Performance Analysis of IEEE 802.11 DCF under Limited Load by broverya76

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									          Performance Analysis of IEEE 802.11 DCF
                   under Limited Load
                                      Yong Shyang Liaw, Arek Dadej, Aruna Jayasuriya
                            Institute for Telecommunications Research, University of South Australia
                   Yong.Liaw@postgrads.unisa.edu.au, Arek.Dadej@unisa.edu.au, Aruna.Jayasuriya@unisa.edu.au


                                                                     presents the model and its analysis. Section III presents
   The goal of effective distributed network control is that each    numerical and simulation results. We conclude in Section IV.
wireless node operates near its maximum capacity, with bounded
queue delay. This “optimum” operating point, where maximum
throughput is achieved, is usually located well below the                 II.   LIMITED LOAD ANALYTICAL MODEL & ANALYSIS
saturation point. The choice of optimum operating point
requires knowledge of MAC performance below saturation,                A. Analytical Model
hence a limited load model of MAC operation needs to be studied      We extend the discrete-time Markov model from [2] by
for useful insights. In this paper, we extend the analytical model   adding another state (i.e. E) to represent the node with empty
of IEEE 802.11 DCF from [2] to study throughput performance          queue, as shown in Fig. 1. The parameters used in the model
under limited load, where the node’s transmit queue may be at
                                                                     are summarised in Table 1. Time is divided into slots of
times empty.
                                                                     duration σ.       A packet is discarded after m failed
                                                                     retransmissions. We assume that unsuccessful transmissions
                       I. INTRODUCTION
                                                                     are solely due to collisions (perfect channel). The model does
    The performance of IEEE 802.11 DCF MAC has been well             not account for packet capture effect. Similarly as in [2], a
studied under the saturation conditions, where each node             key assumption in this model is that collisions happen with the
always has packets in its transmission queue [2, 3, 4, 7]. In [2],   same probability p regardless of the number of retransmission
an analytical Markov model is used to study the IEEE 802.11          attempts.
DCF under saturation. Alternatively, [3, 4] use a model based            When a packet arrives in state bE,, the system enters with
on average values to provide a closed-form solution for              transition probability pe0/W0 one of the back-off states b0,i. The
saturation throughput. These works provide useful insights           pe0 depends on the packet arrival distribution function. At ith
into the IEEE 802.11 performance and in performance tuning           back-off stage, the back-off timer assumes a random value k
of the MAC protocol [7, 8]. However, all are based on the            chosen from {0, 1, 2, …, Wi-1} with uniform distribution,
assumption of saturation, and fail to capture the throughput         where Wi is given by (1). Each time slot, the back-off timer is
behaviour below saturation.                                          decremented by 1, and when zero is reached, the packet is
    A robust analytical model is required to study the DCF           transmitted. When the transmission collides with another
throughput under a full range of traffic loads. The benefits of      transmission (with probability p), the next (i+1)th back-off
such studies include traffic admission policies suitable for use     stage is entered with probability p/Wi+1. There is a maximum of
at individual nodes, under fully distributed network control. It     m back-offs. At mth failed retransmission, the packet is
is with this in mind that our analytical model is built; we are      discarded. After a successful transmission (with probability 1-
particularly interested in expressing the throughput                 p) or after mth failed retransmission, the transmission queue is
performance as a function of queue utilisation measurable            empty with probability q and non-empty with probability 1-q.
locally at every node. Furthermore, we extend our analysis to        If not empty, the system enters state bi,k to serve the next packet
account for the presence of hidden terminals [5, 10].                in the queue. When empty, the system enters state bE, and
                                                                     waits for a packet arrival. As soon as a packet arrives in state
    There are some works on finite load models for IEEE              bE, the system enters b0,k to contend for the channel.
802.11 [6, 9, 11]. However, these models are either based on
simplifying assumptions that limit their use near the optimum            Wi = 2 i W0                                            (1)
throughput point, or use variables not measurable locally at a
node, hence are not suitable for distributed network control.
                                                                       B. Transmission probability in a slot
     The main contribution of our work is the limited load               To solve for p, we balance the equations from the Markov
Markov model of IEEE 802.11, and its analysis based on               chain in Fig. 1 and obtain the following equations for k є {0,
variables observable locally at each node. We offer insight          …, Wi -1}. It shall be noted that y in Fig.1 denotes not a state,
into the throughput behaviour of 802.11 over a wide range of         but the sum of the probabilities of successful transmission and
traffic loads. The paper is organized as follows. Section II         discarding a packet, as given in (4).
                                                                                                                    1 m W i − 1 i Wi − k 1 − ( 2 p ) m +1 1 − p m +1
                                                                                                                      ∑ ∑0 p W =W0 1 − 2 p + 1 − p
                            TABLE I.                      PARAMETERS FOR THE MODEL
                                                                                                              β =
p                   Probability a transmission is unsuccessful/collided.                                            2 i=0 k =         i
q                   Probability an empty queue is found after a packet is served.
bi,k                State being in ith back-off stage, with back-off timer of value k.                                         for 0 ≤ p < 1.
bE                  State being in queue empty state.
Pi,k                Stationary probability being in state bi,k.
                                                                                                              The node will only transmit in state bi,0, hence the probability
PE                  Stationary probability being in state bE.                                                 ptx that a node will transmit in an arbitrary time slot is the
pe0                 Transition probability from state bE to a non-empty state.                                probability of being in any of the states bi,0.
W0                  The initial maximum contention window.
                                                                                                                       m                                  m
                                                                                                                                                                       1 − p m +1
Wi
m
                    Maximum contention window at ith back-off stage.
                    The maximum number of retransmissions allowed.
                                                                                                              p tx =   ∑
                                                                                                                       i=0
                                                                                                                             Pi , 0 =ε P0 , 0 where ε =   ∑
                                                                                                                                                          i=0
                                                                                                                                                                pi =
                                                                                                                                                                         1− p
                                                                                                                                                                                     (8)

            q                              1-Pe0
                           E                                                                                       We note that (6), (7) and (8) are independent of q and pe0,
               (1-q)/W0
                                                                                                              i.e. independent of packet arrival distribution. So far, (6) and
       y
                                Pe0/W0      Pe0/W0                      Pe0/W0             Pe0/W0             (8) are equivalent to expressions obtained (in terms of P0,0) in
                                                                                                              [2] for a saturated system.
            1-p            0, 0                     0,1          ...              0,W0-2             0,W0-1

                                                                                                                C. Solving for probability of collision, p
                                p/W1        p/W1                          p/W1              p/W1                  1) No Hidden Terminals
            1-p                                                                  1,W1-2
                                                                                                                  In a neighborhood with N identical nodes where every node
                           1, 0                    1,1           ...                                 1,W1-1
                                                                                                              can sense other nodes’ transmissions (no hidden terminals), a
                                                                                                              transmission results in a collision when any of the (N-1)
                                p/W2        p/W2                          p/W2                p/W2            neighbors also transmits in the same time slot. Eq. (8) must
            1-p
                                                                                                              satisfy the constraint given by (9), and the system may be
                                                   2,1                           2,W2-2             2,W2-1
                         2, 0                                    ...                                          solved numerically for a given value of PE.

                                                                                                                  p = 1 − (1 − p tx ) N −1 ⇒ p tx = 1 − (1 − p)1 /( N −1) (9)
                            .
            1-p             .                                                                                     2) Hidden Terminals
                            .
                                                                                                                  Let us assume that in addition to the (N-1) direct neighbors,
                                p/Wm        p/Wm                         p/Wm                p/Wm
                                                                                                              there are, on average, Nht hidden terminals in the region, as in
           1            m, 0                       m,1           ...             m,Wm-2             m,Wm-1    Fig. 2. A collision with a neighboring node can occur only
                                                                                                              when the nodes transmit in the same slot, otherwise the nodes
                                                                                                              would be able to sense busy channel. However, a hidden
                   Figure 1. IEEE 802.11 Markov model for limited load                                        terminal cannot sense our node, hence can begin the colliding
                                                                                                              transmission in any slot.        Therefore, for a successful
                           W0 − k
           P0 , k =               ⋅ [ p e 0 . P E + (1 − q ) y ]                                       (2)    transmission, the N-1 neighbors must not transmit in the same
                            W0                                                                                slot, and the Nht hidden terminals must not transmit during the
                                                                                                              entire transmission period.
                          Wi − k                                       for 0 < i ≤ m                   (3)
           Pi , k =              . p i . P0 , 0                                                                   Let Ltx be the average number of slots in a transmission
                           Wi
                                                                                                              period (ttx), and ξ be the average slot duration (i.e. average time
                    m
                                                                                                              spent in a state). We categorise states in the Markov model in
           y=      ∑ (1 −
                   i=0
                                        p ). Pi , 0 + p . Pm , 0 = P0 , 0                              (4)
                                                                                                              Fig. 1 into 3 types, empty-queue state (EQ), transmission state
                                                                                                              (TX) and backoff state (BO). The average time spent in each
           PE ⋅ p e 0 = q ⋅ y                                                                          (5)    state type before a transition occurs is different. Let teq, ttx and
                                                                                                              tbo be the average times spent in states of type EQ, TX and BO
       With (4) and (5), (2) and (3) can be rewritten as (6).                                                 respectively (they are functions of p (and PE), given in the later
               W −k i                                 (6)
                                                                                                              section). Then Ltx and ξ are given by (10), and ptx by (11).
       Pi , k = i  . p . P0 , 0                                                                               Eq. (8) must satisfy constraints given by (10) and (11).
                Wi
                                       for i є {0, …,m}, k є {0,…, Wi -1}
   From the normalising condition and (6), we obtain an
expression for P0,0 in terms of p and PE.
                                                                                                                                                          C
                  W i −1
            m
                                                                         2 (1 − PE )                                                        A      B
           ∑∑P
           i=0 k =0
                                i ,k   + PE = 1 ⇒ P0 , 0 =
                                                                                 β
                                                                                                       (7)
                                                                                                                                                                   B is a neighbor
                                                                                                                                                                   C is a hidden
                                                                                                                                                                   terminal

                                                                                                                    Figure 2. Transmission in the presence of hidden terminal
                 t tx                                                                   period is the transmission period ttx, hence the average time
   L tx =                                                                               spent in a BO state, tbo, is given by (15).
                 ξ
                                                         m W i −1
                                                                                 (10)       t bo = 1 − Pn −1 + Pn −1 t tx
   ξ = PE t eq + p tx t tx + t bo ∑                                                                                                                                (15)
                                                              ∑P
                                                        i = 0 k =1
                                                                     i ,k
                                                                                            Pn −1 = 1 − (1 − p tx ) N −1
                                            α                                               The average time taken to transmit a packet, ts, is the total
   = PE t eq + ( ε t tx +                        t bo ) p 0 , 0
                                            2                                           of the transmission time and MAC access time, as given in
                     W i −1
                                    Wi − k                                              (16), and the average transmission rate (µ) is 1/ts. However,
        1    m
                                                1 − ( 2 p ) m +1 1 − p m +1
α =
        2
            ∑∑                 pi
                                     Wi
                                           =W 0
                                                   1− 2p
                                                                −
                                                                   1− p
                                                                                        only (1-p) proportion of packets transmitted are successful, pdisc
            i=0         k =1                                                            proportion are discarded after maximum number of
                                                                                        retransmissions (m), and the rest (i.e. p-pdisc) result in a
  p = 1 − (1 − p tx ) N − 1 + L tx N ht                                          (11)   collision after which the packet is re-inserted into the queue.
  ⇒ p tx = 1 − (1 − p ) 1 /( N − 1 + L tx N ht )                                        Hence the effective packet service rate (µeff) is a sum of the rate
                                                                                        of successful transmissions (µsucc) and the rate at which packets
    For uniform node distribution over a large area with node                           are discarded (µdisc), as in (17). For small p and m > 1, pdisc is
density D, it can be shown that Nht is given by (12). Eq. (8),                          negligible, hence we may approximate µeff with µsucc.
(10), (11) and (12) may be solved for p iteratively. We note
that Nht = 0 is a special case without hidden terminals.                                    t s = taccess + ttx                                                    (16)
                     3 3 2    3 3                                                (12)       µ eff = µ succ + µ disc                                                (17)
    N ht =              R D =     N
                      4        4π
                                                                                                where µ succ = 1 − p , µ disc = p disc
  D. Packet Service Rate                                                                                         ts              ts
    Next, we find an expression for µ, the packet service rate
(rate at which packets are removed from the queue), in terms of                                                        p . Pm , 0        (1 − p ) p m + 1
PE and p. Let taccess be the average time spent backing off                                                 p disc =                 =                    P0 , 0
                                                                                                                       m
                                                                                                                                           1 − p m +1
before a packet is transmitted (i.e. MAC access time), and ttx be                                                      ∑P     i ,0
the packet transmission time (i.e. the slot time for TX state).                                                        i=0

Since at ith back-off stage the average number of back-off slots
is Wi/2, and the system enters the next back-off stage with                               E. Throughput
probability p, the average access time taccess is given by (13).                            For limited load where the arrival rate λ is less than µeff, the
                                                                                        node’s throughput (Tnode) is the portion of traffic that arrived
                               m
                                          Wi
                          ∑        pi(
                                          2
                                             t bo )
                                                               φ W 0 t bo        (13)
                                                                                        minus the portion that is discarded i.e. λ*(1-Pdisc). To derive
      t access     =       i=0
                                                          =                             the throughput of a node, it is necessary to know the
                                    m
                                                                  2ε                    distribution of packet service time. Here, we assume an
                                    ∑
                                    i=0
                                            pi
                                                                                        idealized M/M/1/kq model for the transmission queue, hence
                                                                                        the packet arrival rate (λ), throughput at a node (Tnode), and the
                                        m
                                                              1 − ( 2 p ) m +1
            where φ =                ∑
                                     i=0
                                             p iW i =
                                                                 1− 2p
                                                                                        total normalized throughput for a neighborhood of N identical
                                                                                        nodes, Ttotal, are given by (18), where ρ is the traffic intensity.
    The transmission time ttx depends on the payload size and                               λ = ρ . µ eff
the access mechanism (i.e. basic access or RTS/CTS). Note
                                                                                                      ⎧ ρ ⋅ µ succ for ρ < 1                                       (18)
that the RTS/CTS mechanism combats the hidden-terminal                                      T node = ⎨
problem, but amplifies the exposed-terminal problem, with a                                           ⎩ µ succ      for ρ ≥ 1
potentially greater performance degradation effect than hidden                                         N ⋅ T node ⋅ PL
terminal [10]. For this reason, we limit our analysis to the                                T total =
basic access mechanism. For basic access mechanism, ttx is                                            ChannelRat e
given by (14), where H is the packet header, PL the payload                                 For M/M/1/kq queue, ρ is related to the queue utilization as
size, τ propagation delay, and tsucc and tcoll the times taken by                       in (19), where kq is the size of the queue buffer. It is noted that
successful and unsuccessful transmissions respectively.                                 the model in Fig. 1 uses time slot of varying duration, and the
                                                                                        probabilities PE and Pi,k are relative frequencies of visiting the
 t tx = (1 − p ) t succ + pt coll                                                       corresponding states. Let SE and Si,k be the equivalent
 t succ = H + PL + SIFS + DIFS + ACK + 2τ (14)                                          probabilities with respect to arbitrary time. They are given by
                                                                                        (20), and specify the distribution of time the node spends in the
 t coll = H + PL + DIFS + τ
                                                                                        corresponding states; in particular SE is the proportion of time
    At the beginning of a BO-slot, the node senses either an                            the queue is empty, hence (1- SE) is the queue utilization.
idle channel with probability Pn-1, or a busy channel followed                                    1− ρ                                              (19)
by an idle slot with probability (1- Pn-1). The average busy                             SE =         k +1
                                                                                                           = 1−Uq
                                                                                               1− ρ q
                       PEteq                                    PEteq                                                                        0.9
SE =                                          =
                              m Wi                                        α ⋅ tbo                                                            0.8
           PEteq + Ptxttx + ∑∑ Pi, k tbo          PEteq + (ε ⋅ ttx +                   ) P0,0
                             i = 0 k =1
                                                                              2                                                              0.7

                                                                                                                                             0.6
                                                                                                          (20)




                                                                                                                    Throughput
                         Pi ,0 t tx                                    Pi ,0 t tx                                                            0.5
S i ,0 =                                          =
                                  m   Wi
                                                                                    α ⋅ t bo                                                 0.4
            PE t eq + Ptx t tx + ∑∑ Pi ,k t bo
                                                                                                                                                                                                                                                 N=4
                                                      PE t eq + (ε ⋅ t tx +                    ) P0,0                                                                                                                                            N=10
                                 i =0 k =1                                             2                                                     0.3
                                                                                                                                                                                                                                                 N=20
                                                                                                                                             0.2                                                                                                 N=30
                         Pi ,k t bo                                   Pi ,k t bo                        for i ≥ 1                                                                                                                                N=40
                                                                                                                                             0.1
S i ,k =                                          =
                                  m   Wi
                                                                                    α ⋅ t bo                                                                                                                                                     N=50
            PE t eq + Ptx t tx + ∑∑ Pi ,k t bo        PE t eq + (ε ⋅ t tx +                                                                   0
                                                                                               ) P0,0
                                 i =0 k =1                                             2                                                           0        0.1         0.2       0.3     0.4    0.5      0.6
                                                                                                                                                                                          Queue Utilization
                                                                                                                                                                                                                    0.7       0.8          0.9          1



    PE.teq is the proportion of time the queue is empty in an                                                                                              Figure 3. System throughput versus queue utilization
average time slot. Eq. (20) gives a unique solution for any
given value of λ, for 0 < teq < 1/λ. Then, for simplicity, a                                                                                 0.9
solution is obtained when PE = SE, and teq is given by (21).                                                                            0.85

                                                α ⋅ t bo                                                                                     0.8
                                  (ε ⋅ t tx +              ) P0 , 0
S E = PE ⇒ t eq =                                 2                                                       (21)                          0.75




                                                                                                                    Throughput
                                             1 − PE                                                                                          0.7

                                                                                                                                        0.65

  F. Queue Delay                                                                                                                             0.6

   For M/M/1/kq queue, the queue delay is given by                                                                                      0.55                                                                               Saturation throughput
                                                                                                                                                                                                                           Max. throughput
                                                                                                                                             0.5
                ⎛ ρ    ( k + 1) ρ
                                                              k q +1
                                                                        ⎞1                                                                             0                10                 20              30                    40                     50

      t queue = ⎜     − q                                               ⎟                                 (22)                                                                              No. of Nodes (N)

                ⎜1− ρ          k +1
                          1− ρ q                                        ⎟λ
                ⎝                                                       ⎠                                                                                       Figure 4. Maximum and saturation throughput

                                                                                                                                              0.3
                              III. SIMULATIONS & RESULTS                                                                                                                                                        N=10
                                                                                                                                                                                                                N=20
                                                                                                                                             0.25
    Table II shows the default parameters used for both                                                                                                                                                         N=30

simulations and numerical evaluation. Uq is approximated by
                                                                                                                     Queue Delay (seconds)




                                                                                                                                                                                                                N=40
                                                                                                                                              0.2
ρ, since a buffer size (kq) of 60 packets is assumed, and Tnode is                                                                                                                                              N=10 (simulation)
                                                                                                                                                                                                                N=20 (simulatoion)
taken as equivalent to λ since the pdisc is negligible in (18).                                                                              0.15
                                                                                                                                                                                                                N=30 (simulation)


                                                                                                                                              0.1
  A. Without Hidden Terminals
    Fig. 3 shows the total system throughput (Ttotal) in respect to                                                                          0.05

queue utilization (Uq) at the node. The point of maximum
throughput is at the saturation point (i.e. Uq = 1) only for less                                                                                  0
                                                                                                                                                       0        1        2         3       4      5       6            7         8          9           10
than 10 nodes in the neighborhood. For N ≥ 10, the maximum                                                                                                                              Node Throughput (pps)

throughput is achieved below the saturation point (i.e. Uq < 1).
                                                                                                                                                                                  Figure 5. Queue delay
The maximum throughput point shifts towards lower Uq with
increasing N. This result agrees with the results in [11].                                                                                   0.6
                                                                                                                                                                                                                                                N=4
   Fig. 4 shows that maximum throughput degrades slower                                                                                      0.5                                                                                                N=10
than the saturation throughput, and approaches an asymptotic                                                                                                                                                                                    N=20

value with large N. Significant benefits can be obtained                                                                                     0.4
                                                                                                                                                                                                                                                N=30
                                                                                                                     prob of collission




                                                                                                                                                                                                                                                N=40
                                                                                                                                                                                                                                                N=50
                                                                                                                                             0.3
   TABLE II.         VALUES FOR SIMULATION & NUMERICAL EVALUATION
   Payload size (PL)                         8184 bits
                                                                                                                                             0.2
   MAC header                                272 bits
   PHY header                                128 bits
                                                                                                                                             0.1
   ACK                                       112 bits + PHY header
   Channel Bit Rate                          1 Mbit/s                                                                                         0
   Propagation delay (τ)                     1 µs                                                                                                  0        2       4         6    8      10   12   14    16        18      20        22     24         26
                                                                                                                                                                                        Node Throughput (pps)
   Slot time (σ)                             20 µs
   SIFS                                      10 µs                                                                                                              Figure 6. Transmission collission probability
   DIFS                                      50 µs
   Initial contention window (W0-1)          31                                                                     through distributed control of queue utilization at each node.
   Maximum number of retransmissions (m)     5
   Uq is very sensitive to the traffic arrival rate near the                                 conditions, in respect to queue utilization before saturation.
maximum throughput point. A small increase in λ will cause a                                 We have shown that for large numbers of nodes the maximum
significant increase in Uq, and consequently a dramatic increase                             throughput is achieved well before the saturation, and have
in queue delay as shown in Fig. 5. This can be explained by                                  validated our model via simulations. Our model offers a useful
sharp increase in collision probability when the throughput                                  method of estimating the optimum operating point for the
exceeds certain threshold, as illustrated in Fig. 6. As a result,                            purpose of distributed network control, based on Uq.
operating at saturation point is not desirable as it dramatically
                                                                                                                                                         m =5
increases queue delay and reduces throughput for larger                                                               0.9

numbers of nodes. For maximum throughput with low delay,                                                              0.8

the “optimum” operating point for IEEE 802.11 MAC is at the                                                           0.7
“knees” of the delay curves shown in Fig. 5.




                                                                                              Normalised Throughput
                                                                                                                      0.6

                                                                                                                      0.5                                                       N=4 (Simulation)
    B. Hidden Terminals                                                                                                                                                         N=10 (Simulation)
                                                                                                                      0.4                                                       N=20 (Simulation)
              0.6
                                                                                                                                                                                N=30 (simulation)
                                                                                                                      0.3
                                                                                                                                                                                N=4 (Analytical)
              0.5                                                                                                     0.2                                                       N=10 (Analytical)
                                                                                                                                                                                N=20 (Analytical)
                                                                                                                      0.1
              0.4                                                                                                                                                               N=30 (Analytical)
 Throughput




                                                                                                                      0.0
                                                                                                                            0       5         10       15           20     25       30              35
              0.3                                                                                                                                    Arrival Rates (pps)
                                                                                N=4

              0.2
                                                                                N=10                                            Figure 8. Throughput performance of IEEE 802.11
                                                                                N=20
                                                                                N=30
              0.1
                                                                                N=40
                                                                                                                                 TABLE III.        ESTIMATED QUEUE ULTILIZATION
                                                                                N=50
               0
                    0.0   0.1   0.2   0.3   0.4   0.5       0.6   0.7   0.8   0.9      1.0
                                                                                             N                                  Uq (Analytical)                  Uq (Simulation)
                                             Queue Utilization                               10                                 0.33±0.04                        0.31±0.09
                                                                                             20                                 0.17±0.04                        0.18±0.05
   Figure 7. Throughput performance in the presence of hidden terminals                      30                                 0.09±0.04                        0.082±0.029

    Fig. 7 depicts throughput performance of the IEEE 802.11
in the presence of hidden terminals. As expected, the                                                                                               REFERENCES
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    Fig. 8 shows the results. For N=4, the maximum                                           [8]                            Haitao Wu, et al, “Performance of Reliable Transport Protocol
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throughput occurs at saturation, and for N ≥ 10 the throughput                                                              Proc. of INFOCOM’02, June 2002.
peaks before the saturation point. The maximum throughput                                    [9]                            O. Tickoo, B. Sikdar, “A Queue Model for Finite Load IEEE
closely matches that from the analytical model. At 90% of                                                                   802.11 Random Access MAC”, Proc. of ICC 2004, June 2004.
maximum arrival rate, queue utilization is relatively low, close                             [10]                           Aruna Jayasuriya, et al, “Hidden vs. Exposed Terminal Problem
to the analytical results, as shown in Table III.                                                                           in Ad hoc Networks”, Proc. of Australian Telecommunication
                                                                                                                            Networks and Applications Conference , Dec 2004.
                                                                                             [11]                           C. L. Fullmer, et al, "Solutions to Hidden Terminal Problems in
                                       IV. CONCLUSIONS                                                                      Wireless Networks", Proc. of ACM SIGCOMM 97, Sept 1997.
    We have presented an analytical model to study the
performance of IEEE 802.11 DCF under a full range of load

								
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