1. TCOM 509 – Internet Protocols (TCP/IP) Lecture 03_a
Instructor: Dr. Li-Chuan Chen Date: 09/15/2003
Based in part upon slides of Prof. J. Kurose (U Mass), Prof. B. Yener (Rensselaer Polytechnic Institute)

2. Outline
Chapter 5 – mapping Internet Addresses to Physical Addresses (ARP)
Chapter 6
Chapter 7

3. Address Resolution Problems
Problem: given an IP address, need to find its equivalent physical address
Sender must map the intermediate router’s IP address and destination IP address to their corresponding physical addresses.
Solutions:
Direct Mapping
Table lookup
Dynamic Binding via ARP (Address Resolution Protocol)

4. Address Resolution Mechanisms
Direct mapping Make the physical addresses equal to the host ID portion.
Mapping is easy.
Only possible if admin has power to choose both IP and physical address or when size of physical address < IP address.
Cannot apply to Ethernet addresses (Ethernet addresses are 48 bits vs. IP addresses which are 32-bits).

5. Address Resolution Mechanisms
Table Lookup Searching or indexing to get MAC addresses
Similar to lookup in /etc/hosts for names
Problem: change Ethernet card => change table
IP Address
MAC Address
0A:4B:00:00:07:08
0B:4B:00:00:07:00
0A:5B:00:01:01:03

6. Address Resolution Mechanisms
Dynamic Binding (ARP)
The host broadcasts a request: “What is the MAC address of ?”
The host whose IP address is replies back: “The MAC address for is 8A-5F-3C ”
Broadcast is expensive.
ARP responses are cached. Issues
Broken hardware: use timer.
Table full: least recently used (LRU)
Each host updates its table when receiving an ARP broadcast.

7. ARP Message Format 8 16 24 31 HW Type Protocol Type HW Len8 16 24 31
HW Type
Protocol Type
HW Len
Protocol Len
Operation
Sender HW Address (6 bytes)
Sender HW Address
Sender IP Address (4 bytes)
Sender IP Address
Target HW Address (6 bytes)
Target HW Address
Target IP Address (4 bytes)
Hardware (HW) Type: 1 for Ethernet
Protocol Type: = IP address
HW Len and Protocol Len allows arbitrary networks to be used
Operation: 1 = ARP Request, 2 = ARP Response = RARP Request, 4 = RARP Response
ARP messages are sent directly to MAC layer
ARP message is 28 octets long.

8. ARP Let Ia = IP address and pa =physical address
To send an internet packet across a physical net, the network software must map Ia to pa and use the pa to transmit the frame.
If Ia < pa, use direct mapping. Else, use ARP to perform dynamic mapping.
Given an IP address, a host uses ARP to find the corresponding hardware address.
To make ARP efficient, all hosts on the network receive the ARP request and update its cache.
The host with the same Ia replies directly to the sender.

9. Outline Chapter 4.1 – 4.18 Chapter 5
Chapter 6 – Determining An Internet Address At Startup (RARP)
Chapter 7

10. RARP Problems: Given pa, how do we find Ia ?
Solution: Reverse ARP (RARP)
Use the same message format as ARP.
Sender broadcast a RARP request (fills its pa in the target field).
Only RARP server replies.
Typically used in Ethernet LAN.
If only one RARP server available on the network, use larger delay time before retransmit another request.
More RARP servers?
Pros – reliable
Cons – overload the network. (assign primary and secondary server to solve this problem)

11. RARP
Diskless host needs to find its IP address at startup before it can communicate using TCP/IP.
Give physical address, a host can use RARP to find its IP address from a RARP server on the network.

12. Outline Chapter 4.1 – 4.18 Chapter 5 Chapter 6
Chapter 7 – Internet Protocol: Connectionless Datagram Delivery

13. Internet Services The three conceptual layers of internet services.
Connectionless Packet Delivery Services
Reliable Transport Services
Application Service

14. Destination IP Address
IP Datagram Format
Vers
HLen
TOS
Total Length
Identification
Flags
Time to live
Protocol
Header Checksum
Fragment Offset
Source IP Address
Destination IP Address
IP Options (if any)
Padding
Data 4 8 16 31

15. IP Datagram Format Internet datagram: basic transfer unit
VERS - Version (4 bits): IPv4
HLEN - Internet header length (4 bits): units of 32-bit words. Min header is 20 bytes or 5 words.
Total Length (16 bits): header + data. Units of bytes. Total must be less than 64 K (216) octets.

16. IP Header TOS - Type of service (8 bits)
precedence (3 bits), delay, throughput, and reliability.
Not widely supported.

17. IP Header How big can a datagram be?
What happens when a datagram is larger than the frame size of the underlying physical network?

18. Maximum Transmission Unit (MTU)
Each subnet has a maximum frame size Ethernet: 1500 octets FDDI: 4470 octets per frame Token Ring: 2K to 4K octets
Transmission Unit = IP datagram (data + header)
Each subnet has a maximum IP datagram length (header + payload) = MTU B
Net 1
MTU = 1500
Net 3
Net 2
MTU = 620 R1 R2 A

19. Fragmentation Datagrams larger than MTU are fragmented
Original header is copied to each fragment and then modified (fragment flag, fragment offset, length,...)
Fragments must be a multiple of 8-octets.
IP Header
Original Datagram
IP Hdr 1
Data 1
IP Hdr 3
Data 3
IP Hdr 2
Data 2

20. Reassembly Reassembly only at the final destination
Partial datagrams are discarded after a timeout
Fragments can be further fragmented along the path. Subfragments have a format similar to fragments.
Minimum MTU along a path  Path MTU B
Net 1
MTU = 1500
Net 3
Net 2
MTU = 620 R1 R2 A

21. IP Header Fragmentation
Identifier (16 bits): used in reassembly to uniquely identify all the pieces of a fragment chain.
Flags (3 bits): more fragments (MF), don’t fragment (DF), and reserved bit.
Fragment offset (13 bits): In units of 8 octets

22. Fragmentation Example
Net 1
MTU = 1500
Net 2
MTU = 620
Net 3
MTU = 1500 A R1 R2 B
Payload size 1400 bytes needs to be transmitted, Packet ID = 2222
Networks: Ethernet (MTU=1500) and Net2 (MTU=620)
Use smallest MTU size (620) to find payload size for the fragment packet.
IP Header = 20 bytes => Payload = MTU – IP Header = 600 bytes
Fragments need to be multiples of 8-bytes.
Nearest multiple to 600 is still 600 bytes
Fragment offset length = 600/8 = 75
Number of fragments = 1400/600 = 2.33 = 3
frag1 = 600, frag2 = 600, frag3 = 200, Packet ID = 2222 for all fragments
Offset1 = 0, Offset2 = 75, Offset3 = 150 MF1 bit = 1, MF2 bit = 1, MF3 bit = 0 H1
frag1 H2
frag2 H3
frag3
Net works
Min MTU = 620 75
150

23. IP Header Time to live (TTL) 8 bits: Protocol (8 bits)
Specifies how long the datagram is allowed to live in the network (in seconds). Typically use number of hops visited.
Protocol (8 bits)
Next level protocol to receive the data, e.g., ICMP (1), IGMP (2), TCP (6), UDP (17).
Header checksum (16 bits)
1’s complement sum of all 16-bit words in the header.

24. IP Header
Source Address (32 bits): Original source. Does not change along the path
Destination Address (32 bits): Final destination. Does not change along the path.
Options (variable length): security, source route, record route, stream id, timestamp recording
Padding (variable length): Makes header length a multiple of 4
Payload Data (variable length): Data + header < 65,535 bytes

25. IP Header Options – for network testing or debugging
Security - for military purpose and is only supported by some products.
Source route – a list of IP address that the datagram must take.
Record route – the nodes in the path must return their IP address.
Stream id - used for voice for reserved resources
Timestamp – the time through the node is returned, so that delays may be measured.
If entries in the options must be recorded by nodes, the sender must reserve sufficient space for the option data.

26. Summary Internetworking Problem
IP header: supports connectionless delivery, variable length pkts/headers/options, fragmentation/reassembly,
Fragmentation/Reassembly, Path MTU discovery.
ARP, RARP: address mapping
Internet architectural principles