EEC484584 Computer Networks Lecture 13 Wenbing Zhao wenbingieee
EEC-484/584 Computer Networks Lecture 13 Wenbing Zhao wenbing@ieee. org (Part of the slides are based on Drs. Kurose & Ross’s slides for their Computer Networking book, and on materials supplied by Dr. Louise Moser at UCSB and Prentice-Hall) Spring Semester 2006 EEC-484/584: Computer Networks
2 Outline • • • Classful IP address allocation CIDR Network address translation ICMP IPv 6 Spring Semester 2006 EEC-484/584: Computer Networks 2
3 IP Addresses • Classful addressing - every host and router has unique IP address consisting of network number and host number (2 level hierarchy) – E. g. , Class A: up to 27 = 128 networks with up to 224 = 16, 777, 216 hosts each • Network numbers are managed by ICANN (Internet Corporation for Assigned Names and numbers) to avoid conflicts • No longer used, but references to it are still common Spring Semester 2006 EEC-484/584: Computer Networks 3
4 IP Addresses Spring Semester 2006 EEC-484/584: Computer Networks 4
5 IP Addresses • IP address are usually written in dotted decimal notation – Each of the 4 bytes is written in decimal, from 0 to 255 – Lowest IP 0. 0, highest 255 • Special IP addresses Spring Semester 2006 EEC-484/584: Computer Networks 5
6 Subnets • Allow a network to be split into several parts for internal use, but to act as a single network to outside world • Take some bits away from host numbers • Subnet mask – needed by the main router. Indicates split between network + subnet number and host – Write the address and the mask as a binary number – If mask bit is 1, then corresponding bit of address matters Spring Semester 2006 EEC-484/584: Computer Networks 6
7 Subnets • E. g. , A class B network can be subnetted into 64 subnets – Originally 16 bits for host info. Now, 6 bits used for subnet and 10 bits for host numbers – Subnet mask can be written as 255. 252. 0 or /22 Subnet 1: 10000010 00110010 000001 00 00000001 Subnet 2: 10000010 00110010 00 00000001 Subnet 3: 10000010 00110010 000011 00 00000001 Spring Semester 2006 EEC-484/584: Computer Networks 130. 50. 4. 1 130. 50. 8. 1 130. 50. 12. 1 7
8 Routing without Subnets • Each router has a table listing two types of entries: – (network, 0): tells how to get to distant networks – (this-network, host): tells how to get to a local host • When an IP packet arrives, its destination address is looked up in the routing table – If the packet is for a distant network, it is forwarded to the next router on the interface given in the table – If it is for a local host, it is sent directly to the destination – If the network is not present, the packet is forwarded to a default router with more extensive tables Spring Semester 2006 EEC-484/584: Computer Networks 8
9 Routing with Subnets • Three-level hierarchy: entries in a routing table take the form – (network, 0) – (this-network, subnet, 0) and – (this-network, this-subnet, host) • A router on subnet k knows how to get to all the other subnets and also how to get to all the hosts on subnet k • If a packet is for this network – Do a Boolean AND of the destination address with the network’s subnet mask to get rid of the host number – Look up the resulting address in the routing table Spring Semester 2006 EEC-484/584: Computer Networks 9
10 Routing with Subnets • For example, a packet address to 130. 50. 15. 6, the subnet mask is 255. 252. 0/22, AND them, we get 130. 50. 12. 0 and this address is looked up in the routing table to find out which output line to use Subnet base address 10000010 00110010 00000100 0000 Output to line A 10000010 00110010 00001000 0000 Output to line B 10000010 00110010 00001100 0000 Output to line C 10000010 00110010 00001111 00000110 (Destination IP Addr: 130. 50. 15. 6) 1111111100 0000 (Subnet mask: 255. 252. 0) 10000010 00110010 00001100 0000 (After AND operation) Spring Semester 2006 EEC-484/584: Computer Networks 10
11 Problems with Classful Addressing • A class is obviously too large for any organization • C class is too small (only 256 addresses available) • B class is requested and allocated, but it is still too large for most organizations Many IP addresses are wasted Shortage of IP addresses Spring Semester 2006 EEC-484/584: Computer Networks 11
12 CIDR – Classless Inter. Domain Routing • For the remaining IP addresses, classless allocation is used – Allocate remaining IP addresses in variable-sized blocks (must be power of 2), without regard to the classes – The starting address must fall on the boundary of the block size – E. g. , if a site needs, say, 2000 addresses, it is given a block of 2048 addresses on a 2048 -byte boundary Spring Semester 2006 EEC-484/584: Computer Networks 12
13 Classless Allocation – Example 5 -59 • Routing tables are updated with the three assigned entries. Each entry contains a base address and a subnet mask (in short: base address/subnet mask) C: 11000010 00011000 00000000 11111111000 0000 E: 11000010 00011000 00000000 1111111100 0000 O: 11000010 00011000 00010000 111111110000 Base address Spring Semester 2006 Subnet mask EEC-484/584: Computer Networks 13
Classless Inter. Domain Routing • Each routing table is extended by giving it a 32 -bit mask • The routing table contains entries of (IP address, subnet mask, outgoing line) triples • When a packet comes in, its destination IP address is first extracted • Then, the routing table is scanned entry by entry, masking the destination address and comparing it to the table entry looking for a match • If multiple entries (with different subnet mask lengths) match, the longest mask is used – E. g. , if there is a match for a /20 mask and a /24 mask, the /24 mask is used Spring Semester 2006 EEC-484/584: Computer Networks 14 14
Classless Inter. Domain Routing: Example 15 • If a packet is addressed to 194. 24. 17. 4, in binary 11000010 00011000 0001 00000100 • First it is Boolean ANDed with the Cambridge mask to get 11000010 00011000 00010000 • This value does not match the Cambridge base address, so next try Edinburgh mask, to get 11000010 00011000 00010000 • This value still does not match, so Oxford is tried, yielding 11000010 00011000 00010000 • This value matches the Oxford base. If no longer matches are found, the Oxford entry is used and the packet is sent along the line named in it C: 11000010 00011000 00000000 11111111000 0000 E: 11000010 00011000 00000000 1111111100 0000 O: 11000010 00011000 00010000 111111110000 Base address Spring Semester 2006 Subnet mask EEC-484/584: Computer Networks 15
16 Classless Inter. Domain Routing • Aggregate entry – all three new entries can be combined into a single aggregate entry 194. 24. 0. 0/19 with a binary address and submask as follows: 11000010 00000000 11111111 11100000 • By aggregating the three entries, a router has reduced its table size by two entries • Aggregation is heavily used throughout the Internet Spring Semester 2006 EEC-484/584: Computer Networks 16
17 NAT – Network Address Translation • Another workaround for the IP addresses shortage problem: network address translation – One public IP address, many private IP addresses – When a packet exits the private network and goes to the ISP, an address translation takes place • Three ranges of IP addresses have been declared as private: – 10. 0 – 10. 255 (16, 777, 216 hosts) – 172. 16. 0. 0 – 172. 31. 255/12 (1, 048, 576 hosts) – 192. 168. 0. 0 – 192. 168. 255/16 (65, 536 hosts) Spring Semester 2006 EEC-484/584: Computer Networks 17
18 NAT – Network Address Translation Placement and operation of a NAT box Spring Semester 2006 EEC-484/584: Computer Networks 18
NAT – What about the Incoming Traffic? 19 • Solution is based on the assumption all traffic is TCP/UDP • TCP/UDP has two port fields, one for source port, the other for destination port, each 16 bits wide • The source port is used as an index to an internal table maintained by the NAT box • The internal sender’s private IP and original port info are stored in the table • When the reply comes back, it will carry the index as the destination port, the NAT box then translates the address back • For both outgoing and incoming address translations, the TCP/UDP and IP header checksums are recomputed Spring Semester 2006 EEC-484/584: Computer Networks 19
20 Drawback of NAT • NAT violates the architectural model of IP, which states that every IP address uniquely identifies a single machine worldwide • NAT box must maintain mapping info for each connection passing through it. This changes the Internet from a connectionless network to a kind of connection-oriented network • NAT violates the most fundamental rule of protocol layering: layer k may not make any assumptions about what layer k+1 has put into the payload field • NAT only support UDP/TCP traffic • NAT has problem supporting apps that include local IPs in payload, such as FTP and H. 323 • Each NAT box can support at most 65, 536 (216) hosts Spring Semester 2006 EEC-484/584: Computer Networks 20
21 Internet Control Message Protocol • ICMP messages are sent using the basic IP header • The first byte of the data portion of the datagram is a ICMP type field – The type field determines the format of the remaining data • Typical format: type, code plus first 8 bytes of IP datagram causing error Destination Unreliable Message Format Spring Semester 2006 EEC-484/584: Computer Networks 21
22 Internet Control Message Protocol • When something unexpected occurs in Internet, the event is reported by routers using ICMP • Principal ICMP message types Spring Semester 2006 EEC-484/584: Computer Networks 22
23 Internet Protocol Version 6 • IPv 4 current version • IPv 5 experimental real-time stream protocol • IPv 6 – Longer addresses than IPv 4 – 16 bytes – Fixed-length 40 byte header • No checksum field • No fragmentation (by intermediate routers) allowed. Only sender is allowed to fragment a packet, using fragment extension – Better support for options Spring Semester 2006 EEC-484/584: Computer Networks 23
24 The Main IPv 6 Header 8000: 0000: 0123: 4567: 89 AB: CDEF Spring Semester 2006 EEC-484/584: Computer Networks 24
25 The Main IPv 6 Header • Version 6 • Priority – 0 -7 slow down in event of congestion – 8 -15 real-time traffic • Flow label – allows source and destination to set up pseudo-connection with particular properties and requirements • Payload length (as opposed to total length in IPv 4) • Next header – additional optional extension header • Hop limit (time to live in IPv 4) • Source/destination address – 128 bits (32 bits in IPv 4) Spring Semester 2006 EEC-484/584: Computer Networks 25
26 IPv 6 Extension Headers Spring Semester 2006 EEC-484/584: Computer Networks 26
27 IPv 6 Extension Headers http: //www. tcpipguide. com/free/t_IPv 6 Datagram. Extension. Headers-2. htm Spring Semester 2006 EEC-484/584: Computer Networks 27
28 IPv 6 Extension Headers • Two types of extension header formats – Fixed format – Variable number of variable-length fields • Each item is encoded as a (type, length, value) tuple • Type: 1 -byte field telling which option this is • Length: 1 -byte field telling how long the value is (0255 bytes) • Value: any info required Spring Semester 2006 EEC-484/584: Computer Networks 28
29 IPv 6 Extension Headers • Hop-by-hop options: used for info that all routers along the path must examine – One option has been defined to support datagrams exceeding 64 K • Next header: 1 -byte field telling type of header • Length field: 1 -byte field telling how long the hop-by-hop header is in bytes, excluding the first 8 bytes, which are mandatory • 1 -byte field indicating that this option defines the datagram size • 1 -byte field telling the size is a 4 -byte number • 4 -byte field: size of datagram Spring Semester 2006 EEC-484/584: Computer Networks 29
30 IPv 6 Extension Headers • Routing option header: lists one or more routers that must be visited on the way to destination – Routing type field: 1 -byte giving format of the rest of the header – Segments left field: 1 -byte keeping track of how many of the addresses in the list have not yet been visited Spring Semester 2006 EEC-484/584: Computer Networks 30
31 Exercise - CIDR • According to Classless Inter. Domain Routing, the remaining IP addresses are allocated in variable-sized blocks, without regard to the classes. However, the starting address must fall on the boundary of the block size allocated. Assuming that a large number of consecutive IP address are available starting at 194. 24. 0. 0. Suppose that three organizations, A, B, and C, request 4000, 1000, and 2000 addresses, respectively, and in that order. For each of these, give the first IP address assigned, the last IP address assigned, both must be in dotted decimal form, and the mask in the w. x. y. z/s notation. Spring Semester 2006 EEC-484/584: Computer Networks 31
32 Exercise - CIDR • A router has just received the following new IP addresses: 57. 6. 96. 0/21, 57. 6. 104. 0/21, 57. 6. 112. 0/21, and 57. 6. 120. 0/21. If all of them use the same outgoing line, can they be aggregated? If so, to what? If not, why not? Spring Semester 2006 EEC-484/584: Computer Networks 32
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