Solutions Manual
Computer Networks: A Systems Approach
5th Edition
By Larry Peterson and Bruce Davie
, Chapter 1 1
Solutions for Chapter 1
3. 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 = 5.458 sec. If we pay more
careful attention to when a mega is 106 versus 220, we get
8,192,000 bits/1,500,000 bps = 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 between 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.
4. The answer is in the book.
5. Propagation delay is 4 103 m/(2 10 × 8 m/s) = 2× 10−5 sec = 20
× µs. 100 bytes/20 µs is 5
bytes/µs, or 5 MBps, or 40 Mbps. For 512-byte packets, this rises to 204.8 Mbps.
6. The answer is in the book.
7. 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 inconsis- tencies. Telephone
numbers, at least those assigned to landlines, are more sim- ilar to network addresses:
they are (geographically) hierarchical, fixed-length, administratively assigned, and in
more-or-less one-to-one correspondence with nodes.
8. 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).
, Chapter 1 2
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.
9. 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. Delivery of video and audio streams for a television
channel only to those households currently interested in watching that channel is another
application.
Trying to reach any of several equivalent servers, each of which can provide the answer
to some query, would be another possible use, although the receiver of many responses
to the query would need to deal with the possibly large volume of responses.
10. 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.
11. 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
12. 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.
13. (a) The minimum RTT is 2 × 385, 000, 000 m / 3×108 m/s = 2.57 seconds.
(b) The delay×bandwidth product is 2.57 s×1 Gbps = 2.57Gb = 321 MB.
(c) This represents the amount of data the sender can send before it would be possible
to receive a response.
, Chapter 1 3
(d) We require at least one RTT from sending the request before the first bit of the
picture could begin arriving at the ground (TCP would take longer). 25 MB is
200Mb. Assuming bandwidth delay only, it would then take 200Mb/1000Mbps =
0.2 seconds to finish sending, for a total time of 0.2 +
2.57 = 2.77 sec until the last picture bit arrives on earth.
14. The answer is in the book.
15. (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.
16. (a) On a 100 Mbps network, each bit takes 1/108 = 10 ns to transmit. One packet
consists of 12000 bits, and so is delayed due to bandwidth (serial- ization) by 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.
17. The answer is in the book.
18. (a) The effective bandwidth is 100 Mbps; the sender can send data steadily at this
rate and the switches simply stream it along the pipeline. We are assuming 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 520 µs as in 16(b) above to be delivered; the 400 bit ACKs
take 4 µs/link to be sent back, plus propagation, for a total of 4 4 µs ×
+4 10 µs = 56 ×µs; thus the total RTT is 576 µs. 12000 bits in 576 µs is
about 20.8 Mbps.
× / 12
(c) 100 4.7 109 bytes × hours = 4.7 1011 bytes/(12
× 3600 s) × MBps ≈
10.9
= 87 Mbps.
19. (a) 100×106bps × 10 × 10−6 sec = 1000 bits = 125 bytes.
Computer Networks: A Systems Approach
5th Edition
By Larry Peterson and Bruce Davie
, Chapter 1 1
Solutions for Chapter 1
3. 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 = 5.458 sec. If we pay more
careful attention to when a mega is 106 versus 220, we get
8,192,000 bits/1,500,000 bps = 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 between 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.
4. The answer is in the book.
5. Propagation delay is 4 103 m/(2 10 × 8 m/s) = 2× 10−5 sec = 20
× µs. 100 bytes/20 µs is 5
bytes/µs, or 5 MBps, or 40 Mbps. For 512-byte packets, this rises to 204.8 Mbps.
6. The answer is in the book.
7. 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 inconsis- tencies. Telephone
numbers, at least those assigned to landlines, are more sim- ilar to network addresses:
they are (geographically) hierarchical, fixed-length, administratively assigned, and in
more-or-less one-to-one correspondence with nodes.
8. 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).
, Chapter 1 2
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.
9. 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. Delivery of video and audio streams for a television
channel only to those households currently interested in watching that channel is another
application.
Trying to reach any of several equivalent servers, each of which can provide the answer
to some query, would be another possible use, although the receiver of many responses
to the query would need to deal with the possibly large volume of responses.
10. 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.
11. 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
12. 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.
13. (a) The minimum RTT is 2 × 385, 000, 000 m / 3×108 m/s = 2.57 seconds.
(b) The delay×bandwidth product is 2.57 s×1 Gbps = 2.57Gb = 321 MB.
(c) This represents the amount of data the sender can send before it would be possible
to receive a response.
, Chapter 1 3
(d) We require at least one RTT from sending the request before the first bit of the
picture could begin arriving at the ground (TCP would take longer). 25 MB is
200Mb. Assuming bandwidth delay only, it would then take 200Mb/1000Mbps =
0.2 seconds to finish sending, for a total time of 0.2 +
2.57 = 2.77 sec until the last picture bit arrives on earth.
14. The answer is in the book.
15. (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.
16. (a) On a 100 Mbps network, each bit takes 1/108 = 10 ns to transmit. One packet
consists of 12000 bits, and so is delayed due to bandwidth (serial- ization) by 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.
17. The answer is in the book.
18. (a) The effective bandwidth is 100 Mbps; the sender can send data steadily at this
rate and the switches simply stream it along the pipeline. We are assuming 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 520 µs as in 16(b) above to be delivered; the 400 bit ACKs
take 4 µs/link to be sent back, plus propagation, for a total of 4 4 µs ×
+4 10 µs = 56 ×µs; thus the total RTT is 576 µs. 12000 bits in 576 µs is
about 20.8 Mbps.
× / 12
(c) 100 4.7 109 bytes × hours = 4.7 1011 bytes/(12
× 3600 s) × MBps ≈
10.9
= 87 Mbps.
19. (a) 100×106bps × 10 × 10−6 sec = 1000 bits = 125 bytes.
