Medium Access Control

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					   Medium Access Control

       Tanenbaum (Chapter 4)
Others References:
Walrand J., Communication Networks: A first Course
Bertesekas and Gallager, Data Networks
Where in the OSI Reference Model ?
            Application
            Presentation

            Session

            Transport

            Network                       Medium Access Control

            Link Layer
                                           Data Link Layer
            Physical
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Why Do We Need a MAC Layer ?
• Let us consider different topologies
     – Point-to-point channel
            • full duplex
            • half duplex


     – Broadcast channel



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

                                              Servers

 • Contention: let stations compete for the ALL
     channel
                                                 One server

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        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

6/22/2010          Medium Access Control            5
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



  6/22/2010               Medium Access Control   6
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



  6/22/2010                    Medium Access Control   7
       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
                                                 Buffer
                                                          Packet
                                                          Switch
                                                          PS
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            Queues Everywhere
•   Processes waiting for CPU
•   I/O Disk requests
•   Network interface card
•   IP input (output) queues
•   Events (mouse click, keyboard…)
•   ………………………

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               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

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            Queueing Theory
                                         Servers

                jobs
  Job
Sources
              Or customers




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                  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



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                        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

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              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
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Queueing System: Variables of Interest

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

                    jobs
  Job
Sources
                  Or customers
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           A Key Result : Little’s Law
               Arrivals                     Departure
                          Black Box


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

   6/22/2010                Medium Access Control        18
        A Key Result : Little’s Law
                Example
A monitor on an HTTP server showed that the average time to
satisfy a request was about 50 milliseconds. The requests
arrival 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) = l. E(w)
   Here l =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
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               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 l packets per second
  – Prob(n arrivals during T) = [(l.T)n e-lT]/n!


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                          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, l = 100/h
Prob(n arrivals during T) = [(l.T)n e-lT]/n!
Prob(10 arrivals during 0.1 h)= (100x0.1)10xe-(100x0.1)/10!=0.12
Prob(10 arrivals during 15 mn) = 0.003
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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) = [(l.t)0 e-lt]/0!= e-lt
  • So, Prob(IA <= t) = 1 - e-lt
  6/22/2010             Medium Access Control            22
 InterArrival Times of Poisson Process (2)

• The cumulative distribution function (CDF) is:
  – Prob(IA <= t) = 1 - e-lt
• The probability distribution function is the
  derivative of CDF, i.e., PDF = l.e-lt
• This what is called the exponential distribution

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

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Properties of Poisson Process (PP)
 • Merging of K P.P with mean rate li results
   in a P.P with mean rate the sum of the li’s.
        l1
                           S li
        l     2
             li
             ln
 • Splitting (randomly) a P.P with mean rate l
   with probabilities pi results in P.Ps with
   mean rates pi. l.     p1.l1
                             p2.l2
              l
                           pi.li
                    pn.ln
 6/22/2010         Medium Access Control          26
            Analysis of an M/M/1
                   Queue
 Jobs
Source                                          Server

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


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             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)
     –      l= arrival rate
     –      m = service rate (service time = 1/ m)



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  Markov Chain for M/M/1 System
              l.d       l.d          l.d              l.d       l.d       l.d
          0         1         2             3               4         5
1-l.d         m.d       m.d          m.d              m.d       m.d       m.d

  • 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) = l.d, and Proj(j,j-1) = m.d

  6/22/2010                   Medium Access Control                             29
            Remarks about M/M/1
        l       l       l             l             l       l
 0          1       2          3                4       5
        m       m       m             m             m       m

• l > m, otherwise the system will be instable
• After some time in operation, an M/M/1
  gets into some equilibrium.
                                         l
• When in equilibrium l.Pi = m.Pi+1  i
                                                        m
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                           Determining Pn
          l            l             l              l                l             l
  0            1              2             3                4             5
          m            m             m              m                m             m
 l.Po = m.P1                      P1 = (l\m).P0
 l.P1 = m.P2                      P2 = (l\m).P1                          Pn = (l\m)n.P0
 …………..                           …………..
 l.Pn = m.Pn+1                    Pn+1 = (l\m).Pn
Using the equations above and the sum                        SP =i       1, we can derive Pi’s

SP = S (l\m) .P
      i
                   i
                           0 = P0. S (l\m) = P /(1-(l\m)) = P /(1-r) = 1
                                                i
                                                        0                      0

              P0 =(1-r)                     Pi = ri.(1-r)
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 Mean Queue Length E(N) = r/(1-r)
            r =l/m is the traffic intensity
• The average number of customers in the
  system is E(N)
• E(N) = Si.ri.(1-r) = r/(1-r)
• E(N) = (1-r).Si.ri
• E(N) = r/(1-r)
• E(N) = l/(m-l)

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Mean Waiting Time E(W) = 1/(1-r)m

• What is the average time in the system?
  (queueing delay + service time)
• We use Little’s formula: E(N) = l.E(W)




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            Prob{Server is busy} ?




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            Prob{Server is idle} ?




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             Analysis of M/M/m (m=3)
         l        l        l             l             l        l
  0           1        2          3                4        5
         m        2m       3m            3m            3m       3m
 lPo = mP1
 lP1 = 2mP2
 …………..
 lPn = 3mPn+1

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

 6/22/2010                 Medium Access Control                     36
                  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 l
• Job’s computation time is exponentially
  distributed 1/m
• Source partition is m M/M/1 while global FCFS
  may be viewed as M/M/m with Arrival rate n.l.




6/22/2010          Medium Access Control             37
            A Fundamental Result In
               Queueing Theory
• One powerful server for all is better than
  one weak server for each one !


                                One server


• Why ? Better utilization, as a dedicated
  channel may stay IDLE.
6/22/2010          Medium Access Control       38
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
6/22/2010                   Medium Access Control   40
               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%




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            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



6/22/2010           Medium Access Control              42
            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


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            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



6/22/2010     t0               +Tr
                             t0Medium Access Controlt0+2Tr   t0+3Tr   44
              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



  6/22/2010    t0          +Tr
                         t0Medium Access Controlt0+2Tr   t0+3Tr   45
                     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



6/22/2010   t0           +t
                       t0Medium Access Controlt0+2t   t0+3t   46
      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
              t
  successful.


6/22/2010     t0              +t
                            t0Medium Access Controlt0+2t   t0+3t   47
    Carrier Sense Multiple Access
               CSMA
• 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?



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             Collisions with CSMA
 • Key: signal takes time tp to propagate
 • Problem 1: If two stations are listening to grab
   the channel…
 • Problem 2: If S1 starts transmitting, S2 may well
   send during tp.
 • Problem 3: S1 is not aware of the collision
Sender S1                                    Sender S2

                    tp
 6/22/2010           Medium Access Control             49
            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%.
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         Problem 2 and 3:
  Vulnerability Period with CSMA
 • At time t, S1 sends a frame f1
 • At time tp-e, S2 may well send a frame f2 because
   S2 does not hear yet first bit of f1
 • At time 2tp-e, S1 does not hear yet first bit of f2
 • Vulnerability period is 2tp

Sender 1                                        Sender 2

                    tp
 6/22/2010            Medium Access Control              51
            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

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            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 (tmax)
 • The sender listens while sending: it takes Tr
   to send a frame.
 • To detect collision: Tr > 2.tmax
6/22/2010           Medium Access Control     53
      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 tmax?
 • The bandwidth is 10 Mbps
 • What should be the smallest Tr?
 • What is the minimum size of a frame?




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       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.tmax? Or 8.tmax?
6/22/2010       Medium Access Control   55
   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.tmax) 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.tmax) before trying again


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            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.tmax) 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)

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              Efficiency of Ethernet (2)
• Efficiency = Tr/(Tr+E(X) .(2.tmax)). 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?

  6/22/2010            Medium Access Control        58
              Efficiency of Ethernet (3)
• 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?
  6/22/2010            Medium Access Control             59
             Efficiency of Ethernet (3)
• Efficiency = Tr/(Tr+E(X) .(2.tmax)). What is
  E(X)?

• We found that E(X) = 1/A

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

 6/22/2010            Medium Access Control      60
             Efficiency of Ethernet (4)
• Efficiency = Tr/(Tr+(2.tmax)/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.


 6/22/2010            Medium Access Control       61
              Efficiency of Ethernet (5)
• Efficiency = Tr/(Tr+(2.tmax)/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.

  6/22/2010            Medium Access Control      62
              Efficiency of Ethernet (6)
• Efficiency = Tr/(Tr+(2.tmax).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.tmax is the is the smallest time to send a
  frame

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             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 q (10ms in 802.5)
  – Release the token

 6/22/2010           Medium Access Control           64
             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 q (10ms in 802.5)
  – Release the token

 6/22/2010           Medium Access Control           65
            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


6/22/2010           Medium Access Control   66
      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 q.
     –      Station 2 gets the token and sends for q.
     –      Station i gets the token and sends for q.
     –      Station N gets the token and sends for q.
• The duration of one cycle is then S=N.q + t.
  (t is the propagation time around the ring)
6/22/2010                  Medium Access Control        67
      Efficiency of Token Ring (3)
• Recall that the cycle duration is S=N.q + t
• The question now is: how long was the
  medium used during a cycle
• The answer is N.q.
• Efficiency = N.q/S = 1/(1+t/(N.q))
• Example: N= 40 stations, Ring of 2500m,
  what is the efficiency?
6/22/2010         Medium Access Control         68
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).q)
     – Wait for the token (t)
• What is the maximum with the previous
  example?


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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

6/22/2010            Medium Access Control   71
            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
6/22/2010                Medium Access Control              72
        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
6/22/2010               Medium Access Control    73
            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)

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                   MACAW
• Improved version of MACA
     – Adds acknowledgements for successful data
       frames
     – Use of CSMA
     – Exponential Backoff mechanism




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            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


6/22/2010             Medium Access Control          76
            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
6/22/2010           Medium Access Control     77
            Delay After Sensing Idle
                                   DIFS


                       PIFS


                SIFS
Medium Busy
               ACK        PCF          DCF      time
               CTS



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