中科大計算機網絡05課件

Chapter5Objectives

Uponcompletionyouwillbeableto:IPAddresses:

ClasslessAddressing

UnderstandtheconceptofclasslessaddressingBeabletofindthefirstandlastaddressgivenanIPaddressBeabletofindthenetworkaddressgivenaclasslessIPaddressBeabletocreatesubnetsfromablockofclasslessIPaddressesUnderstandaddressallocationandaddressaggregation1TCP/IPProtocolSuite5.1VARIABLE-LENGTHBLOCKSInclasslessaddressingvariable-lengthblocksareassignedthatbelongtonoclass.Inthisarchitecture,theentireaddressspace(2^32addresses)isdividedintoblocksofdifferentsizes.Thetopicsdiscussedinthissectioninclude:Restrictions可以解決很多問題，整個體系能夠工作得有些限制FindingtheBlock尋找地址塊GrantedBlock分配到的地址塊2TCP/IPProtocolSuiteFigure5.1

Variable-lengthblocks可變長度塊不同大小的地址塊塊地址數：一個地址塊中的地址數必須是2的整數次冪開始地址：開始地址必須能被地址數整除3TCP/IPProtocolSuiteWhichofthefollowingcanbethebeginningaddressofablockthatcontains16addresses?a.205.16.37.32 b.190.16.42.44

c.17.17.33.80 d.123.45.24.52Example1Solution

Onlytwoareeligible(aandc).Theaddress205.16.37.32iseligiblebecause32isdivisibleby16.Theaddress17.17.33.80iseligiblebecause80isdivisibleby16.4TCP/IPProtocolSuiteWhichofthefollowingcanbethebeginningaddressofablockthatcontains256addresses?a.205.16.37.32 b.190.16.42.0

c.17.17.32.0 d.123.45.24.52Example2Solution

Inthiscase,theright-mostbytemustbe0.AswementionedinChapter4,theIPaddressesusebase256arithmetic.Whentheright-mostbyteis0,thetotaladdressisdivisibleby256.Onlytwoaddressesareeligible(bandc).5TCP/IPProtocolSuiteWhichofthefollowingcanbethebeginningaddressofablockthatcontains1024addresses?a.205.16.37.32 b.190.16.42.0

c.17.17.32.0 d.123.45.24.52Example3Solution

Inthiscase,weneedtochecktwobytesbecause

1024=4×256.Theright-mostbytemustbedivisibleby256.Thesecondbyte(fromtheright)mustbedivisibleby4.Onlyoneaddressiseligible(c).6TCP/IPProtocolSuiteFigure5.2

Formatofclasslessaddressingaddressn的意義：表示該地址塊中的每一個地址都有n個比特是相同的（從左至右，稱地址前綴）無分類標志斜線記法也叫CIDR

記法7TCP/IPProtocolSuiteTable5.1Prefixlengths8TCP/IPProtocolSuiteClassfuladdressingisaspecialcaseofclasslessaddressing.Note:9TCP/IPProtocolSuiteWhatisthefirstaddressintheblockifoneoftheaddressesis167.199.170.82/27?Example4Addressinbinary:101001111100011110101

Keeptheleft27bits:101001111100011110101010

01000000

ResultinCIDRnotation:167.199.170.64/27Solution

Theprefixlengthis27,whichmeansthatwemustkeepthefirst27bitsasisandchangetheremainingbits(5)to0s.Thefollowingshowstheprocess:10TCP/IPProtocolSuiteWhatisthefirstaddressintheblockifoneoftheaddressesis140.120.84.24/20?Example5SeeNextSlideSolution

Figure5.3showsthesolution.Thefirst,second,andfourthbytesareeasy;forthethirdbytewekeepthebitscorrespondingtothenumberof1sinthatgroup.Thefirstaddressis140.120.80.0/20.11TCP/IPProtocolSuiteFigure5.3

Example512TCP/IPProtocolSuiteFindthefirstaddressintheblockifoneoftheaddressesis140.120.84.24/20.Example6SeeNextSlideSolution

Thefirst,second,andfourthbytesareasdefinedinthepreviousexample.Tofindthethirdbyte,wewrite84asthesumofpowersof2andselectonlytheleftmost4(mis4)asshowninFigure5.4.Thefirstaddressis140.120.80.0/20.13TCP/IPProtocolSuiteFigure5.4

Example614TCP/IPProtocolSuiteFindthenumberofaddressesintheblockifoneoftheaddressesis140.120.84.24/20.Example7Solution

Theprefixlengthis20.Thenumberofaddressesintheblockis232?20or212or4096.Notethat

thisisalargeblockwith4096addresses.15TCP/IPProtocolSuiteUsingthefirstmethod,findthelastaddressintheblockifoneoftheaddressesis140.120.84.24/20.Example8SeeNextSlideSolution

Wefoundinthepreviousexamplesthatthefirstaddressis140.120.80.0/20andthenumberofaddressesis4096.Tofindthelastaddress,weneedtoadd4095(4096?1)tothefirstaddress.16TCP/IPProtocolSuiteTokeeptheformatindotted-decimalnotation,weneedtorepresent4095inbase256(seeAppendixB)anddothecalculationinbase256.Wewrite4095as15.255.Wethenaddthefirstaddresstothisnumber(inbase255)toobtainthelastaddressasshownbelow:Example8(Continued)

140.120.80.0

15.255

-------------------------

140.120.95.255Thelastaddressis140.120.95.255/20.17TCP/IPProtocolSuiteUsingthesecondmethod,findthelastaddressintheblockifoneoftheaddressesis140.120.84.24/20.Example9SeeNextSlideSolution

Themaskhastwenty1sandtwelve0s.Thecomplementofthemaskhastwenty0sandtwelve1s.Inotherwords,themaskcomplementis0011111111

or0.0.15.255.Weaddthemaskcomplementtothebeginningaddresstofindthelastaddress.18TCP/IPProtocolSuite

140.120.80.0

0.0.15.255

----------------------------

140.120.95.255Example9(Continued)Weaddthemaskcomplementtothebeginningaddresstofindthelastaddress.Thelastaddressis140.120.95.255/20.19TCP/IPProtocolSuiteFindtheblockifoneoftheaddressesis190.87.140.202/29.Example10SeeNextSlideSolution

Wefollowtheprocedureinthepreviousexamplestofindthefirstaddress,thenumberofaddresses,andthelastaddress.Tofindthefirstaddress,wenoticethatthemask(/29)hasfive1sinthelastbyte.Sowewritethelastbyteaspowersof2andretainonlytheleftmostfiveasshownbelow:20TCP/IPProtocolSuite202 ?128+64+0+0+8+0+2+0Theleftmost5numbersare?128+64+0+0+8Thefirstaddressis190.87.140.200/29Example10(Continued)Thenumberofaddressesis232?29or8.Tofindthelastaddress,weusethecomplementofthemask.Themaskhastwenty-nine1s;thecomplementhasthree1s.Thecomplementis0.0.0.7.Ifweaddthistothefirstaddress,weget190.87.140.207/29.Inotherwords,thefirstaddressis190.87.140.200/29,thelastaddressis190.87.140.207/20.Thereareonly8addressesinthisblock.21TCP/IPProtocolSuiteShowanetworkconfigurationfortheblockinthepreviousexample.Example11SeeNextSlideSolution

Theorganizationthatisgrantedtheblockinthepreviousexamplecanassigntheaddressesintheblocktothehostsinitsnetwork.However,thefirstaddressneedstobeusedasthenetworkaddressandthelastaddressiskeptasaspecialaddress(limitedbroadcastaddress).Figure5.5showshowtheblockcanbeusedbyanorganization.Notethatthelastaddressendswith207,whichisdifferentfromthe255seeninclassfuladdressing.22TCP/IPProtocolSuiteFigure5.5

Example1123TCP/IPProtocolSuiteInclasslessaddressing,thelastaddressintheblockdoesnotnecessarilyendin255.Note:24TCP/IPProtocolSuiteInCIDRnotation,theblockgrantedisdefinedbythefirstaddressandtheprefix（前綴）length.Note:25TCP/IPProtocolSuite5.2SUBNETTINGWhenanorganizationisgrantedablockofaddresses,itcancreatesubnetstomeetitsneeds.Theprefixlengthincreasestodefinethesubnetprefixlength.Thetopicsdiscussedinthissectioninclude:FindingtheSubnetMaskFindingtheSubnetAddressesVariable-LengthSubnets26TCP/IPProtocolSuiteInfixed-lengthsubnetting,thenumberofsubnetsisapowerof2.Note:27TCP/IPProtocolSuiteAnorganizationisgrantedtheblock130.34.12.64/26.Theorganizationneeds4subnets.Whatisthesubnetprefixlength?Example12Solution

Weneed4subnets,whichmeansweneedtoaddtwomore1s(log24=2)tothesiteprefix.Thesubnetprefixisthen/28.28TCP/IPProtocolSuiteWhatarethesubnetaddressesandtherangeofaddressesforeachsubnetinthepreviousexample?Example13SeeNextSlideSolution

Figure5.6showsoneconfiguration.29TCP/IPProtocolSuiteFigure5.6

Example1330TCP/IPProtocolSuiteThesitehas232?26=64addresses.Eachsubnethas232–28=16addresses.Nowletusfindthefirstandlastaddressineachsubnet.Example13(Continued)SeeNextSlide1.Thefirstaddressinthefirstsubnetis130.34.12.64/28,usingtheprocedureweshowedinthepreviousexamples.Notethatthefirstaddressofthefirstsubnetisthefirstaddressoftheblock.Thelastaddressofthesubnetcanbefoundbyadding15(16?1)tothefirstaddress.Thelastaddressis130.34.12.79/28.31TCP/IPProtocolSuiteExample13(Continued)2.Thefirstaddressinthesecondsubnetis130.34.12.80/28;itisfoundbyadding1tothelastaddressoftheprevioussubnet.Againadding15tothefirstaddress,weobtainthelastaddress,130.34.12.95/28.3.Similarly,wefindthefirstaddressofthethirdsubnettobe130.34.12.96/28andthelasttobe130.34.12.111/28.4.Similarly,wefindthefirstaddressofthefourthsubnettobe130.34.12.112/28andthelasttobe130.34.12.127/28.32TCP/IPProtocolSuiteAnorganizationisgrantedablockofaddresseswiththebeginningaddress14.24.74.0/24.Thereare232?24=256addressesinthisblock.Theorganizationneedstohave11subnetsasshownbelow:

a.twosubnets,eachwith64addresses.

b.twosubnets,eachwith32addresses.

c.threesubnets,eachwith16addresses.

d.foursubnets,eachwith4addresses.Designthesubnets.Example14SeeNextSlideForOneSolution33TCP/IPProtocolSuiteFigure5.7

Example1434TCP/IPProtocolSuite1.Weusethefirst128addressesforthefirsttwosubnets,eachwith64addresses.Notethatthemaskforeachnetworkis/26.Thesubnetaddressforeachsubnetisgiveninthefigure.2.Weusethenext64addressesforthenexttwosubnets,eachwith32addresses.Notethatthemaskforeachnetworkis/27.Thesubnetaddressforeachsubnetisgiveninthefigure.Example14(Continuted)SeeNextSlide35TCP/IPProtocolSuite3.Weusethenext48addressesforthenextthreesubnets,eachwith16addresses.Notethatthemaskforeachnetworkis/28.Thesubnetaddressforeachsubnetisgiveninthefigure.4.Weusethelast16addressesforthelastfoursubnets,eachwith4addresses.Notethatthemaskforeachnetworkis/30.Thesubnetaddressforeachsubnetisgiveninthefigure.Example14(Continuted)36TCP/IPProtocolSuiteAsanotherexample,assumeacompanyhasthreeoffices:Central,East,andWest.TheCentralofficeisconnectedtotheEastandWestofficesviaprivate,point-to-pointWANlines.Thecompanyisgrantedablockof64addresseswiththebeginningaddress70.12.100.128/26.Themanagementhasdecidedtoallocate32addressesfortheCentralofficeanddividestherestofaddressesbetweenthetwooffices.Figure5.8showstheconfigurationdesignedbythemanagement.Example15SeeNextSlide37TCP/IPProtocolSuiteFigure5.8

Example1538TCP/IPProtocolSuiteThecompanywillhavethreesubnets,oneatCentral,oneatEast,andoneatWest.Thefollowingliststhesubblocksallocatedforeachnetwork:Example15(Continued)SeeNextSlidea.TheCentralofficeusesthenetworkaddress70.12.100.128/27.Thisisthefirstaddress,andthemask/27showsthatthereare32addressesinthisnetwork.Notethatthreeoftheseaddressesareusedfortheroutersandthecompanyhasreservedthelastaddressinthesub-block.Theaddressesinthissubnetare70.12.100.128/27to70.12.100.159/27.NotethattheinterfaceoftherouterthatconnectstheCentralsubnettotheWANneedsnoaddressbecauseitisapoint-to-pointconnection.39TCP/IPProtocolSuiteExample15(Continued)SeeNextSlideb.TheWestofficeusesthenetworkaddress70.12.100.160/28.Themask/28showsthatthereareonly16addressesinthisnetwork.Notethatoneoftheseaddressesisusedfortherouterandthecompanyhasreservedthelastaddressinthesub-block.Theaddressesinthissubnetare70.12.100.160/28to70.12.100.175/28.NotealsothattheinterfaceoftherouterthatconnectstheWestsubnettotheWANneedsnoaddressbecauseitisapoint-to-pointconnection.40TCP/IPProtocolSuiteExample15(Continued)c.TheEastofficeusesthenetworkaddress70.12.100.176/28.Themask/28showsthatthereareonly16addressesinthisnetwork.Notethatoneoftheseaddressesisusedfortherouterandthecompanyhasreservedthelastaddressinthesub-block.Theaddressesin.thissubnetare70.12.100.176/28to70.12.100.191/28.NotealsothattheinterfaceoftherouterthatconnectstheEastsubnettotheWANneedsnoaddressbecauseitisapoint-to-pointconnection.41TCP/IPProtocolSuite5.3ADDRESSALLOCATIONAddressallocationistheresponsibilityofaglobalauthoritycalledtheInternetCorporationforAssignedNamesandAddresses(ICANN).ItusuallyassignsalargeblockofaddressestoanISPtobedistributedtoitsInternetusers.42TCP/IPProtocolSuiteAnISPisgrantedablockofaddressesstartingwith190.100.0.0/16(65,536addresses).TheISPneedstodistributetheseaddressestothreegroupsofcustomersasfollows:Example16SeeNextSlidea.Thefirstgrouphas64customers;eachneeds256addresses.

b.Thesecondgrouphas128customers;eachneeds128addresses

c.Thethirdgrouphas128customers;eachneeds64addresses.43TCP/IPProtocolSuiteDesignthesubblocksandfindouthowmanyaddressesarestillavailableaftertheseallocations.Example16(Continued)SeeNextSlideSolution

Figure5.9showsthesituation.44TCP/IPProtocolSuiteFigure5.9

Example1645TCP/IPProtocolSuiteGroup1

Forthisgroup,eachcustomerneeds256addresses.Thismeansthesuffixlengthis8(28=256).Theprefixlengthisthen32?8=24.Theaddressesare:Example16(Continued)SeeNextSlide1stC