Chapter 5
Bidirectional Traffic Flow
Fig.5-1 and Fig.5-2 show the scenarios of the bidirectional traffic flow. In the scenario of the unidirectional traffic flow, if a frame is transmitted successfully, the backoff counter will be reset. But in the scenario of the bidirectional traffic flow, when a frame is transmitted successfully, the backoff counter may not be reset. The key point is if the frame in the front of the buffer or not. There are examples as below.
(1) One station sends an aggregated data frame to the AP. The AP will return the Ack+AggregatedData frame to the station. If the AggregatedData frame is at the head of buffer in the AP, the backoff counter of AP will be reset by the piggyback. If not, the counter will not be reset.
(2) The AP sends an aggregated data frame to some station. The station will return the Ack+AggregatedData frame to the AP. The backoff counter of the station must be reset by the piggyback.
Fig.5-1: WLAN Scenario – Bidirectional traffic flow
Fig.5-2: Bidirectional traffic flow with aggregation
Chapter 5 Bidirectional Traffic Flow
5-1 Our Method for Calculating the Channel Utilization
If a collision happens in the channel, the channel will pay for this collision. The is the same as the
If a successful transmission happens in the channel, the channel will pay for this successful transmission as shown in
_
channel success EmptySlots OurAnalysis RTS CTS phy up data
phy down data ACK
T W T T T
T T SIFS τ DIFS
= × + + +
+ + +
Fig.5-3 Successful transmission state for bidirectional traffic flow
channel
p is the probability that a collision happens to the medium. is the probability that a collision happens to a station. The channel utilization can be represented as below:
channel mac up data mac down data OurAnalysis
channel channel collision channel channel success
mac up data mac down data
p T T
TRTS,TCTS,TACK,SIFS DIFS and , τ are deterministic values. Please reference the table 1 in the chapter 6. Next and will be calculated as the following sections. From the result in [5] and the equation (5.17),
channel
p WEmptySlots
p satisfies
(5.2) 1 1
1 (1 )
p N
Wbi
= − − −
Similar to the evaluation of (4.11), pchannel is
(5.3) 2
collision channel
success collision
r p
p = =
r +r − p
Chapter 5 Bidirectional Traffic Flow
5-2 The Proposed Method of Paper [4] for Calculating the Channel Utilization
Fig.5-4 shows a virtual transmission time. The evaluation is the same as the chapter 4.
Fig.5-4 A virtual transmission time
So tv can be expressed as below:
TTXOP is the time duration for a TXOP in a virtual transmission time.
(5.5)
TXOP RTS CTS up down ACK
T T
Colli
T is the duration of the i-th collision in a virtual transmission time.
(5.6)
From the result of [4], tv can be expressed as below:
where and are the average transmission time for a physical data frame in the uplink and downlink directions respectively
Tup Tdown
(5.8) where and are the average transmission time for a MAC data frame in the uplink and downlink directions respectively
_ mac up
T Tmac_down
The utilization in [4] can be expressed as:
where p satisfies p N in the
p CW N
Chapter 5 Bidirectional Traffic Flow
5-3 How to Calculate WEmptySlots
5-3-1 Our Analytic Method
Suppose there are stations (including the AP) in the saturated WLAN. They have the equal probability
N 1
N to contend for the channel successfully. Fig.5-5shows a possible transmission pair in the saturated WLAN that supports bidirectional traffic flow. represents that the AP contends for the channel successfully to send a frame to and will return a Ack+AggregatedData frame to the AP. So the backoff counter of the can be reset by the piggyback.
represents that contends for the channel successfully to send a frame to the AP and the AP will return a Ack+AggregatedData frame to the . If this frame is in the head of the buffer, the backoff counter of the AP can be reset by the piggyback..
Fig.5-5 A possible transmission pair in the saturated WLAN that supports bidirectional traffic flow for the AP
Next we will extend the concept from the unidirectional traffic flow to the bidirectional traffic flow. is the average number of backoff slots experienced by a packet until it is transmitted successfully or discarded in the saturated WLAN with unidirectional traffic flow. From the equation (4.19) and
Wx
Fig. 5-5, we approximate the
time interval that the AP contends for the channel successfully by . Next we will discuss the average time interval after which the backoff counter of the AP can be reset through the piggyback.
Wx
So the average time interval after which the backoff counter of the AP is reset by the piggyback isWx
2 . If the backoff counter of the AP is not reset by the piggyback, the average time interval after which the backoff counter is reset isW .x
Case Ⅰ: The backoff counter can be reset by the piggyback. The average time interval isWx
Let y1 be the average number of transmission tries for the AP during 2 Wx
and we can get the average time between two successive transmissions. Let’s recall the calculation (4.20), if is small, it will be smaller than 1.5. Since is smaller than , we approximate by 1.
x p y1
x y1
Chapter 5 Bidirectional Traffic Flow
Case Ⅱ: The backoff counter can not be reset by the piggyback. The average time interval isWx.
The probability of this case is:
(5.11)
We suppose that the AP in a saturated WLAN that supports bidirectional traffic flow will make a transmission for every slot in average: We estimate
by Next we will discuss the case that a station will make a transmission for every
slot in average. As shown in
_ th
bi STA
W Fig.5-6, we approximate the time interval that
contends for the channel successfully by .
STA1 Wx
Fig.5-6 A possible transmission pair in the saturated WLAN that supports bidirectional traffic flow for STA1
1
Let denote the event that AP contends for the channel successfully and sends a frame to AP by the piggyback. For an example, the pair
(AP,STA )i
STAi
{
(AP,STA )| i = 1 indicates the backoff counter of i}
is reset by the piggyback.The average time interval after which the backoff counter of is reset by the piggyback is
STA1
STA1
Wx
2 . Otherwise, the average time interval after which the backoff counter of STA1 is reset isW .x
Case Ⅰ: The backoff counter is reset by the piggyback. The average time interval isWx
Case Ⅱ: The backoff counter is not reset by the piggyback. The average time interval isWx.
The probability of this case is:
(5.15)
We suppose that one station in a saturated WLAN that supports bidirectional traffic flow will make a transmission for every slot in average, we estimate
by So the AP or stations in a saturated WLAN that supports bidirectional traffic flow
Chapter 5 Bidirectional Traffic Flow
will make a transmission for every Wbith slot in average:
(5.17) bi_AP ( 1) bi_STA
bi
W N
W W
N
+ −
=
Similar to the analysis in chapter 4, WEmptySlots_OurAnalysis can be expressed as:
(5.18) W _ N
bi EmptySlots OurAnalysis
= W
5-3-2 The Proposed Method of Paper [4]
There are two possible ways to piggyback it
(1) Each station always has at least one frame to the AP. If the AP contends for the channel successfully and send a frame to a particular station (with probability 1
1
N− ). The frame in the front of the buffer in the particular station can be piggybacked to the AP.
(2) The AP always has at least one frame to every station. If one station contends for the channel successfully (with probability 1
1
N− ) and send a frame to the AP. The frame in the AP to the station can be piggybacked to the station.
In these two cases, the backoff counter is half of . Therefore, the actual average backoff counter for a station or AP satisfies:
bidirectional
The average period of virtual transmission time for the AP and each STA is (N-1)
× . In this paper, it gets a conclusion that the total empty slots in a virtual transmission time as below:
tv
(5.21) Random variable is the number of collisions in a virtual transmission time and it is a geometrical distribution. So its mean can be calculated as below:
Nc
Chapter 5 Bidirectional Traffic Flow
The backoff slots between two transmissions in the medium in the saturated WLAN that support bidirectional traffic flow can be calculated from the equation as below:
Chapter 6
Simulate and Numerical Results
Table 1 lists the parameters of the 802.11 WLAN shared by the simulate environment and analytic equation.
PLCP header bit rate 1Mbps
PLCP overhead 192 µ sec
Basic bit rate 5.5Mbps
Data bit rate 11Mbps
TRTS 20 8
PLCP overhead + 221 sec
5500000× = µ
TCTS 14 8
PLCP overhead + 212 sec
6000000× = µ
TACK 14 8
PLCP overhead + 212 sec
6000000× = µ Table 6-1 Input parameters and values
From (4.11) (4.4) (5.2), we can give the values of CWmin and M to calculate the analytic collision probabilitiesp by bisection method.