Download 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.