peterson davie

7/28/2019 peterson davie

1/48

Log InSign Up

Browse

Download

Add Note

Link

Embed

Save

of 122

Readcast
https://www.scribd.com/loginhttps://www.scribd.com/loginhttps://www.scribd.com/loginhttp://www.scribd.com/explorehttp://www.scribd.com/explorehttps://www.scribd.com/loginhttps://www.scribd.com/login


2/48


3/48

Chapter 1730. (a) A file server

needs lots of peak

bandwidth. Latencyis relevant only if it

dominates bandwidth;

jitter and averagebandwidth are

inconsequential.No

lost data is acceptable,


4/48

but without real-time

requirements we cansim-ply retransmit lost

data.(b) A print server

needs less bandwidth

than a file server

(unless imagesareextremelylarge).

We may be willing to

accept higherlatency


5/48

than (a), also.(c) A

file serverisa digital library of a

sort, but in general the

world wide webgets

along reasonably wellwith much less peak

bandwidth than most

fileservers provide.(d)Forinstrumentmonito


6/48

ringwedontcareabou

tlatencyorjitter.Ifdatawerecontinually

generated, rather than

bursty, we might be

concerned

mostlywith averagebandwidthratherthanpea

k, andif the datareally

were routinewe mightjust accept a certain


7/48

fraction of loss.(e) For

voice we needguaranteed average

bandwidth and bounds

on latencyand

jitter. Some lost data

might be acceptable;e.g. resulting in

minordropouts many

secondsapart.(f) Forvideowea


8/48

reprimarilyconcerned

withaveragebandwidth. Forthesim-ple

monitoring application

here, relatively modest

video of Exercise

28(b)mightsuffice; we could

even go to

monochrome


9/48

(1bit/pixel), at

whichpoint 160

120

5frames/sec requires12KB/sec. We could

tolerate multi-secondlatency delays; the


10/48

primary restriction is

that if the monitoringre-vealed a need for

intervention then we

still have time to

act. Considerableloss,

even of entire frames,would be

acceptable.(g) Full-

scale televisionrequires massive


11/48

bandwidth. Latency,

however, couldbehours. Jitter would be

limited only by our

capacity absorb the

arrival-time variations

by buffering. Someloss would

be acceptable, but

largelosses would bevisually annoying.31.


12/48

In STDM the offered

timeslices are alwaysthe same length, and

are wasted if they are

unused by

the assigned

station. The round-robin access

mechanismwould

generally give eachstation only as much


13/48

time as it needed to

transmit,or none if thestation had nothing to

send, and so network

utilization would

beexpected to be

much higher.32. (a) Inthe absence of

any packet losses or

duplications, when weare expectingthe


14/48

N

th packet weget

theNth packet, and so we

can keep track ofNlocally at the

receiver.(b) The


15/48

scheme outlined here

is the stop-and-waitalgorithm of Section

2.5;as is indicated

there, a header with at

least one bit of

sequence numberisneeded (to

distinguish between

receiving a new packetand a duplication


16/48

of the previous

packet).(c) With out-of-order delivery

allowed, packets up to

1 minute apart must

bedistinguishable via

sequence number.Otherwise a very old

packet might

Chapter 1


17/48

8arrive and be

accepted ascurrent. Sequence

numbers would have

to countas high as

bandwidth

1minute/packet size


18/48

33. In each case we

assume the local clockstarts at

1000.(a) Latency: 100.

Bandwidth: high

enough to read the

clock every 1unit.1000 11001001

11011002 1102100

3 1104 tiny bit of jitter: latency = 101


19/48

1004 1104(b) Latenc

y=100;bandwidth: only

enough to read the

clock every 10

units.Arrival times

fluctuate due tojitter.1000 11001 0 2

0 1 1 1 0 l a t e n c y =

9 0 1040 11451 0 6 01 1 8 0 l a t e n c y = 1


20/48

2 0 1080 1184(c) Lat

ency = 5; zero jitterhere:1000 10051001

10061003 1008

w e l o s t 1 0 0 2 1004

10091005 101034.

Generally, with MAXPENDING =1, one or

two connections will

be acceptedandqueued; that is, the


21/48

data wont be

delivered to the server.The others will

beignored; eventually

they will time

out.When the first

client exits, anyqueued connections

are processed.36. Note

that UDP accepts apacket of data from


22/48

any source at any

time; TCP re-quires anadvance connection.

Thus, two clients can

now talk

simultaneously;their

messages will beinterleaved on the

server.


23/48

Solutions for

Chapter 21.

BitsNRZClockManchesterNRZI1 0


24/48

0 1 1 1 1 1 0

0 0 1 0 0 0 1 2. See the figure

below.B i t s 1 1 10 0 0 1 0 1 11 1 1 1 0 1 01 0 1 N R Z I3. T

he answer is in thebook.4. One can list

all 5-bitsequencesand

count,but here is


25/48

anotherapproach:

there are23

sequencesthatstartwit

h00,and

23thatendwith00. There

aretwosequences,00000and 00100,that do


26/48

both. Thus, the

numberthat do eitheris

8+82 = 14,and finally the

number that do neither

is


27/48

32

14 = 18. Thus there wouldhavebeen enough 5-

bit codes meeting thestronger requirement;

however,

additionalcodes are

needed for control


28/48

sequences.5. The

stuffed bits (zeros) arein bold:1101 0111 1100 1011 1110

1010 1111 1011 06. The


29/48

marks each position

where a stuffed 0 bitwas removed. There

were nostuffing errors

detectable by the

receiver; the only such

error the receivercouldidentify would be

seven 1s in a row.1101

0111 11


30/48

10 1111 1

010 1111 1

1107. The answer is in

the book.8. ..., DLE,

DLE, DLE, ETX,

ETX9. (a) X DLE Y,where X can be

anything besides DLE

and Y can beanythingexcept DLE


31/48

or ETX. In other

words, each DLE mustbe followed by

eitherDLE or ETX.(b)

0111 1111.9


32/48

10. (a) After 48


33/48

8=384 bits we can be

off by no more than

1/2 bit, which isabout

1 part in 800.(b)

Oneframeis 810bytes; atSTS-

151.8Mbpsspeedwea

resending51.8


34/48

106

/(8

810)= about 8000frames/sec, or about

480,000frames/minute. Thus,


35/48

if stationBs clock ran

faster than station Asby one part in

480,000, A would

accu-mulate about one

extra frame

per minute.11. Supposeanundetectablethre

e-

biterroroccurs. Thethreebadbitsmustbespre


36/48

adamong one, two, or

three rows. If thesebits occupy two or

three rows, then

somerow must have

exactly one bad bit,

which wouldbe detected by the

parity bit forthat

row. But if the threebits are all in one row,


37/48


38/48

be

correct. Furthermore,if four bits changeand

no error is detected,

then the bad bits must

form a rectangle: in

order for theerror to goundetected, each row

and column must have

no errors or exactlytwoerrors.13. If we


39/48

know only one bit is

bad, then 2-D paritytells us which row and

column itis in, and

we can then flip it. If,

however, two bits are

bad in the same row,thenthe row parity

remains correct, and

all we can identify isthe columns in


40/48

whichthe bad bits

occur.14. We need toshow that the 1s-

complement sum of

two non-0x0000

numbersis non-

0x0000. If nounsigned overflow

occurs, then the sum

is just the 2s-complementsum andc


41/48

antbe 0000withoutov

erflow;in the absenceofoverflow,addition is

monotonic. If overflow

occurs, then the result

is at least 0x0000

plusthe addition of acarry bit, i.e.


42/48

0x0001.15. Lets

define swap([A,B]) =[B,A], where A and B

are one byte each. We

onlyneed to show [A,

B] + [C, D] =

swap([B, A] + [D,C]). If both (A+C)

and(B+D) have no

carry, the equationobviously holds.If


43/48

A+C has a carry and

B+D+1 does not,[A,B] + [C, D] = [(A+C)

& 0xEF,

B+D+1]swap([B, A]

+ [D, C]) =

swap([B+D+1, (A+C)& 0xEF]) = [(A+C)&

0xEF, B+D+1](The

case where B+D+1 hasalso a carry is similar


44/48

to the last case.)If

B+D has a carry, andA+C+1 does not,[A,

B] + [C, D] =

[A+C+1, (B+D) &

0xEF].swap([B, A] +

[D, C]) =swap([(B+D) & 0xEF,

A+C+1]) =

[A+C+1,(B+D) &0xEF].10


45/48

Recommended

107 p.

Peterson Solutions

nikhat88

8282 Reads

108 p.

Pd3esolutions Manual

api_user_11797_P Doshi9827 Reads
http://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/nikhat88http://www.scribd.com/nikhat88http://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/api_user_11797_P%20Doshihttp://www.scribd.com/api_user_11797_P%20Doshihttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/111352315/Solution-Manual-for-Computer-Networks-by-Larry-L-Peterson-Bruce-S-Daviehttp://www.scribd.com/api_user_11797_P%20Doshihttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/doc/7026088/Pd3esolutions-Manualhttp://www.scribd.com/nikhat88http://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/20472262/Peterson-Solutionshttp://www.scribd.com/doc/20472262/Peterson-Solutions


46/48


47/48


48/48

About

About Scribd

Team

Blog

Join our team!

Contact Us

Premium Premium Reader

Scribd StoreAdvertise with us

Get started

AdChoicesSupport

Help

FAQ

PressPartners

Developers / APILegal

Terms Privacy

CopyrightGet Scribd Mobile

Copyright 2013 Scribd Inc.

Language:

English
http://www.scribd.com/abouthttp://www.scribd.com/about/teamhttp://blog.scribd.com/http://www.scribd.com/jobshttp://www.scribd.com/contacthttp://www.scribd.com/archive/plans?metadata=%7B%22platform%22%3A%22web%22%2C%22logged_in%22%3Afalse%2C%22context%22%3A%22landingpage%22%2C%22action%22%3A%22footer%22%7Dhttp://www.scribd.com/explore/storehttp://www.scribd.com/advertisehttp://www.scribd.com/adchoiceshttp://support.scribd.com/loginhttp://www.scribd.com/faqhttp://www.scribd.com/static/presshttp://www.scribd.com/developershttp://www.scribd.com/termshttp://www.scribd.com/privacyhttp://www.scribd.com/copyrighthttp://www.scribd.com/doc/45905524/Computer-Networks-Peterson-and-Davie-4e-Solution-Manualhttp://www.scribd.com/doc/45905524/Computer-Networks-Peterson-and-Davie-4e-Solution-Manualhttps://play.google.com/store/apps/details?id=com.scribd.app.reader0&hl=enhttps://itunes.apple.com/us/app/scribd-worlds-largest-online/id542557212?mt=8https://play.google.com/store/apps/details?id=com.scribd.app.reader0&hl=enhttps://itunes.apple.com/us/app/scribd-worlds-largest-online/id542557212?mt=8http://www.scribd.com/doc/45905524/Computer-Networks-Peterson-and-Davie-4e-Solution-Manualhttp://www.scribd.com/copyrighthttp://www.scribd.com/privacyhttp://www.scribd.com/termshttp://www.scribd.com/developershttp://www.scribd.com/static/presshttp://www.scribd.com/faqhttp://support.scribd.com/loginhttp://www.scribd.com/adchoiceshttp://www.scribd.com/advertisehttp://www.scribd.com/explore/storehttp://www.scribd.com/archive/plans?metadata=%7B%22platform%22%3A%22web%22%2C%22logged_in%22%3Afalse%2C%22context%22%3A%22landingpage%22%2C%22action%22%3A%22footer%22%7Dhttp://www.scribd.com/contacthttp://www.scribd.com/jobshttp://blog.scribd.com/http://www.scribd.com/about/teamhttp://www.scribd.com/about

peterson davie

Documents

Transcript of peterson davie