### Introduction

**Note: This is the translation of an old paper from a research activity prior to IEEE 802.11n standard publication, so all calculations are based on IEEE 802.11g standard, as well as IEEE 802.11a, that were the fastest physical layers for the IEEE 802.11 family at that time.**

This paper will use IEEE 802.11g standard, which defines the highest data rates within IEEE 802.11 set of standards. IEEE 802.11a runs at the same data rate but in 5 GHz band instead of 2.4 GHz. The throughput and delay limits will be calculated within an ideal scenario. The paper will focus on maximum throughput and minimum delay that can be achieved by the MAC protocol.

### Hypothesis

This is the scenario and the hypothesis used in order to calculate the throughput limits for IEEE 802.11 MAC protocol:

**Collision free scenario**: Only a wireless device transmits at a time. The remaining devices just listen and answer to control frames when they are required to do it.**Error free scenario**: The wireless channel is free of errors and interferences. That is, when a wireless device starts a frame transmission, it will be received by every wireless device within the range of the radio.**Always data ready**: The source wireless device always has a packet, or MSDU, ready to send at any time. The packets are never queued, so there is no waiting time for none of them.**No signal delay**: The wireless devices are close enough one from another, so the signal delay time will be close to 0 (δ=0).**The slot time length is 20 µs**. The IEEE specification sets the standard length of every slot time in 20 µs, however there is also an optional 9 µs length that can be used when the BSS is made of wireless devices that are able to use ERP modulation and implement this functionality, although within a IBSS the slot time length must be established in 20 µs. For this reason, this one is the chosen length for this analysis.

### Using basic DCF schema

The Transmission Cycle (TC) for the first packet from the buffer, with L bytes length using the M_{d} transmission mode for data frames and the M_{c} mode for control frames, can be calculated for the DCF basic schema by means of the next equation:

\begin{eqnarray}

TC_{Basic}(L_{MSDU},M_{d},M_{c}) & = & T_{DIFS}+T_{Contention}+ \nonumber \\

& & +T_{PPDU}(L_{PSDU},M_{d})+ \nonumber \\

& & +T_{SIFS}+T_{ACK}(M_{c})+2 \delta

\label{eqn:CTDCFBasico}

\end{eqnarray}

Parameters for this calculation are available on table 1. A packet length can vary between 0 and 2312 bytes, or until 2304 when WEP is used. The MAC protocol adds a 24 bytes length header, even though when an access point is configured in wireless bridge mode, a fourth address field is added to the header, so in this case the header length would be 30 bytes long. Finally, 4 more bytes are added for the FCS field that contains a 32 bit CRC.

T_{SIFS} |
20 µs |

T_{DIFS} |
50 µs |

T_{Slot} |
20 µs |

CW_{Min} |
15 |

T_{Preamble} |
16 µs |

T_{Signal} |
4 µs |

T_{Symbol} |
4 µs |

Signal extension | 6 µs |

Propagation delay(δ) | 0 µs |

The average contention time (T_{Contention}) can be calculated from the probability distribution linked to the chosen backoff counter which is an integer uniform distribution. So when time tends to infinite, the contention time will tend to:

\begin{equation}

\lim_{t \rightarrow \infty} T_{Contention} = \lceil

\frac{CW_{min}}{2} \rceil *T_{Slot}

\label{eqn:TContienda}

\end{equation}

The IEEE 802.11g standard establishes the minimum contention window value (CW_{min}) in 15 slots. Because of backward compatibility of IEEE 802.11g with IEEE 802.11b devices, there is an alternate value for this parameter, that is 31, although this value only is used when the transmission rate set for each device is composed of 1, 2, 5.5 and 11 Mbps. For this reason, for this analysis, the selected value will be 15.

[HERE A FIGURE...]

The figure {ref} shows the structure of a PPDU, or physical frame. With the next equation, transmission time for a physical frame can be calculated depending on the length, in bytes, of the frame, or PSDU (which also is equivalent to the MPDU) and the used transmission mode:

\begin{eqnarray}

T_{PPDU}(L_{PSDU},M) & = & T_{Preamble}+T_{Signal} + \nonumber \\

& & + T_{Sym}*\lceil \frac{16+8*L_{PSDU}+6}{N_{DBPS}} \rceil + \nonumber \\

& & +Signal\ extension

\label{eqn:TPPDU}

\end{eqnarray}

The term L_{PSDU} is the length, in bytes, for the packet, the MAC header and the FCS field. The term NDBPS is the number of coded bits per OFDM symbol, and it depends on the mode M used in order to transmit the physical frame. The table {reference} shows the relation of transmission rates next to their related parameters: Modulation, rate, date bits per OFDM symbol, etc.

R1-R4 |
Rate (Mbps) |
Modulation |
Code rate (R) |
Coded bits per subcarrier (NBPSC) |
Coded bits per OFDM symbol (NCBPS) |
Data bits per OFDM symbol (NDBPS) |

1101 | 6 | BPSK | 1/2 | 1 | 48 | 24 |

1111 | 9 | BPSK | 3/4 | 1 | 48 | 36 |

0101 | 12 | QPSK | 1/2 | 2 | 96 | 48 |

0111 | 18 | QPSK | 3/4 | 2 | 96 | 72 |

1001 | 24 | 16-QAM | 1/2 | 4 | 192 | 96 |

1011 | 36 | 16-QAM | 3/4 | 4 | 192 | 144 |

0001 | 48 | 64-QAM | 2/3 | 6 | 288 | 192 |

0011 | 54 | 64-QAM | 3/4 | 6 | 288 | 216 |

The equation can be used for calculating the duration of control frames. These frames length are 14 bytes for ACK and CTS, an 20 bytes for RTS .

The IEEE 802.11 standard specifies that the control frames can be transmitted at any rate included in the BSS basic rate set. Since every IEEE 802.11g device must be able of transmit and receive frames at 6, 12 and 24 Mbps, control frames must be transmitted at one of these three rates. In the current analysis, it has been decided to use the highest rate lesser or equal to the lowest data transfer rate, in the same way as Xiao {cite} did in his paper. By using the following notation: {data frame rate, control frame rate}, the valid combinations are: {54,24}, {48,24}, {36, 24}, {24,24}, {18,12}, {12,12}, {9,6}, {6,6}.

#### Throughput limit

The throughput (S) of a communications protocol is defined as the number of information bits transmitted per transmission cycle (TC). It can be calculated , in bps, by means of the next equation:

\begin{equation}

S_{Basic}(L_{MSDU},M_{d},M_{c}) = \frac{8*L_{MSDU}}{TC_{Basic} \

(L_{MSDU},M_{d},M_{c})}

\label{eqn:SDCFBasico}

\end{equation}

In figure {cite}, upper limits for throughput have been drawn related with the packet length and the transmission rate.

As the figure {} shows, the longer the packet lenght the higher throughput, this is because there is a direct relationship between the packet payload length related with preambles and headers size. Logically, the higher the transmission rate, the higher the throughput too. Even though with an hypothetical infinite transmission rate, the throughput reaches a finite value! This behaviour is due to the inter-frame spaces. These spaces limit the theoretical maximum value the throughput could reach.

In this analysis, the infinite transmission rate is got by letting the physical frame data field (DATA) be coded with an infinite number of bits per OFDM symbol, this way, the DATA field duratin will be 0 µs, and the total physical frame duration will be the one for PLCP preamble and SIGNAL field.

#### Transmission delay limit

A packet transmission delay (D) is calculated from the transmitting device point of view. Taking into account this premise, a packet of L bytes length will be transmitted using the Md transmission mode, its required transmission time can be calculated by means of the next equation:

\begin{equation}

D_{Basic}(L_{MSDU},M_{d})=T_{DIFS}+T_{Contention}+ \

T_{PPDU}(L_{PSDU},M_{d})+\delta

\label{eqn:DDCFBasico}

\end{equation}

The figure {reference} shows that the transmission delay increases with the packet length, which is logical: The greater the length the greater, the transmission delay. Also the transmission delay increases when transmission rate decreases. Again, there is a lower limit for this delay, even though we would use an infinite transmission rate, this is due to DIFS, the content time and physical layer protocol overload.

It must be pointed out that, the transmission in one way is taken into account for calculating the delay: from the transmitter device to the receiver, or receivers. This is because of the hypothesis 2, or channel error free hypothesis. So, when frames arrive at receiving device, they are delivered to upper layers in order to be processed. In the meanwhile, an ACK frame is sent back to the transmitter, if necessary.

[HERE A FIGURE...]

### Using DCF schema with RTS/CTS

The transmission cycle for the first frame from the buffer, with a length of L bytes, using transmission mode M_{d} for data frames and mode M_{c} for control frames, and using DCF schema with CTS/RTS, can be calculated by means of the following equation:

\begin{eqnarray}

CT_{RTS}(L_{MSDU},M_{d},M_{c}) & = & T_{DIFS}+T_{Contention}+T_{ACK}(M_{c})+ \nonumber \\

& & +T_{RTS}(M_{c})+T_{CTS}(M_{c})+ \nonumber \\

& & +T_{PPDU}(L_{PSDU},M_{d})+ \nonumber \\

& & +3*T_{SIFS}+4\delta

\label{eqn:CTDCFRTS}

\end{eqnarray}

#### Throughput limit

#### Transmission delay limit

### An example

Here you have an example of use of the equations previously exposed. In this example, the transmitting IEEE 802.11a, or IEEE 802.11, device sends data packets with an average length of 1024 bytes, and selects a transmission mode that lets it send data at 54 Mpbs ans control frames at 24 Mbps. The example will illustrate both Basic DCF and RTS/CTS schemas.

In the first place, we are going to calculate the transmission time for the physical frames, both data and control frames. For data frames this time is:

\begin{eqnarray}

T_{PPDU}(1024+28,54) & = & 16+4+4* \lceil \frac{16+8*(1024+28)+6}{216} \

\rceil+6 = \nonumber \\

& = & 186\ \mu s \nonumber

\end{eqnarray}

The transmission time for ACK and CTS frames, which length are 14 bytes, is:

\begin{eqnarray}

T_{PPDU}(14,24) & = & 16+4+4* \lceil \frac{16+8*14+6}{96} \

\rceil+6 = \nonumber \\

& = & 34\ \mu s \nonumber

\end{eqnarray}

The transmission time for RTS frames, which length are 20 bytes, is:

\begin{eqnarray}

T_{PPDU}(20,24) & = & 16+4+4* \lceil \frac{16+8*20+6}{96} \

\rceil+6 = \nonumber \\

& = & 34\ \mu s \nonumber

\end{eqnarray}

The cycle time for both es will be calculated with the equations {ref1} and {ref2}

\begin{eqnarray}

CT_{Basic}(1024,54,24) &=& 50+ \lceil \frac{15}{2} \rceil *20+186+10+34+0 \

= \nonumber \\

& = & 440\ \mu s \nonumber

\end{eqnarray}

\begin{eqnarray}

CT_{RTS}(1024,54,24) &=& 50+\lceil \frac{15}{2} \rceil *20+34+34+34+186+ \

\nonumber \\

& & +3*10+0 = 528\ \mu s \nonumber

\end{eqnarray}

Now, the upper limit for throughput for both schemas will be calculated with the equations {ref3} and {ref4}:

\begin{eqnarray}

S_{Basic}(1024,54,24) = \frac{8*1024}{440\ \mu s} = 18,62\ Mbps \nonumber

\end{eqnarray}

\begin{eqnarray}

S_{RTS}(1024,54,24) = \frac{8*1024}{528\ \mu s} = 15,52\ Mbps \nonumber

\end{eqnarray}

The relative values are easily calculated:

\begin{eqnarray}

S_{Basic}(1024,54,24)\ (\% ) = \frac{18,62\ Mbps}{54\ Mbps} = 34,48\ \% \

\nonumber

\end{eqnarray}

\begin{eqnarray}

S_{RTS}(1024,54,24)\ (\% ) = \frac{15,52\ Mbps}{54\ Mbps} = 28,74\ \% \

\nonumber

\end{eqnarray}

Finally, the transmission delay limits for both schemas will be calculated with equations {ref5} and {ref6}:

\begin{eqnarray}

D_{Basic}(1024,54) = 50 + \lceil \frac{15}{2} \rceil * 20 +186+0 = \

396\ \mu s \nonumber

\end{eqnarray}

\begin{eqnarray}

D_{RTS}(1024,54,24) & = & 50+ \lceil \frac{15}{2} \rceil * 20+34+34+186+ \nonumber \\

& & +2*10+0 = 484\ \mu s \nonumber

\end{eqnarray}

**THIS ARTICLE IS STILL UNDER CONSTRUCTION...**