IP

Internet Protocol (IP) Universal service in a heterogeneous world IP over everything Globally unique logical address for a host

IP Addressing A 32-bit number that uniquely identifies a location Written using dotted decimal notation

IP Addressing IP address is assigned to each network interface Routers connect two or more physical networks Each interface has its own address Multi-homed host A host having multiple connections to Internet Multiple addresses identify the same host Does not forward packets between its interfaces

IP Packet

IP “Classful” Addressing Scheme Three unicast address classes: A, B, and C One multicast: class D network host 10 110 1110 multicast address A B C D class 1.0.0.0 to 127.255.255.255 128.0.0.0 to 191.255.255.255 192.0.0.0 to 223.255.255.255 224.0.0.0 to 239.255.255.255 32 bits

IP Addressing first 24 bits are network address LAN 223.1.1.1 223.1.2.1 223.1.1.2 223.1.1.4 223.1.2.9 223.1.2.2 223.1.1.3 223.1.3.27 LAN 223.1.3.1 223.1.3.2

Classless Inter-Domain Routing Classful addressing scheme wasteful IP address space exhaustion A class B net allocated enough for 65K hosts Even if only 2K hosts in that network Solution: CIDR Eliminate class distinction No A,B,C Keep multicast class D

Classless Addressing Addresses allocated in blocks Number of addresses assigned always power of 2, and always on the boundary. That is, if 2048 addresses, it will start with some address with all lower 11 bits being 0. Network portion of address is of arbitrary length Address format: a.b.c.d/x x is number of bits in network portion of address 11001000 00010111 00010000 00000000 network part host 200.23.16.0/23

Allocating Addresses Assume abundant addresses are available starting at 194.24.0.0. Cambridge university needs 2048 addresses, it is given 194.24.0.0 to 194.24.7.255. Mask 255.255.248.0. Oxford need 4096 addresses. Because the requirement is that must be on the boundary, it is given 194.24.16.0 to 194.24.31.255. Mask 255.255.240.0. Edinburg needs 1024 addresses, is given 194.24.8.0 to 194.24.11.255. Mask 255.255.252.0.

CIDR A router keeps routing table with entries IP address, 32-bit mask, outgoing line When an IP packet arrives, the router checks its routing table to find the longest match. Match means anding the IP address with the network address mask (1111…10000), and check if the result is the same as the network address.

CIDR Example. Cambridge 194.24.0.0/21 194.24.0.0 -- 194.24.7.255 Edinburgh 194.24.8.0/22 194.24.8.0 -- 194.24.11.255 (Available) 194.24.12.0/22 194.24.12.0 -- 194.24.15.255 Oxford 194.24.16.0/20 194.24.16.0 -- 194.24.31.255 When a packet addressing to 194.24.17.4 arrives, where should it be sent to? And with all masks, find one that matches the longest.

CIDR – Entry aggregation How does a router in Tallahassee route packet to C,E and O, assuming that he has only two outgoing links? All to New York. Can it reduce the size of its routing table? C E N O H T

CIDR Entry Aggregation From 194.24.0.0 to 194.24.31.255, all to N. So aggregate the three entries into one 194.24.0.0/19. The N router can do the same thing. C E N O H T

CIDR If later the free address space 194.24.12.0/22 194.24.12.0 -- 194.24.15.255 is assigned to Pittsburgh and has to go through Houston, what should the router at Tallahassee do? C E N P O H T

CIDR When a packet arrives addressing 194.24.15.8, the router checks the routing table and there will be two matches: 194.24.12.0/22 and 194.24.0.0/19. Pick the longest match.