fall 2005 by: h. veisi computer networks course olum-fonoon babol chapter 4 medium access control...

Fall 2005By: H. Veisi

Computer networks course

Olum-fonoon Babol

Chapter 4

Medium Access Control (MAC)

Computer networking, Olum-Fonoon Babol H. VeisiH. Veisi Fall 2005

Overview

In broadcast networks (Multi-access/random-access channels)

The key issue is how to determine who gets to use the channel when there is competitioncompetition for it.

MAC=Protocol to determine who goes next on channel

It’s important for LANs, WANs are point-to-point.

The Channel Allocation Problem: StaticStatic Channel Allocation in LANs and MANs

FDM and TDM

DynamicDynamic Channel Allocation in LANs and MANs


Static channel allocation (1)

The traditional (phone) way of allocating a single channel is Frequency Division Multiplexing. FDM works fine for limited and fixed number of users.

Inefficient to divide into fixed number of chunks. May not all be used, or may need more. Doesn't handle burstly traffics of computer systems.

From queuing theory (Poisson distribution for C and T): T = mean time delay for a channel C = capacity (Bits/Sec.) λ = arrival rate (Frames/Sec.) 1/μ = mean length of a frame

1T = ---------- C -


Static channel allocation (2)

Example: C=100Mbps, 1/μ =10000 Bits, λ =500 Fps

T=200 Micro Sec.

If divide this channel into N sub-channels, each with capacity C/N. Input rate on each of the N channels is λ/N. So:

N times worse for FDM In example for N=10 => T=2 Mili Sec.

Same arguments can apply for TDM

1 NT(FDM) = ----------------- = ------------ = NT

μ(C/N) - λ/N μC - λ


Dynamic channel allocation (1)

Assumptions (1): Station Model: Assumes that each of N "stations"

(packet generators, Terminal) independently produce frames. The probability of producing a packet in the interval Δt is λ.Δt where λ is the constant arrival rate. That station generates no new frame until that previous one is transmitted.

Single Channel Assumption: There's only one channel; all stations are equivalent and can send and receive on that channel.

Collision Assumption: If two frames overlap in any way time-wise, then that's a collision. Any collision is an error, and both frames must be retransmitted. Collisions are the only possible error.


Dynamic channel allocation (2)

Assumptions (2): Continuous/ Slotted Time: There's no "big clock

in the sky" governing transmission. Time is not in discrete chunks. Frame transmission can begin at any instant. Alternatively, in slotted, frame transmissions always begin at the start of a time slot. Any station can transmit in any slot (with a possible collision.)

Carrier/No-Carrier Sense: Stations can tell a channel is busy before they try it. NOTE - this doesn't stop collisions. LANs have this, satellite networks don't.


Multiple Access Protocols: ALOHA

ALOHA Developed in Hawaii in the 1970s.

PURE ALOHA: Every station transmits whenever it wants to. Colliding frames are destroyed. The sender

knows if its frame got destroyed using feedback property, and if so waits a random time and then retransmits.

ANY overlap is a collision. Best efficiency if frames are same size. A contention system: Multiple users share a

common channel that can lead to conflict.


PURE ALOHA (2)

Pure ALOHA: frames are transmitted at completely arbitrary times.

What is the efficiency of ALOHA?


PURE ALOHA (3)

Frame time: Time needed to transmit a standard, fixed-length frame = Frame length divided by bit-rate

N= Mean No. of frame per frame-time and infinite users if N>1: All frames suffer a collision if <0N<1: Expected frame rate

G = Load= N + frames retransmitted due to previous collisions. P0 = Transmission succeeding = probability that a frame does

NOT suffer collision. Throughput, S = P0 . G


PURE ALOHA (4)

Vulnerable period for the shaded frame:


PURE ALOHA (5)

Probability that k frames are generated during a given frame time (Poisson distribution):

Probability of no traffic initiated during the vulnerable period: P0 = e-2G so: Throughput per frame time is: S = G . e-2G

Max. channel utilization: G=0.5, S=1/2e=18%

G k . e-G Pr[k] = --------------

k!


SLOTTED ALOHA

Slotted ALOHA: Doubles efficiency by dividing time into discrete

intervals. Sends occur only at the start of a slot time. Vulnerable period is 1/2 of pure Aloha case, so:

S = G . e-G

Best throughput is at G = 1 , S = 37%


ALOHA

Throughput versus offered traffic for ALOHA systems


CSMA Protocols (1)

Carrier Sense Multiple Access (CSMA) 37% utilization is low yet! Stations can listen for a carrier and there is no

transmission send it’s data.

Persistent and non-persistent: persistent: When channel is found to be busy, keep

monitoring to find THE instant when it becomes free.

non-persistent: When channel is found to be busy, don't keep monitoring to find THE instant when it becomes free. Instead, wait a random time and then sense again.


CSMA Protocols (2)

1-persistent CSMA Station listens. If channel idle, it transmits. If collision,

wait a random time and try again. If channel busy, wait until idle.

If station wants to send AND channel == idle then do send.

Propagation delay has an important effect on transmission.

Success here depends on transmission time - how long after the channel is sensed as idle will it stay idle (there might in fact be someone else's request on the way.)

Non-persistent CSMA (equivalent to 0-persistent CSMA) Same as above EXCEPT, when channel is found to be busy,

don't keep monitoring to find THE instant when it becomes free. Instead, wait a random time and then sense again.

Leads to better utilization


CSMA Protocols (3)

p-persistent CSMA For slotted time channels If ready to send AND channel == idle then send with probability p and with prob. q = 1 -

p defers to the next slot.

Example: 0.5-persistent , 0.01-persistent

Lower probability is better for higher frame transmission per frame-time


CSMA Protocols (4)

Comparison of the channel utilization versus load for various random access protocols.

Higher load


CSMA with Collision detection

CSMA with Collision detection (CD) = CSMA/CD CSMA protocols are clearly improved over ALOHA CSMA protocols can improve if stations abort

their transmission as soon as they detect a collision.

Used with LANs.

In CSMA/CD, when a station detects a collision, it stops sending, even if in mid-frame. Waits a random time and then tries again.


CSMA/CD (2)

CSMA/CD can be in one of three states: 1-Contention, 2-transmission, 3-idle.


CSMA with Collision detection

CSMA with Collision detection (CD) = CSMA/CD What is contention interval- how long must

station wait after it sends until it knows it got control of the channel? It's twice the time to travel to the furthest station.

Collision-free protocols: Bit-map protocol Binary countdown


IEEE standards for LAN

IEEE 802.2: Describes the upper part of the data link layer, the LLC (Logical Link Control). Hide the differences between the various kinds of MAC

protocols by providing a single format to network layer IEEE 802.2 standard

Descriptions of the physical and lower part of the DLL are (MAC): IEEE 802.3 CSMA/CD LAN IEEE 802.4 Token Bus IEEE 802.5 Token Ring


Ethernet (1)

Ethernet LAN standard (IEEE 802.3) 1-persistent CSMA/CD + Binary Exponential

Back-off

Architecture of the original Ethernet.


Ethernet (2)

Ethernet Cabling (1)

10Base=10 Mbps, Base-band signaling


Ethernet (3)

Ethernet Cabling (2)

Transceiver: Electronic circuits that handle carrier detection and collision detection

(a) 10Base5,

(b) 10Base2,

(c) 10Base-T.


Ethernet (4)

Ethernet Cabling (3) In Binary coding there's no way to distinguish a 0 bit

from nothing-happening. Need to know when is middle of bit WITHOUT a clock => ManchesterManchester Encoding

In Manchester: Bit 1= High-Low Bit 0= Low-High Used in Ethernet

Differential Manchester Bit 1 indicated by absence of transition at the start of

interval Better noise immunity Used in Token ring


Ethernet (5)

Ethernet Cabling (4) ManchesterManchester Encoding


Ethernet (6)

Ethernet MAC Protocol

Preamble == 7 bytes of 10101010 for synchronization SOF: Start of Frame == 1 byte of 10101011 for compatibility with

802.4 and 802.5 Dest. Add. == 6 bytes of mac address

multicast == (47th bit=1) sending to a group of stations. Broadcast == (dest. = all 1's) to all stations on network

Source Ad. == 6 bytes of MAC address Length/Type == number of bytes of data/Type of protocol Data == comes down from network layer Pad == ensures 64 bytes from Dest. Add. addr thru

checksum. Checksum == 4 bytes of CRC.

DIX Ethernet

IEEE 802.3.


Ethernet (7)

1500 data byte restrict Max. frame length to 1526 bytes Also need Min. frame lengthMin. frame length! Why?

To distinguish valid frames from garbage (Made by collision) need at-least 64 byte frame.

If data in frame is less than 46 [=64-18] the Pad field is used to filled out frame to Min. frame length.

To prevented a station from completing the transmission of a short frame before the first bit has reached the far-end of the cable, where it may collide

Transmitter need 2τ time to detect noise of collision, where τ is propagation time of a frame to reach another end of cable.


Ethernet (8)

Collision detection can take as long as 2τ

For a 10Mbps LAN with max. length of 2500 meters and 4 repeaters the round-trip time= τ ≈ 50 μsec. So Min. frame length=500 bits => 64 bytes

As the network speed goes up, the Min. frame length must go up or Max. length of cable come down.


Ethernet (9)

Binary exponential back-off algorithm: Determine how randomize is done when collision occurs. After a collision, station waits 0 or 1 slot. If it collides

again while doing this send, it picks a time of 0,1,2,3 slots. If again it collides the wait is 0 to 23 -1 times.

In general after i collisions an random number between 0 and 2i-1 is chosen and that number of slots is skipped.

Max time is 210 -1 (or equal to 10 collisions.) After 10 collisions, an error is reported.

Slot is determined by the worst case times; 500 meters + 4 repeaters = 512 bit times = 51.2 μsec.

Algorithm adapts to number of stations.


Ethernet (10)

Ethernet Performance (1) Channel efficiency depends on

F: Frame length, B: Network Bandwidth, L: Cable Length, C: Speed of signal propagation, e: optimal number of contention slots per frame.

T= Time for transmission a frame = F/B

τ = is propagation time of a frame thru cable = L/C A= Probability that a station acquires the channel in a slot with

contention, Optimum=1/e

T 1channel efficiency = ------------------ = --------------------------- T+2τ/A 1 + 2 B L e / c F


Ethernet (10)

Ethernet Performance Efforts focus on improving both B and L, both of which

will decrease efficiency. In all theatrical researches on performance evaluation

of Ethernet, it’s assumed that traffic is Poisson but in real data they are not Poisson, but self similar.

Efficiency of Ethernet at 10 Mbps with 512-bit slot times:


Fast Ethernet

IEEE 802.3u: An addendum of existing 802.3 (1995)

Idea: keep all old frame formats, interfaces and procedural rules but just reduce bite time from 100 nsec. to 10 nsec.

Commonly use twisted pair cabling 100Base-T4: Use Ternary (3-level) signaling and UTP-Cat3

cabling 100Base-TX: Use 4B/5B (use 5 bits for transmit 4 bits)

signaling and UTP-Cat5 cabling


Gigabit Ethernet (1)

IEEE 802.3z: Another addendum of existing 802.3 (1998) Goal: make 10 times faster + Compatible with existing

Ethernet standard Support unacknowledged datagram service in both

unicast and multicast Configuration is point-to-point rather than multi-drop

(Use Hub or Switch)



Two different modes: Full-Duplex:

Allows traffic in both directions at same time Is used when there is a switch connect to computer or

other switches All lines are buffered Contention is impossible

CSMA/CD is not used Half-Duplex

When computers connect to Hub rather than Switches Don’t buffer frames, is just like classic Ethernet

Need CSMA/CD protocol Reduce length of cable Carrier Extension: Tell hardware to add its own

padding to extend frame length to 512 bytes.



Frame Bursting: Allow a sender to transmit a concatenated sequence of frames in a single transmission. This is efficient and perfect over carrier extention

Cabling

Use 8B/10B signaling for Fiber optic Use different encoding for 1000Base-T


Repeaters, Hubs, Bridges, Switches, Routers and Gateways (1)

User generate some data: Data are passed to Transport layer and adds a header, i.e. TCP

header The resulting unit passes down to Network layer, adds headers, IP

packet Then goes to DLL which adds its own header (CRC) Resulting frame given to Physical layer

These devices operate in different layers



Physical layer: Repeaters:

Analog devices that are connected two cable segments, Amplifying incoming signal from one side and send out other.

Don’t understand frames, packets or headers, Just understand Volt

In classic Ethernet, to extend max. cable length from 500 meters to 2500 meters, 4 repeaters was allowed.

Hubs: Has a number of input lines that it joins electrically. Frames

arriving on any of the lines are send out on all others. If two frames arrive at the same time Collision occurs Are like repeater and don’t understand frames, but usually

don’t amplify incoming signal



Data Link Layer: Bridges:

Connect two or more LANs. Use des. Add. In frame header to determine destination

Switches: Are like Bridges and use des. Add. To find the route. Used to connect individual computers need more number of

ports, each port has own collision domain. Store & Forward: Get entire a frame then transmit Cut-through-switches: start forwarding the frame as soon as

the des. Add. Field has come in.



Network layer: Routers:

When a packet comes into, the frame header and trailer are stripped off and packet located in the frame’s payload field is passed routing software.

Transport and Application layer: Transport gateway:

Connect two computers that use different connection-oriented transport protocols.

Ex. TCP/IP and ATM

Application gateway: Understand the format and content of data and translate

message from one format to another one.


Virtual LAN (1)

Logical rather than physical configuration in hubbed or switched Ethernet

Reasons? Security, Load, Broadcast,


Virtual LAN (2)

In VLAN: How many VLANs there will be? Which computer will be on which VLAN? What the VLANs will be called? Four physical LANs organized into two VLANs, gray and

white, by (a) two bridges. (b) by switches.


Summary


ATM (1)

ATM (Asynchronous Transfer Mode) Is underlying mechanism. Transmits in small

fixed-size cells. A connection-oriented network Use virtual circuits and small, fixed-size

packets (Cells)


ATM (2)

Packet (cell) switching is dramatic change for phone companies.

ATM is connection oriented; make connecting request first; then all cells follow the same path.

Target is 155 Mbps and 622 Mbps. Allows TV transmission.


ATM (3)

ATM Reference Model: 3 layers:

1- Physical layer :Physical medium (voltage, bit timing, ….)

2- ATM layer : deal with cells and transports it + establish/release

virtual circuits + congestion control

3- ATM adaptive layer : Segment large cells and resemble after transmission


ATM (4)

ATM Reference Model:


ATM (5)

Comparisons to other models:

fall 2005 by: h. veisi computer networks course olum-fonoon babol chapter 4 medium access control...

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