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21-IP addressing IPv4

21-IP addressing IPv4. Dr. John P. Abraham Professor UTRGV. Lecture notes. I would strongly suggest you pay close attention to my lecture and take good notes. I go far beyond the textbook on this topic. Addressing. Physical

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21-IP addressing IPv4

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  1. 21-IP addressing IPv4 Dr. John P. Abraham Professor UTRGV

  2. Lecture notes • I would strongly suggest you pay close attention to my lecture and take good notes. I go far beyond the textbook on this topic.

  3. Addressing • Physical • MAC addresses do not suffice because the Internet can include multiple network technologies and each technology defines its own MAC addresses. • Will Discuss more later • Logical – see next slide

  4. Logical Address • The advantage of IP addressing lies in uniformity: an arbitrary pair of application programs can communicate without knowing the type of network hardware or MAC addresses being used

  5. IP address • 32 bits • Hierarchy – prefix and suffix • Prefix identifies the network – given by IANA In classful addressing, the network address (the first address in the block) is the one that is assigned to the organization. The range of addresses can automatically be inferred from the network address. • Suffix identifies the computer –given locally • No two computers can have the same public IP

  6. Octets • IP address is written in 4 octets with separating dots. • 10000000.00001011.00000011.00011111 • 128.11.3.31 • One of the UTRGV CS address is: • 10000001.01110001.10000010.01101110 • What is this address in decimal notation?

  7. Classful IP addressing • Divided IP address space into three primary classes A, B, C and also there exist class D (multicasting) and E. • First four (MSB) bits will determine its class

  8. Finding the address class

  9. Figure 4.5Finding the class in decimal notation

  10. Each student must do these independantly. Find the class of each address: a. 227.12.14.87 b.193.14.56.22 c.14.23.120.8d. 252.5.15.111 e.134.11.78.56 Solutiona. The first byte is 227 (between 224 and 239); the class is D.b. The first byte is 193 (between 192 and 223); the class is C.c. The first byte is 14 (between 0 and 127); the class is A.d. The first byte is 252 (between 240 and 255); the class is E.e. The first byte is 134 (between 128 and 191); the class is B.

  11. LAN The network portion of all IP addresses should be the same on all machines within a LAN You can use VLAN to put machines from different networks together artificially. All computers within a LAN can be made visible to others in the LAN Machines in two different LANs can communicate using a Bridge.

  12. LAN One should be able to share files with others within the LAN. Note: a virtual LAN can be created with differing IPs. An user of the file should be able to MAP the directory and give a drive letter (in windows) Anyone wanting to connect to machine that does not belong the LAN must go through a routing table (usually located within the Router)

  13. Transition to IPv6 • When IP4 was defined only few computer networks existed • Each year the networks are nearly doubling • Changing IP addressing scheme is very costly. So, IPV4 is still used.

  14. Authority for IP addresses • The IP address space is managed globally by the Internet Assigned Numbers Authority (IANA), and by five regional Internet registries (RIRs) • Internet Corporation for Assigned Name and Numbers (ICANN) performs the actual technical maintenance work of the Central Internet Address pools and DNS root zone registries pursuant to the Internet Assigned Numbers Authority (IANA) function contract. • As of Oct 1, 2016 IP distribution authority no longer controlled by USA (President Obama ended it)

  15. Subnet Addressing • The network address is the beginning address of each block. It can be found by applying the default mask to any of the addresses in the block (including itself). It retains the netid of the block and sets the hostid to zero.

  16. Subnets • Makes smaller networks for better security and easy management • Mask is used by routers to find the network portion of the address. AND operation.

  17. Creating subnet masks • Determine how many subnets you need • How many bits would be required to create that many subnets? • For example you need 8 subnets, need 3 bits set aside. Certain addresses are reserved, so all subnets can’t be used. • Turn those bits (msb) to 1. • For example class C, default subnet is 255.255.255.0, if you decided to subnet it to 8 subnets, then you would use the subnet 255.255.255.224. I will explain in class.

  18. Example 12 Given the address 23.56.7.91, find the beginning address (network address). SolutionThe default mask is 255.0.0.0, which means that only the first byte is preserved and the other 3 bytes are set to 0s. The network address is 23.0.0.0.

  19. Example 15 What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0? SolutionWe apply the AND operation on the address and the subnet mask. Address ➡ 11001000 00101101 00100010 00111000 Subnet Mask ➡ 11111111 11111111 11110000 00000000 Subnetwork Address ➡ 11001000 00101101 00100000 00000000.

  20. Figure 4.25Comparison of a default mask and a subnet mask

  21. Private IPs Table 4.3 Special addresses

  22. CIDR notation • Classless Inter-Domain routing (CIDR) • Ddd.ddd.ddd.ddd/m 192.5.48.69/26 • Used for subnetting and supernetting • An ISP can now subdivide a larger blocks and hand out 1, 2, 4, 8,.. (power of 2) to the customers. These are called variable length bocks and we use a slash notation.

  23. Slash notation • In classless addressing we need to include the prefix length to each address if we need to find the block of the address. • Given the address 167.199.170.82/27 the netmask is 255.255.255.224 (27 ones).

  24. Calculations • Given the address 110.23.120.14/20 find the number of address, the first and last address in the block • There are 20 1s. So 255.255.?.0 11110000 is 240. The mask is 255.255.240.0 • Number of addresses is 32-20, so 212 or 4096 • To find the first address AND the given address with the mask. 110.23.120.14 AND 255.255.240.0 = 110.23.112.0 (third octet 01110000 AND 11110000 • To find the last address: OR the address with the complement of the mask. 110.23.127.255 (also we can keep all the 32-20 left most bits of the IP and change the 12 rightmost to 1s.

  25. 10.4.12.0/22 • Mask is 255.255.252.0 • First IP 10.4.00011000.0 AND 255.255.11111100.0 =10.4.00011000.0 = 10.4.12.0 • Number of IP is 32-22 = 10. 210 =1024 • Last 10.4.15.255

  26. Block Assignment • The number of IP addresses requested should be power of 2 • The requested block needs to be allocated where there are a contiguous number of unallocated addresses in the address space. • Suppose a block of 1000 IP requested. • Power of 2 is 210 which is 1024. • Prefix length is 32-10 which is 22. • From the available address space, say 18.14.12.0/22 (which is divisible by 1024) is given. Last address is 18.14.15.255

  27. Special addresses • All zeros – When a device does not know its on IP address, it requests one using this address as the source and sends a limited broadcast asking for an IP. Therefore, all zeros means, me. • All ones. Limited broadcast address in the current network. • Loopback address. 127.0.0.0 is localhost. Used to test a software on a machine.

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