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by: Ellsworth O'Hara

Internetworking EE281

Ellsworth O'Hara
GPA 3.9


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This 41 page Class Notes was uploaded by Ellsworth O'Hara on Tuesday October 20, 2015. The Class Notes belongs to EE281 at San Jose State University taught by JalelRejeb in Fall. Since its upload, it has received 33 views. For similar materials see /class/225336/ee281-san-jose-state-university in Electrical Engineering at San Jose State University.


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Date Created: 10/20/15
Internetworking Network Layer HT TP FJPwsMTP N V TcP UDP g l i i Data link iayer protocols P Myra l yef ruLuDDlS Introduction amp Outline 0 In the previous chapter we discussed techniques to overcome some ofthe limitations ofLAN ie usage of bridges to extend LAN packets can travel over geographically larger areas with better use of bandwidth learning bridges spanning tree But still we had 7 Heterogeneity LANs of similar networks of similar addresses Billion hosts 9 addressing problem logical addresses In this chapter we examine some approaches to overcome these problems This leads to the design of internetworking internet arbitrary collection of networks or LANS interconnected by routers and other devices not to be confused with Internet We will discuss the Juou aspects and relevant protocols wwc mam the building of the internet feasible and efficient o This chapter consists of two main parts 7 IP Protocol Subnetting and IPv6 7 Routing and Control Protocols EE 2811ntemetworking FARTI IP PROTOCOL amp SUBNETTING EE 2811nta networking Router vs Switch I A ke A1a er in the imA lemeniztion of the internet is the router I Router is a networklayer device Its job is to move data between different network segments by looking into packet headers amp determine the best path for the packet to travel Nmm a NakwmkA Network a EE 2811nta networking lP Internet o Noterouters not like bridges can look into a packet header Example of Internet Using IP Internet Protocol that runs over anything routers can connect network segments that use different technologies It s a key tool to build scalable h rl Hetewge euus netwu ks Journey of Pkt from H1 to H8 Note IP protocol makes it all possible by being platform independent TCP R R2 R3 III I ETH EE 281 Internetworking Pop Quiz Example True or False Routing occurs exclusively at 051 layer 3 and above While switching occurs exclusively at OSI layer 2 and above a True b False Hint39While in routing datagram forwarding decisions are based on OS layer 3 information and in switching datagram forwarding decisions are based on OS layer2 information the information still needs to be passed down to OS layer 1 and transmitted over the physical media as a stream of bits EE 2811ntemetworking Router Hardware 39 Arouter can also be viewed as a s ecial rocessor CPUMemoryBusOS that forwards network traf c along optimized paths It determines these optimized routes throughthe assistance of routing protocols In this chapter many of these protocols are discussed EE 2811ntemetworkmg Internet Architecture TCPIP Suite I De ned by Internet Engineering Task Force IETF I Hourglass Architecture With four layers EE 2811ntemetworkmg TCPIP vs OSI Model BB 281 lntanetwurkmg Internet Protocol IP 0 IP is the primary network protocol of TCPIP suite o All applications depend on IP and IP runs over all networks 39 quot L 39 39 uuueto rw AI39 k Al39 a I o IntheIP layer 7 addressing and subnetting are de ned 7 network routing takes place 7 Fragmentation based no on MTU no reassembly 7 Type of service is speci ed EE 281 lntanetwurkmg IP Datagram Format 32 bin Identi cation 16 bits Fragment Offset 13 bits Header Checksum 16 bits Source IP address 32 bits Destination IP address 32 bits Ethernet Header TCP Header Applica ion data Ethernet Traiier EE 281 Internetworking IP Datagram Format 0x00 44m ooooooooooooo2 Ebff Applicaion data I EthernetTraiier I EE 2811ntemetworking IP Datagram Format 0 15 16 31 version P3348 bit ngsfferm 16 bit total length bytes 16 bit identification I 13 bit fragment offset 8 bit TTL I 8 bit protocol 16 bit header checksum 20 bytes 32 bit source IP address 32 bit destinatioan address Options if present Data EE 281 Internetworking IP Header Version IPv4 for nowother Pv6 larger no need ur 39 header length in 32bits words range 5 words 20B to max 60B total length total length of IP datagram 7 max 216 l 65535 bytes ofwhich 20 to 60 bytes are for the header typeofservice speci es to the router how datagram should be handled 39 39 39 e ay maximize throughput maximize reliability based on the application needs T TL Time to live the maximum number of routers the datagram can traverse Default value is 64 7 each router decrements the eld by l 7 When it reaches 0 the router discards it and sends an error message to the source identification flags fragment o fret control the fragmentation reassembly process reassembly is done at the host more next header checksum only IP header ones complement sum EE 2811ntemetworking IP FragmentationReassembly o Fragmentation is the 7 rocess of dividing a datairam messai e into smaller separate packets based on the MTU Maximum Transmission Unit MTU is a network dependent parameter Its value is set by the datalink layer and passed to IPlayer Example 48B 1500B39 and 4464B are the MIUs of ATM Ethernet and FDDI networks respectively 0 In IP network Fragmentation is done at the router and reassembly the reverse process is done the host 7 is there an advantage or having this setting In IP network an identification number is assigned to all fragments of the same packet indicated in the identi cation ags fragment a set elds 15 EE 2811ntemetworking Question Host H1 sends a packet X with ident x and size 1400B to H8 through the network shown below the MTU in PPP El EEEE segment networks are as indicated in the Figure Illustrate the fragmentation process using a diagram Assume header size is 20B Answer Note DataSize 532B20B512B Offset field all fragments must be multiple of 8 Bytesexcept s st so how many fragments are allowed Check out 8192fragmentsdatagram man I Iii mam fragmentation only at router R2 Reassembly at H8 maulll l 12 data bytes tart ofh ade II E m e we llama E est of heade 76 data bytes EE 2811nternetworking What is an IP Address 0 An IP address is a mi iue ilobal address for a network interface 7 TheIP address or Internet address is 32bit long version4 4 Billions 7 IP addresses are unique and universal 7 however a machine can have more than one IP address IP address is a logical address not the physical address MAC 7 use ARP to map the IP address to MAC address 0 Each IP address consists oftwo elds netid and hartid 7 the netid de nes the network and the hostid de nes the host on that network All hosts in the same network must have the same netid 17 EE 281 Internetworking Dotted DeCI mal N otatl on o For readability LP addresses are often written in dotteddecimal format 7 offour IP address elds each eld runs from 0 to 255 2551111 1111Z 0 Example 1ooooooo 10001111 10001001 1001oooo 1 5 Byte 2nd Byte 3rd Byte 4th Byte 128 143 137 144 128143137144 18 EE 2811nternetworking Internet Address Classes There are five types called classes for IP addressing All are 32bits IclassAIIoI I 14 bits 16 bits 128000 to IClassB 1 I 0 I HCtid I hOStid I mm 7 a 21 bits 8 bits IClassC 1 I 1 I 0 I netid I hostid I 1920oo to 223255255255 28 bits I Class D 1 I 1 I 1 I 0 I Multicast group id I 27 bits 24oooo to I Class E 1 I 1 I 1 I 1 I 0 I reserved for future use I 19 EE 2811nternetworking Problem s With Internet Address Classes Three Big Bear Problem 0 Based on these classes the IP address allows us to have networks from the following combinations of valid network and host addresses class Possible Possible Total x Percent networks Hosts ofA 11 addresses A 126 16777214 2113928964 50 Toofew Hawks B 16382 65534 1073709056 25 Depleted C 2097150 254 532676608 125 Too few hosts 0 To determine the number of possible hosts use the formula 2n 7 2where n is the number of bits in host eld The subtraction of 2 is for special address the local network id and broadcast address 20 EE 2811nternetworking Problems with Classful IP Addresses o The original classful address scheme had a number of problems Problem 1 Too few network addresses for large networks 7 Class A and Class B addresses are gone Problem 2 Twolayer hierarchy is not appropriate for large networks with Class A and Class B addresses Problem 3 Inflexible Assume a company requires 2000 addresses 7 Class A andB addresses are overkill 7 Class C address is insuf cient requires 8 Class C addresses Problem 4 Exploding Routing Tables Routing on the backbone Internet needs to have an entry for each network address In 1993 the size of the routing tables started to outgrow the capacity of routers EE 281 Internetworking Solution to Classful IP addresses The BIG Picture To overcome the IPv4 addressing problems several approaches were developed as summarized below I l I IPv6 l LNATandDHCPj Subnetting lFixedlengthj Variablelength VLSM Next will discuss most of these approaches except the NAT Network Address Translation and DHCP Dynamic Host Configuration Protocols are not part of our IP layer discussion but in later chapters 22 EE 2811ntemetworking Subnetting 0 Problem Organizations have multiple networks which are independently managed University Network Engineering Medical School School Solution 1 Allocate one or more Class C address for each network Difficult to manage From the outside of the organization each network must be addressable Solution 2 Add another level of hierarchy to the 1P addressing structure gt Subnetting EE 281 Internetworking BaSIC Idea of Subnetting 0 Split the host number portion of an 1 address into a subnet number and a smaller host number 0 Result is a 3layer hierarchy hiding local structure network ID host ID network ID subnet ID host ID 0 Then extended network ID 0 Subnets can be freely assigned within the organization 0 Internally subnets are treated as separate networks 0 Subnet structure is not visible outside the organization 0 In xedsubnet all subnets in a given network have the same size number of hosts while in variablelength can have different sizes EE 2811nternetworking Subnet Mask 0 Recall given a host IP address one can determine the corresponding network id by looking at the MSB bits of its IP address 7 for example ifthe first 2 bits are quot10 in the 1P host address then the rst 16 bits of the address indicate its networkid and the last 16 bits indicate its host dress this can e viewe also as natural or defau su ne o In subnetting one needs to look beyond its class to determine its subnet and host id This is done using a SubnetMask o A subnet mask is a 32bit binary number whose bits are set to 7 1 if it corresponds to a bit in the IP address that should be considered as part of the extended network id 7 0 if the system should treat the bit as part of the host id 0 It is the job of subnet mask to extract mask 011 the network and host segments of the address 25 EE 2811nternetworking Example Subnet Mask Two examples of subnets using class B Example 1 8 bits are used to assign hosts 16 bits 8 bits 8 bits 1 Class B Netid Subnetid Hostid Subnetmask 11llllllllllllll llllllll 00000000 Decimal notatio 255 255 255 0 Example 2 6 bits are used to assign hosts 16 bits 10 bits 6 bits 1 Class B Netid Subnetid Hostid SubnetmaskHlllllllllllllllllllllllllll 000000H 255255255192 26 EE 2811nternetworking Example 1 Fixedlength subnetting A that is all subnem have the same number him for the subnetiid and hostiid elds on a network ofl39P network address 1411400 Observe 7 Without subnetting the router needs to store all hostLP addresses which can in the range of 21565536 entries 7 With subnetting only 4 entries 7 We now have different type of routes edge router backbone router m N 1111402 m Il lZ2 llllquot n3 4sts25a r F 1quot erwurk39 MLHILU m M 2m 4 m in llw wst ur Ilw mm 27 BB 281 lntEmetwurkmg Netwo k out Sub emng 1401 in mm land ll in III27M U 3 1 T i SuImLt 5 SnlmM m mm 5 lrllJ LSrLD 1an m n 1 slumc M m2xu HI Nth m u imzn 39iilr I n 110 391 u m w ul Klu lmrnwt twork thubne g 28 BB 281 lntEmetwurkmg Subnetting 0 In ieneral to determine the subnet network address we siin 17 perform bitAND operation of the IP address and subnet mask that is to determine boundary between subnetiid and hostiid Subnet Mask 2552551920 1411141224 AND MilHMO 1 address erwurk nddrvss 0 To determine the range of IP addresses in a given subnet we need to know the rst address of the subnet and the subnet mask EE 2811ntemetworl ng Example 2 o What is the network subnetaddress if the desh39natt39on address is 200 45 3456 and the subnet mask is 25525524007 0 Solution We apply the AND operation on the address and the subnet mask Address 391100100000101101 0010001000111000 SubnetMash 111111111111000000000000 7mm y L 1114 139 139 139 139 EE 2811ntemetworkmg Exa m p e 3 Questions Class B network on the lntemet has a subnet mask of2552552400 What is the maximum number of hosts per subnet assume xed subnetting Answer The lower 16 bits in B class are for subnet and host elds Since the most signi cant 4bits are 1111 240 27 26 25 24 so mask is binary 1111 1111 1111 1111 1111 0000 0000 0000 So that leaves 12 bits for host or total of 212 host 72 4094 hosts 31 EE 2811ntemetworking Subnet Mask and IP address 0 Given its l39P address a host can determine which class it belongs to and the bOLmdary between hostiid and networkiid A B or C using leading bits 9 That is with IP address one can determine the host network 0 Given the subnet mask host can determine boundary between subnetiid and hostiid 9 that is with its own the IP address and subnet mask a host can determine 1 similar hosts in its own subnet 2 a host on a different subnet but on its own network 3 a host on a different network 32 EE 2811ntemetworking Example 4 Assume a host EE2817student was given the IP address l40252ll and the subnet mask 2552552550 What is the class of EE2817student IP address How many bits are assigned for its subnetiid Determine whether the following IP addresses are on the same subnet or network as of the host EE2817student a 14025245 b 140252123 c 1924311 P N Answer EE 281 Internetworking EE 2811ntemetworking Subnet Mask and Routing Therefore network devices need two pieces of information in order to calculate what its host address is and what its subnet and netvvor are l the actual address amp 2 the mask A network address is obtained from applying the bitwise AND operation on the IP address and the mask after determining its class When a host wishes to send a packet it rst performs the AND operation to check if the receiving host is in the sam bne 7 if so the the packet is sent directly to the receiving host 7 otherwise the datagram is sent to the default router Sending HostIP 159215640 SM 2552552550 LAN39 15921560 DeSlIIP 1592156 42 S Desan 159215142 M 2552552550 SM39 552552550 LAN 15921560 LAN 15921570 sending amp receiving hosts are sending amp receiving hosts are 39 e same AN in dif erem LANs EE 2131 Internetworking Fixedlength Subnetting Example 0 Problem Consider an university campus that is granted a class C 1 address sayXYZO needs to arrange 5 departments 5 subnets with the following number of hosts 60 60 60 30 30 Provide a possible subnetting arrangement for this campus 0 Solution 7 in class C we have only 1 byte the 4Lh byte for the host This byte would be splitted into host and subnet elds 7 Using xedlength subnetting that is all subnets have the same number of hosts it would require at least 3bits for the subnet eld 238 to accommodate 5 subnets 7 Which leaves 5 bits for the host eld so each subnet would have only 30 hosts 252 9 not enough for the 60 host department One approach probably the best also too overcome this problem is using VariableLength Subnet Masking VLSM RFC 1009 7 A Later note Rva2 and OSPF both support variablelength subnets but RIP does not EE 2811ntemetworking VariableLength Subnet Masking VLSM Example 0 In VLS con guration we can allow one subnet eld of2 bits 7 So we divide the network into 4 subnets 21 each of 52 hosts 262 3 t rm 7 me 4m by using 3 bits in the Subnet eld so 4 bits rot host 0r242 hosts see gure in the next slide 0 How about the mask address In VLS routers usually needs to 2 or more masks one applied after the other 7 in this example it need 15 one mask to divide the network into four Mas consistson ones 249m oftheclassc defauithtask andl mm for the Subnet mask 11111111 11111111 11111111 11000000 255 255 255192 Maskz consists of 27 ones24 om ofthe class c defauitmask ands ones for the Subnet mask 11111111 11111111 11111111 11100000 255 255 255 224 37 EE 281 lntEmetwurkmg quotquotsi quotquotquotquotquotquotquot quotquotquotquotquotquotquot 39 XYZ0 I Router 1 SE EE 281 intEmetwurkmg Pop Quiz a You have the following IP addresses 1716833 and 17168231fthe subnetmask is 2552552550 these addresses belong to the same subnet a True b False b Now you have the following IP addresses 1716833 and 1716823 1fthe subnet mask is 25525500 these addresses belong to the same subnet a True b False EE 2811ntemetworking Example 2 VariableLength Subnets with noncontiguous mask M The network designer in Example 1 is given class C network with 20011 address and wants to form the two subnets for the departments SA with 72 hosts and SB with 35 hosts Devise a possible an39angement of subnet mask to make this possible Answer in class C the most lower 8 bits are for subnet and host fields A possible arrangement SA set rst bit to 0 to identify the subnet SA the remaining 7 bits are for hosts 272 hosts large enough to accommodate 72 hosts so the mask 25525525597 bits for host address notation 0 SB Q6 bits for the host address notation 10 Note a 01 or OOarenot possible mm for SB why The essennal requzremem I that any mm subnet numbers remam mam when me anger 0712 I truncated to me length ufthz shmter o How about ifSB is as big as SA Use two or more smaller subnets for SB EE 2811ntemetworking 20 S u pe rn ettl n g 0 Problem Consider an university campus that is granted a class C P address say XYZ0 ut needs to accommodate its 1000 hosts users Also assume that class C LP addresses are only the still available IP address classes class A B addresses were all depleted 0 Solution in class c we have only 1 byte for the host So the maximum number of hosts is 254 282 lt 1000 not enough 7 Apply for more blocks of class c IP addresses then form one subnet How a use supemetn39ng o Supernettjng convert some ofls for Netld to OS 7 Note this opposite to what we have been doing with subnetung where were converting ls to OS but in the HostIdportjon see the next gure 7 In our example We to convert 2 1 s into 2 OS to have to Hostld 0f10 bits which enough to accommodate 1000 host in order 2 7 That is 4 blocks of class C IP addresses and then connect them together to m a single subnetsupemet for the university EE 281 lntEmetwurk ing 4i Nclid Slihnclid Hmlid Subnet mask a Subuetting Netill Hosrid Defaultmask llliilll 1111111 1111111 00000000 Slipcmclid Hoslid gamma 00 00000000 Note It can be viewe subnettjng regular one is just moving more ls to right of the default mask 1 supernettjng is is just moving more ls to left ofthe default mask 1 42 EE 281 lntEmetwurk ing Supernetting mask 0 How about the supemet mask 0 As in any other subnet to rest of the world the university supemet should look as one single networ o In this example we need to convert 2 ls from the NetID into 2 OS 1111111111111111111111110 9 11111111 11111111 111111000 2552552550 defaultmask 2552552520 0 similar to subnet in supernetting we need the rst lowest address of the supernet and the supernetmask to de ne the range of addresses In this example assume we choose to start with XY321 so network id for this supemet is XY320 other possible choices that we could also have started with are XY41 or XY81 or also XY1281 but not with XY35 1 set to 0s because the last two bit in 3 byte read from leftare now part of the host hence are XY zzzzzz0000000000 z can be any choice 43 BB 281 Internetworking First class C address Secoan class C nddrcs I I E E XY32254 XYu XY322 Cl CI XY33253 7 quot Cl XY33 754 To the rest of the Internet I XY35254 43p XY341 I XY35253 39 Q Q XY342 7 XY3Sl XY34254 1 y i I 1 Fourth chm C address Third class C address Supernetting with 4 blocks of class C NOTE you need the mask 2552552520 and the XY320 id to uniquely define this supernet 44 BB 281 Internetworking CIDR Classless Interdomain Routing o In supernetting blocks of C class IP addresses were used to create a single network where we moved bits from Netld field to increase host field 0 CIDR is an extension of C class supernet to any IP address not only C class 0 CIDR similar to subnetting CIDR was developed to solve 1 the three bigbear problem but also 2 to address the increasing sizes of routing tables 7 IP backbone routers have one routing table entry for each network address a With subnetting a backbone router only needs to know one entry for each Class A B or C networks 0 This is acceptable for Class A and Class B networks 7 27 128 Class A networks 7 214 16384 Class B networks 0 But this is not acceptable for Class C networks 7 221 2097152 Class C networks 0 In 1993 the size of the routing tables started to outgrow the capacity of routers 45 EE 281 Internetworking CIDR o Consequence The Classbased assignment of IP addresses had to be abandone o CIDR abandons the notion of classes and instead it views 32bit TP address as 2 fields 7 network id field the x most significant bits of the IP address called the network pre x A key concept of CIDR is that the prefix is of an arbitrary length flexible All hosts in the same network or organization must have network prefix 7 The remaining 32x bit address is used to distinguish hosts within subnets 46 EE 2811ntemetworking 23 CIDR Example 0 CIDR notation of a network address 19202018 o quot18quot says that the first 18 bits are the network part of the address and 14 bits are available for speci c host addresses 0 Assume that a site requires a network address with 1000 addresses 0 With CIDR the network is assigned a continuous block of 1024 addresses with a 22bit long pre x EE 281 Internetworking CIDR and Address Assignments Backbone ISPs obtain large block of IP addresses 5 ace and then reallocate portions of their address blocks to their customers In CIDR notation the black granted is de ned by the rst address and the pre x length Example Assume that an ISP owns the address block 206064018 which represents 16384 2 IP addresses Suppose a client requires 800 host addresses With classful addresses need to assign a class B address and waste N64700 addresses or four individual Class Cs and introducing 4 new routes into the global Internet routing tables With CIDR Assign a 22 block eg 206068022 and allocated a block of1024 210 IP addresses EE 2811nternetworking 24 Example Problem a Find the block network address if one of the addresses is 1908714020229 Find the number of addresses The I address that is the network address The last address that is the broadcasting address Show a network con guration for the block in these subnets Solution 7 the network mask is 255255255 248 writing 29 1s 7 AND operation 19087140202 1908714020029 255255255 248 222 sse This the network address EE 281 Internetworking Example observing 39 r 5 bitsofthegiven 1 is 1 5 L39 L L I b the number of addresses 232 or 8 c The I address is 1908714020029 in this course we will denote the 1 address network address special address with all zeros inhost suffix Note 190140201 is the 15 h0st address in this network d To nd the last address we use the complement of the mask The mask has 29 Is the complement has 29 0s and three Is The complement is 000 7 Then we ADD this to the network address we get 908714020729 e The network con guration is as shown in the next gure Note that the last address is usually set as the special address by the router broadcasting as all hostbits are 111 similar to subnetting but no needfor 255 other approaches are possible shortcut for instance please check that the last address is also the bit0R of the given address with the mask complement 9087140202 0R 0007 19087140207 EE 2811ntemetworking 25 Ne twork Organization 1908714020129 1908714020229 1908714020329 39 190 14020429 Q7 87 i e Block 8 Addresses 1908714020029 I a c 1908714020729 Network Address 190 37140 2 00 29 1908714020529 1908714020629 19087140 20729 Special Address KB 281 Intemetworking CIDR and Routing Abstraction ernet tckbone anization z1 EE 2811ntemetworking 26 CIDR and Routing Information Backbone routers do not know anything about Company X ISP Y or Organizations 21 22 ompany X i ISPy sends everything which 8390 2 matches the pre x 2098823719226 to Organizations 21 20988237026 to Organizations 22 ISP X does not know about Organizations 21 22 I 1 1 I ISP X sends everything which matches the prefix SP y 206068022 to Company X 20988237024 to ISPy 0 24 l m Backbone sends everything which matches the pre xes 206064018 2041880015 gan39zatmquot Z1 39 0 ganization 22 20988232021 to ISP x 2 2 EE 281 Internetworking You can find about ownership of EP addresses in North America via httpWWWarinnetWhois Exa m pl e o The IP Address 207 2 88 13970 88 1397 0 I 11001111 I 00000010 I 01011000 I 10101010 I Belongs to CityofCharlot tesville VA 2072880 207292255 I 11001111 I 00000010 I 01011000 I 00000000 I belongsto CableampWireessUSA 207000 2073255255 I 11001111 I 00000000 I 00000000 I 00000000 I 54 EE 2811ntemetworking 27 CIDR and Aggregation 0 An entry in a router look up table ltip7address mask nextihopgt lt1291213gt lt1283215161gt o Aggregation or routing table entrles 7 1281430016 and 1281440016 are represented as 1281420015 o Longest pre x match Routing table lookup finds the routing entry that matches the longest prefix What is the outgoing interface for 128143137024 more examples in review problems EE 281 Internetworking General Rules for CIDR Subnet Design Example From our previous discussion of subnetting we can easily observe andverify the following which will be helpful in solving designing subnetting problems if N the number of addresses granted by CIDR then N is a power 2 N 2quot if n size of the pre x then n32lag2N 32k the ISt address which is often the network address of CIDR subnet is divisible by the number of addresses in that subnet Forn the size of the pre x of the network NET which can consist ofmany subnets andN is the total number of its addresses IfNETl is a subnet ofNET with N addresses then nswj the size of the pre x ofNET is given by n IcyWM Let us illustrate some of these rulesobservations in the following problem Question An organization is granted the block 13034126426 The organization needs four subnetworks each with an equal number of hosts Design the subnetworks and find the information about each network Answer done in class 56 EE 2811ntemetworking 28 EE 281 Internetworking EE 2811ntemetworking 29 Example Binary Structure and Look up Table Example 1 Find the set of prefixes 2 Construct the tree using unique prefix EE 281 Internetworking Answer Binarytree Structure 7 Bit Address 0011mm 001 0100011 00000000 00000000 00000000 01010110 0000000 00000000 0000000 0110000 00000000 00000000 00000000 011 1010mm 10 1011000 00000010 00000000 00000000 10110 10111011 0001010 00000000 00000000 10111 A set of 327bit IP addresses and the corresponding set of pre xes that uniquely identify each of them EE 2811nternetworking 30 Internet Protocol Version 6 IPv6 0 Motivation The use of Subnetting and ClIDR strategies have helped to slow down the process of running out of IP addresses however with the continuing exponential growth in the number of Internet hostssers the IP in it current version IPv4 won ent the high demands and interest in the Internet recall Intemet initially g mm it was not was used by few universities and hightech industry and ove meant to accommodate anyone who wishes toJoin To remedy this weakness of addressing in IPv4 IETF Internet Engineering Task Force organization responsible for providing standards and solutions for intemet defined a new version for IP IPv6 It consists of 128bits 16B for addressing space four times the size ofIPv4 7 So we can have in the order of2128 IPaddresses 100 ef ciency that is using all bits for addressing which is about 7 10 23 IP addresses per square meter of earth s surface land water 7 Even with an ef ciency less than 100 the most pessimistic estimates indicate that there are 1000 addresses per square meter In short it seems unlikely that we will run out of addresses with IPv6 As of now we are using only about 30 of the total address space EE 281 Internetworking IPv6 Other Important Features wishlist 0 Besides address space IETF uses the IPv6 to correct other weaknesses in IPv4 of which the most important i Simplify the protocol to allow routers to process packets faster Pay more attention to type of service in particular realtime services 19 9 Provide security gt Allow protocols to evolve39 exible Lquot Permit the old amp new protocols to coexist needed at least during transition period Next let us examine how IPv6 supports these features EE 2811nternetworking 31 IPv6 Packet Header h h to the header in 0 IPv6 header has 6 elds IPv4 has 10 9 Simpler format easy to process by the router faster These elds are de ned as follows Version size 4 bit alwa s equal to 6 we c remove this eld if all agree to use IPv6 H gt It is instructive to compare IPv6 header 5 own ere IPv4 Traf cClass used to hel con estion b 0 4 12 16 24 Versiorl T f CCIES4 FlowLabel PayloadLen NextHeaderl HopLimit SourceAddress Y c assi ymg the type 0 traf 1cs So traffic with 0 11 Wed to tr 39t at slower rate Whll 1 traf c with values 8 to 15 can keep constant D t t Add rate ofIIansmission for instance audio an es mi 39 quot ress FlowLabel still experimental but some are using it Wit Traf cClass for ToS PayloadLen length of the packet in Bytes exc u mg the header header is fixed 40B Next header or data NextHeader Clever wa to include Option header and Protocol field 0 IPv4 ifneeded fr mentation ifneeded 65 535B 9 exible and lets old version to coexist up to 63 EE 2811ntemetworking IPv6 Packet Header HopLimit replaces TTL in IPv4 Also observe 0 fields relating to fragmentation in IPv4 were removed in TPv6 So what to do with large packets IPv6 hosts and routers have to conform with a maximum packet size of 1280 B MTU default value increased from 576 B which is large enough for most cases If host sends a larger packet IPv6 issues an error message asking the sender to fragment its larger packet 9 Faster routing How about with stubborn hostrouter ie who insists to use IPv4 and Iragment large pkt IPv6 sets the NextHeader 44 indicating that there is fragmentation and puts all the related information in the extension header eld NextHeader in data eld I The checksum eld is removed checksum reduces performance plus Data Link and Transport layers do checksum so another one was not worth it not to forget the header is 40B larger than IPv4 20B for which we need to do the checksum 64 EE 2811ntemetworking 32 IPv6 Addressing 8 Notation 0 IPv6 defines three types of addressing Unicast Anycast Multicast Unicast and Multicast are as in va4 Anycast is similar to multicast in that the destination is a group of addresses members but instead of delivering the packet to each one of them it tries to deliver to just one of them any member of the group possibly the closet Example of applications client wants to access information from servers any server will be ne Mobile lP 0 Notation a new notation has been devised to represent the 16B address of va6 It consists of 8 groups of hexadecimal digits with colons in between as illustrated in the figure below 28 ltils lblxylus 321m digits I I illllllttl l l 101 IUO l l l lllll 65 BB 281 lntemetworkjng Abbreviation 0 Since the address is too long it has many zeros not used yet it is commonly abbreviated Below two examples of abbreviation U11 I I I I I I I FDEC I BA98 74 32 OOOF BBFF 0000 FFFF KG A I I I I I I FDEC BA9SI74I321gF BBFF 0FFFF Abbreviated Abbreviatpd FDEC00BBFF0FFFF More Abbriated BB 281 lntemetworking 33 Address Space Assignment 0 IPv6 has no classes no A B C as in IPv4 we want to be more efficient but instead it divides the address bits 128 bits into 2 fields 7 Typeprefix 7 Address I28 bits 4 Variable Variable Type prefix Rest of address I o The Type prefix has variable length it defines the purpose of the address the table in next slide lists the value and purposes of each typei see pg321320 table 411 for complete list 67 BB 281 Intemetworking I I Lqul Lignta hi 2 Iml L u awh IJqucgtxgi 011 I IIIVzvsluru ll 11 1111 1 1111 11n 11 1 1 11 I 111 lnr 14 C Xr 68 BB 281 Intemetworking 34 IPv6 addressing for IPv4 0 Address beginning with 80 zeros are reserved for IPv4 addresses IPv6 supports two type of formats distinguished by the next 16 bits ompatr39ble used when computers using va6 want to sendreceive packets over va4 network this tunneling as we discussed earlier 33 hits 2 hm Hl ls I ll v4 address nunrmurml A n u umpulilvlc ilLIdI39CM ll vn ll rt 4 h U20Dl IOE r 2l3i74 l4 1 An example nl address lrumlurmulmn Mapping used when sender and networks use va6 but the receiver uses va4 8 bits 72 bits 32 bits All Os u All Is IPv4 address 00000000 a Mapped address IPv4 va6 0FFFF1020D 1 0E 2131714 b An Example of address transformation BB 281 lntemetworking Review Problems for FARTI BB 281 lntemetworking 35 Practice Problems 0 In this course routing is done using IPv4 and CIDR unless otherwise specified 0 Probleml hint for HW A router has built the table shown state to what next hop the following packets will be delivered l0 a C45El387 b C46B3l2E c 0000 note longestmatch pre x is implied why C1DR NetMachength Nexthop Also recall hexadecimal notation C4500012 A C45El0020 B C4600012 C C46800l4 D 800001 E w 400002 F 00002 G 71 EE 2811ntemetworking 72 BB 281 Intemetworking 36 Problem 1 BB 281 Intemetworking Problem 2 Question An organization is granted a block of addressquot m quotginning address 142474024 The organization needs to have 3 subblocks of addresses to use in its three subnets as shown below CI One subblock of 120 addresses CI One subblock of 60 addresses CI One subblock of 10 addresses 1 Design the subnetworks and nd the information about each networ 2 Are they unused addresses in your setting Answer done in class BB 281 Intemetworking m2 EE 281 Internetworking Problem 3 0 on Consider the Wireshark trace for IPv6 packet as shown in next slide A er consulting 39 v6 header format ll out the corresponding elds of IPv6 of the captured packet Insert NA Whenever information is not provi e 2 4 12 16 24 1 Versiolll Trafficmask FlowLabel PayloadLen I NextHead4r HopLimit SourceAddress Next header or data WW 38 E s gun 3mm Bisplay 1w EM an n q ry r spans yahooltmx Neighbor solicitation if D onssfFfensmona oo97 r Neighbor advermsanent F ffe u 207692A Peso ng w Response so e 1 26 o12 Beef 6 a so ans 1 104 1 s a f I Hammer 1 Src nnnnnaas I 39 A Emtarnet Protocol versiun a Version a N w h x14 Yraffiz cuss 0x00 0 ere 19 Wide 39 Next heads un om o O Hop Hm 6A Sour the upper protocol is UDP This used to replace the protocol in IPv4 It illustrates the payim length othon 15 used to indicate that 0039120056FFFEDSE013 AZ 01 4819 User Datagram Protocol Sr pm 2397 2397 051 Port domain 5 on am fl xibility ofl39Pv n o Authority m o momma ans 0 am type Mx class in w yahoocuxn m Man exchange Class 39Inet so no 97 o7 at as no no as os 50 an as an an on no no no 27 11 4o 21 f2 os o7 no no on o1 oz on no oo oo oo oo 42 09 so on n1 no no 01 no no no no no 51 as 61 of 03 a m 1 g m 3 new 25 4m s oo o 77 7 3 of on oo on of no o1 Problem 4 discussion 0 uestion It was mentioned before IPv6 does not support fragmentation and suggested value of MTU size 1280 B Why you think network designers did not increase this size to say to 1600 B 0 Answer 78 EE 2811ntemetworking HE Problem 5 Discussion 6 o uestion Consider IPv6 addresses if block of 1 million addresses is allocated every picosecond 1012 how long will the address last The age of the universe is 1010 years a Assume 100 ef ciency b Assume 01 ef ciency Answer a There are 2128 addresses Locating them at rate million 103912 1018sec There are 360024365 31536 k106 secondsl year Therefore they will last 212831536 l 106 1018 1013years This number is 1000 times the age of the universe b Of course the address space is not at so they are not located linearly 11010 but still with as low as 01 efficiency will not run out of a resses BB 281 Intemetworking BB 281 Intemetworking 40 Problem extra without formula nor calculator 0 Question In this problem you are to allocate network addresses using CIDR instead of class subnetting for a router with the following three interface constrains Subnetl has 125 hosts Subnet2 and Subnet 3 have each 60 hosts Provide your hosts address of the form abcdx to satisfy this setting assume that ISP provided you with the prefix 2231l724 Answer 81 EE 281 Internetworking 41


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