Transcript Week10_2
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 class A 0 network B 10 C 110 D 1110 1.0.0.0 to 127.255.255.255 host network 128.0.0.0 to 191.255.255.255 host network multicast address 32 bits host 192.0.0.0 to 223.255.255.255 224.0.0.0 to 239.255.255.255 IP Addressing first 24 bits are network address 223.1.1.1 223.1.2.1 223.1.1.2 223.1.1.4 223.1.1.3 223.1.2.9 223.1.2.2 223.1.3.27 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 network part 11001000 00010111 00010000 00000000 200.23.16.0/23 host part 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 Edinburgh 194.24.8.0/22 (Available) 194.24.12.0/22 Oxford 194.24.16.0/20 194.24.0.0 -- 194.24.7.255 194.24.8.0 -- 194.24.11.255 194.24.12.0 -- 194.24.15.255 194.24.16.0 -- 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 P E N 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.