Network La (or 8 Prot, SE Se \ Network Laye ~The network layer is responsible for carrying the packet from the source all the way to destination. In short it is responsible for host-to-host delivery. ~The network layer has a higher responsibility than the data link layer, because the data link layer is only Supposed to move the frames from one end of the wire to the other end, ~ Thus network layer is the lowest layer that deals with the end to end transmission, 1d Position of Network Layer : ~ Fig. 1.1. shows the position of network layer in the 5 layer internet model. Its the third layer. [Network layer provides services to the transport layer Network layer thes serves fm he data nk yor (6-435) Fig. 1.1.1: Position of network layer ~ _Tereceives services from the data link layer and provides ‘Services to the transport layer. 1.1.2 Network Layer Duties ; Fig. 1.12 shows the set of duties ofthe network layer, Duties ofthe network layer Intemetworking Addressing Routing Packatizing Fragmenting (6-496 Fig. 1.2 : Duties of the network layer 1. Internetworking : This is the main duty of network layer. It provides the 'ogical connection between different types of Networks, 2 Addressing : ~ _ Addressing is necessary to identity each device on the Internet uniquely, This is sir imilar to a telephone ‘system. ~The addresses used in the able to uniquely define computer to the Intemet un Network layer should be the connection of a iversally 3. Routing: In a network, there are multiple roots avai, 1 from a source to a destination and one oF then. tobe chosen. = The network layer decides which root is to taken. Ths is called as routing and it depends cy various criterions. 4. Packetizing : - As discussed earlier, the network layer receives the packets from upper layer protocol ang ‘encapsulates them to form new packets. ~ This is called as packetizing. A network layer protocol called IP (Interetworking Protocol), does the job of packetizing. 5. Fragmenting : The sent datagram can travel through different ‘networks. Each router decapsulates the IP datagram from the received frame. Then the datagram is Processed and encapsulated in another frame. Other issues : ‘The other issues which are not directly related to the Cuties of network layer but need to be discussed are : 1. Address resolution. 2. Multicasting. 3. Routing protocols. Other supporting protocols ; The Internetworking Protocol (P) needs the Support of ‘another protocol ICMP or ARP etc. in the network layer, How to achieve the goals 7 the network layer must the communication subnet It also should choose. ‘communication,ACN (Comp. /Sem. S/MSBTE) 13 Network Layer & Protocols 1.2 IPv4 Addresses : Each computer connected to the Internet should be identified uniquely. The identifier used for this purpose is called as the Internet address or IP address. The hosts and routers on the Internet have unique IP addresses. The current version of IP (Internet Protocol) is IPv4 whereas the advanced version is IPv6. The IPv4 address is a 32-bit address and it is used for defining the connection of a host or router to the Internet. Thus an IP address is an address of the interface. 1.2.1 Uniqueness of IP Addresses : The IP address is unique and universal. That means each IP address defines only one connection to the Internet. ‘At any given time, no two devices connected to the Internet can have the same IP address. But if a device is connected to the Intemet via two connections through two different networks, then it can have two different IP addresses. All the IPv4 addresses are 32 bit long and they are used in the source address and destination address fields of the IP header. ‘The IP addresses for hosts are assigned by the network ‘administrator. For Internet it has to be obtained from the network information center. 1.2.2 Address Space : ‘The IPv4 protocol has an address space. It is defined as, the total number of addresses used by the protocol IFN number of bits are used for defining an address then the address space will be 2" addresses. For IPv4, Nis 32 bits. Hence its address space is 2" or 4, 234, 967, 296 (more than 4 billion). So. theoretically more than 4 billion devices could be connected to the Internet. Thus the address space of Pv is 2°. 1.2.3 Notation : The IPv4 addresses can be shown use three different notations as follows 1. Binary notations (base 2) 2. Dotted decimal notation (base 256) 3. Hexadecimal notation (base 16) = Out of these the dotted decimal notation is most commonly used Dotted decimal notation : = This notation has become popular because of the two advantages it offers. This notation makes the IPv4 address more compact and easy to read. ~The 32 bit IPV4 address is grouped into groups of 8-bits each separated by decimal points (dots). = Each B-bit group is then converted into an equivalent decimal number as shown in Fig. 1.2. = Each octet (byte) can take a value between 0 and 255. Therefore the IPv4 address in the dotted decimal ‘notation has a range from 010.00 to 255.255.255.255. ~ For example the IPv4 address of 1001 0001.00001010 00100010 00000021 is denoted in the dotted decimal form as 14510343. ° 1 “Qe Dated decal [5-10 +94 +9 (G-2001) Fig. 12.2 : Dotted decimal notation 1.2.4 IPv4 Address Format : = A32bit IPv4 address consists of two parts. The first part is called as net id ie. network identification which identifies a network on the Internet and the second part is called as the host id which identifies @ host on that network, = Fig. 12.2 shows the IPv4 address format. Note that the net id and host id are of variable lengths depending on the class of address. ~ Note that class D and & addresses are not divided into net id and host id for the reasons discussed later on k-— bits —— [netid [host] IP v4 address (G-2002) Fig. 1.2.2 : IPv4 address formatcn (Comp. /Sem. S/MSBTE) The concept of IP addresses is few decades old. It uses {he concept of classes. This architecture is called as the classful addressing. Later on in mid 1990s a new architecture of addressing was introduced which was known as classless ‘addressing. This new architecture has superseded the Original architecture. Ii this section we are going to discuss the classful addressing, 1.3.1 IPv4 Address Classes : In the classful addressing architecture, the IP address Pace has been divided into five classes : A,B, C, D and E Fig. 13.1 shows the percentage of occupation of the Address space by each class. ~The number of class A addresses is the highest ie. 50% ‘and those of classes D and E is the lowest ie. 6.25%. A BL | 25% c | | 126% D | | 625% Ee | 2" | e25% (G-2003) Fig. 1.3.1 : Classful addressing occupation of ‘address space 1.3.2 Formats of Various Classes : bits, 7 24 bite: (Px rer Ta] (G-531) Fig, 13.2(0) : Class A IPv4 address formats 14 NOWeH L0" 8 Fog, Class A format The formats used for IPv4 address are as sho, © Fig. 132. The IPv4 address for class A nena” ® shown in Fig. 13.2(0). is 7 bit long as shown in Fig. 23, = The network field is 7 bit “ and the host field is of 24 bit length. So the netae field can have numbers between 1 to 126. = But the host numbers wil range from 0000 ,, 127.255.255.255. = Thus in class A, there can be 126 types of networks ang 17 million hosts. ~The “0” in the first field identifies that it is a class 4 network address. Class B format : ~The class B address format is shown in Fig. 1.3.26). — The first two fields identify the network, and the ‘number in the first field must be in the range 128 - 191 14 bits 16 bits (6-532) Fig. 13.2(b) : Class B format ~ Class B networks are large. Host numbers 0.0 and 255.255 are reserved, so there can be upto 65,534 (216-2) hosts in a class B network. Most of the 16,382 class B addresses have been allocated. The first block covers address from 128.0.0.0 to 128.255.255.255 and the last block covers from 191.255.00 to 191.255.255.255. ~ Example : 128.89,0.26, for host 0.26 on net 128.89, Class C format : ~ The class C address format is shown in Fig. 13.2(c) a te 2 (6-533) Fig. 1.3.2(c) : Class C format The first block in class C covers addresses from 192.0.0.0 to 192.0.0.255 and the last block covers addresses from 223.255.2550 to 223.255.255.255, Class D format :ACN (Comp. /Sem_ SMSSTE) 15 Network Layer & Protocols = The dass format allows for upto 2 milion networks with upto 254 hosts each and dass D format allows the multicast in which a datagram is directed to multiple hosts. Class E address format : = Fig. 132(e) shows the address format for 2 class & address. This address begins with 11110 which shows that itis reserved for the future use. Fig. 1.3.2(¢) : IP address for class E network ~The 32 bit (4 byte) network addresses are usually written in dotted decimal notation. In this notation each of the 4-bytes is written in decimal from 0 to 255. = So the lowest IP address is 0.0.0.0 ie. all the 32 bits are 1 zero and the highest IPv4 address is 255.255.255.255. 3.3 How to Recognize Classes ? ‘When an IPv4 address is given to us either in the binary or dotted decimal notation, we can find the class of the address. If the given address is in the binary notation then we can identify its class by inspecting the first few bits of the address. This is as shown in Fig. 1.3.32). Byto2 Bytes Byte 4 Class A Byto 1 | Class B Bye caseo Bye coo omy TT 1 Byte | Class E {6-2006 Fig. 1.3.3(0) : Finding the address class If the given address is in the dotted decimal notation then we can identify the address class by inspecting the first byte of the address. Ths is as shown in Fig. 1.3.3). Bye, Clase A Owe [ems [ome] eye cane [ies to] Set [es Lowe Byo2 usec [fse-225] Ove? | Ores | Broa Bye t cased [Bes -255] ere? | eves | eves | eye! Classe [ao - 288] Bye2 | Byes | Byes {6-2005) Fig. 1.3.3(b) : Finding the address class = is important to note here that there are some special addresses which fall in class A or E. These special addresses are to be treated as the exceptions to the dlassful addressing. = In computers, the IPv4 addresses are generally stored in the binary notation format. Therefore it is possible to write an algorithm which can identify the address class by using the continuous checking process. = The principle of such an algorithm has been shown in Decision box to ‘check the next bit (6.2006) Fig. 1.3.4 : Algorithm to identify address class 1.3.4 Two Level Addressing : = The IPv4 addressing is used for defining @ destination for an Intemet packet at the network layer. — At the time when classful addresses were designed, the Internet was considered as the network of networks. In other words the whole Intemet was divided into a umber of smaller networks with many hosts connected to each network. — Normally an organization which wants to connect to the Internet creates a network and the Internet authorities allocate @ block of address to the organization. These addresses can be in class A, B or C = Allthe addresses allotted to an organization belong to 2 single block. Therefore each IPv4 address in classful ‘TechKnowledgsNetwork Layer & Pr [ACN (Comp. (Sem. SMSBTE) 16 = _ eee 3, Last address In the block addressing system is made up of two parts namely net Geameunaeenagise #32 bts 4 ret aeons reer ta] Fn bits ate (2 m) Bi (G-2007) Fig. 1.3.5 : Two level addressing eens ~ ‘The job of the met id is to define a network and that of Frat tees areata eter - ‘As shown in Fig. 135 if n bits define net Id then the ring Gao dene host = Trrtas erie note ora be cases fc Taepend onthe cases thown n Tale 3 Table 13.1 pa) A n=8 B | n=16 n=26 1.3.5 Extracting Information In a Bloc! = A block is nothing but a range of addresses. For any given block we would be interested to extract the following three pieces of information : 1. The total number of addresses in the block. 2. The first address ofthe block 3. The last address in the block. ‘extracting all this information, we have to identify the class of the address as discussed earlier. ‘Once we find the dass of the block, we will have the values of ‘n* (the length of net id in bits) and (32 - n) i.e. the length of the host id in bits. It is now possible to obtain the three pieces of information mentioned above as shown in Fig. 13.6. ‘Total number of addresses In the block : The total number of IPv4 addresses in the given block will be equal to, N= 20° 3.2) 2. First address in the block : ‘The first address in the given block can be obtained by keeping the leftmost “n* bits in the address as it is and setting all the (32 ~ n ) rightmost bits to 0 as shown in Fig. 136. the given block can be obtained g, raping te lefmest “0” its i the adres 95 ang reeretting al the 20) rightmest Bits £0235 shown in Fig. 136. (32-7) bits 1.3.6 Network Address : The network address is an address that defines the network itself. It cannot be assigned to a host. Fig. 13.7 shows the examples of network addresses for different classes. ‘Not 1D Host ID 6 E20) 115.837.90 —Neto that 115 is common inthe address ofall ‘machinos in class ‘Anetwork 18.90.1214 Network aderoos (cass A network address 151.1500 151.156769 Note that 151.15 ia common inthe address ofall machines in lass B notwork 151.189091 Network address (b) Class B network address (6-536 Fig. 1.3.7 (Contd.)ACN (Comp. ‘Sem. SMSBTE) s Network Layer & Protocols 58 Note that 22321.70 is ‘common in the address of ‘ll machines in class. +, Cretwork 223.21.70.126 Network address (©) Class C network address (6536 Fig. 13.7 ~The following examples will enable you to find the network address. Ex.1.3.1: For the address 24.46.8.95 identity the type of network and find the network address. Soin. : — _Bramine the first byte. Its value is 24 ie. ‘and 127. So itis a class A network. = So only the first byte defines the Net id. So we can find the network address by replacing the host id with 0s. = The process of obtaining the network address is shown between 0 in Fig. P.132. fe Not kts} host ld ——a] be ee | Replace host id by o's Netw wereen —+ (6-537) Fig. P. 2.3.2 So the network address is 24.0.0.0. Ex.132: For the address 132.7.21.84 find the type of network and the network address. - Sola, Examine the first byte. It is 132 ie. between 128 and 192. So it is a class B network. = Sothe first two bytes define the net id. Replace the host id with 0's to get the network address as shown in Fig. P.13.2. Not ip —+}e— host id —+} ie ‘replace host id by O's neworcacroe [eT [0-0] (@-538) Fig. P. 1.3.2 So the network address is 132.7.0.0. Ex. 1.3.3: Find the class of the network if the address is 221.46.75.64, Soin. : The first byte is 221i. between 192 and 255. So this is a dass C network. The net id and host id are as shown in Fig. P.133. J+ Not ig) ——sf host (6-530 Fig. P.13.3, What is the difference between net id and network address ? The network address is different from a net id. A network address has both net id and host id, with Os for the host id. Where to use the network address 7 ‘The network address is used to route the packets to the desired location. 1.3.7 Network Mask or Default Mask : — Earlier we have discussed the methods for extracting different pieces of information. But all these methods are theoretical methods which are useful in explaining the concept. — But practically these methods are not used. When a packet arrives at the input of the router in the Internet, it uses an algorithm to extract the network address from the destination address in the received packet. is can be achieved by using a network mask. Definition of default mask : ‘Anetwork mask or default mask in classful addressing is defined as a 32-bit number obtained by setting all the “n” leftmost bits to 1s and all the (32 ~ n) rightmost bits to 0. 1.3.8 Default Masks for Different Classes : = We know that the value of n is different for different classes. Therefore their default masks also will be different. v Tectaewtedyt\CN (Comp. Sem. 5/MSBTE) ~The default masks for class A, B and C addresses are as shown in Fig. 1.3.8, Class A:nas bens 6s (22 - n= 24 iso} [iti e0000090 Teoooea00 [55500000] 255.00. ‘MI (22—) nghemost bats eottoos Alot most n* ta sao is (0) Default mask for class A address Clase Bn = 16 be nw 16 ig (22-0) = 18 too} CET TAT SATTTTAT 17777 [06000000] 258.255.00 (©) Defautt mask for class B address (22-n) = bits Sy] LLAITHTAT 00000000] 255.255.256,0 (©) Default mask for class C address (6-200 Fig. 13.8 ~ Table 13.2 enlists the default masks of the three classes Of IPv4 addresses, Table 1.3.2 : Default masks A 255.0.0.0 B 255.256.0.0 c 255.255.255.0 1.3.9 Finding Network Address using Default Mask : ~ The router uses the AND operation for extracting the ‘network address from the destination address of the received packet. ~The router ANDs the destination address with the default mask to extract the network address as shown inFig. 139, ~ _ tis possible to use the defauit mask to find the number of addresses and the last address in the block. dross Dofauit mask address (©2019 Fig. 1.3.9: Finding a network address using the default mask (0 Three Level Addressing : Subnetting ‘As discussed earlier, the originally designey addresses were with two level addressing with ne and host id. ; The two level addressing is based on the principle thy, in order to reach a host on the Internet, we have reach the network frst and then the host. But very soon it became evident that the two ley addressing would not be sufficient for the following two reasons : 1. First it was needed to divide a large network of an ‘organization (to which a block in class A or B is allotted) into many smaller subnets (subnetworks) for improved management and security 2. Second reason is more important. The blocks in class A and B were almost depleted and the blocks in class C were smaller than the needs of most ‘organization. Therefore the organizations had to divide their allotted class A or B block into smaller subnetworks and share them. Definition of subnetting : ~ We can define the subnetting as the principle of splitting a block of addresses into smaller blocks of addresses. ~ _In the process of subnetting we divide a big network into smaller subnetworks or subnets. ~ Each such subnet has its own subnet address, Subnet mask : - The network mask or default mask that we discussed cari is used when the given network is not to be ivided into smaller subnetworks ie. when Subnetting is not to be done.ACN (Comp. /Sem. SMSBTE) = In Fig. 13.10, we have shown the default mask and subnet mask when a cass C network is to be divided into 8 subnets. (32-7) =8 ery Dotaun mask [FrarnssT ersasnte] erate [00000000] Tass C network fein 29 +3 = 27 a} Suorot mack [FATT | a aTT | FTTH O00] -——— No change Shera te (G-2011) Fig. 1.3.10 : Default and subnet masks 1.3.11 Special IP Addresses : «ag 1st nena ec se eae as as ee (at: oe (G-540 Fig, 1.3.21 : Special IP addresses = All zeros means this host or this network and all 1s, means broadcast address to all hosts on the indicated network. ~The IP address 0.0.00 is used by the hosts when they are being booted but not used afterward. ~The IP addresses with 0 as the network number refer to their own network without knowing its number as shown in Fig.1.3.11(b). ‘The address having all ones is used for broadcasting on the local network such as a LAN as shown in Fig. 13.110). Refer Fig. 1.3.11(d). This is an address with proper network number and all 1s in the host field. This address allow machines to send broadcast packets to distant LANs anywhere in the Internet. If the address is “127. Anything® as shown in Fig. 13.11(e) then it is a reserved address loopback testing. This feature is also used for debugging network software 1.3.12 Limitations of IPv4 : ‘The most obvious limitation of IPv4 is its address field. TP relies on network layer addresses to identify ‘end-points on networks, and each networked device has a unique IP address. IPv4 uses a 32-bit addressing scheme, which gives it 4 billion possible addresses. With the proliferation of networked devices including PCs, cell phones, wireless devices, et, unique IP addresses are becoming scarce, and the world could theoretically run out of IP addresses. If a network has slightly more number of hosts than a particular clas, then it needs either two IP addresses of that class or the next class of IP address. For example, let use say a network has 300 hosts, this network needs either a single class B IP address or two class C IP addresses. If class B address is allocated to this network, as the number of hosts that can be defined in a class B network is (2"° ~ 2), a large number of host IP addresses are wasted. If two class C IP addresses are allocated, as the number of networks that can be defined using a class C address is only (2%), the number of available class C networks will quickly exhaust. Because of the above two reasons, 2 lot of IP addresses are wasted and also the available IP address space is rapidly reduced. Other identified limitations of the IPv4 protocol are: Complex host router configuration, hierarchical addressing, difficulty in re-numbering addresses, large routing tables, non-trivial implementations in providing security, QoS (Quality of Service), mobility and multi-homing, multicasting etc. and ron- To overcome these problems the intemet protocol version 6 (IPV6) which is also known as internet protocol, next generation (IPng) was proposed. In IPV6 the internet protocol was extensively modified for accommodating the unforeseen growth of the internet. The format and length of the IP addresses has been changed and the packet format also is changed. ‘TedsNetwork Layer & W cn (comp. rSem. SMSBTE) val Restrictions Ex.1.3.4: A router inside an organization receives the same packet with a destination address 190.240.34.95. If the subnet mask is /19 (first 19-bits are 18 and following bits are Os). Find the subnet address. Soin. : ~ To find the subnet address, AND the destination ‘dress withthe subnet mask as shown in Fig. P. 134 fe—— 1019 —— v4 100 ——¥ S.t0 a] as. ey ce EI ore eon 2 wre es = comnass Thus the subnet address is 190.240.32.0. 1.3.13 Classless Addressing : Eventhough the number of actual devices connected to Intemet is much less than 4 billion, the address depletion has taken place due to flaws in the dlassful addressing scheme. We have run out of class A and B addresses. To overcome these problems, the classless addressing is now being tried out. In the classless addressing, there are no classes but the ‘address generation take place in blocks. Address blocks : ‘Address block is defined as the range of addresses. In the classless addressing, when an entity wants to get connected to the internet, a block (range) of addresses is granted to it. The size of this block ie. number of addresses depends on the size of the entity as well as its nature. That means for a small entity such as @ household only ‘one or two addresses will be given whereas for a larger entity like an organization, thousands of addresses can be allotted. ici address blocs, restriction on classless 7 a simplify the process of address handling. The addresses in a block should be continuo, i. serial in manner. The total number of addresses in BlOCk has to by ‘equal to some power of 2 Le. 2.27.2) ete. ‘The first address should be evenly divisible by the number of addresses. 1.3.14 Supernetting : The class A and class 8 addresses are almost depleted, But class C addresses are still available. But the size of class C address with a maximum number of 256 addresses does not satisfy the needs of an ‘organization. More addresses will be required. — The solution to this problem is supernetting. = In superetting an organization combines several class C blocks to create a large range of addresses i.e. several networks are combined to create a supernetwork. - By doing this the organization can apply for a set of dass C blocks instead of just one. Example of supernetting : ~ If an organization needs 1000 addresses, they can be ‘obtained by using four C blocks one C block corresponds to 256 addresses). ~ The organization can then use these addresses as one ‘supernetwork as a whole. 1 1.3.15 Who Decides the IP Addresses ? ~ Ne two IP addresses should be same. This is ensured by 2 central authority that issues the prefix or the network number portion of the IP address, ~ Locally an ISP is to be contacted in order to get 2 Unique IP address prefix “ ~ At the ‘global level the Internet Assig ned Number Authority (IANA) allots an IP address prefix to the ISP. Thus it is ensured that the IP addresses are not# ACN (Comp. (Sem, SMSBTE) 1 Network Layer & Protocols Conceptually IANA is a wholesales and ISP is a retailer of the IP addresses because ISP purchases IP addresses, from LANA and sells them to the customers. 3.16 Registered and Unregistered Addresses : computer that is connected to the Intemet. other technologies for protecting the computers. Internet to access them. ~ These workstations are given the unregistered private IP addresses. These addresses are assigned by the network administrator without obtaining them from an ISP (internet Service Provider) or IANA. = These are special network addresses in each class as shown in Table 13.3. These addresses are to be used for | ~ are called unregistered private networks and addresses. ~ We can choose any of these unregistered address while building our own private network. ‘Table 1.3.3: IP addresses for private networks ‘A. | 10.0.0.0 through 10.255.255.255 Registered IP addresses are required for computers which are accessible from the Intemet but not every For security reasons, networks use firewalls or some The firewalls will enable the workstations to access the Intemet but do not allow the other systems on the Soin. ‘Step 1: To find the subnet address : In order to find the subnet address we have to AND the IP address and the mask as follows 120 14 2 19 : IP address 255 255. 12 o ayy it111111 . 10000000. MASK (6-559 Fig. P.1.3.5(0) So the subnet address is 1201400. Similarly we can find the other subnet addresses. Step2: Host id: ~ Examine the first byte of the subnet address. It is 120 which is between 0 and 127. Hence this is a class A network. So only the first byte corresponds to the net id and the remaining three bytes correspond to the host id as shown in Fig. P.135(b). a Netid Host id (@-554) Fig. P. 1.3.5(b) So the host id is 14.00, Similarly we can find the other host id. B__| 172,16.0.0 through 172.31.255.255 Ex. 13.6: The IP address of a host on class C network is 496.123.46.237. Four networks are allowed for | 102.168,0.0 through 192.168.255.255 this network. What is subnet mask ? Soin 1.3.17 Solved Examples : The default mask for a class C network is, £x.1.35: Find the sub-network address and the host Id feared " In order to have four networks, we must have two extra for the folowing: 1s, Hence the default mask and subnet mask are shown in Fig, P.136. 255. 255. 256 . 0 (a) | 120,14.22.16 | 255.255.1280 Ld (o) | 140.11.96.22 | 256.255.256.0 255 . 255.286. 192 Subnet (| 141.181.14.16 | 256.255.224.0 mask (6) _| 200.34.22.166 | 265.255.256.240 ce) (6-555) Fig. P.13.6 ‘Thus the required subnet mask is 255.255.255.192.Wercmenaeng Tan ACN (( ‘Sem. 5SMSBTE) ibnets are othe various addresses of 6 $4 85 shown g What is the subnet address if the destination address is 200,45.34.56 and subnet mask is 255.255.240.07 En 137: Soin, To find the subnet address we have to AND the IP ‘address and the subnet mask as shown in Fig. P. 13.7. Eotetate Ld ooo. or10110r 790010 OO ee aors et Joo ane Stet [io wore oom (6-556 Fig. P.1.3.7 Thus the required subnet address is 200.45.32.0. Ex.1.38: A company is granted a site address 201.70.64.0. The company needs six subnets. Design the subnets. Soin. : = Thisis a class C network. So the default mask is, 255.255.2550 ‘As we need 6 subnets, we need three extra 1s. So the subnet mask is, 255.255.255.200 In the binary form the subnet mask is as shown in Fig. P.138. (6-557) Fig. P.138 In order to have six subnets, we can have 6 different combinations of the 3-extra 1s as shown in Table P. 1.38(@). Table P, 1.3.8(a) ao eee y 201,70.64.32 to 201.70.64.63 201.70.64.64 to 201.70.64.95 201.70.64.96 10 201.70.64.127 201, 70.64.128 to 201.70.64.159 '201.70,64.160 to 201.70.64.191 in For a given class C network 195.168.65.0 design equal subnets in such a way tha each subnet has atleast 60 nodes. E139: Soin. : Fig. P. 139(2) shows the structure of a class C address in which 3-bytes are reserved for net ID and 1-byte for host ID. 3 byte ————e — byte —t ‘bits (6-550 Fig. P. 1.3.9(0) We are expected to design equal subnets such that each subnet has atleast 60 nodes (i.e. 60 users). In order to identity at least 60,users we need 6-bits in the host ID. The remaining 2-bits are assigned for subnetting as shown in Fig. P.13.9(0). (6-559) Fig. P.1.3.9(6) This shows that there will be four equal subnets each ‘one having at least 60 nodes. ec Ex.1.3.10: Show by caloulations how many network each son. IP address class can have with one example ? Number of networks in different IP address : Class A address : ~The format of dass A address is shown ir Fig. P1310). Here one byte defines the seswort ID don | Subnet number | 000 Subnet 1 oot Subnet2 010 Subnet 3 o1t ‘Subnet 4 100 ‘Subnet § 101 Subnet 6 ‘and three bytes define the host ID. WetACN (Comp. /Sem. SMSBTE) [Network Layer & Protocols fe} byte ——Sbyies—$—$<4 (6-560) Fig. P.1.3.10(a) : Class A address = The MSB in the network field is reserved, So actually there are only 7-bits in the network fields. = So the number of networks in class A address will be 128. Class B address : - The format of class B address is shown in Fig, P, 13.10(b). Here 2-bytes are reserved for network field and remaining two bytes are forthe host fel. = Out of 26-bits in the network field the first two bits (MBs) are reserved. So actually 14 bits are available in the network field (@-56n Fig, P. 13.10(b) : Class B address = So the number of networks in class B address is 2 = 16, 368. Class C address : = The format of class C is shown in Fig. P. 1.3.10(0. Here S+bytes are reserved for network field and only one byte for the host field = Out of 24-bits in the network field 3-bits are again reserved. So actually only 21-bits are available. (6-562 Fig. P. 1.3.10(¢) : Class C address ~ So the number of networks in class C addresses is 2, 097, 152. Ex.1.3.11: How many host per network in each IP address class can exist, show with example ? Soln, : Number of hosts in different IP addresses : Class There are 3-bytes (24-bits) in the host field. Hence the number of hosts in class A address will be 2 = 16, 7772, 16. Class B: There are 2-bytes (16-bits) in the host field. So the ‘number of hosts in class B address will be 65536 i. 2" per network. Class C: There is 1-byte (B-bits) in the host field. So number of hosts in class C address will be 2° = 256 per network. Ex. 1.3.12: Convert the IP address whose hexadecimal representation is C22F15B2 to dotted decimal (6-563) Fig. P. 1.3.12 . The IP address in the dotted decimal notation is as follows : 194,79.21.226 ‘A class B network on intemet has a subnet mask of 255.255.240.0. What is the maximum, ‘number of hosts per subnet ? Ex. 1.841 The structure of class 8 address is as shown in Fig. P.1.3.13(0). 4614 bits —rhe—=16 bits —o1 (G-564) Fig. P, 1.3.13(a) : Class B address The given subnet mask is 255.255.2400. So it is as shown in Fig. P.1.3.13(b). 12bits tor ost ID (G-565) Fig. P. 1.3.13(b) : Subnet mask Thus there are 4 extra 1s as shown in Fig. P. 1.3.13(b). So there will be 16 subnets and each subnet can have 2 = 4096 hosts. PeterietePerform the subnetting of the following IP ‘Address 160.111. X.X Original subnet mask 255,255.0.0 Number of subnets 6 (six) Soin. : ~The original subnet mask indicates that we are dealing with a class B address. ~ Marder to have six subnets we need to use 3 extra bits from the bits that are reserved for host ID. So the subnet mask is as shown in Fig. P. 13.14 3 bis for ‘brat 2852s *3bie [eu sevafisssnes [ifeoeo foc coo (Net ID +e Host ID» (6-366) Fig. P. 2.3.14 ~The bits reserved for subnetting will have 8 combinations from 000 to 111 out of which any six ‘combinations can be used for 6 subnets. = Let us decide that the combinations 000 to 001 are not to be used. Then the subnet masks for the 6 possible subnets will have the following addresses. 285.255.224.0 1.4 Classless Addressing In IPv4 = Eventhough the number of actual devices connected to Internet is much less than 4 billion, the address depletion has taken place due to flaws in the classful ‘addressing scheme. ~ We have run out of class A and B addresses. To ‘overcome these problems, the super netting and subnetting has been tried as discussed earlier. - But subnetting and supernetting also could not solve the problem of address depletion in IPv4, Network Layer & p 114 W cw comp. /som. susers) of Internet users Due to increased number se it evident that a larger address space ead begs Fr ong term solution to this Problem. For q, th of the IP address should be increaseg wy, ree the IP packet itself must be changed 1g term solution is to switch to IPV6. But ay, Seales ses the same address space, Aa i fr tet Bi KOM 3 cag addressing. In the classless ada address generation take place in blocks. ‘The classless addressing was announced by the Inter authorities in 1996 in which blocks of variable leng, which do not belong to any class are used. essing, there are no classes but y, 1.4.1. Varlable Length Blocks : ‘Address block is defined as the range of addresses. In the classless addressing, when an entity wants to ge, connected to the internet, a block (range) of addresse, is granted tot. = The size of this block ie, number of addresses depends on the size of the entity as well as its nature. = That means for a small entity such as a household only fone or two addresses will be given whereas for a large entity like an organization, thousands of addresses can be allotted. - Fig. 14.1 shows how the address space is divided int non overlapping address blocks. (6-1806 Fig. 1.4.1; Variable length blocks in classless addressing ‘Two level addressing : ~ _ We have discussed the two level addressing for classtul addressing which divided an address into two part ‘namely : net id and host id. Defines the network. Defines the host (6-805) Fig. 1.4.2 : Two layer addressing in classfull addressing wrestACN (Comp. S/MSBTE) _ 116, Network Layer & Protocols = The net id and host id define the network and host respectively. It is possible to use the same idea in the classless addressing as well. = A block of addresses granted to an organization is divided into two parts called as the prefix and the suffix. = The role of prefix is same as that of the net id whereas as the role of suffix is same as that of the host id. Thus in a block granted to an organization, al the addresses will have the same prefix but each address will have @ different suffix. = Thus the prefix defines the network (organization to Which the address block has been granted) while the suffix defines individual hosts on the network, = The concept of two level addressing in classless addressing using the prefix and suffix is as shown in Fig. 143. = The IPv4 address is 32 bit long out of which the prefix will be of length “a” which can take any value from 0 to 32 and the length of the suffix will be (32 n) bits. = Note that the value of “n* ie. length of the prefix depends on the length of the: address block allotted (granted) to an organization. '}-——— 22 bits —— fe nia —efe— (82-1) bits —f (6-006 Fig. 1.4.3: Two level addressing using prefix ‘ond suffix for classless addressing Ex.1.4.1: Find out the values of prefix and suffix lengths in classless addressing if all the available addresses in IPv4 is to be considered as one single block. Soln.: = The total addresses in v4 is 2° = 4,294,967,296. We have to consider this as one block hence the prefix length n = 0. Whereas all the hosts will have their individual addresses. So all the 32 bits willbe allotted to the suffix length Ex.14.2: For the same data of the previous example find out the values of prefix and suffix lengths if all the available IPv4 addresses are divided into 4,294,967,296 blocks with each block having only one host. Soln.: = Here the prefix length for each block is n = 32, and the suffix length would be (32-n) = 0. The address of the single host in each block will be same as its block address itself. 1.4.2 The Slash Notation (CIDR Notation) : = fan address (classful or classless) is given to us and we want to extract information from it, then the net id in classful addressing or the prefix in classless addressing are extremely important and useful to us. = However it is not easy to identity the prefix bits in a given classless address. It is easy to identify the net id from the given classful address. = For a given classless address it is not possible to find the prefix length because the given address can belong to a block with any prefix length. ~ Therefore, in classless addressing jit is essential to include the prefix length to each address if the block of the given address is to be found. = Hence the prefix length “n* is added to the classless address separated by a slash and the notation is known as the slash notation. — Fig. 144 demonstrates a classless address with slash notation, Classloss accross (ve TT eve TT eve TT eve VET] PREFIX longi SLASH notation (G-1007) Fig. 1.4.4 Slash notation - The slash notation is also called as Classless Interdomain Routing or CIDR notation.14.3 Network Mask : ~ We have discussed the concept of network maskin the ‘dassful addressing. The same concept is also applicable. in the classless addressing as well. ~ A network mask in classless addressing is a 32 bit ‘number. With its “n* left most bits (corresponding to the prefix) all set to 2s and the remaining (32-0) bits Corresponding to the suffix all set to Os. Ex.14.3: For the following addresses identity the ‘number of prefix bits and write down the network mask : 1 12.26.25.79/8 2. 130.12.230.156 / 16 Soln, 1. Classless CIDR address : 12.26.25.70/8 AS per the slash notation we have n = 8 ie. number of prefix bits is 8 ‘Therefore the number of suffix bits = 32-8 = 24. ~ In order to obtain the network mask the prefix bits all set to 18 and the suffix bits all set to zero as shown in Fig. P. 2. Classless CIDR Address : 130.12290.156/16 AAs per the slash notation, n= 16 ie. number of prefix bits is 16. Number of suffix bits = 32-16 = 16 ~ In order to obtain the network mask, set all the prefix bits to 1s and set all the suffix bits to 0s as shown in Fig. P.1.43(b). (6-109) Fig. P. 14.300) = Thus the network mask = 255,255.00 4.44. Extracting the Block Information : = We can extract all the required information from the given classless address in the CIDR notation. The information that we can obtain is as follows: 1. The first address (network address) 2. The number of addresses. 3. Thelastaddress. We can obtain the number of addresses in a block as follows : Number of addresses in a block N Where n = Number of prefix bits. The fist address or network address in block can be obtained by ANDing the address with the network mask. First address = (Any address) AND (Network mask) (14.2) - OR what we can do is keep the “n” leftmost bits of any address as it is and set the remaining (32-n) bits to Os. This is equivalent to the ANDing operation mentioned above. = _Tnorder to obtain the last address in the block we have to add the first address with the number of addresses in the block directly. «+ Last address = Fist address + Number of addresses in the bed (1.43) ~ tis also possible to obtain the last address by ORing the address with complement of the network mask. + Last address = (Any address) OR [NOT (Network Mask] 44) One more way of obtaining the last address of the block isto keep all the ‘n' left most bits (prefix bits) 35 aoe (4a) itis and set al the (32-n) bits (sufix bits) to 1s.ACN (Comp. ‘Sem. SMSETE) Ex. 144: It an address in a block is given in CIDR Classless notation as 64.32.16.8 / 27 then find the following : 1. Number of addresses in the block (N) 2. The first address and 3. The last address. ‘Soin. : Step1: Findn: Given address = 64.32.68 /27 Hence n = 27 from the slash notation. cons D7bits, ». Prefix bits = 27, suffix bits = 32-27=5 ‘Step2: | Number of addresses in the block (N) : N= 25927232 ‘Stop3: Find the first address : - Refer Fig. P. 1.44(a) to obtain the first address in the block For this we have to AND the given address with the network mask. Network Layer & Protocols n_@@-n) ‘Network mask = 27 ones | 5 zer08 cs Network mask = 255.255.255.224 = For ANDing write the given address and network mask in their binary notations as shown in Fig. P. 144(0). “x. From Fig, P. 14.4(@) we get the first address in the block as: Ans. First address = 6432160 ‘Step 4: Find the lest address : To obtain the last address in the block, we have to keep the left most 27 bits in the given address as itis and set the remaining 5 bits to 1s as shown in Fig. P. 1.4.4(0). ‘From Fig. P. 144(b) we get the last address in the block as follows Last address = 64.32.1631W cn Ex14.5: For the classless address 129.65.33.01 / 24 find the following : 1. Number of addresses in the block (N) 2. The first address. 3. The last address. /Sem. SMSBTE) Soin, : Step1: Find: Given address = 129.65.33.01 / 24 hence n = 24 from the slash notation. n= 24bits = Prefix bits = 24, suffix bits = 32-24 =8 Step2: Number of addresses In the block (N) : N = 27222 256 ww Ans. ‘Step 3: Find the first address : ~ Refer Fig. P. 1.45(a) to obtain the first address in the block. For this we have to AND the given address with the network mask. a (2-n) + Network mask = 255.255.255.0 - For ANDing write the given address and network mask in their dotted decimal notations as shown. (6-112) Fig. P. 1.4.5(a) : First address in the block From Fig. P. 1.4.5(a) we get the first address in the block as : First address = 129.65.33.0 Ans. ‘Step Find the last address : To obtain the last address in the block, we have to keep the left most 24 bits in the given address as it is and set the remaining 8 bits to 1s as shown in Fig. P. 1.4.5(b). (G-1813) Fig. P. 1.4.5(b) : Last address in the block = From Fig. 1A5{b) we get the last adressin he is as follows : Last address = 129.65.33.255 an 1.4.5 Block Allocation : Now let us understand how to allocate the BIOCKS in ng classless addressing. The global authority for the bloc, allocation is ICANA means Internet Corporation fg, ‘Assigned Names and Addresses. But the individual addresses of the Internet users isnot allotted by the ICANA. Instead ICANA will assign large blocks of addresses to various ISPs or large organizations. These ISPs or organization will assign addresses to the individual Internet users from thei allotted blocks. Restrictions Some of the restriction on classless address blocks have been imposed by the internet authorities in order to simplify the process of address handling. 1. The addresses in a block should be continuous, ie. serial in manner. 2, The total number of addresses in a block has to be equal to some power of 2 Le. 24, 2", 2° ete. 3. The first address should be evenly divisible by the number of addresses. 1.4.6 Relation to Classful Addressing : ~The classful addressing may be imagined as the special ‘case of classless addressing such that the blocks of addresses in class A, B and C type addresses will have the prefix lengths ng = 8, ng = 16 and ne = 24, Table 14.1 lists the prefix lengths for class A to F lassful addresses and using this information we can change a block in classful addressing to a block in classless addressing. Table 1.4.1 : Prefix lengths for classful addressingACN (Comp. /Sem. 5/MSBTE) 14.7 Subnetting : = The concept of subnetting in dassless addressing domain is similar to that discussed for the classful addressing, = The subnetting is used for creating a three level hierarchy in the classless addressing domain. = An organization or an ISP have a biock of addresses granted to them. It can divide these addresses into several subgroups and each subgroup of addresses is assigned to a subnetwork or subnet. ~The subnetworks may be subdivided further if the organization want it that way. 1.4.8 Designing Subnets : LetN = Total number of addresses granted to an organization. n= Prefix length Assigned number of addresses to each ‘subnetwork Prefix length for each subnetwork Total number of subnetworks. New = ss ~ Now follow the steps given below to ensure that the subnetworks operate propery. Steps to follow : = The number of addresses in each subnetwork should always be equal to a power of 2. ie. 2°, 2',2%...ete. = We can use the following expression to find the prefix length of each sunetwork. nev net (45) = The starting address in each subnet should be divisible by the number of addresses in that subnework, To achieve this we need to first assign address to larger 1.4.9 Finding Information about Each Network > After designing the subnetworks, we can find the information about the subnets such as starting and last addresses, we can use the same procedure that was Network, sed to find the information about each network in the Internet. Ex 146 ‘A block of addresses granted to an ISP is ‘given by 130.34.13.64 / 26. These addresses fare to be divided into four subnetworks with equal number of hosts. Design the subnetworks and obtain all the information ‘about each subnet. Soln.: Step 1: Find total number of addresses (N) : = From the given address we get n = 26 (prefix length). = Hence the number of addresses in the whole network will be N= 288% 2). a 6 = The first address in this block will be 130.34.13.64 / 26 whereas the last address will be 130.34.13.127 / 26. ‘These values have been obtained using the procedure that we have discussed earlier. ‘Subnet design : ‘Step2: = There are four subnetworks with equal number of guests. Find number of hosts per subnetwork : Number of hosts per subnetwork is given by, N84 N= N=N=Ne=g 2g 216 Ans. ~ Note that the first requirement that 64 / 16 should be a power of 2 has been satisfied here. Step 3: Find the prefix lengths of the subnets : ~The pref lengths of the four subnets are given by, N Ay = ngeny=ny=n-+ loge] 5 — 6 + log, 4 wns. Starting and ending addresses of all the subnets : ~ Refer Fig. P. 146 which shows all the starting and ending addresses of the 4-subnets. Step 4: It should be noted from Fig. P. 1.46 that all the starting addresses should be divisible by the number of addresses in the subnet ie, by 16.1-20 I Se Tonwarang ager n we 8 cee sytem ee | maps fond 9 the OE SAH Srey Ea 20 tee see ey pret forwarding packets i a5 O1OWS | f saa ere soon asa packet aves 2.2 TOURE 2p, eee pee rea dress was shied HOht BY 28 bry a “eee ‘obtain a 4 bit class number. sonaner Me 2, A é-way branch then sorts packets into class» COL eed at supported) with eight of the cass ‘Address aggregation is considered to be one of the cass A four of the cases for class B, 1WO of thy cases for class Cand one each for Dand E. advantages of CIDR architecture. As we know, ICANN assigns then masked off the & 3 2a large block of addresses to an ISP which is divided into smaller subnets and assigned to the customers by the ISPs. ‘Thus many blocks of addresses are aggregated in one block and assigned to one ISP. peas ee caeeaes Ree ‘The code for each class 16-, or 24- bit network n\ itina 32 bit word. sic number was then searched in the A yumber and right alignes 4, The networ BorC table. Ex 147: routing table : Address/Mask Next Hop 195.46,56.022 Interface 0 135.46.60.0/22 Interface 1 192.53.40.0/28 Router 1 Default Router 2 For each of the following IP addresses, what does the router do if a packet with that address: arrives ? 1. 195.46.63.10 2, 192.53.56.7 Soin. : CIDR - Classless inter Domain Routing + ‘A router has following CIDA entries in its IP is being heavily used for decades. However, due to the exponential growth of intemet, IP is running out of addresses. This is a potential disaster and the internet community has begun discussion over it. In this section we are going to discuss one of the solutions to this problem. One of the solutions is CIDR (Classless Inter Domain Routing). The CIDR is based on the principle of allocating the remaining IP addresses in variable-sized blocks regardless of the class. Ifa site needs say 2000 addresses, then a block of 2048 addresses on the 2048 byte boundary is given to it. However the classless routing makes forwarding of 'As soon as the entry was found, the outgoing line was decided and the packet was forwarded upon it Forwarding with CIDR : ‘The simple forwarding algorithm explai not work with CIDR. Instead now each router table entry is extended by siving if a 32 bit mask So now there is a single routing table for all networks (no different tables for class A. B, C ete) which consists of an array of triples. Each triple consists of an IP address, subnet mask and outgoing earlier does line. When a packet arrives at the input, the router first extracts its destination IP address. Then the routing table is scanned entry by entry to look for a match. It is possible that different entries with different subnet ‘mask lengths match. In such a case the longest mask is used. For example if there is a match for a/20 mask and 2/24 mask then /24 entry is used. Solution of problem : = Convert the IP address to bits and then AND it with the subnet mask of the interface whose address is closest to that of the IP addresses. = The result of the ANDing will give you the network address and the interface to send the packet to. packets more complicated.IP = 138.46.63.10: The interface whose address is closest to this IP is interface 1. This interface uses a 22 bit mask. So AND the given IP address with a 22 bit mask as follows : IP = 136.4669.10 = 10000111.00101110.00111111.00001010 22 bt mask = 255.256 25200 = 11911191.11111911.11111100,00000000 1P AND Mask = 10000111.00101110.00111100.00000000 TP ANO Mask = 135.46.0.0 (61575) This result of ANDing matches with the network address Of interface 1. Hence the router will forward this packet to interface 1. 2 P= 102.53.56.7: The interface whose address is closest to this IP is interface 2. This interface uses a 23 bit mask. So AND the packet IP address with a 23 bit mask as follows : IP 19288567 = 11000000.00110101.00111000.00000111, Za bit mask = 255.255.2540 0 1448111.41111111.11111190.00000000 JP AND Mask = 11000000.00110701.00111000,00000000 = 19253560 e974) This result of ANDing does not match with the network addresses of interface 0 or 1. Hence the packet will forwarded to the default ie. Router 2. 1.5 Special Addresses : In the dlassful addressing, some addresses were reserved for special purpose. Similarly in the classless addressing as well some addresses are reserved. 1.5.1 Special Blocks : Some address blocks have been reserved for special Purpose. 1.5.2 All Zeros Address : = The block (20.0.0 / 32 contains only one address. It is called as the all zero address and has a prefix length of n= 32, ~ This address has been reserved for communication when a host has to send an IPv4 packet but it does not know its own address. ~ In such situations, the host sends an IPv4 packet to a DHCP server using this all zero address as the source [ACN (Comp /Sem. SMSSTE) 4.21 Network Layer & Protocols Address and a limited broadcast address (all one address) as the destination address, so as to find its own address. 1.5.3 All One Address-Limited Broadcast = The block 255.255.255.255 / 32 contains only one address. It is called as an all one address and has a prefix length of n = 32. = This all one address has been reserved for limited broadcast address ie. if a host wants to send message to all the hosts simultaneously then the sending host ‘can use all one address as a destination address inside the IPv4 packet. = Such a broadcasting is confined to the network only because routers do not allow the all one packet to pass through them. = The datagram sent with the all zero address as destination will be received and processed by all the hosts on the network. 1.5.4 Loopback Address : = Alloopback address is the address which is used to test the software on a machine, The block 127.0.0.0 / 8 with 2 prefix length of 8 is used for the loopback address. = On using this address, a packet does not leave the machine at all but it returns to the protocol software. It can be used for testing the IPv4 software. 1.5.5 Private Addresses : = The address blocks that are not recognized globally still ‘assigned for private use are known as private addresses. ~ These addresses are neither connected to nor isolated from the Network Address Translation (NAT) techniques. ~ Table 15.1 depict such address blocks. Table 1.5.1: Private addresses Block | Number of Et aiiteoenen 10.00.0/8 | 16,777,216 |192.168.00/16| 65,536 172.16.0.0/12| 1,047,584 |169.254.0.0/16| 65536Network Layer & 122 \CN (Comp. /Sem. SMSBTE) tion to this problem iS NAT ie. neg ut a 1.5.6 Multicast Addresses : The block 2260.00 / 4 with a prefix length of n = 4 has been reserved for the multicast IP communication. 1.5.7 Special Addresses in Each Block : ~The usage of some address in each block for special addresses has been recommended. But it has not been ‘made mandatory. These addresses are not assigned to any host. One important point to be remembered is that a very small block of addresses should not be used as special addresses. 1.5.8 Network Address : ~The network address is defined as the first address (with the suffix set all to Os) in a block. It is used for defining the network itself. It does not define any host in the network. ~ With the same principle, the first address in a ‘Subnetwork is called as the subnetwork address. 1.5.9 Direct Broadcast Address : ~ We can use the last address in a block or subblock (with the suffix part set to all 1s), as a direct broadcast address for that block or subblock, wick sol ; e translation. It is described in RFC 3022 address the basic idea in NAT is that each COMPANY is asi, e ingle P address or at the most a small number g, a sing! addresses so as to access the Internet. Within the company, every computer gets 2 unique address which is used for routing the internal trafic gy the office But when a packet goes out of the company, and gos, to ISP, the translation of IP address takes place there, In order to make this scheme work, three ranges of p ‘addresses have been declared as private. Companie; can use these addresses internally as per ther requirement. However no packet containing these addresses is allowed to appear on the Internet. The three reserved ranges are as follows : 10.0.0.0 to 10.255.255.255/8 | 16777216 Hosts 1048 576 Hosts | ~ A router generally uses this address for sending a Packet to all the hosts connected to a specific network. This address is used as the destination address in the Pv packet and all the hosts will accept and process the datagram which has this destination address, 1.6 NAT — Network Address Translation mn Address Translation : ~ The problem that existing number of IP addresses is tess than the actully required ones is practically important, ~ Along term solution to this problem is thatthe whole Intemet should be migrated from IPv4 to IP. Ths has begun, but will take year to get complete. (That means all the computers should have IPV6 addresses instead of v4 addresses), _| 173.31.255.255/12 | | Ranges | 192.168.0.0 to 65536Hosts 5 | 102.168.255.258116 | Generally most companies choose the addresses from the first range. Refer Fig. 1.6. which explains the operation of NAT. It Shows that within the company premises, every Machine has a unique address of the form 12abe. But when @ packet leaves the company premises, it Passes through the NAT box This box converts the internal IP address 120.02 in Fig. 161 to the ComPany/s true IP address 198.6049 19, The NAT box is Senerally combined with a firewall Iti 3150 possible to integrate the NAT box into company’s ‘outer.ACN (Comp. /Sem. S/MSBTE) 1.7__Internet Protocol Version 4 (IPv4) — We have already discussed the addressing mechanism, for the IP packets. — Now we will discuss the format of IP packet in the next few sections. = _ In the discussion we will see that an IP packet consists of a base header and options which are sometimes Useful in controlling the packet delivery. 1.7.1 Position of IP : = The main protocols corresponding to the network layer in the TCP/IP suite as well as Intemet layer are : ARP, RARP, IP, ICMP and IGMP. This is as shown in Fi - Out of these protocols IP is the most important protocol. It is responsible for host to host delivery of datagrams from a source to destination. But IP needs to take services of other protocols. ~ IP takes help from ARP in order to find the MAC (physical) address of the next hop. 1.7.2 Network Layer & Protocols, IP uses the services of ICMP during the delivery of the datagram packets to handle unusual situations such as presence of an error. IP is basically designed for unicast delivery. But some new Internet applications as well as multimedia need multicast delivery. S0 for multicasting, IP has to use the services of another protocol called IGMP. IPv4 is the current version of IP whereas IPv6 is the latest version of IP. Internet Protocol (IP) : The Internet Protocol is the host to host delivery protocol which belongs to the network layer and is designed for the Internet. IP is used as the transmission mechanism by the TCP / IP protocols. That means the TCP or UDP packets ‘are encapsulated in the IP packet and the IP carries it from source to destination. PP is a connectionless datagram protocol with no guarantee of reliability. It is an unreliable protocol because it does not provide any error control or flow control. IP can only detect the error and discards the packet if it is corrupted If IP is to be made more reliable, then it must be paired with a reliable protocol such as TCP at the transport layer. Each IP datagram is handled independently and each one can follow a different route to the destination. So there is a possibility of receiving out of order packets at the destination. Some packets may even be lost or corrupted. IP relies on a higher level protocol to take care of all these problems, The version of IP that we are going to discuss is called as IPv4 ie. IP version 4, IP is also called as a best effort delivery protocol. The meaning of the term best effort delivery is that the IP Packet can get lost or corrupted or delayed. They may arrive out of order at the destination or may create congestion in the network. TochKnewlodgsACN (Comp. ‘Sem. SMSBTE) 1.7.3 Datagram: = Packets in IP layer are called datagrams. Fig. 17.2 shows yer 9 the typical format of an IP packet A datagram has two parts namely the header and data as shown, The length of datagram is not fied. It varies from 20 bytes to 65536 bytes. ~The length of the header is 20 to 60 bytes. The information necessary for the routing and delivery of the datagram has been stored in the header = The other part of the datagram is the data field which is of variable length 720 to 65596 byles (6-525) Fig. 1.7.2: IPv4 datagram format — Its a custom in TCP/IP to show the header in 4-byte (22 bit) sections. 1.7.4 IPv4 Header Format : ~The IP frame header contains routing information and control information associated with datagram delivery. The IP header structure is as shown in Fig. 1.7.3. 31 Destination IP address ‘Options + Padding (0 - 40 bytes) (6-2082) Fig. 1.7.3 :1Pv4 header format = _ Various fields in the header format are as follows : 1. VER (Version) : — This is a 4 bit field which is used to define the version of IP protocol. The current version of IP is 4 i.e. IPv4 but in future it may be completely replaced by the latest version of IP i. IPv6. = This field will indicate the IP software running on the processing machine that this datagram belongs to IPv4 version, 4-24 Network Layer & Pio, essing machine gram some If the proc 0 of IP, then the ersio il be discardes HLEN (Header length) : ‘This 4-bit long field is of the datagram header i! this field is multiplied by 4 to get, Pv header which varies between z, sed f0r defining the ln, - in 4-byte words = The value of length of the and 60 bytes. there are no ~ inet he header length is 5x 4 = 20 bytes, men the value of option field i maximum the aan of HLEN field is 15 and the corresponding rane length is maximum ie. 15 x 4 = 60 bytes, options, the value of this Fela, ‘Service type = In the earlier designs of IP header, this field was called as Type of Service (TOS) field and its job was to define how the datagram should be handled. = At that time, a part of this field used to define the precedence of datagram and the remaining part used to define the type of service out of different possible services such as low delay, high throughput etc. = But now the interpretation of this field has been changed by IETF. This field is now supposed to define a set of differential services. Fig. 17.4 illustrates the new interpretation of the service type field TTT fo Procedon0e epee Pt intorrotation = epee b fot Differential service interpretation (G-2083) Fig. 1.7.4 : New interpretation of service type field ~ As seen in Fig. 1.7.4, in the new interpretation, the ae ‘ye fed is divided into two subfields paaree ee codepoint subfield and a 2 bit We can use the 6-bit codepoint subfield in two different ways, as follows 1 For the interpretation, Purpose of precedence 2. For the differential service interpretation. Technolog4 wv ACN (Comp. ‘Sem. SMSBTE) Network Layer & Protocols Wren corp Som swseTe) Oe Precedence Interpretation : = If the three right most bits are zeros, then the three leftmost bits are interpreted the same as the precedence bis in the service field (old interpretation) ‘That means itis compatible with the old interpretation of this fila = The precedence interpretation is used for defining the priority level of this datagram (from 0 to 7) in the situations lke congestion. In the event of congestion, the datagrams with lowest precedence (0) will be discarded first. \tferential service interpretation : When the three rightmost bits are not all zeros, the 6 bit codepoint subfield is used for differential service interpretation. In that case these 6 bits can be used for defining a total of 56 (64 - 8) services, on the basis of the priorities assigned by the Intemet or local authorities as per Table 17.1. Table 1.7.1 : Values of codepoints 1 xxxxx0 | Internet 2 | xxxxt1 | Local 3, | xxxx01 | Temporary or Experimental The first, second and third categories contain 24, 16 and 16 service types respectively. The Intemet authorities assign the first category. The local authorities assign the second while the third one is temporary and can be used for experimental purposes. Total length : = This 16 bit field is used to define the total length of the IP datagram. The total length includes the length of header as well as the data field - The field length of this fields is 16 bits so the total length of the IP datagram is restricted to (2° - 1) = 65535 bytes out of which 20 to 60 bytes constitute the header and the remaining bytes are reserved to carry data from upper layers. This field allows the length of a datagram to be upto 65,535 bytes, although such long datagrams are impractical for most hosts and networks. = Alllhosts must be prepared to accept datagram of upto 576 bytes, regardless of whether they arrive whole or in the form of fragments. ~The hosts are recommended to send datagram larger than 576 bytes only if the destination is prepared to accept larger datagram. = We can find the length of data by subtracting the header length from the total length. = As stated earlier the header length can be obtained bby multiplying the contents of HLEN field by four. Length of data = Total length— header length = The total length (maximum value) of 65,535 bytes might seem to be large but in future the size of IP datagram is likely to increase further because the improvement in technology will allow more bandwidth, Why do we need the total length field ? We might feel that the total length field is not at all required because the host or router will drop the header and trailer when it receives a frame. Then why to include this field ? The answer to this question is that in many situations ‘we do not need this field at all. But in some special situations, only the datagram is not encapsulated in the frame but there are some padding bits as well that are included. In such situations, the machine (host or router) that decapsulates the datagram, needs to check the total length field so as to understand how much is the data and how much is the padding ? Identification : This field is used to identify the datagram ‘originating from the source host. When a datagram is fragmented, the contents of the identification field get copied into all fragments. This identification number is used by the destination to reassemble the fragments of the datagram, W eaten,ACN (Comp. /Sem. SMSBTE) 6 Flags: ~ Flags : This is a three bit field. The 3 bits are as shown in Fig 175. Con oe] Ln treme This is do not fragment bit (©-2nFig, 17.5: lag bits ~ First bits reserved, and it should be 0. - ‘The second bit is known as the “Do Not Fragment” bit If this bit is “1” then machine understands that the datagram is not to be fragmented. - But if the value of this bit is 0 then the machine should fragment the datagram if and only if necessary ~ The third bit is known as “More Fragment Bit” (M). M = 1 indicates that the datagram is not the last fragment and M = 0 indicates that this is the last or the only fragment. Fragmentation offeet : - This is a 13 bit field which is used to indicate the relative position of this fragment with respect to ‘the complete datagram. ~ _ Itis the offset of the data in the original datagram ‘Measured in units of 8 bytes. ~ Tounderstand this refer Fig. 1.7.6. — The original IP packet (datagram) contains 4000 bytes numbered from 0 to 3999, It is fragmented into three fragments. ~ The firs fragment contains 1400 bytes numbered from 0 to 1399, The offset for this fragment is 0/8 = 0. Similarly the offsets for the other two fragments are 1400/8 = 175 and 2800/8 = 350 respectively as shown in Fig. 1.76, ~The offset is measured in units of 8 bytes. Because fragments should be of size such that first byte ‘umber is divisible by 8, This is an 8-bit field which controls the maximum number of routers visited by the datagram during its lifetime. A datagram has a limited lifetime for travelling through an Internet. = Originally the TTL field was designed to hold the timestamp. This timestamp value was decremented by one, everytime the datagram visits arouter. ~ AS soon as the timestamp value reduces to zero the datagram is discarded. But for this scheme to become successful, all the machines must have synchronized clocks and they must know the time taken by a datagram to travel from one router to the other. ~ Today the TTL field is used to control the maximum number of hops ie. router by a datagram. ~ _Atthe time of sending a datagram, the source host wil store a number in the TT field. This number is approximately twice the maximum number of ‘outers present between any two hosts. ~ Everytime this datagram visits a outer, this value is decremented by one. If after decrementing, the value of TTL field reduces to Zero then that router discards the datagram, Need of TTL feld : the length of the offet field is 13 bits, so the ~AGN (Comp. ‘Sem. SMSBTED - Network Layer & Protocols, - ‘The TTL field is needed in such situ ions for limiting the lifetime of a datagram. = The TTL field is also used to iit the Journey of packet intentionally. For exemple if @ packet is to be confined to a local network ‘only then 2 1 is stored in the TTL field of this packet. = As soon as it reaches the first router, then TTL field value is decremented from 1 to 0 and the packet will be discarded 9. Protocol : = This is an 8-bit field which is used for defining the higher level protocol which uses the services of IP layer. _ The data from different high level protocols can be encapsulated into an IP datagram. These protocols: could be UDP, TCP, ICMP, IGMP etc. The protocol field contents would tell the name of the protocol at the final destination to which this IP datagram is to be delivered. — At the destination, the value of this field helps in the process of demultiplexing. = Table 1.7.2 shows some of the values of this field corresponding to different high level protocols. Table 172 value | ) | 1 ICMP 17 UDP 2 IGMP 89 OSPF 6 | Top 10, Header checksum : ‘A checksum in IP packet covers on the header only. ‘Since some header fields change, this field is recomputed and verified at each point that the Internet header is processed, 11. Source address : This field is used for defining the IP address of the source. Itis a 32 bit field. 12, Destination address This field is used for defining the IP address of the destination. Its also 2 32 bit field. 13. Options Options are not required for every datagram. They are used for network testing and debugging, We have discussed all the options in detail, later in this chapter, Fragmentatio! 1.8 Fragmentation: = Inthe Internet, a datagram sent by a host has to travel through different networks before it is delivered to the destination host. _ at every router, the received frame is decapsulated, the 1p datagram is extracted and processed and encapsulated in another frame. = The size and format of the frame received by @ router depends on the protocol used by the previous physical network to the router. — Asan example, imagine that a router connects a LAN to a WAN, Then the frame received by the router is in the LAN format and the one forwarded by itis in the WAN format. 4.8.1. Maximum Transfer Unit (MTU) : = The frame format of each data link layer protocol is different in its own way. One of the important field in the frame format is the maximum size of data field. = Therefore when we encapsulate an IP datagram in a frame, the datagram size should be less than the maximum data size specified by the maximum size field. = The concept of MTU has been illustrated in Fig. 1.8.1. (6-2084) Fig. 1.8.1 : Concept of MTU = Now the problem is that the value of MTU changes from one protocol to the other used for the physical network. - We have to make the IP protocol independent of the physical network. In order to do so the maximum length of IP datagram was decided to be equal to 65,535 bytes.\CN (Comp. ‘Sem. /MSBTE) ~ we use a physical network protocol which has MTU = 65,535 bytes, then the transmission will become more efficient. For the other protocols having MTU smaller than 65,535 bytes, the IP datagram is divided into small parts called So that they can pass through the physical etworks successfully, This processes of dividing the IP datagram in smaller Parts is called as fragmentation. ~ The fragmentation generally does not take place at the ‘Source because the transport layer there will adjust the ‘Segment size in such a way that they will fit in the IP datagrams and data link layer frames. ~ After fragmentation, each fragment will have its own hheader. Most of the fields of the original header are Copied into the fragment header but some fields are changed. ~ Such a fragmented datagram can be fragmented further if it comes across a network with even ‘smaller MTU. ~The fragmentation of a datagram can be carried by the ‘source host or any router on the route of the datagram. — But the process of reassembly of all the fragments will be carried out only by the destination host. - All the fragments of a datagram are free to take any route and we do not have any control over them. In short each fragment acts as an independent datagram. = The reassembly of fragments is not done during the transmission because of the loss of efficiency associated with it. — At the time of fragmentation, all the required parts of the header are copied into the fragments. But the ‘options field may or may not be copied as discussed later on. = The following three fields are altered when the host or router fragments a datagram : 1 Flags. 2. Fragmentation offset. ; Network Layer & p 1.8.2. Flelds Related to Fragmentation The following three fields in an TP datagram hea, related tothe fragmentation and reassembly oy, datagram. 1. Identification. 2. Flagsand 3. Fragmentation offset field. 49 Options: = Inthe IP header there are two parts : A fixed part ang, variable part. We have already discussed the fixed pay of 20 byte length, = At the most 40 byte long variable part consists of ‘options which we are going to discuss inthis section ~ Options as the name suggests are not required for datagram. Their main application is for network testng ‘and debugging. = Options are not a required part of a datagram but ‘option processing is very much a required part of the software. ~ This implies that if the options are present in the header, then all the implementations should be able to handle them. 1.9.1 Format: = The format of an option has been shown in Fig. 1.9.1. As shown, it consists of three fields. namely, a type field (1- byte), length field (1-byte) and a variable length value field. Je byle—vie— 1 byto—sle Variable —o Copy L cass L Number (G-2009) Fig. 1.9.1: Option format 3. Total length. = The remaining fields in the IP header are copied as its. The value of checksum should be calculated again regardless of fragmentation. ~ And the final point about fragmentation is that only data in a datagram is fragmented. ~ _Letus discuss these fields one by one. 1. Type: ~ As shown in Fig. 1.9.1, the type field is an B-bit field 4nd it contains three subfields as follows 1 Copy (bit, 2 Class @ bits), 3. Number (5 bits)W 40 comp. som. SMSBTE) (@) Copy This is a 1 bit subfield. So it can have only two possible values, 0 or 1. If copy = 0, then the option ‘must be copied only into the frst fragment. ‘Whereas if copy = 1, then the option field must be copied into all the fragments. 0 _| Copy option field only in first fragment. 1._ | Copy option field in all fragments (>) Class: This 2-bit subfield is used to define the purpose of ‘option. It has four possible values, out of which ‘only two (00 and 10) are defined right now. The ‘ther two possible values (01 and 11) are not yet defined. If dass = 00, it indicates that the option is being used for datagram control, Whereas if copy = 10 then the option is used for debugging and 10_| Debugging and management. 11_| Not defined or reserved. (©) Number; This 5-bit subfield is used for defining the type of ‘option. This subfield has 32-possible values (types), but currently only 6-types are defined as shown in Table 191. Table 19.1 (00000 _| End of option. 00001 | Nooption. 00011 | Loose source route 00100 _| Timestamp 00111 | Record root 01001 _| Strict source route 4.29 2 3. Network Layer & Protocols ‘We will discuss these later inthis chapter. Length : ~ This 8-bit field is used for defining the total length Cf the option with the type field and the length field included. = The length field will not be present in al the option types. Value : = This is variable length field which contains the specific data which is required by that option. = Similar to the length field, the value field also will not be present in all the option types. 10 Option Type ‘As we started earlier, only six options are being used currently. Fig. 1.10.1 shows the classification of these options. (62086 Fig, 1.10.1 : Categories of options Options are classified into two option types is. single byte options and multiple byte options. There are two single byte options which do not require the data or length fields. ‘The remaining four options are multibyte options which the data and length fields. Let us now discuss these options one by one. 1.10.1 No Operation Option : This is a single byte option which is being used as a filler between options. ‘As shown in Fig. 1.102, we can use the no operation ‘option to align the next option on a 16 bit or 32 bit boundary. eee[No-oP ] ‘An 11 - byte option (©) NO-OP Is being used to align _(<) NO-OP is being used beginning of an option to align the next option (G-2087) Fig. 1.10.2 : No operation option 1.10.2 End of Option Option : ~ The second one byte option is the end of option Option. It finds its application in padding at the end of the option field. ~ Two important points about this option are as follows : 1. We can use it only as the last option. 2. We can use only one end of option. That means after this option, the receiver should expect the Payload data to arrive, ~ There if we need more than 1 byte to align the option field, then we must use more than one no-operation options and after that only one end-of-operation option as shown in Fig. 1.10.3. Type 0 190000000. (@) End-of-option (b) Used for padding (G-2088) Fig. 1.10.3 1.10.3 Record-Route Option : ~The record route option is a multiple byte option and it is used for recording the Intemet routers which handle the datagram, Since the maximum size of the header is 60 bytes, including 20 bytes of base header, this option can list upto 9-1P addresses of the routers. ~ So actually only 40 bytes are left for the option part. The format of the record-root option is as shown in Fig. 1.104. The source creates fields that are to be filled by each router visited by the datagram, Network Layer & Pry, 1:30 ‘ At ‘only Hot addros908, omy ‘can be ben stored ate {(¢-2089) Fig. 1.10.4 : Round trip option ‘The pointer field is an offset integer field which cont, the byte number of the frst empty entry. That means, 1 points towards the first available entry. : All the empty fields for the IP address are empty whe the datagram leaves the source. The value of pointy field is 4 which points to the first empty field. ‘When the datagram starts travelling, each router visite, by this datagram, will insert its outgoing IP address in the next empty field and increments the value of pointer by 4 : 4 Strict-Source-Route Option : = This is also @ multi byte option which is used by the source to determine the route in advance for the datagram travelling over the Internet. ~ Due to this it becomes possible for the sender to choose root to get a specific type of service (ie. minimum delay, maximum throughput etc). ~ _ Itis also possible for a sender to choose a safer and more reliable root. — Ifa datagram specifies a strict source route, then the datagram must visit all the routers which are defined in the option, ~ It should not visit any router whose IP address is not sted in the detagram. If it does so then that datagram willbe discarded and an error message wil be issued. However the strict source routing is not generally Preferred even by the regular users of the Intemet, as they are not much aware of the phys pology wysical toy of the Internet. :(6-2090) Fig, 1.10.5 : Format of strict source root option 1.10.5 Loose-Source-Root Option : This option is similar to the strict source root option discussed earlier. However this option is not as strict as the strict source root option, itis more relaxed. Here each router whose IP address is mentioned in the list must be visited by the datagram as before but the l ‘Unused (All zeros) ‘tthe rocolved IP datagram including ‘of datagram data (G-2110) Fig. 1.12.6 : Format of parameter problem error message 1.12.5 Redirection Error Messagi If a router or host wants to send a packet to another network then it should know the IP address of the next router. = The routers and hosts must have a routing table to find ‘the address of the next router and the routing table has to be updated automatically on a continuous basis. The redirection message is used for such updating. TechKaowledsNetwork Layer & p, ACN ( /Sem. SMSBTE) oe, Code = 2: Redirection f0€ dome fr ang ~The ICMP sends a redirection message back to its host to.carry out an automatic periodic updating ~ In order to ensure higher efficiency, the hosts do not Participate in the process of routing table update. This {5 because the number of hosts in the Internet is much higher than the number of routers. ~ If the routing tabies of hosts are updated dynamically then it creates an unwanted trafic. ~ Generally the static routing is used by the hosts. That means the routing table of a host contains limited ‘number of entries. Generally a host knows the IP ‘address of only one router that isthe default router. = Due to this, a host can send a datagram which is destined for another network, to a wrong router. ~ Here the datagram receiving router will route the datagram the correct router. However it sends a redirection message to the host to update the routing table of the host. = Fig. 1.12.7 shows the format of the redirection error message, fea iB ho [Tyee 18 [cose “Oto EE Ghecksum 7] (6-211) Fig. 1.12.7 : Format of the redirection message ‘As shown in Fig. 1127, the second row of the redirection message contains the IP address of the appropriate target router. It is important to understand that the redirection message is different from the other error message eventhough it is considered as an error reporting message. What is the difference ? In this case the router does not discard the erroneous datagram. Instead itis sent to the ‘appropriate router. This process of redirection is narrowed down by the contents of the code field as follows: 1 Code = 0 : Redirection will be for a network specific route. 2. Code = 1: Redirection is to be done for a host specific route. specific route and based UPON 8 specifc yo." service Code = 3 : Redirection is to be done for sj, speci rote on the Bass of SPecied yg, service. “A route sends the redirection message back 1.13 Query Messages. (ICMPv4) : Query ‘The ICMP can diagnose some of the network problem, This is in addition with the error reporting feature. Suc, 1a diagnosis is done through the query messages. ‘The query messages is a group of five different pais 1 messages as shown in Fig. 1.13.1. 4. Echo request and reply 2, Time stamp request and reply 8, Address mask request and reply 4. Router solatation and advertisement 65. Information request and reply messages (G-2112) Fig. 1.13.1 : Query messages However out of these five pairs of messages, only two pairs are being used today. They are : 1. Echo request and reply. 2. Timestamp request and reply. Let us discuss them one by one. 1.13.1 Echo Request and Reply : This pair of query messages has been designed for the diagnostic purpose. This pair of messages is utilized by the network managers and users for identifying the network problems, This pair of query messages would determine whether the two given systems (either hosts or routers) can communicate with each other or not. ‘The communication will take place as follows 1. Abost or router sends the echo-request message to another host or router it wants to communicate to. 2. The host or router which receives the echo request message will create an echo-reply message and sends it back to the original sender.ACN (Comp. ‘Sem. SMSBTE) 1 = We can also use the echo-request echo-reply pair to determine if the IP level communication is present or not = The network managers can use the echo request and echo reply pair of messages to check the operation of P protocol = Ahost can also use this message pair to see if another host is reachable or not. At the users level, this is done by invoking the packet Internet groper command ing). = Now a days a version of ping command is provided by most systems which can create a string of echo-request ‘and echo-reply messages for providing statistical information. = It is also possible to check whether a node is functioning properly or not with the help of the echo- request echo reply pair of messages. The format of the echo request echo reply pair of messages is as shown in Fig. 1132. mf 00m he 10th (G-2119 Fig. 1.13.2: Echo request and echo reply messages = _InFig. 1132, the protocol does not formally define the identifier and sequence number fields. Therefore the sender can use them in an arbitrary manner. 1.13.2 Timestamp Request and Reply : = This pair of messages can be used by the hosts and routers to find out the round trip time that an IP datagram needs to travel between them. = Itcan also be used for synchronizing the clock signals sed in the two machines (hosts or routers). ~ Fig. 1133 shows the format ofthese two messages. | (G-2116 Fig. 1.13.3 : Format of timestamp request and timestamp reply messages Network Layer & Protocols = As shown in Fig. 1.133, there are three timestamp fields ‘and each field is 32-bit long. The number in each of these fields represents time in milliseconds from the ‘midnight in Universal time = Eventhough, the 32 bit field can represent a number between 0 and 4,294,967,295 but @ timestamp in this ‘case can have the maximum value of 86,400,000 = 24 x 60 x 60 x 1000. = The timestamp request message is created by the source. It fills the original timestamp field at departure time, and fills the other two timestamp fields will zeros. — The timestamp reply message is created by the destination host. The original timestamp value from the timestamp request message is copied as itis into the ‘original timestamp field in the timestamp reply message, by the destination. = The destination then fills up the receive timestamp field by the time at which the request was received. — At the end the destination fils up the transmit timestamp field with the departure time of the reply message. ‘Computation of one way or round trip time (RTT) : — We can use the pair of timestamp messages to compute the one way or RTT i.e. the time required by the datagram to travel from source to destination and then come back to source again, as follows : Sending time = receive timestamp - original timestamp. Receiving time = retumed time - transmit timestamp. Rount trip time = sending time + receiving time. = If we want the calculations of the sending time and receiving time to be accurate, then the two clocks in the source and destination computers should be synchronized. = But the calculation of RTT will be correct even if the locks at the source and destination machines are not synchronized. = We can calculate the one way time duration by dividing the RTT by two. 1.13.3 Deprecated Messages : IETF has declared the following three pairs of query messages as obsolete : 1L.__ Information request and reply messages. TecNetwork Layer & p,, ‘ACN (Comp. /Sem. SMSBTE) 138 ; Obtain the checksum by complementing the.” 2. Address mask request and reply messages. 3. Router solicitation and advertisement. 1. The Information request and reply messages = — These messages are not used now a days because the Address Resolution Protocol (ARP) is doing their duties. 2. Address mask request and reply : - The IP address of a host contains a network ‘address, subnet address and host identifier. - A host may know its full IP address but may not know it is divided into three parts mentioned above. — Sot can send an address mask request message to the router. The router then sends back the address mask reply message. — These messages are not being used today because their duties are done by the Dynamic Host Configuration Protocol (DHCP). 3. Router solicitation and advertisement : ‘A host that wants to send data to a host on another network must know the address of routers connected to its own network. In such situations the router solicitation and advertisement messages can help. ‘A host can broadcast or multicast a router solicitation message. The routers receiving this message can broadcast their routing information using the router advertisement message. ‘These messages are not being used today because their duties are done by the DHCP. 1.13.4 Checksum : Earlier we have discussed the concept of checksum. In ICMP, the entire message (including the header and data) is considered for calculation of checksum. Checksum calculation : ‘The checksum calculation is done at the sending end by following the steps given below : 1. Set the checksum field to zero. Calculate the sum of al the 16 bit words including header and data. 2 calculated in step 2 ‘4. Store the checksum in the checksum fil Checksum testing = the folowing steps are followed by the receiver yi, {Vs complement arithmetic the sum of all words (header and data 1. Calculate 2. Complem 3 Accept the message if the result obtained inst, 2 is 16 zeros. Otherwise the message is rejected, 1.14 Debugging Tools In the Internet many tools can be used for debugging ility of @ router or host. The rent the sum calculated in step 1. We can decide the feasi route of a packet can be traced. Following are the tools in which ICMP is used for debugging : 1 Ping 2. Trace route or Tracert 1.14.1 Ping: Ping program is used to search whether a host is fing or alive. To the destination, the source host respor sends ICMP echo-request message if it is alive destination responds with ICMP echo-reply messages. Inthe echo-request and echo-reply messages the ping program sets the identifier fields and it starts with the sequence number 0. When a new message is sent, at every time this number is incremented by 1. = Ping can compute the round trip time. In the data section of message it adds the sending time in it. - After arriving the packet, the round trip time can be \ calculated as follows : Round trip time = Departure time of packet - Arrival time of packet - To understand the concept of ping, consider the following example, which shows how we send a ping ‘message to particular sit.W ACN (Comp. Sem. SMSBTE) Example We will use ping to test the server xyz edu having IP ‘address 192,181.84 with 64 bytes of data. $ Ping xyz-edu ing xyz edu (192.181.8.4)-56 (84) bytes of data a 62 6 62 a e 6 3 | 62|1.00ms a 4 _|62|191ms Cc 8 _| 62 | 200ms “4 6 _|62| 201ms ~ x72: edu ping statistics is as follows : 7 packets are transmitted. 7 packets are received. (0% packet loss. Minimum rtt (round trip time) = 1.90 mS Average rtt = 1.95 mS awe ene Maximum rtt = 2.04 mS ‘The ping starts sending messages from sequence number 0 which gives us RTT time for each probe. - In the IP datagram which encapsulates an ICMP message has been set to 62 it means that the packet can travel only upto 62 hops. Ping defines 56 number of data bytes at the beginning and 84 total number of bytes (56 + 8 bytes of ICMP header + 20 bytes of IP header = 84 bytes). The ping program continues message sending with 64 bytes in each probe (56 + 8 = 64 bytes). With interrupt key (ctrl + C) we can stop ping program. The statistics of probe is printed after interruption. The statistics includes the number of sent and received 1-39 Network Layer & Protocols, Packets, minimum, maximum and average round trip time, packet loss etc. 1.14.2 Traceroute or Tracert : — From a source to the destination to trace the path of a packet in UNIX, the traceroute program and in Windows, tracert program can be used. = Traceroute / tracert can search the IP addresses of all the routers which are visited in that path. = Usually the program is set for checking of maximum 30 routers (hops) to be visited = Inthe internet normally the number of hops is less than this. Traceroute : = The ping and traceroute programs are different. To get the help the ping program uses two query messages whereas traceroute program uses two error reporting messages namely time-exceeded and destination- unreachable. This is application layer program in which only the client program is required. There is no server traceroute program because in the destination host, client program never reaches the application layer. ~ Ina UDP user datagram the traceroute program is encapsulated. A port number which is not available at ‘the destination is intentionally used by traceroute program. = The traceroute sends (n + 1) messages if there are n routers are present in the path. Each router discards one message that means the first n messages are discarded and the destination host discards the last message. ~The (n + 1) ICMP error reporting messages received are used by the tracceroute client program for searching the path between the routers. = Fig. 1:14. shows the use of ICMPv4 in traceroute program in which the value of n is automatically found, there is no need to the value of TecknowtndstNetwork Layor 4 1-40 See, ACN (Comp. /Sem. 5MSBTE) rs ene message Wl EE 1 row gy ~ From the host A with TTL value 1 the first traceroute message is sent at the first router (RI) this message is discarded which sends ICMP time exceeded error message from which the traceroute program will come 0 know the IP address of the fist router and the name of router, ~ With TTL value 2, second traceroute message is sent which will find the name and the IP address of second router (RZ) Similarty the third and fourth message can find information about R3 and Ré respectively. ‘The fifth message reaches the destination host is also dropped but for some another reason. ie. the Gestination host B cannot be able to find the port ‘number which is specified in the UDP datagram. Then ICMP sends the destination-unreachable message with ‘ode 3 which shows that the port number isnot found. = __The traceroute program after receiving destination- destination is reached. To find the name of the final destination ang tp Xe, it uses the data in the received message, i, fig, 1.14.1 value of nis 4. To find the destination and the roundtrip time og, router the traceroute program sets a timer. To find better estimate for round trip time most, traceroute programs sends three messages wit, . same TTL value to each device. Tracert : In Windows, the tracert program is used which behay, differently than ping and traceroute. In IP datagran the tracert program is directly encapsulated. = Similar to traceroute, the tracert sends echo reque; ‘messages to routers, when last echo request reacte the destination an echo reply message is received,141 W A0N (comp. ‘Sem. SMSBTE) — ri Layer & Protects, 1.15 Mobile IP : Mobile IP is the extension of IP protocol. It has been developed for the mobile and personal computers such as notebook. = Mobile IP allows the mobile computers to get connected to the Internet at any location. 1.15.1 Addressin = Addressing is a very important problem in providing mobile communication using IP protocol We will discuss its solution in this section. 1.15.1.1 Addressing In Stationary Hosts : ~The original IP addressing was designed on the basis of two assumptions 1. The host is stationary. 2. The host is connected to only one network. = _AnIP datagram is routed by the routers on the basis of the IP address. As discussed earlier in this chapter, an IP address is made of two parts: a prefix and a suffix. = Abhost gets associated with a network due to the prefix part ofits IP address. That means a host cannot carry its IP address with itself from one place to the other. ~ That means with change in place, the network changes and so does the IP address of the host. = Routers use the fixed association between a host and its ‘network for routing the packets to the network to which the host is attached. 1.15.1.2 Mobile Hosts : ~The IP addressing structure needs to be changed when ‘host moves from one network to the other. To achieve this, various solutions have been suggested. - Two of them are as follows : |. Changing the address : - One of the solutions is to allow the mobile host to change its IP address as it changes the network. = This can be achieved by using DHCP. The mobile host can obtain a new IP address using DHCP and {get associated with the new network = But this technique has many drawbacks. Some of them are as follows Drawbacks 1. We need to change all the configuration files. 2. The mobile host would need rebooting, everytime it ‘moves from one network to the other. 3, It would be necessary to revise the DNS table everytime so that all the other hosts on the Internet are aware of this address change. 4, If the mobile host moves from one network to the other when transmission is taking place, then the exchange of data will be interrupted because during the transmission, the client and server cannot change their port and IP addresses. 2. Twoaddresses : = Due to all the drawbacks of the first approach, the second approach of using two IP addresses for a mobile host is tried out and it is found to be a more feasible approach. = The two IP addresses assigned to a mobile host are: 1L.Home address and 2. Temporary address = The home address is the original IP address of the mobile host, and the temporary address is called as the care of address. ~The home address associated the host with its home network (ie. the network which is permanent home of the host and it is its permanent IP address. ~ When the host moves to the other network, its temporary (care-of) address changes. This care-of address associates the host with the foreign network. 1.15.2 Agents : ~ Alhome agent and a foreign agent are required for ‘making the change of address transparent to the rest of Intemet. TechKnowtedgs(G-2257) Fig. 1.15.1 : Home agent and foreign agent In Fig. 1.15.1 the home and foreign agents have been shown as routers. However actually they act as a router as well as a host. Home Agent : - A home router is basically a router attached to the home network of a mobile host. When a remote host sends a packet to the mobile host, the home agent acts on behalf of the mobile host, receives the packet and sends it to the foreign agent. Foreign Agent : A foreign agent is a router connected to the foreign network. The packets sent by the home agent are received by the foreign agent and delivers them to the mobile host. Sometimes, a mobile host itself can act as foreign ‘agent. Then there is no need of using a separate foreign agent. ~ For thi, the mobile host should have the ability to receive a care-of address on its own. This can be done using DHCP. ~ _In addition to this @ special software needs to be installed at the mobile host to enable it to communicate with the home agent and to have the two addresses (home and temporary) Network Layer & p at keep the dual % ACN (Comp. /Sem. 5/MSBTE) Lis necessary 10 ceep ua san - I parent to the application programs, trans me The position of home agent with respect to the ho it to ‘network and that of the foreign agent with respect the foreign network are shown in Fig. 1.15.1. led as coll ‘The care-of-address is call located - mobile host itself is acting. address if 2 "9 as 5, foreign agent. The use of collocated care-of address ha, advantage that the mobile host can move ig, foreign network without even thinking about, availability of the foreign agent. However its disadvantage is that an extra softy, needs to be installed with the mobile host. 1.15.3 Three Phases : The communication of a mobile host with 2 remo, host goes through the following three phases : 1 2 3 Agent discovery Registration Data transfer. All these phases are shown in Fig. 1.15.2. Phase-|: Agent Discovery (Steps 1 to 4) : This is the first phase in mobile communication. t Consists of the following two subphases : 1. Agent solicitation and 2. Agent advertisement. ‘A mobile host must leam the address of (discover) its home agent before moving to any foreign ‘network (Steps 1 and 2). The mobile host must also lear the address of (discover) the foreign agent ‘once it moves to a foreign network (Steps 3 and 4) This process of address leaming includes leaming Of both the care-of address and the foreign agents address, The agent discovery phase involves the discovery of home and foreign agents, This Process requires {he use of two messages namely : 1. Advertisement ‘Message and 2. Solicitation method. CO143 Network Layer & Protocols, ACN (Comp. Sem. SMSBTE). cae The Tine (¢-z2snFig. 1.15.2 : Communication between mobile host and remote host 2. Phased! : Regletration (Steps & to 8): 3. Phase-ll: Data Transfer : ~ _Thisis the second phase of mobile communication. The mobile host frst moves to the foreign network ‘and discovers the foreign agent (Phase-D. = This is the third phase in mobile communication after the agent discovery and registration. In this phase the mobile host can communicate with the remote host as shown in Fig. 1.15.2. 1. From Remote Host to Home Agent : ~ Ifa packet is to be transferred from the remote host to mobile host, then the remote host uses its agent (Sep). address as the souree address and home address 2. Registration of mobile host with its home of mobile host as destination address. agent. This is normally done by the foreign ‘agent on behalf of mobile host (Step 6). After this it must undergo the registration phase, which comesponds to steps 5 to 8 in Fig. 115.2. = The four aspects of registration are as follows : 1. Registration of mobile host with the foreign ~ But practically the home agent is pretending as the mobile. So it will intercept the packet with the help 3 The mobile host must renew its registration if of proxy ARP. the registration has expired. ~ Thus the communication from remote host to 4, The mobile host is supposed to cancel its mobile host actually takes place between the registration when it retums back to its home remote host and home agent as shown in network. Fig. 1.15.2. ~The registration request and registration reply ~The mobile communication between the Remote messages are used as shown in Fig. 1152 for Host and Home agent has been marked by @ thick registration of mobile host with the home agent path marked by °1* in Fig. 125.2(a). and foreign agent.4 Pesce 2. From Home to Foreign Agent : 3. From Foreign Agent to Moblie Host : 4. From Mobile Host to Remote Host : destination address. (G-2259)Fig. 1.15.2(a): Data transfer from remote host to home agent ~ AS the packet is received by the home agent it sends the packet to the foreign agent using the concept of tunneling. ‘The home agent encapsulates this received IP Packet into a new IP packet by using its own address as the source address and foreign agents ‘address as the destination address and sends this new IP packet to the foreign agent as shown by the thick path marked by °2" in Fig. 115.2(a, From the IP packet received the foreign agent will recover the original packet by decapsulation Process, However the recovered original packet has the hhome address of mobile host as its destination address. The foreign agent will refer to a registry table and finds the eare-of-address of the mobile host. The Original packet is then sent to the care-of-address 35 shown by the thick path marked by °3* in Fig. 1152). If @ mobile host wants to send a packet to a remote hes it does itn anormal way To do this the mobile host creates a packet wih ts home address (and not the Care-of-address) as Source address and remote host's address ag the = Its very important to note that eventhough ty packet originates from the foreign network, ith, the home address of the mobile host. This communication has been shown by the thik path “4” in Fig. 115.2(a). 1.15.4 Transparency : In the entire data transfer Process, the remote host absolutely does not know anything about the Movement of the mobile host. Because, the remote host uses the home address as the destination address when sending a packet to the mobile host. ‘Similarly the mobile host uses its home address as the Source address while sending a packet to the remote host, mobile host is totally transparent because the Fest of the {ntemet has absolutely no idea about the Movement of the mobile host, 1.15.5 Inefficiency in Mobile Ip ; The communication done with the help of mobile IP can be moderately to severly inefficient1 ACN ( ‘Sem. 5MSBTE) 145 twork Layer & Protocols, (G-2260)Fig. 1.15.3. Double Crossing or 2X : Now consider a situation in which a remote host wants to communicate with a mobile host which has moved to the same network as that of the remote host as shown in Fig. 1.15.3. This is called as a double crossing or 2X case ie. the case of severe inefficiency. ‘As discussed earlier, a mobile host can send a packet directly to the remote host. Therefore there {sno loss of efficiency in this communication. However if the remote host wants to send a packet to the mobile host then it cannot do so directly (via the dotted direct path in Fig. 125.3). Instead the remote host has to send the packet first home agent (path-1 in Fig. 1.15.3) and the home agent will route the packet to the mobile host (path-2 in Fig. 115.3). In this the packet has to cross the Internet twice. Thus the resources are used twice unnecessarily in this communication which reduces the efficiency severely. Hence the double cross case is called as the case of agent Double crossing (G-2260Fig. 1.15.4 : Triangle routing = In this situation as well if a mobile host wants to send a packet to @ remote host it can do so directly without any loss of efficiency. = But when a remote host wants to send a packet to a mobile host the packet has to first travel to the home agent and then to the mobile host as shown in Fig. 1154, ~ Thus the packet has to travel along two sides of @ triangle instead of only one whichis the direct path shown by a dotted line in Fig. 1.15.4 1.15.6 Remedy : severe inefficiency. Trlangle Routing or Dog Leg Routing : ‘triangle routing or dog leg routing is a case of moderate inefficeny. It occurs when a remote host wants to send a packet to the mobile host which is Not attached to its own (remote) network. This situation is illustrated in Fig. 1.15.4. Binding the care-of-address to the home address of ‘mobile could be one of the solutions to the problem of inefficiency. That means when the home agent receives the first packet from the remote host and sends it to the foreign ‘agent it should also send an update binding packet to the remote host. By doing this itis ensured that all the future packets to this mobile host can be sent to the ‘care-of-address rather than home address. The remote host can save this information in a cache However this remedy also has an inherent flaw. The cache entry would become outdated as the mobile host moves to a new network.\CN (Comp. /Sem. S/MSBTE) To avoid this the home agent must send a warning packet to the remote host to inform that the mobile host has moved to a new network. 6 __Virtual Private Networking (VPN) a3 a4 Due to Internet software, it appears that the Internet is 2 single, seamless system of communication to which lots of networks containing a large number of computers are connected. The internal details of these real or actual networks get hhidden when they become a part of the Internet. Every computer connected to the Intemet has it own unique address assigned to it. The users of the Internet do not have to bother about the internal structure of the physical networks and the details related to them. Thus the user is a part of @ virtual network. Intemet is thus the best example of virtual networks. The concept of virtual networks states that in such types of networks, different computer networks are not only connected together but you feel that they are a part of 2 big single network. The concept of virtual networks is illustrated in Fig, 1.16.1. Explain the connection oriented and connectionless services. Why modem computer use dynamic routing ? Explain with example how distance vector routing is used to route the packet and why count-to-nfinty problom arises and how does it get solved. What is fragmentation ? White short notes on : Hierarchical routing, 1-46. as a6 a7 a8 ag a.10 an a2 13 a4 15 a6 ai 0.18 a9 20 21 a2 23 24 27 Q.28 Q.29 routing. port notes O° we ork layer, waite s! nt protons ine nem Name aire P. ‘of ARI Explain ue es proadast but ARP 1g, why is unicast ? wit ceo in IP. mentation in IP. Explain fragt packet in IP? Whatis the name of & Explain the IP header. how is fragmentation related tj write @ note on ICMP. Name and describe three types of IPv6 addresses, What is unicast routing ? What is multicast routing ? Write a note on mobile IP. What is fragmentation ? Explain how is it suppor in IPv4 and IPv6. Explain the addressing scheme in IPv4 and IPt When IPv6 protocol is introduced, does the ARF protocol have to be changed ? Explain. What is fragmentation ? Explain how it is supporte: in IPv4 and IPv6. Given an IP address, how will you extract its net « and host id. What is PING utility ? How many ways are there t implement PING ? Explain steps. What is subnetting in IP network, explain wit suitable examples. Why is an ARP Query sent within a broadcas frame ? Why is an ARP response sent within ¢ {frame with a specific destination LAN address ? ‘A network on the intemet has a subnet mask o (255.255.240.0. What is the maximum number ¢ hosts it can handle ? ‘An IP datagram using the strict source routing opi" has to be fragmented. Do you think the option # copied into each fragment, or is it sufficient 10 Put itis the fist fragment ? Explain your answer aa Wy eaten