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We are providing instructors with a generic password that will work past this 6-

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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 solutions;

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 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 +

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