Effective Goodput Computation

In this section, we focus our analysis on the effective goodput of the RTS/CTS access method. Our objective is to calculate the maximum throughput achievable with the IEEE 802.11a WLAN at the MAC layer for a given SNR by taking into consideration of the MAC, PHY, and retransmission overheads. In our analysis, we make following assumptions: i) the transmitter generates, at an infinite rate, l-octet long data payload (MSDU); ii) the MSDU is not fragmented; iii) the propagation delays are neglected; iv) the constant wireless channel condition throughout the entire frame delivery period. After a station makes the PHY mode selection and starts transmitting, the selected PHY mode will be used for all the potential retransmissions.

In brief, a frame with l-octet data payload is to be transmitted using the PHY mode m

over the wireless channel with condition γ . Let m’ denote the PHY mode used for the

corresponding ACK frame transmission and it is determined based on m according to the rule specified in Section 3.2.

To simplify the analysis, we separate the transmission phases of the RTS/CTS access method into two stages: i) channel reservation ii) data transmission. The first stage, channel reservation, includes the DIFS deferral phase, backoff phase, RTS transmission phase, SIFS deferral phase, CTS transmission phase, and SIFS deferral phase. The remained phases belong to the data transmission stage. Now, let’s consider the entire delivery progress of the frame transmission. Since the maximum numbers of transmission attempt to deliver the RTS frame and the data frame are specified by ShortRetryLimit, denoted as n_s, and LongRetryLimit, denoted as n_l, respectively.

Recalling Section 2.1.1, every station maintains an SSRC as well as an SLRC.

Whenever a CTS frame is received in response to the RTS frame, the SSRC is reset to 0. On the other hand, SLRC is reset to 0 when an ACK is received in response to a data frame. That is to say two counters are independent to each other. An example will make it clearer. Let n_s be 7 and n_l be 4. At beginning, we send the RTS frame but unfortunately it fails, i.e. it fails at the channel reservation stage, for first 6 times. At the 7th attempt, it succeeds. So there are 7 retries in total so far. Now the data frame is sent afterwards, but the transmission fails at the data transmission stage. Then one retry count for the data frame is consumed, i.e. there are 3 more left. Again, the RTS frame is sent, and it will get 7 more chances. Therefore, for a single data frame, there may be totally 28 (7×4) RTS frames and 4 data frames sent in the transmission progress in the worst-case scenario. We illustrate all discussions above with Figure 3.6.

stands for the probability of a successful channel reservation within the

, ( ,

succ ch

P γ n)

retry limit and it can be calculated by

is the probability of a successful channel reservation. represents the probability of a successful data transmission and it can be computed by

, , ( ,

By referring to Figure 2.2(a), a successful channel reservation duration is equal to a backoff delay, plus the RTS transmission time, plus an SIFS time, plus the CTS transmission time, and plus an SIFS time. However, whenever the channel reservation fails, the station has to wait for a CTS_Timeout period or an EIFS interval, and then execute a backoff procedure before the retransmission (see Figure 2.2(b) and (d)).

According to the IEEE 802.11 MAC standard, an EIFS interval is equal to an SIFS time plus a DIFS time plus the ACK transmission time at the most robust 6 Mbps and a CTS_Timeout is equal to an SIFS time plus a CTS transmission time plus a Slot_Time.

Therefore, the average transmission duration of the channel reservation can be calculated by

bkoff rts wait ch bkoff rts

i

where is the conditional probability that the channel is successfully reserved at the kth transmission attempt. It can be computed by

[ | ]( P k succch γ,n)

[ ]( ) ( ) ( )

T i is the average backoff interval before the ith transmission attempt or,

equivalently, the (i -1)th retransmission attempt. It can be calculated by

( ) ^{min 2 1}

(

)

On the other hand, the average time wasted in attempting reservation of the channel n_s times in error is given by

( ) ( ) , ( )

We have discussed the average duration the channel reservation stage takes no matter it succeeds or not. Using this statistics, we can compute the duration for a data frame transmitted using the RTS/CTS access method. By referring to Figure 3.6, the probability of the successful frame transmission within the retry limit is

( ) ( ) ( )

succ succ ch s xmit data

n i

succ ch s xmit data

i

A successful exchange of the frame transmission must complete both channel reservation and data transmission stages. Therefore, the overall duration of the frame transmission is equal to a channel reservation duration, plus the data transmission

duration which include the data transmission time, plus an SIFS time, plus the ACK transmission time, and plus a DIFS time. Once the data transmission fails, the station has to wait for an ACK_Timeout period or an EIFS interval, and then perform another channel reservation again (see Figure 2.2(c) and (e)). Therefore, the average transmission duration of the whole frame transmission can be calculated by

( ) [ ]( )

wait data succ ch data

i

where is the conditional probability that the data is successfully transmitted at the kth attempt and it can be computed by

[ | ](, , ,

succ ch s xmit data s xmit data

succ

On the other hand, the average time wasted in attempting transmission of the data frame is given by

( )

succ ch s xmit data succ ch

succ

wait data

succ ch data fail ch

succ ch s xmi

succ ch data fail ch

succ ch s xmit data succ ch

succ

succ ch s xmit data

wait data

The expected effective goodput can then be calculated by

( )

which is based on the fact that, with probability , there is a successful data frame delivery

within the retry limit after dropping the previous k frames. It can also be interpreted as follows: the expected effective goodput is equal to the ratio of the expected delivery data payload to the expected transmission time. Note that, under the constraint of the

( )

[

]

frame retry limit, the successfully delivered data payload is no longer a fixed value of l.

It is actually a two-value random variable and can take the value of l if delivery succeeds, with probability , or 0 if delivery fails. So,

is the expected data payload. Similarly, is the expected transmission time spent on the frame delivery attempt, irrespectively of whether it is successful or not.

(, , ,