計算機網(wǎng)絡(luò)雙語課件第4章答案

AnswerofHWChapter4AnswerofHWChapter41R12:Whatisthe32-bitbinaryequivalentoftheaddress223.1.3.27?11011111000000010000001100011100R12:Whatisthe32-bitbinary2R13:DoroutershaveIPaddresses?Ifso,howmany?Yes.TheyhaveoneaddressforeachinterfaceR13:DoroutershaveIPaddres3R18:Yes,becausetheentireIPv6datagram(includingheaderfields)isencapsulatedinanIPv4datagramR18:Yes,becausetheentireIP4R21:Linkstatealgorithms:Computestheleast-costpathbetweensourceanddestinationusingcomplete,globalknowledgeaboutthenetwork.Distance-vectorrouting:Thecalculationoftheleast-costpathiscarriedoutinaniterative,distributedmanner.Anodeonlyknowstheneighbortowhichitshouldforwardapacketinordertoreachgivendestinationalongtheleast-costpath,andthecostofthatpathfromitselftothedestination.R21:Linkstatealgorithms:Com5AnswerofHWP447P8a)PrefixMatch LinkInterface111000000000000001110000000000001 111100000 2111000013otherwise 4b)Prefixmatchforfirstaddressis4thentry:linkinterface4Prefixmatchforsecondaddressis2ndentry:linkinterface2Prefixmatchforfirstaddressis3rdentry:linkinterface3AnswerofHWP447P86AnswerofHWP448P9DestinationAddressRange LinkInterface0000~~001100100~~011111000~~101121100~~11113numberofaddressesineachrange=AnswerofHWP448P97AnswerofHWP44810Addressinterface10000000through10111111(64addresses) 011000000through11011111(32addresses)111100000through11111111(32addresses)200000000through01111111(128addresses)3PrefixMatchInterface101111112otherwise3AnswerofHWP448108P448P11Subnet1:2000211

=2048Subnet2:1000210=1024Subnet3:1000210

=1024220.2.11110000.00000000220.2.11110000.00000000throughsubnet1220.2.11110111.11111111220.2.11111000.00000000throughsubnet2220.2.11111011.11111111220.2.11111100.00000000throughsubnet3220.2.11111111.11111111P448P119P448P12DestinationAddressLinkInterface200.23.16/210200.23.24/241200.23.24/212otherwise3P448P12DestinationAddress10P448P13Destinationaddressrangeinterface224.0/160224.0/161224/82225/83otherwise4P448P13Destinationaddress11P448P14101.101.101.64/26101.101.101.01000000through

101.101.101.01111111AnyIPaddressinrange101.101.101.64to101.101.101.127insubnetwithprefix101.101.101.64/26Fourequalsizesubnetsfromtheblockofaddressesoftheform101.101.128.0/17101.101.10000000.0000000022

=4subnets

101.101.10000000.00000000101.101.128/19

101.101.10100000.00000000101.101.160/19

101.101.11000000.00000000101.101.192/19

101.101.11100000.00000000101.101.224/19

P448P14101.101.101.64/2612P449Problem15214.97.254/23

214.97.11111110.00000000SubnetA:214.97.254.0/24(28=

256addresses)

214.97.11111110.00000000

SubnetB:214.97.255.0/25-214.97.255.120/29(27=128addresses-8=120)

214.97.11111111.00000000214.97.11111111.01111000through-through214.97.11111111.01111111214.97.11111111.01111111SubnetC:214.97.255.128/25(27=128addresses)214.97.11111111.10000000SubnetD:214.97.255.120/31(2addresses)214.97.11111111.01111000SubnetE:214.97.255.122/31(2addresses)214.97.11111111.01111010SubnetF:214.97.255.124/30(4addresses)214.97.11111111.01111100

P449Problem15214.97.254/2313Tosimplifythesolution,assumethatnodatagramshaverouterinterfacesasultimatedestinations.Also,labelD,E,Ffortheupper-right,bottom,andupper-leftinteriorsubnets,respectively.Forthesecondschedule,wehave:Router1LongestPrefixMatch OutgoingInterface110101100110000111111111SubnetA1101011001100001111111100000000SubnetD110101100110000111111110000001SubnetF

Router2LongestPrefixMatch OutgoingInterface110101100110000111111111000001SubnetF1101011001100001111111100000001SubnetE1101011001100001111111101SubnetCRouter3LongestPrefixMatch OutgoingInterface1101011001100001111111110000000SubnetD1101011001100001111111100 SubnetB1101011001100001111111100000001 SubnetETosimplifythesolution,assu14計算機網(wǎng)絡(luò)雙語ppt課件第4章答案15計算機網(wǎng)絡(luò)雙語ppt課件第4章答案16P449Problem18a)Homeaddresses:192.168.0.1,192.168.0.2,192.168.0.3withtherouterinterfacebeing192.168.0.4b)NATTranslationTableWANSideLANSide128.119.40.86,4000192.168.0.1,3345128.119.40.86,4001192.168.0.1,3346128.119.40.86,4002192.168.0.2,3445128.119.40.86,4003192.168.0.2,3446128.119.40.86,4004192.168.0.3,3545128.119.40.86,4005192.168.0.3,3546

P449Problem18a)Homeaddress17P449Problem19Itisnotpossibletodevisesuchatechnique.InordertoestablishadirectTCPconnectionbetweenArnoldandBernard,eitherArnoldorBobmustinitiateaconnectiontotheother.ButtheNATscoveringArnoldandBobdropSYNpacketsarrivingfromtheWANside.ThusneitherArnoldnorBobcaninitiateaTCPconnectiontotheotheriftheyarebothbehindNATs.

P449Problem19Itisnotpos18P22Step N’ D(s),p(s) D(t),p(t) D(u),p(u) D(v),p(v) D(w),p(w) D(y),p(y) D(z),p(z)1x∞

∞∞8,x6,x6,x∞2xy∞

15,y∞7,y6,x18,y3xyw∞

15,y14,w7,y18,y4xywv∞

11,v10,v18,y5xywvu14,u

11,v18,y6xywvut12,t

16,t7xywvuts16,t