large Retailer, Any Of Dozens Of Phones May
Answer. Arguably, Then, All These Phones Have The Same Non-Unique “Address”. A
More Traditional Application For Non-Unique Addresses Might Be For Reaching Any Of
Several Equivalent Servers (Or Routers).
11. Video Or Audio Teleconference Transmissions Among A Reasonably Large Number Of
Widely Spread Sites Would Be An Excellent Candidate: Unicast Would Require A Separate
Connection Be- Tween Each Pair Of Sites, While Broadcast Would Send Far Too Much
Traffic To Sites Not Interested In Receiving It.
Trying To Reach Any Of Several Equivalent Servers Or Routers Might Be Another Use For
Multicast, Although Broadcast Tends To Work Acceptably Well For Things On This Scale.
12. Stdm And Fdm Both Work Best For Channels With Constant And Uniform Bandwidth
Require- Ments. For Both Mechanisms Bandwidth That Goes Unused By One Channel Is
Simply Wasted, Not Available To Other Channels. Computer Communications Are Bursty
And Have Long Idle Periods; Such Usage Patterns Would Magnify This Waste.
Fdm And Stdm Also Require That Channels Be Allocated (And, For Fdm, Be Assigned
Band- Width) Well In Advance. Again, The Connection Requirements For Computing
Tend To Be Too Dynamic For This; At The Very Least, This Would Pretty Much Preclude
Using One Channel Per Connection.
Fdm Was Preferred Historically For Tv/Radio Because It Is Very Simple To Build Receivers;
It Also Supports Different Channel Sizes. Stdm Was Preferred For Voice Because It Makes
Somewhat More Efficient Use Of The Underlying Bandwidth Of The Medium, And
Because Channels With Different Capacities Was Not Originally An Issue.
13. 1 Gbps = 109 Bps, Meaning Each Bit Is 10−9 Sec (1 Ns) Wide. The Length In The Wire Of
Such A Bit Is 1 Ns × 2.3 × 108 M/Sec = 0.23 M
14. X Kb Is 8 × 1024 × X Bits. Y Mbps Is Y × 106 Bps; The Transmission Time Would Be 8 × 1024 ×
X/Y × 106 Sec = 8.192x/Y Ms.
15. (A) The Minimum Rtt Is 2 × 385, 000, 000 M / 3×108 M/Sec = 2.57 Sec.
(b) The Delay×Bandwidth Product Is 2.57 Sec×100 Mb/Sec = 257mb = 32 Mb.
(c) This Represents The Amount Of Data The Sender Can Send Before It Would Be
Possible To Receive A Response.
(d) We Require At Least One Rtt Before The Picture Could Begin Arriving At The
Ground (Tcp Would Take Two Rtts). Assuming Bandwidth Delay Only, It Would
Then Take 25mb/100mbps = 200mb/100mbps = 2.0 Sec To Finish Sending, For A
Total Time Of
2.0 + 2.57 = 4.57 Sec Until The Last Picture Bit Arrives On Earth.
16. The Answer Is In The Book.
17. (A) Delay-Sensitive; The Messages Exchanged Are Short.
(b) Bandwidth-Sensitive, Particularly For Large Files. (Technically This Does Presume
That The Underlying Protocol Uses A Large Message Size Or Window Size; Stop-
And-Wait Transmis- Sion (As In Section 2.5 Of The Text) With A Small Message Size
Would Be Delay-Sensitive.)
Computer Networks: A Systems Approach
Author: Larry L. Peterson,Bruce S. Davie
6th Edition
,Chapter 1 1
Solutions For Chapter 1
3. Success Here Depends Largely On The Ability Of Ones Search Tool To Separate Out
The Chaff. I Thought A Naive Search For Ethernet Would Be Hardest, But I Now Think
It’s Mpeg.
Mbone Www.Mbone.Com
Atm
Www.Atmforum.Co
m
Mpeg Try Searching For “Mpeg Format”, Or (1999)
Drogo.Cselt.Stet.It/Mpeg Ipv6 Playground.Sun.Com/Ipng,
Www.Ipv6.Com
Ethernet Good Luck.
5. We Will Count The Transfer As Completed When The Last Data Bit Arrives At Its
Destination. An Alternative Interpretation Would Be To Count Until The Last Ack Arrives
Back At The Sender, In Which Case The Time Would Be Half An Rtt (50 Ms) Longer.
(a) 2 Initial Rtt’s (200ms) + 1000kb/1.5mbps (Transmit) + Rtt/2 (Propagation)
≈ 0.25 + 8mbit/1.5mbps = 0.25 + 5.33 Sec = 5.58 Sec. If We Pay More Careful
Attention To When A Mega Is 106 Versus 220, We Get 8,192,000 Bits/1,500,000
Bits/Sec = 5.46 Sec, For A Total Delay Of 5.71 Sec.
(b) To The Above We Add The Time For 999 Rtts (The Number Of Rtts Between When
Packet 1 Arrives And Packet 1000 Arrives), For A Total Of 5.71 + 99.9 = 105.61.
(c) This Is 49.5 Rtts, Plus The Initial 2, For 5.15 Seconds.
(d) Right After The Handshaking Is Done We Send One Packet. One Rtt After The
Handshaking We Send Two Packets. At N Rtts Past The Initial Handshaking We
Have Sent 1 + 2 + 4 +
+2· ·n· = 2n+1 1 Packets.
− At N = 9 We Have Thus Been Able To Send All 1,000 Packets;
The Last Batch Arrives 0.5 Rtt Later. Total Time Is 2+9.5 Rtts, Or 1.15 Sec.
6. The Answer Is In The Book.
7. Propagation Delay Is 2× 103 M/(2× 108 M/Sec) = × 1 10−5 Sec = 10 µs. 100 Bytes/10 µs
Is 10 Bytes/µs, Or 10 Mb/Sec, Or 80 Mbit/Sec. For 512-Byte Packets, This Rises To 409.6
Mbit/Sec.
8. The Answer Is In The Book.
9. Postal Addresses Are Strongly Hierarchical (With A Geographical Hierarchy, Which
Network Ad- Dressing May Or May Not Use). Addresses Also Provide Embedded “Routing
Information”. Un- Like Typical Network Addresses, Postal Addresses Are Long And Of
Variable Length And Contain A Certain Amount Of Redundant Information. This Last
Attribute Makes Them More Tolerant Of Minor Errors And Inconsistencies. Telephone
Numbers Are More Similar To Network Addresses (Although Phone Numbers Are
Nowadays Apparently More Like Network Host Names Than Ad- Dresses): They Are
(Geographically) Hierarchical, Fixed-Length, Administratively Assigned, And In More-Or-
Less One-To-One Correspondence With Nodes.
,2 Chapter 1
10. One Might Want Addresses To Serve As Locators, Providing Hints As To How Data
Should Be Routed. One Approach For This Is To Make Addresses Hierarchical.
Another Property Might Be Administratively Assigned, Versus, Say, The Factory-Assigned
Ad- Dresses Used By Ethernet. Other Address Attributes That Might Be Relevant Are
Fixed-Length V. Variable-Length, And Absolute V. Relative (Like File Names).
, Chapter 1 3
If You Phone A Toll-Free Number For A Large Retailer, Any Of Dozens Of Phones May
Answer. Arguably, Then, All These Phones Have The Same Non-Unique “Address”. A
More Traditional Application For Non-Unique Addresses Might Be For Reaching Any Of
Several Equivalent Servers (Or Routers).
11. Video Or Audio Teleconference Transmissions Among A Reasonably Large Number Of
Widely Spread Sites Would Be An Excellent Candidate: Unicast Would Require A Separate
Connection Be- Tween Each Pair Of Sites, While Broadcast Would Send Far Too Much
Traffic To Sites Not Interested In Receiving It.
Trying To Reach Any Of Several Equivalent Servers Or Routers Might Be Another Use For
Multicast, Although Broadcast Tends To Work Acceptably Well For Things On This Scale.
12. Stdm And Fdm Both Work Best For Channels With Constant And Uniform Bandwidth
Require- Ments. For Both Mechanisms Bandwidth That Goes Unused By One Channel Is
Simply Wasted, Not Available To Other Channels. Computer Communications Are Bursty
And Have Long Idle Periods; Such Usage Patterns Would Magnify This Waste.
Fdm And Stdm Also Require That Channels Be Allocated (And, For Fdm, Be Assigned
Band- Width) Well In Advance. Again, The Connection Requirements For Computing
Tend To Be Too Dynamic For This; At The Very Least, This Would Pretty Much Preclude
Using One Channel Per Connection.
Fdm Was Preferred Historically For Tv/Radio Because It Is Very Simple To Build Receivers;
It Also Supports Different Channel Sizes. Stdm Was Preferred For Voice Because It Makes
Somewhat More Efficient Use Of The Underlying Bandwidth Of The Medium, And
Because Channels With Different Capacities Was Not Originally An Issue.
13. 1 Gbps = 109 Bps, Meaning Each Bit Is 10−9 Sec (1 Ns) Wide. The Length In The Wire Of
Such A Bit Is 1 Ns × 2.3 × 108 M/Sec = 0.23 M
14. X Kb Is 8 × 1024 × X Bits. Y Mbps Is Y × 106 Bps; The Transmission Time Would Be 8 × 1024 ×
X/Y × 106 Sec = 8.192x/Y Ms.
15. (A) The Minimum Rtt Is 2 × 385, 000, 000 M / 3×108 M/Sec = 2.57 Sec.
(b) The Delay×Bandwidth Product Is 2.57 Sec×100 Mb/Sec = 257mb = 32 Mb.
(c) This Represents The Amount Of Data The Sender Can Send Before It Would Be
Possible To Receive A Response.
(d) We Require At Least One Rtt Before The Picture Could Begin Arriving At The
Ground (Tcp Would Take Two Rtts). Assuming Bandwidth Delay Only, It Would
Then Take 25mb/100mbps = 200mb/100mbps = 2.0 Sec To Finish Sending, For A
Total Time Of
2.0 + 2.57 = 4.57 Sec Until The Last Picture Bit Arrives On Earth.
16. The Answer Is In The Book.
17. (A) Delay-Sensitive; The Messages Exchanged Are Short.
(b) Bandwidth-Sensitive, Particularly For Large Files. (Technically This Does Presume
That The Underlying Protocol Uses A Large Message Size Or Window Size; Stop-
And-Wait Transmis- Sion (As In Section 2.5 Of The Text) With A Small Message Size
Would Be Delay-Sensitive.)
