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Computer Networks: A Systems ApproachFourth EditionSolutions ManualLarry Peterson and Bruce Davie20071Dear Instructor:This Instructors’ Manual contains solutions to most of the exercises in the fourth edi-tion of Peterson and Davie’s Computer Networks: A Systems Approach.Exercises are sorted (roughly) by section, not difficulty. While some exercises aremore 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 sidebarsummarizing much of the applicable basic probability theory in Chapter 2.An occasional exercise is awkwardly or ambiguously worded in the text. This manualsometimes suggests better versions; also see the errata at the web site.Where appropriate, relevant supplemental files for these solutions (e.g. programs) havebeen placed on the textbook web site, www.mkp.com/pd4e. Useful other materialcan also be found there, such as errata, sample programming assignments, PowerPointlecture slides, and EPS figures.If you have any questions about these support materials, please contact your Mor-gan Kaufmann sales representative. If you would like to contribute your own teach-ing materials to this site, please contact our Associate Editor Rachel Roumeliotis,[email protected] welcome bug reports and suggestions as to improvements for both the exercisesand the solutions; these may be sent to [email protected] PetersonBruce DavieFebruary, 2007Chapter 11Solutions for Chapter 13. Success here depends largely on the ability of one’s search tool to separate outthe chaff. The following are representative examplesMbone http://antc.uoregon.edu/MBONED/ATM http://www.mfaforum.org/MPEG http://www.m4if.org/IPv6 http://www.ipv6forum.org/Ethernet http://standards.ieee.org5. 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 arrivesback 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 (propaga-tion)≈ 0.25 + 8Mbit/1.5Mbps = 0.25 + 5.33 sec = 5.58 sec. If we pay morecareful attention to when a mega is 106versus 220, we get8,192,000bits/1,500,000bits/sec = 5.46 sec, for a total delay of 5.71sec.(b) To the above we add the time for 999 RTTs (the number of RTTs betweenwhen 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 thehandshaking we send two packets. At n RTTs past the initial handshakingwe have sent 1 + 2 + 4 + ···+ 2n= 2n+1−1 packets. At n = 9 we havethus 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× 103m/(2×108m/sec) = 1×10−5sec = 10µs. 100bytes/10µsis 10 bytes/µs, or 10MB/sec, or 80 Mbit/sec. For 512-byte packets, this rises to409.6 Mbit/sec.8. The answer is in the book.9. Postal addresses are strongly hierarchical (with a geographical hierarchy, whichnetwork addressing may or may not use). Addresses also provide embedded“routing information”. Unlike typical network addresses, postal addresses arelong 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 (althoughphone numbers are nowadays apparently more like network host names than ad-dresses): they are (geographically) hierarchical, fixed-length, administrativelyassigned, and in more-or-less one-to-one correspondence with nodes.Chapter 1210. One might want addresses to serve as locators, providing hints as to how datashould 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 berelevant are fixed-length v. variable-length, and absolute v. relative (like filenames).If you phone a toll-free number for a large retailer, any of dozens of phones mayanswer. Arguably, then, all these phones have the same non-unique “address”. Amore traditional application for non-unique addresses might be for reaching anyof several equivalent servers (or routers).11. Video or audio teleconference transmissions among a reasonably large numberof widely spread sites would be an excellent candidate: unicast would require aseparate connection between each pair of sites, while broadcast would send fartoo much traffic to sites not interested in receiving it.Trying to reach any of several equivalent servers or routers might be another usefor multicast, although broadcast tends to work acceptably well for things on thisscale.12. STDM and FDM both work best for channels with constant and uniform band-width requirements. For both mechanisms bandwidth that goes unused by onechannel is simply wasted, not available to other channels. Computer communi-cations are bursty and have long idle periods; such usage patterns would magnifythis 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 muchpreclude using one channel per connection.FDM was preferred historically for TV/radio because it is very simple to buildreceivers; it also supports different channel sizes. STDM was preferred for voicebecause it makes somewhat more efficient use of the underlying bandwidth ofthe medium, and because channels with different capacities was not originallyan issue.13. 1 Gbps = 109bps, meaning each bit is 10−9sec (1ns) wide. The length in thewire of such a bit is 1 ns × 2.3 × 108m/sec = 0.23 m14. x KB is 8 × 1024 × x bits. y Mbps is y × 106bps; the transmission time wouldbe 8 × 1024 × x/y × 106sec = 8.192x/y ms.15. (a) The minimum RTT is 2 × 385, 000, 000 m / 3×108m/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 bepossible to receive a response.Chapter 13(d)
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