These slides are used in the presentation at https://vimeo.com/156386656 . In that video, Daan Pareit (iMinds / Ghent University) explains how to calculate Wi-Fi throughput ("your Wi-Fi speed") based on the theory for WLAN medium access. It is a good starting point before doing their online lab which uses live actual Wi-Fi hardware remotely and which is explained at https://vimeo.com/152678614. That online lab itself is accessible at forge.test.iminds.be/wlan . More information about the FORGE project which enabled the succeeding lab session: at ict-forge.eu .

1. Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN throughput:
calculation exercises
Daan Pareit, Ph.D.
daan.pareit@intec.ugent.be
www.ibcn.intec.ugent.be
www.iminds.be

2. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Overview
1. Short MAC theory recapitulation
2. Short PHY theory recapitulation
3. Example exercise explained
4. Other exercise assignments

3. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Recapitulation of MAC layer theory

4. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 – DFWMAC-DCF using CSMA/CA
DFWMAC-DCF using CSMA/CA
 station has to wait for DIFS (+ random back-off time if medium is busy)
before sending data
 receivers acknowledge at once (after waiting for SIFS) if the frame was
received correctly (CRC)
 automatic retransmission of data frames in case of transmission errors
(but new random back-off time)
t
SIFS
DIFS
data
ACK
waiting time
other
stations
receiver
sender
data
DIFS
contention

5. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 - MAC layer
 Different IFS = different medium access priority
 SIFS (Short Inter Frame Spacing)
 highest priority, for ACK, CTS, polling response
 length of SIFS determined by PHY
 DIFS (DCF IFS)
 lowest priority, for asynchronous data service
 DIFS = SIFS + 2 time slots (length of time slot determined by PHY)
t
medium busy
SIFS
DIFS
next framecontention

6. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 – DFWMAC-DCF using CSMA/CA
 Contention Window
 Back-off time = random value between 0 and CW
 Low CW  many collisions
 High CW  high delays
 Exponential back-off: adaptation to load of medium
 If collision: CW doubles
 Example for HT PHY (802.11n):
CWmin = 15 and CWmax= 1023
then CW = (15, 31, 63, 127, 255, 511, 1023) depending on load
t
medium busy
DIFSDIFS
next frame
contention window
(randomized back-off
mechanism)
slot time

7. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 – DFWMAC-DCF with RTS/CTS
 DFWMAC-DCF with RTS/CTS
 RTS/CTS: Request to Send / Clear to Send
t
SIFS
DIFS
data
ACK
defer access
other
stations
receiver
sender
data
DIFS
contention
RTS
CTS
SIFS SIFS
NAV (RTS)
NAV (CTS)

8. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 – DFWMAC-DCF with RTS/CTS
 Fragmentation
t
SIFS
data
ACK1
other
stations
receiver
sender
frag1
DIFS
contention
RTS
CTS
SIFS SIFS
NAV (RTS)
NAV (CTS)
NAV (frag1)
NAV (ACK1)
SIFS
ACK2
frag2
SIFS
DIFS

9. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
MAC headers: data
Frame
Control
Duration/
ID
Address
1
Address
2
Address
3
Sequence
Control
Address
4
payload
MSDU
CRC
2 2 6 6 6 62 40-2312bytes
MPDU = MAC Protocol Data Unit
(= PSDU = PLCP Service Data Unit)
data
QoS
Control
2
sender

10. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11 – MAC address format
DS: Distribution System
AP: Access Point
DA: Destination Address
SA: Source Address
BSSID: Basic Service Set Identifier
RA: Receiver Address
TA: Transmitter Address
Address 1: physical receiver
Address 2: physical transmitter
Address 3: logical receiver/sender/BSSID
Address 4: logical sender
 Filtering on address 1
 ACK to address 2
scenario address 1 address 2 address 3 address 4
ad-hoc network DA SA BSSID -
infrastructure network,
from AP
DA BSSID SA -
infrastructure network,
to AP
BSSID SA DA -
infrastructure network,
within DS
RA TA DA SA

11. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
MAC headers: ACK
receiver
ACK
Frame
Control
Duration
Receiver
Address
CRC
2 2 6 4bytes

12. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
MAC headers: RTS/CTS
sender
RTS
receiver
CTS
Frame
Control
Duration
Receiver
Address
CRC
2 2 6 4bytes
Frame
Control
Duration
Receiver
Address
Transmitter
Address
CRC
2 2 6 6 4bytes

13. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11n
 Frame aggregation
 A-MSDU  A-MPDU

14. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11e
 Special control packets
 Block ACK Request
 Block ACK
Frame
Control
Duration
Receiver
Address
CRC
2 2 6 4bytes
BlockAck Start
Seq Control
2
Transmitter
Address
6
BlockAckReq
Control
2
Frame
Control
Duration
Receiver
Address
CRC
2 2 6 4bytes
BlockAck Start
Seq Control
2
Transmitter
Address
6
BlockAck
Control
2
BlockAck
bitmap
128

15. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Recapitulation of PHY layer theory

16. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11a: PHY frame format
rate service PSDU
Variable [bits]
6 Mbit/s
PLCP preamble signal
symbols12 1 variable
reserved length tailparity tail pad
616611214 variable
6, 9, 12, 18, 24, 36, 48, 54 Mbit/s
PLCP header
16 µs
data
sender
PLCP preamble

17. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Payload
MSDU
CRC
40-2312
IEEE 802.11a
SERVICE PSDU Tail PadPHY
MAC
LLC
Frame
Control
Duration/
ID
Address
1
Address
2
Address
3
Sequence
Control
Address
4
2 2 6 6 6 62bytes
4 µs16 µs 16 bits 6 bits
LLC
header
Payload
IP IP packet
8 bytes
@ PHY data rate
PLCP preamble signal

18. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11a: OFDM
 Rate dependent parameters
 250 000 OFDM symbols/s (symbol duration: 4 µs)
Data rate
[Mbit/s]
Modulation Coding rate
Coded bits
per
subcarrier
Coded bits
per OFDM
symbol
Data bits
per OFDM
symbol
6 BPSK 1/2 1 48 24
9 BPSK 3/4 1 48 36
12 QPSK 1/2 2 96 48
18 QPSK 3/4 2 96 72
24 16-QAM 1/2 4 192 96
36 16-QAM 3/4 4 192 144
48 64-QAM 2/3 6 288 192
54 64-QAM 3/4 6 288 216

19. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11n: PHY frame
format
PLCP preamble signal data
service PSDU
Variable [bits]
tail pad
616 variable

20. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Payload
MSDU
CRC
40-2312
QoS
Control
2
IEEE 802.11n
L-STF L-LTF L-SIG HT-SIG HT-STF HT-LTF HT-LTF SERVICE PSDU Tail PadPHY
MAC
LLC
Frame
Control
Duration/
ID
Address
1
Address
2
Address
3
Sequence
Control
Address
4
2 2 6 6 6 62bytes
8 µs 8 µs 4 µs 4 µs 4 µs 4 µs8 µs 16 bits 6 bits
LLC
header
Payload
IP IP packet
8 bytes
@ PHY data rate

21. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
IEEE 802.11n: data rate
 Modulation and Coding Schemes (MCS)
 Symbol duration with long GI = 4 µs, with short GI = 3.6 µs after training fields

22. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Example exercise explained

23. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 1
 What is the maximum throughput for IEEE 802.11a
DFWMAC-DCF using CSMA/CA?
 Assumptions:
 There is only one UDP sender (= STA) occupying the
wireless medium
 The UDP receiver (= AP) is close enough to the sender so
that data can be sent at maximum bit rate and no
transmission errors occur
 Propagation delay can be neglected
 Parameters
 TSIFS = 16 s and Tslot = 9 s
 CWmin = 15 and CWmax = 1023
 IP packet length = 1500 bytes & 500 bytes

24. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)

25. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Exercise 1: solution method
 Maximum throughput calculation (DFWMAC-DCF using CSMA/CA)
Throughput = 8 * payload (bytes) / (TDIFS + CWaverage + Tdata + TSIFS + TACK + 2 * )
TDIFS = TSIFS + 2 * Tslot = 16 s + 2 * 9 s = 34 s
CWaverage = CWmin/2 * Tslot if there is only 1 client (medium is always free)
Tdata = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + MAC-header + LLC-header + payload)
= 16 s + 4 s + 1 / R * (16 bits + 6 bits + MAC-header + LLC-header + payload)
= 20 s + 1 / R * (22 bits + 8 bits/byte * (28 byte + 8 byte + payload (byte))
TSIFS = 16 s
TACK = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + ACK-message)
= 16 s + 4 s + 1 / R * (16 bits + 6 bits + 8 bits/byte * 14 byte)
= 20 s + 1 / R * (22 bits + 112 bits)
= 20 s + 1 / R * 134 bits
: propagation delay
R: physical data rate
padding: granularity for Tdata and TACK is 1 symbol = 4 s  roundup

26. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Exercise 1: answer 1
 Maximum throughput calculation (DFWMAC-DCF using CSMA/CA) for 1500 bytes
Throughput = 8 * payload (bytes) / (TDIFS + CWaverage + Tdata + TSIFS + TACK)
TDIFS = TSIFS + 2 * Tslot = 16 s + 2 * 9 s = 34 s
CWaverage = CWmin/2 * Tslot = 15/2 * 9 s = 67.5 s
Tdata = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + MAC-header + LLC-header + payload)
= 16 s + 4 s + 1 / (54*106 bits/s) * (16 bits + 6 bits + 8 * 28 bits + 8 *8 bits + 8 *1500 bits)
= 20 s + 1 s / 54 bits * (22 bits + 8 * 28 bits + 8 * 8 bits + 8 * 1500 bits)
= 247.96  248 s
TSIFS = 16 s
TACK = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + ACK-message)
= 16 s + 4 s + 1 s / 54 bits * (16 bits + 6 bits + 8 * 14 bits)
= 22.5  24 s
Throughput = 8 * 1500 bits / (34 s + 67.5 s + 248 s + 16 s + 24 s )
= 8 * 1500 / 389.5 Mbps = 30.8 Mbps

27. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Exercise 1: answer 2
 Maximum throughput calculation (DFWMAC-DCF using CSMA/CA) for 500 bytes
Throughput = 8 * payload (bytes) / (TDIFS + CWaverage + Tdata + TSIFS + TACK)
TDIFS = TSIFS + 2 * Tslot = 16 s + 2 * 9 s = 34 s
CWaverage = CWmin/2 * Tslot = 15/2 * 9 s = 67.5 s
Tdata = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + MAC-header + LLC-header + payload)
= 16 s + 4 s + 1 / (54*106 bits/s) * (16 bits + 6 bits + 8 * 28 bits + 8 *8 bits + 8 *500 bits)
= 20 s + 1 s / 54 bits * (22 bits + 8 * 28 bits + 8 * 8 bits + 8 * 500 bits)
= 99.8  100 s
TSIFS = 16 s
TACK = Tpreamble + Tphy + 1 / R * (bitsservice + bitstail + ACK-message)
= 16 s + 4 s + 1 s / 54 bits * (16 bits + 6 bits + 8 * 14 bits)
= 22.5  24 s
Throughput = 8 * 500 bits / (34 s + 67.5 s + 100 s + 16 s + 24 s )
= 8 * 500 / 241.5 Mbps = 16.6 Mbps

28. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Other exercise assignments

29. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 2
 What is the maximum throughput for IEEE 802.11a
DFWMAC-DCF using RTS/CTS?
 Assumptions:
 There is only one UDP sender (= STA) occupying the
wireless medium
 The UDP receiver (= AP) is close enough to the sender
so that data can be sent at maximum bit rate and no
transmission errors occur
 Propagation delay can be neglected
 Parameters
 TSIFS = 16 s and TSlot = 9 s
 CWmin = 15 and CWmax = 1023
 IP packet length = 1500 bytes

30. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 3
 What is the throughput for IEEE 802.11a DFWMAC-DCF
using RTS/CTS?
 Assumptions:
 There is only one UDP sender (= STA) occupying the
wireless medium
 The receiver (= AP) is far away from the sender: data can
still be sent at maximum bit rate, but packets are
fragmentized (at MAC layer) to a maximum size of 500
bytes to limit transmission errors (which may be further
neglected)
 Propagation delay can be neglected
 Parameters
 TSIFS = 16 s and TSlot = 9 s
 CWmin = 15 and CWmax = 1023
 IP packet length = 1500 bytes

31. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 4
 What is the throughput for IEEE 802.11a DFWMAC-
DCF using RTS/CTS?
 Assumptions:
 There is only one UDP sender (=STA) occupying the
wireless medium
 The receiver (= AP) is far away from the sender: data
cannot be sent at maximum bit rate, but at a lower bit
rate of 36 Mbps
 Propagation delay can be neglected
 Parameters
 TSIFS = 16 s and TSlot = 9 s
 CWmin = 15 and CWmax = 1023
 IP packet length = 1500 bytes

32. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 5
 What is the maximum throughput for IEEE 802.11n DFWMAC-
DCF using CSMA/CA?
 Assumptions:
 There is only one UDP sender (= STA) occupying the wireless
medium, sending best effort (BE) traffic
 The UDP receiver (= AP) is close enough to the sender so that data
can be sent at maximum bit rate and no transmission errors occur
 The network is set up for a mixed 802.11g/n environment and
supports QoS
 The used hardware supports maximum 2 space-time streams
 Propagation delay can be neglected
 Parameters
 TSIFS = 10 s, Tslot = 20 s
 Channel: 20 MHz @ 2.4 GHz, short guard interval
 AIFSN[BE] = 2, CWmin[BE] = 15 CWmax[BE] = 1023
 IP packet length = 1500 bytes
 No frame aggregation

33. Ghent University - Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
WLAN Throughput
 Exercise 6
 What is the maximum throughput for IEEE 802.11n DFWMAC-
DCF using CSMA/CA with frame aggregation?
 Assumptions:
 There is only one UDP sender (= STA) occupying the wireless
medium, sending best effort (BE) traffic
 The UDP receiver (= AP) is close enough to the sender so that data
can be sent at maximum bit rate and no transmission errors occur
 The network is set up for a mixed 802.11g/n environment and
supports QoS
 The used hardware supports maximum 2 space-time streams
 Propagation delay can be neglected
 Parameters
 TSIFS = 10 s, Tslot = 20 s
 Channel: 20 MHz @ 2.4 GHz, short guard interval
 AIFSN[BE] = 2, CWmin[BE] = 15 CWmax[BE] = 1023
 IP packet length = 1500 bytes
 aggregation of 3 frames: A-MPDU versus A-MSDU

34. Department of Information Technology – Internet Based Communication Networks and Services (IBCN)
Contact
Daan Pareit, Ph.D.
daan.pareit@intec.ugent.be
www.ibcn.intec.ugent.be
Internet Based Communication Networks and Services research group (IBCN)
Department of Information Technology (INTEC)
Ghent University - iMinds