medium access control tanenbaum (chapter 4) others references: walrand j., communication networks: a...

Medium Access Control

Tanenbaum (Chapter 4)

Others References:Walrand J., Communication Networks: A first Course

Bertesekas and Gallager, Data Networks

04/18/23 Medium Access Control 2

Where in the OSI Reference Model ?

Session

Transport

Network

Link Layer

Physical

Presentation

Application

Data Link Layer

Medium Access Control


Why Do We Need a MAC Layer ?

• Let us consider different topologies– Point-to-point channel

• full duplex

• half duplex

– Broadcast channel


Management of Broadcast Channels

• Contention free: allocating statically shares of the channel to all stations (by TDM, FDM, or other)

• Contention: let stations compete for the ALL channel

Servers

One server


Objectives of This Chapter:Establish A Key Result

• 1) Efficiency: with heavy load, contention free systems are better than contention systems.

• 2) For average delay: contention systems are better than contention free (Whatever is the protocol! ) Introduction to Queueing Theory


Objectives of This Chapter (cont’d):Establish some Properties

• Relationship between a MAC protocol and physical properties of network:– Upperbound on efficiency– Maximum length of medium– Bandwidth– Minimum packet size


Objectives of This Chapter (cont’d):Techniques of Analysis

• Initiate students to the techniques used to analyze the performance of MAC protocols:– Aloha protocol– Slotted Aloha– CSMA– CSMA/CD

Objective 1: Establish A Key Result on

Average Delay

Queueing Theory

Introduction to Analytical Modeling

L.Kleinrock, “Queueing Systems”

R.Jain, “The Art of Computer Systems Performance Analysis”

E. Modiano, MIT course on Communications


Packet Switched Networks

PS

PS

PS

PS

PS

PS

PS

PacketSwitch

Buffer

PS


Queues Everywhere

• Processes waiting for CPU

• I/O Disk requests

• Network interface card

• IP input (output) queues

• Events (mouse click, keyboard…)

• ………………………


Queueing Theory

• What creates queues?– Randomness!!!

• Used for analyzing network performance• In packet networks, randomness comes from:

– random packet arrival– Random packet length

• Information of interest:– Delay in buffer (queueing delay)– Buffer size


Queueing Theory

JobSources

Servers

jobs

Or customers


A Queueing System A/S/m/B/K/SD

• A : Interarrival time of jobs distribution• S : The service time distribution• m : Number of servers• B : Capacity of the system (max # of jobs allowed)• K : population size• SD : Service discipline

– Default values B = ∞; K=∞; and Z=FCFS– For A and B

• G : General distribution• M : Exponential Distribution (memoryless property)• D : Deterministic


Example

• M/D/2/15/20000/FCFS– Time between arrivals exponentially distributed

– Service time constant (no variation)

– 2 servers

– System capacity is 15 (2 places for currently served + 13 waiting)

– Population is 20000 (20000 customers will ever come to the system)

– Service discipline is first come first served


Queueing Models

• Model good for– Requests received by Google– Customers waiting in line– Packets waiting to be transmitted over a line

• Information of interest– Average number of customers in the system– Average delay experienced by a customer


• Mean arrival rate: • Mean service rate: (service time per job)• Number of customers in system: n• Waiting time (in queue+service): w

Queueing System: Variables of Interest

JobSources

Servers

jobs

Or customers


A Key Result : Little’s Law

Black BoxArrivals Departure

Mean # in system = Mean arrival rate X mean time spent in systemE(n) = .E(w)

Remarks :- Result independent of A/B distributions or SD- Can be applied to all system or part of it- Crowded system long delays


A Key Result : Little’s LawExample

A monitor on an HTTP server showed that the average time to satisfy a request was about 50 milliseconds. The requestsarrival rate is 200 requests per second.What should be the buffer size (unit is requests) at the http server For the requests ?

Little’s law states E(n) = . E(w)

Here =200 req/s and E(w) = 0.050 s

The expected number of customers in the system is E(n)

E(n) = 200 x 0.050 = 10

It would be safe to have a buffer size of 20


Arrival Process

• Packets arrive according to some random process

• There are many stochastic processes.

• A nice stochastic process is the Poisson process– Mean arrival rate of packets per second– Prob(n arrivals during T) = [(.T)n e-T]/n!


Example• The number of phone calls arriving to a

switch can be closely modeled as a Poisson process. Suppose that the mean arrival rate is 100 per hour.

• What is the probability to receive 10 calls in 6 minutes?

T = 0.1 hour, = 100/h

Prob(n arrivals during T) = [(.T)n e-T]/n!Prob(10 arrivals during 0.1 h)= (100x0.1)10xe-(100x0.1)/10!=0.12Prob(10 arrivals during 15 mn) = 0.003


InterArrival Times of Poisson Process

• This is the time between consecutive arrivals. IA is a continuous random variable

• What is its probability distribution function?

• Prob(IA <= t) = 1 – Prob(IA > t)

• = 1 - Prob(0 arrivals within t)• Prob(0 arrivals during t) = [(.t)0 e-t]/0!= e-t

• So, Prob(IA <= t) = 1 - e-t


InterArrival Times of Poisson Process (2)

• The cumulative distribution function (CDF) is:– Prob(IA <= t) = 1 - e-t

• The probability distribution function is the derivative of CDF, i.e., PDF = .e-t

• This what is called the exponential distribution

• This distribution is largely used to model the service times, time between error losses, ..


Memoryless Property

• Def: A random variable X is said to be without memory, or memoryless, – P(X>s+t|X>t) = P(X>s) for all s, t ≥ 0

• In words: “When I get to the bus station, I am told that the probability that the bus comes within the next 10 minutes is 0.90. After one hour waiting, I am told that the probability that the bus comes within the next 10 minutes is still 0.90 ”

• Important result : X is memoryless iff it is exponentially distributed


More Examples

• Suppose that the time to graduate from AU is exponentially distributed with mean 4 years.

• Given that a student already spent 3 years at AU, what is the expected remaining time before he graduates?


• Merging of K P.P with mean rate i results in a P.P with mean rate the sum of the i’s.

• Splitting (randomly) a P.P with mean rate with probabilities pi results in P.Ps with mean rates pi. .

Properties of Poisson Process (PP)

1

2

i

n

i

p1.1

p2.2

pi.ipn.n


Analysis of an M/M/1

• Interarrival exponentially distributed (Memoryless)• Service time exponentially distributed (Memoryless)• One server• Infinite capacity of system• Infinite population• First come, first served

JobsSource Server

Queue


Analysis of an M/M/1 (2)

• We are interested in the average number of jobs in the system, the average waiting time in the system, or in the probability to have a given number of customers in the system.

• Notations– N(t): number of jobs in system at time t– Pn(t) = Prob{N(t)=n}– Pn = lim t-->∞ Pn(t)– = arrival rate– = service rate (service time = 1/ )


Markov Chain for M/M/1 System

• Circle i = state i means there are i customers in the system

• What is the probability Prob(i,j), i.e, the probability of transition from state i and j?

• Prob(j,j+1) = and Proj(j,j-1) =

0

1

2

3

4

5


Remarks about M/M/1

• > , otherwise the system will be instable

• After some time in operation, an M/M/1 gets into some equilibrium.

• When in equilibrium Pi = Pi+1

0

1

2

3

4

5

i


Determining Pn

0

1

2

3

4

5

Po = P1

P1 = P2

…………..Pn = Pn+1

Using the equations above and the sum Pi = 1, we can derive Pi’s

P1 = (\P0

P2 = (\P1

…………..Pn+1 = (\Pn

Pn = (\nP0

Pi = (\iP0 = P0.(\i = P0/(1-(\= P0/(1-

P0 =(1- Pi = i(1-


Mean Queue Length E(N) = /(1-) =/ is the traffic intensity

• The average number of customers in the system is E(N)

• E(N) = ii(1- = (1-

• E(N) = (1-.ii

• E(N) = (1-

• E(N) = (-


Mean Waiting Time E(W) = (1-)

• What is the average time in the system? (queueing delay + service time)

• We use Little’s formula: E(N) = l.E(W)


Prob{Server is busy} ?


Prob{Server is idle} ?


Analysis of M/M/m (m=3)

0

1

2

3

4

5

Po = P1P1 = 2P2…………..Pn = 3Pn+1

Using the equations above and the sum Spi = 1, we can derive Pi’s


Example

• Comparison source partition vs global FCFS• Let the system have n sources and m servers.Jobs

generate at each source as P.P with rate • Job’s computation time is exponentially

distributed 1/• Source partition is m M/M/1 while global FCFS

may be viewed as M/M/m with Arrival rate n..


A Fundamental Result In Queueing Theory

• One powerful server for all is better than one weak server for each one !

• Why ? Better utilization, as a dedicated channel may stay IDLE.

One server

Analysis of MAC Protocols

Read Tanenbaum Chapter 4

Key sections (Intro, 4.1, 4.2.1,4.2.2, 4.3)


Multiple Access Protocols

• Competing stations (possible collisions)– Aloha– Slotted Aloha– CSMA– CSMA/CD

• Collision-free– Bit-map protocol– Binary countdown

• Limited contention– Adaptive tree walk protocol


Pure Aloha• Designed by Abramson (wireless)

• A station emits whenever it has something to send

• If other station emits, a collision happens

• If collision, frame must be resent

• Best possible utilization at high load 18%


Analysis of Pure Aloha• Assume infinite population of users• Let Tr=time to trasmit a frame (“frame time”)• The population generates a traffic that is Poisson

with mean N per “frame time” (new frames)• Since there are also retransmissions, the total

traffic generated is Poisson with mean G• What is the throughput S?• S = G.Po where Po is the probability that a frame

does not suffer a collision


Analysis of Pure Aloha (2) (Example)

• Suppose that the bandwidth is 10 Mbps, and packet size is 1500 bytes

• Tr= 1500.8/10 Mbps = 1.2 ms• Possible values for N (mean number of frames

generated per time frame): between 0 and 1• Values for generated traffic (G > N)

– No retransmissions at all G = N– Low load G~N– High load G > N

• Po will be higher at low load


Analysis of Pure Aloha (3)• What is the probability to generate k frames

during a “frame time”?– Prob(k arrivals during 1) = [(G.1)k e-G.1]/k!

(page 20)

• What is the probability Po that that a frame does not suffer a collision?

Tr

t0 t0+2Trt0+Tr t0+3Tr


Analysis of Pure Aloha (4)• Po is the probability that ZERO frame is generated during the

vulnerable period

– Po = Prob(0 arrivals during 2) =

– = [(G.2)0 e-G.2]/0!

– Po = e-G.2

– S = G.Po = G. e-G.2 (When is S maximum?)

– The maximum throughput S is 1/(2.e) = 18.4%

Tr

t0 t0+2Trt0+Tr t0+3Tr


Slotted Aloha• Designed by Roberts (wireless)

• Requires synchronization and division of time in discrete slots

• A station emits whenever it has something to send AND must wait for beginning of slot

• Best possible utilization at high load 37%t

t0 t0+2tt0+t t0+3t


Analysis of Slotted Aloha (2)• The key: slotted time reduces the vulnerable

period to Tr (instead of 2Tr).– Po = Prob(0 arrivals during 1) =– = [(G.1)0 e-G.1]/0!– Po = e-G.1

– S = G.Po = G. e-G (When is S maximum?)– The maximum throughput S is 1/e = 36%

• Exercise: derive the average number of transmissions of one frame before being successful.

t

t0 t0+2tt0+t t0+3t


Carrier Sense Multiple AccessCSMA

• Designed by Metcalfe, analyzed by Kleinrock and Tobaggi

• A station listens to the channel before sending.

• If channel busy, wait until it becomes idle

• When channel free, send with probability 1

• Are collisions still possible?


Collisions with CSMA

• Key: signal takes time p to propagate• Problem 1: If two stations are listening to grab the channel…• Problem 2: If S1 starts transmitting, S2 may well send during

p.• Problem 3: S1 is not aware of the collision

Sender S1 Sender S2

p


Solving Problem 1: p-persistent CSMA

• 1-persistent: after collision, waits a random time and starts over (slide 48)

• Non persistent: if channel busy, the station does not keep listening, but rather waits for a random time before listening again.

• p-persistent: for slotted, if idle, send with probability p, otherwise defers to next slot

• Much better utilization than Aloha, may go beyond 95%.


Problem 2 and 3:Vulnerability Period with CSMA• At time t, S1 sends a frame f1• At time p-, S2 may well send a frame f2 because

S2 does not hear yet first bit of f1• At time 2p-, S1 does not hear yet first bit of f2• Vulnerability period is 2p

Sender 1 Sender 2

p


Solving Problem 2 and 3: Collision Detection (1)

• Objective:

1) Sender must detect collision

2) Stop transmission of damaged frames to avoid waste of medium

• CD: while sending, a sender must keep listening to detect a collision

• Can the sender always detect collisions?

• Relationship with medium length


Solving Problem 2 and 3: Collision Detection (2)

• Collision detection imposes a minimum frame length based on bandwidth and maximum medium length

• Sender must listen for at least the time it takes for the signal to tral between the farthest points on the medium (max)

• The sender listens while sending: it takes Tr to send a frame.

• To detect collision: Tr > 2.max


Solving Problem 2 and 3: Collision Detection (Example)

• A medium has a length of 1 km, the speed of light on the medium is 2/3 the speed in free space. What is max?

• The bandwidth is 10 Mbps

• What should be the smallest Tr?

• What is the minimum size of a frame?


Binary Exponential Backoff

• Problem: CSMA/CD does not adapt to the number of competing stations.

• What should be INTERVAL of the random time to wait after a collision occurs?

• Should it be 1 second?

• Should it be 1ms?

• Should it be 2.max? Or 8.max?


Binary Exponential Backoff (2)

• Problem: If the INTERVAL I is constant, there is NO ADAPTATION to the number of competing stations.

• How to solve the problem?– Use (2.max) as the unit of time

– At the ith collision for the same frame, choose randomly a number B beween 0 and 2i-1. (Max is i=10)

– Wait B.(2.max) before trying again


Efficiency of Ethernet

• Efficiency = Tr/T• Tr: transmission of one frame• T: T is the average time it takes to get a frame

from a sender to a receiver

• T= Tr + X.(2.max) where X is the number of slots we “waste” before sending out a frame successfully

• X is a random variable: we can compute E(X)


Efficiency of Ethernet (2)

• Efficiency = Tr/(Tr+E(X) .(2.max)). What is E(X)?

• We have to find the average number of collisions before a frame makes it.

• In others words, we have to find the average number of attempts before we get a frame through

• Suppose that I know the probability A that a frame makes it through.

• What would be E(X), based on A?



• Recall that A is the probability for a packet to get transmitted successfully

• X takes the value 1 with probability A• X takes the value 2 with probability (1-A).A• X takes the value 3 with probability (1-A)2.A• ………………………………………………• X takes the value i with probability (1-A)i-1.A

• What is E(X), based on A?



• Efficiency = Tr/(Tr+E(X) .(2.max)). What is E(X)?

• We found that E(X) = 1/A

• Then, Efficiency = Tr/(Tr+(2.max)/A).


Efficiency of Ethernet (4)• Efficiency = Tr/(Tr+(2.max)/A).

• Nice! But, what is the value of A?

• Recall: the probability A that a frame makes it through

• In other words: A is the probability that one station attempts to transmit while the other stations do not.


Efficiency of Ethernet (5)• Efficiency = Tr/(Tr+(2.max)/A). • Suppose that there are k stations.• Each station tries to send in each slot with

probability p.• Question 1: What is the probability that SOME

station acquires the medium?• A = k.p.(1-p)k-1

• Do we want A to be high or low?• A is maximal for p = 1/k. What is Amax?• Amax = 1/e when k gets large.


Efficiency of Ethernet (6)• Efficiency = Tr/(Tr+(2.max).e).

• We could express as a function of the bandwidth Bw, the maximal length of the medium L, and the packet size F.

• What is the expression of the Efficiency?

• Hint: 2.max is the is the smallest time to send a frame


Collision Free MAC• Token Ring MAC Protocol (IEEE 802.5)• N nodes are connected in a ring topology• A special packet (token) is circulated

periodically: only a station possessing the token can transmit:

• IEEE 802.5 Strategy:– Wait for the token and grab it– Transmit packets for at most (10ms in 802.5)– Release the token


Efficiency of Token Ring

• Suppose, we have N stations

• We want to establish the proportion of time the medium is used to send frames

• We assume that all stations have a frame to send.

• We neglect the transmission time of a token


Efficiency of Token Ring (2)

• The function of a Token Ring is periodical• Let us consider one cycle:

– Station 1 gets the token and sends for .– Station 2 gets the token and sends for .– Station i gets the token and sends for .– Station N gets the token and sends for .

• The duration of one cycle is then S=N. + . ( is the propagation time around the ring)


Efficiency of Token Ring (3)

• Recall that the cycle duration is S=N. + • The question now is: how long was the

medium used during a cycle

• The answer is N.q.

• Efficiency = N./S = 1/(1+/(N.))

• Example: N= 40 stations, Ring of 2500m, what is the efficiency?


Maximum Access time of T.Ring

• If all stations have packets to send, a station may well have to wait for:– All other stations to send their packets ((N-1).)– Wait for the token ()

• What is the maximum with the previous example?

Problem of CSMA on Wireless Networks


Air Is A Broadcast Medium

• Can we use CSMA/CD ?

– Talking/Listening problem

– Hidden Terminal problem

– Exposed Terminal Problem


Hidden Terminal Problem

A B C D

•A wants to send to B•AND•C wants to send to B•A and C do not hear each other, they cannot detect collisions


Exposed Terminal Problem

A B C D

•B wants to send to A•AND•C wants to send to D•B and C believe they are bothering each other


MACA (Phil Karn)

• When X wants to send to Y:– X sends an RTS frame– if Y gets RTS frame, Y sends a CTS frame

• If you hear an RTS, you should keep quiet (to let the CTS come back)

• If you hear a CTS, keep quiet (to let the incoming frame data)


MACAW

• Improved version of MACA– Adds acknowledgements for successful data

frames– Use of CSMA– Exponential Backoff mechanism


MAC Layer for 802.11

• Two kinds of policies– Distributed (with contention) : Distributed

Coordination function– Centralized contention free: Point coordination

function

• The two may be used simultaneously


Distributed Coordination function

• Check first a logical maintained variable (NAV: Network allocation vector)

• If NAV not null wait

• if NAV null, then sense carrrier

• if idle, transmit (if collision, random exponential backoff)

• otherwise Random exponential backoff


Delay After Sensing Idle

Medium Busy

time

SIFS

PIFS

DIFS

ACKCTS

PCF DCF

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