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SOLUTIONS MANUAL
INSTRUCTOR PASSWORD FOR NETWORK SIMULATION EXPERIMENTS MANUAL
Anyone who purchases the 3rd edition of Computer Networks: A Systems Approach
has access to the online Network Simulation Experiments Manual
(http://www3.us.elsevierhealth.com/MKP/aboelela) for 6 months.
We are providing instructors with a generic password that will work past this 6-
month period of time under the condition that this password is not distributed
by instructors to students or professionals.
We appreciate your discretion. Note: a print version of the Manual is available
from the publisher for purchase with unexpiring access to the simulation
software for $19.95 (ISBN: 0120421712).
Password: CONE3INST007 (second character is a letter "O", middle character is a
letter "I")
Dear Instructor:
This Instructors’ Manual contains solutions to most of the exercises in the third edition of Peterson
and Davie’s Computer Networks: A Systems Approach.
Exercises are sorted (roughly) by section, not difficulty. While some exercises are more difficult than
others, none are intended to be fiendishly tricky. A few exercises (notably, though not exclusively,
the ones that involve calculating simple probabilities) require a modest amount of mathematical
background; most do not. There is a sidebar summarizing much of the applicable basic probability
theory in Chapter 2.
An occasional exercise is awkwardly or ambiguously worded in the text. This manual sometimes
suggests better versions; also see the errata at the web site.
Where appropriate, relevant supplemental files for these solutions (e.g. programs) have been placed
on the textbook web site, www.mkp.com/pd3e. Useful other material can also be found there,
such as errata, sample programming assignments, PowerPoint lecture slides, and EPS figures.
If you have any questions about these support materials, please contact your Morgan Kaufmann
sales representative. If you would like to contribute your own teaching materials to this site, please
contact Karyn Johnson, Morgan Kaufmann Editorial Department, kjohnson@mkp.com.
We welcome bug reports and suggestions as to improvements for both the exercises and the solu-
tions; these may be sent to netbugs@mkp.com.
Larry Peterson
Bruce Davie
May, 2003
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.com
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 (50ms) 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 10
6
versus 2
20
, we get 8,192,000 bits/1,500,000 bits/sec = 5.46sec,
for a total delay of 5.71sec.
(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
= 2
n+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 × 10
3
m/(2 × 10
8
m/sec) = 1 × 10
−5
sec = 10 µs. 100bytes/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.
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).
2
Chapter 1
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 = 10
9
bps, meaning each bit is 10
−9
sec (1ns) wide. The length in the wire of such a
bit is 1ns × 2.3 × 10
8
m/sec = 0.23 m
14. x KB is 8 ×1024 ×x bits. y Mbps is y ×10
6
bps; the transmission time would be 8 ×1024×
x/y × 10
6
sec = 8.192x/y ms.
15. (a) The minimum RTT is 2 × 385, 000, 000 m / 3×10
8
m/sec = 2.57 sec.
(b) The delay×bandwidth product is 2.57sec×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.)
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