Solution Manual
Computer Networks: A Systems Approach
By Larry L. Peterson,Bruce S. Davie
6th Edition
,Solutions For Chapter 1
4. We Will Count The Transfer As Completed When The Last Data Bit Arrives At
Its Desti- Nation. 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 (25 Ms) Longer.
(a) 2 Initial Rtt’s (100ms) + 1000kb/1.5mbps (Transmit) + Rtt/2 (Propaga-
Tion = 25ms)
≈ 0.125 + 8mbit/1.5mbps = 0.125 + 5.333 Sec = 5.658 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 = 5.461 Sec, For A Total Delay Of
5.586 Sec.
(b) To The Above We Add The Time For 999 Rtts (The Number Of Rtts
Be- Tween When Packet 1 Arrives And Packet 1000 Arrives), For A Total
Of 5.586 +
49.95 = 55.536.
(c) This Is 49.5 Rtts, Plus The Initial 2, For 2.575 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 +− + 2n = 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 .575 Sec.
5. The Answer Is In The Book.
6. Propagation Delay Is 4× 103 M/(2× 108 M/Sec) = 2× 10−5 Sec = 20 µs. 100 Bytes/20
µs Is 5 Bytes/µs, Or 5 Mb/Sec, Or 40 Mbit/Sec. For 512-Byte Packets, This
Rises To
204.8 Mbit/Sec.
7. The Answer Is In The Book.
8. Postal Addresses Are Strongly Hierarchical (With A Geographical Hierarchy,
Which Network Addressing May Or May Not Use). Addresses Also Provide
Embedded “Routing Information”. Unlike Typical Network Addresses, Postal
Addresses Are Long And Of Variable Length And Contain A Certain Amount Of
Redundant Informa- Tion. This Last Attribute Makes Them More Tolerant Of
Minor Errors And Incon- Sistencies. 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.
9. 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 Addresses Used By Ethernet. Other Address Attributes That
Might Be Relevant Are Fixed-Length V. Variable-Length, And Absolute V.
Relative (Like File Names).
, 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). Non-
Unique Addresses Are Also Useful When Global Reachability Is Not Required,
Such As To Address The Computers Within A Single Corporation When Those
Computers Cannot Be Reached From Outside The Corporation.
10. 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 Between 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.
11. Stdm And Fdm Both Work Best For Channels With Constant And Uniform
Band- Width Requirements. For Both Mechanisms Bandwidth That Goes
Unused By One Channel Is Simply Wasted, Not Available To Other Channels.
Computer Communi- Cations 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
As- Signed Bandwidth) Well In Advance. Again, The Connection Requirements
For Com- Puting 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.
12. 10 Gbps = 1010 Bps, Meaning Each Bit Is 10−10 Sec (0.1 Ns) Wide. The Length
In The Wire Of Such A Bit Is .1 Ns × 2.3 × 108 M/Sec = 0.023 M Or 23mm
13. 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.
14. (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×1 Gb/Sec = 2.57gb = 321 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/1000mbps = 200mb/1000mbps = 0.2 Sec
To Finish Sending, For A Total Time Of 0.2 + 2.57 = 2.77 Sec Until The
Last Picture Bit
Arrives On Earth.
, 15. The Answer Is In The Book.
16. (A) Delay-Sensitive; The Messages Exchanged Are Short.
(b) Bandwidth-Sensitive, Particularly For Large Files. (Technically This Does
Pre- Sume That The Underlying Protocol Uses A Large Message Size Or
Window Size; Stop-And-Wait Transmission (As In Section 2.5 Of The
Text) With A Small Mes- Sage Size Would Be Delay-Sensitive.)
(c) Delay-Sensitive; Directories Are Typically Of Modest Size.
(d) Delay-Sensitive; A File’s Attributes Are Typically Much Smaller Than The
File Itself (Even On Nt Filesystems).
17. (A) One Packet Consists Of 12000 Bits, And So Is Delayed Due To
Bandwidth 120 µs Along Each Link. The Packet Is Also Delayed 10 µs On
Each Of The Two Links Due To Propagation Delay, For A Total Of 260 µs.
(b) With Three Switches And Four Links, The Delay Is
4 × 120µs + 4 × 10µs = 520µs
(c) With Cut-Through, The Switch Delays The Packet By 200 Bits = 2 µs.
There Is Still One 120 µs Delay Waiting For The Last Bit, And 20 µs Of
Propagation Delay, So The Total Is 142 µs. To Put It Another Way, The
Last Bit Still Arrives 120 µs After The First Bit; The First Bit Now Faces
Two Link Delays And One Switch Delay But Never Has To Wait For The
Last Bit Along The Way. With Three Cut-Through Switches, The Total
Delay Would Be:
120 + 3 × 2 + 4 × 10 = 600 µs
18. The Answer Is In The Book.
19. (A) The Effective Bandwidth Is 10 Mbps; The Sender Can Send Data Steadily
At This Rate And The Switches Simply Stream It Along The Pipeline. We
Are As- Suming Here That No Acks Are Sent, And That The Switches Can
Keep Up And Can Buffer At Least One Packet.
(b) The Data Packet Takes 2.04 Ms As In 17(B) Above To Be Delivered; The
400 Bit Acks Take 40 µs/Link For A×Total Of 4× 40 µs+4 10 µs = 200
µsec = 0.20 Ms, For A Total Rtt Of 2.24 Ms. 5000 Bits In 2.24 Ms Is
About 2.2 Mbps, Or 280 Kb/Sec.
(c) 100 × 6.5× 108 Bytes / 12 Hours = 6.5× 1010 Bytes/(12× 3600 Sec)≈ 1.5 Mbyte/Sec
= 12 Mbit/Sec
20. (A) 1×107bits/Sec × 10−6 Sec = 100 Bits = 12.5 Bytes
(b) The First-Bit Delay Is 520 µs Through The Store-And-Forward Switch, As In
17(A). 107bits/Sec × 520 ×10−6 Sec = 5200 Bits. Alternatively, Each Link
Can Hold 100 Bits And The Switch Can Hold 5000 Bits.
(c) 1.5×106 Bits/Sec × 50 × 10−3 Sec = 75,000 Bits = 9375 Bytes
Computer Networks: A Systems Approach
By Larry L. Peterson,Bruce S. Davie
6th Edition
,Solutions For Chapter 1
4. We Will Count The Transfer As Completed When The Last Data Bit Arrives At
Its Desti- Nation. 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 (25 Ms) Longer.
(a) 2 Initial Rtt’s (100ms) + 1000kb/1.5mbps (Transmit) + Rtt/2 (Propaga-
Tion = 25ms)
≈ 0.125 + 8mbit/1.5mbps = 0.125 + 5.333 Sec = 5.658 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 = 5.461 Sec, For A Total Delay Of
5.586 Sec.
(b) To The Above We Add The Time For 999 Rtts (The Number Of Rtts
Be- Tween When Packet 1 Arrives And Packet 1000 Arrives), For A Total
Of 5.586 +
49.95 = 55.536.
(c) This Is 49.5 Rtts, Plus The Initial 2, For 2.575 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 +− + 2n = 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 .575 Sec.
5. The Answer Is In The Book.
6. Propagation Delay Is 4× 103 M/(2× 108 M/Sec) = 2× 10−5 Sec = 20 µs. 100 Bytes/20
µs Is 5 Bytes/µs, Or 5 Mb/Sec, Or 40 Mbit/Sec. For 512-Byte Packets, This
Rises To
204.8 Mbit/Sec.
7. The Answer Is In The Book.
8. Postal Addresses Are Strongly Hierarchical (With A Geographical Hierarchy,
Which Network Addressing May Or May Not Use). Addresses Also Provide
Embedded “Routing Information”. Unlike Typical Network Addresses, Postal
Addresses Are Long And Of Variable Length And Contain A Certain Amount Of
Redundant Informa- Tion. This Last Attribute Makes Them More Tolerant Of
Minor Errors And Incon- Sistencies. 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.
9. 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 Addresses Used By Ethernet. Other Address Attributes That
Might Be Relevant Are Fixed-Length V. Variable-Length, And Absolute V.
Relative (Like File Names).
, 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). Non-
Unique Addresses Are Also Useful When Global Reachability Is Not Required,
Such As To Address The Computers Within A Single Corporation When Those
Computers Cannot Be Reached From Outside The Corporation.
10. 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 Between 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.
11. Stdm And Fdm Both Work Best For Channels With Constant And Uniform
Band- Width Requirements. For Both Mechanisms Bandwidth That Goes
Unused By One Channel Is Simply Wasted, Not Available To Other Channels.
Computer Communi- Cations 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
As- Signed Bandwidth) Well In Advance. Again, The Connection Requirements
For Com- Puting 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.
12. 10 Gbps = 1010 Bps, Meaning Each Bit Is 10−10 Sec (0.1 Ns) Wide. The Length
In The Wire Of Such A Bit Is .1 Ns × 2.3 × 108 M/Sec = 0.023 M Or 23mm
13. 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.
14. (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×1 Gb/Sec = 2.57gb = 321 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/1000mbps = 200mb/1000mbps = 0.2 Sec
To Finish Sending, For A Total Time Of 0.2 + 2.57 = 2.77 Sec Until The
Last Picture Bit
Arrives On Earth.
, 15. The Answer Is In The Book.
16. (A) Delay-Sensitive; The Messages Exchanged Are Short.
(b) Bandwidth-Sensitive, Particularly For Large Files. (Technically This Does
Pre- Sume That The Underlying Protocol Uses A Large Message Size Or
Window Size; Stop-And-Wait Transmission (As In Section 2.5 Of The
Text) With A Small Mes- Sage Size Would Be Delay-Sensitive.)
(c) Delay-Sensitive; Directories Are Typically Of Modest Size.
(d) Delay-Sensitive; A File’s Attributes Are Typically Much Smaller Than The
File Itself (Even On Nt Filesystems).
17. (A) One Packet Consists Of 12000 Bits, And So Is Delayed Due To
Bandwidth 120 µs Along Each Link. The Packet Is Also Delayed 10 µs On
Each Of The Two Links Due To Propagation Delay, For A Total Of 260 µs.
(b) With Three Switches And Four Links, The Delay Is
4 × 120µs + 4 × 10µs = 520µs
(c) With Cut-Through, The Switch Delays The Packet By 200 Bits = 2 µs.
There Is Still One 120 µs Delay Waiting For The Last Bit, And 20 µs Of
Propagation Delay, So The Total Is 142 µs. To Put It Another Way, The
Last Bit Still Arrives 120 µs After The First Bit; The First Bit Now Faces
Two Link Delays And One Switch Delay But Never Has To Wait For The
Last Bit Along The Way. With Three Cut-Through Switches, The Total
Delay Would Be:
120 + 3 × 2 + 4 × 10 = 600 µs
18. The Answer Is In The Book.
19. (A) The Effective Bandwidth Is 10 Mbps; The Sender Can Send Data Steadily
At This Rate And The Switches Simply Stream It Along The Pipeline. We
Are As- Suming Here That No Acks Are Sent, And That The Switches Can
Keep Up And Can Buffer At Least One Packet.
(b) The Data Packet Takes 2.04 Ms As In 17(B) Above To Be Delivered; The
400 Bit Acks Take 40 µs/Link For A×Total Of 4× 40 µs+4 10 µs = 200
µsec = 0.20 Ms, For A Total Rtt Of 2.24 Ms. 5000 Bits In 2.24 Ms Is
About 2.2 Mbps, Or 280 Kb/Sec.
(c) 100 × 6.5× 108 Bytes / 12 Hours = 6.5× 1010 Bytes/(12× 3600 Sec)≈ 1.5 Mbyte/Sec
= 12 Mbit/Sec
20. (A) 1×107bits/Sec × 10−6 Sec = 100 Bits = 12.5 Bytes
(b) The First-Bit Delay Is 520 µs Through The Store-And-Forward Switch, As In
17(A). 107bits/Sec × 520 ×10−6 Sec = 5200 Bits. Alternatively, Each Link
Can Hold 100 Bits And The Switch Can Hold 5000 Bits.
(c) 1.5×106 Bits/Sec × 50 × 10−3 Sec = 75,000 Bits = 9375 Bytes
