cse401n: computer networks
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CSE401N: COMPUTER NetworkS
LAN address & ARPEthernet Basics
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LAN technologies
Data link layer so far: services, error detection/correction, multiple
access
Next: LAN technologies addressing Ethernet hubs, bridges, switches 802.11 PPP ATM
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LAN Addresses and ARP
32-bit IP address: network-layer address used to get datagram to destination IP network
(recall IP network definition)
LAN (or MAC or physical or Ethernet) address:
used to get datagram from one interface to another physically-connected interface (same network)
48 bit MAC address (for most LANs) burned in the adapter ROM
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NIC or Network Adaptor
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LAN Addresses and ARPEach adapter on LAN has unique LAN address
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LAN Address (more)
MAC address allocation administered by IEEE manufacturer buys portion of MAC address space
(to assure uniqueness) Analogy: (a) MAC address: like Social Security Number (b) IP address: like postal address MAC flat address => portability
can move LAN card from one LAN to another
IP hierarchical address NOT portable depends on IP network to which node is attached
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Recall Earlier Routing Discussion
223.1.1.1
223.1.1.2
223.1.1.3
223.1.1.4 223.1.2.9
223.1.2.2
223.1.2.1
223.1.3.2223.1.3.1
223.1.3.27
A
BE
Starting at A, given IP datagram addressed to E:
look up net. address of E, find C
link layer send datagram to C inside link-layer frame
C’s MACaddr
A’s MACaddr
A’s IPaddr
E’s IPaddr
IP payload
datagramframe
frame source,dest address
datagram source,dest address
C
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ARP: Address Resolution Protocol
Each IP node (Host, Router) on LAN has ARP table
ARP Table: IP/MAC address mappings for some LAN nodes
< IP address; MAC address; TTL>
TTL (Time To Live): time after which address mapping will be forgotten (typically 20 min)
Question: how to determineMAC address of Cknowing C’s IP address?
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ARP protocol A knows B's IP address, wants to learn physical
address of B A broadcasts ARP query pkt, containing B's IP
address all machines on LAN receive ARP query
B receives ARP packet, replies to A with its (B's) physical layer address
A caches (saves) IP-to-physical address pairs until information becomes old (times out) soft state: information that times out (goes
away) unless refreshed
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Address Resolution Protocol (ARP)
1. With TCP/IP networking, a data packet must contain both a destination MAC address and a destination IP address.
2. Some devices will keep tables that contain MAC addresses and IP addresses of other devices that are connected to the same LAN.
3. These are called Address Resolution Protocol (ARP) tables. 4. ARP tables are stored in RAM memory, where the cached
information is maintained automatically on each of the devices. 5. Each device on a network maintains its own ARP table. 6. When a network device wants to send data across the network, it
uses information provided by the ARP table. 7. When a source determines the IP address for a destination, it
then consults the ARP table in order to locate the MAC address for the destination.
8. If the source locates an entry in its table, destination IP address to destination MAC address, it will associate the IP address to the MAC address and then uses it to encapsulate the data.
>arp -a
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Address Resolution Protocol (ARP) 1. The computer that requires an IP and
MAC address pair broadcasts an ARP request.
2. All the other devices on the local area network analyze this request, and if one of the local devices matches the IP address of the request, it sends back an ARP reply that contains its IP-MAC pair.
3. Another method to send data to the address of a device that is on another network segment is to set up a default gateway.
4. If the receiving host is not on the same segment, the source host sends the data using the actual IP address of the destination and the MAC address of the router.
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ARP conversation
HEY - Everyone please listen! Will 128.213.1.5 please send me his/her Ethernet address?
not me
Hi Green! I’m 128.213.1.5, and my Ethernet address is 87:A2:15:35:02:C3
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RARP conversation
HEY - Everyone please listen! My Ethernet address is 22:BC:66:17:01:75.Does anyone know my IP address ?
not me
Hi Green! Your IP address is 128.213.1.17.
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Getting a datagram from source to dest.
IP datagram:
223.1.1.1
223.1.1.2
223.1.1.3
223.1.1.4 223.1.2.9
223.1.2.2
223.1.2.1
223.1.3.2223.1.3.1
223.1.3.27
A
BE
miscfields
sourceIP addr
destIP addr data
datagram remains unchanged, as it travels source to destination
addr fields of interest here
Dest. Net. next router Nhops
223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2
routing table in A
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Getting a datagram from source to dest.
Starting at A, given IP datagram addressed to B:
look up net. address of B find B is on same net. as A link layer will send datagram
directly to B inside link-layer frame B and A are directly connected
Dest. Net. next router Nhops
223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2
miscfields223.1.1.1223.1.1.3data
223.1.1.1
223.1.1.2
223.1.1.3
223.1.1.4 223.1.2.9
223.1.2.2
223.1.2.1
223.1.3.2223.1.3.1
223.1.3.27
A
BE
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Getting a datagram from source to dest.
Dest. Net. next router Nhops
223.1.1 1223.1.2 223.1.1.4 2223.1.3 223.1.1.4 2
Starting at A, dest. E: look up network address of E E on different network
A, E not directly attached routing table: next hop router
to E is 223.1.1.4 link layer sends datagram to
router 223.1.1.4 inside link-layer frame
datagram arrives at 223.1.1.4 continued…..
miscfields223.1.1.1223.1.2.3 data
223.1.1.1
223.1.1.2
223.1.1.3
223.1.1.4 223.1.2.9
223.1.2.2
223.1.2.1
223.1.3.2223.1.3.1
223.1.3.27
A
BE
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Getting a datagram from source to dest.
Arriving at 223.1.4, destined for 223.1.2.2
look up network address of E E on same network as
router’s interface 223.1.2.9 router, E directly
attached link layer sends datagram to
223.1.2.2 inside link-layer frame via interface 223.1.2.9
datagram arrives at 223.1.2.2!!! (hooray!)
miscfields223.1.1.1223.1.2.3 data network router Nhops interface
223.1.1 - 1 223.1.1.4 223.1.2 - 1 223.1.2.9
223.1.3 - 1 223.1.3.27
Dest. next
223.1.1.1
223.1.1.2
223.1.1.3
223.1.1.4 223.1.2.9
223.1.2.2
223.1.2.1
223.1.3.2223.1.3.1
223.1.3.27
A
BE
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Routing to another LANwalkthrough: routing from A to B via R
In routing table at source Host, find router 111.111.111.110
In ARP table at source, find MAC address E6-E9-00-17-BB-4B, etc
A
RB
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A creates IP packet with source A, destination B A uses ARP to get R’s physical layer address for
111.111.111.110 A creates Ethernet frame with R's physical address as
dest, Ethernet frame contains A-to-B IP datagram A’s data link layer sends Ethernet frame R’s data link layer receives Ethernet frame R removes IP datagram from Ethernet frame, sees its
destined to B R uses ARP to get B’s physical layer address R creates frame containing A-to-B IP datagram sends to B
A
RB
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Ethernet (IEEE 802.3)“dominant” LAN technology: first widely deployed LAN technology simpler, cheaper than token ring, FDDI, and
ATM Lesson learned: KISS (Keep It Simple, Stupid)
kept up with speed race: 10, 100, 1000 Mbps
Metcalfe’s Ethernetsketch
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Ethernet(2)
First widely used LAN Technology Simpler than token ring, FDDI, or ATM Comply with new Technology and Speed
Can run over coaxial cable,or twisted pair, or fiber optics or radio link
10Mbps, 100Mbps, 1Gbps, 10Gbps. Ethernet Hardware(Hub/Bridge/Switch)
is widely available and cheap
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Ethernet and the OSI Model
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Ethernet and the OSI Model
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Repeaters: Layer1 Device
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Ethernet and the OSI Model
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Naming(MAC Address)
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802.3 Frame Structures
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Ethernet Frame Structures
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Ethernet Frame Structures
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Ethernet Frame Structure
Sending adapter encapsulates IP datagram (or other network layer protocol packet) in Ethernet frame
Preamble: 7 bytes with pattern 10101010 followed
by one byte with pattern 10101011 used to synchronize receiver, sender
clock rates
8 6 6 2 46-1500 (including padding) 4
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Ethernet Frame Structure(2) Addresses: 6 bytes each
if adapter receives frame with matching destination address, or with broadcast address (e.g. ARP request), it passes data in frame to network layer
otherwise, adapter discards frame Type (2 bytes): indicates the higher layer
protocol, mostly IP but others may be supported (such as Novell IPX and AppleTalk)
Data (46-1500 bytes): MTU is 1500 bytes, MIN frame size = 46 + 18 = 64 bytes = 512 bits
CRC (4 bytes): checked at receiver, if error is detected, the frame is simply dropped
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Unreliable, connectionless service
Connectionless: No handshaking between sending and receiving adapter.
Unreliable: receiving adapter doesn’t send ACKs or NAKs to sending adapter stream of datagrams passed to network layer can have
gaps gaps will be filled if app is using TCP otherwise, app will see the gaps
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Ethernet: From Bit to Electrical Signal
Use Manchester encoding One voltage change per bit
For a “1”, a voltage from 1 to 0 For a “0”, a voltage from 0 to 1
Example
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MAC Rules & Collision Detection/Backoff
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MAC Rules and Collision Detection/Backoff
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Ethernet Timing
Slot time: amount of time required to travel between
the furthest points of the collision domain, collide with another transmission at the last possible instant, and then have the collision fragments return to the sending station and be detected.
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Interframe Spacing and Backoff
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The Basic MAC Mechanisms of Ethernet
get a packet from upper layer;K := 0; n :=0; // K: random wait time; n: no. of
collisionsrepeat: wait for K * 512 bit-time; while (network busy) wait; wait for 96 bit-time after detecting no signal; transmit and detect collision; if detect collision stop and transmit a 48-bit jam; n ++; m:= min(n, 10), where n is the number of
collisions choose K randomly from {0, 1, 2, …, 2m-1}. if n < 16 goto repeat else giveup
CSMA/CD + Exponential backoff
Question: Why exponential backoff?
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Ethernet: uses CSMA/CD
A: sense channel, if idle then {
transmit and monitor the channel; If detect another transmission then { abort and send jam signal;
update # collisions; delay as required by exponential backoff algorithm; goto A}
else {done with the frame; set collisions to zero}}
else {wait until ongoing transmission is over and goto A}
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Ethernet’s CSMA/CD (more)Jam Signal: make sure all other transmitters are aware
of collisionBit time: 0.1 microsec for 10 Mbps Ethernet; for
K=1023, wait time is about 50 msecExponential Backoff: Goal: adapt retransmission attempts to estimated
current load heavy load: random wait will be longer
first collision: choose K from {0,1}; delay is K x 512 bit transmission times
after second collision: choose K from {0,1,2,3}… after ten or more collisions, choose K from
{0,1,2,3,4,…,1023}
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Why 64 bytes min frame length?
10Base5 Ethernet: 10Mbps, max segment 500m, max 4 repeaters, max network diameter 2500m Repeater: physical layer device that amplifies and retransmits
bits it hears on one interface to its other interfaces, used to connect multiple segments
Round trip time (worst case collision detection time) about 50 microsec All frames must take more than 50 microsec to send so that
transmission is still taking place when the noise burst gets back to sender
With 10Mbps bandwidth, 1 bit time = 0.1 microsec minimum frame size at least 500 bits, choose 512 bits to
add some margin of safety As network speed goes up, the minimum frame length
must go up or the maximum cable length must come down E.g. for a 2500m 1Gbps LAN, minimum frame size should be
6400 bytes
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CSMA/CD efficiency
Tprop = max prop. time between 2 nodes in LAN ttrans = time to transmit max-size frame
Efficiency goes to 1 as tprop goes to 0 Efficiency goes to 1 as ttrans goes to infinity Much better than ALOHA, but still decentralized, simple, and cheap
transprop tt /51
1efficiency
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Parameters for 10Mbps Ethernet
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