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Consider an CSMA/CA 802.11 WLAN with three stations: A, B, and C. Node B is the

access point and is in full wireless contact with both A and C. The following are the

appropriate parameter values for this WLAN (note: artificial values are used for the

WLAN parameters to simplify your calculations)

Time slot = 1, DIFS = 3, SIFS = 1

Length of RTS, CTS, or ACK = 1

Initial backoff range for a frame is [0,7]

At time t= 0, the application layer of station A generated data to be transmitted as a

frame of length 10 destined to B. At that same time (t=0), the application layer of

station C generated data to be transmitted as a frame of length 10 destined to B. Before

transmitting a frame, both node A and node C must wait for DIFS, followed by a

random backoff period. The hidden node problem does not exist in this WLAN

Note: Clearly indicate the backoff random counter selected by each node and give time

charts supporting your answer.

Problem 1

This is an exercise on the standard 802.11 protocol,

not 802.11e

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a) Assume the RTS/CTS mechanism is used by all nodes. What is the earliest

time T at which both data frames are successfully received and

acknowledged by node B? What is the probability that this earliest time is

achieved?

b) Assume the RTS/CTS mechanism is not used in this WLAN. What is the

maximum time T at which both data frames are successfully received

without any collision and acknowledged by node B? What is the

probability that this maximum time occurs?

Problem 1

This is an exercise on the standard 802.11 protocol,

not 802.11e

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Problem 1 Part a- extension to 802.11e

There are two scenarios for the minimum time

Scenario 1: ra =1, rc=2


Finish Time = 38

Probability {Time = 38} = Prob {ra=1 & rc=2 OR ra=2 & rc=1 }

= (1/8)

(1/8) + (1/8)

(1/8) = 1/32

For 802.11e, the backoff mechanism is different. The backoff counter in

802.11e is chosen randomly from the range [1, CW+1] and is frozen

during channel usage periods but is activated one slot time before the

AIFS expiration. If the standard 802.11e backoff method is applied in

part a assuming the two frames are of the same traffic category TC1.AIFS(TC1) = DIFS, CWmin(TC1) = 7

3 5 7 17 19 22 24 26 36 38 time

t=19

rc= 1

t=3

ra = 1

rc = 2

DIFS DIFS

ACKACK

t=22

rc = 0

SIFSSIFS

CTSRTS RTS CTS


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Problem 1 Part a-Comparison

3 5 7 17 19 23 25 27 37 39 time

t=19

rc= 1

t=3

ra = 0

rc = 1

DIFS DIFS +1

ACKACK

SIFSSIFS

CTS

t=22

rc = 1

RTS RTS CTS

802.11


3 5 7 17 19 22 24 26 36 38 time

t=19

rc= 1

t=3

ra = 0

rc = 1

DIFS DIFS

ACKACK

SIFSSIFS

CTSRTS RTS CTS

t=22

rc = 0

802.11e


Finish Time = 38

Finish Time = 39

t=0

ra = 1

rc = 2

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Problem 1 Part b - Solution

3 9 19 21 25 35 37 time

b) Maximum time is achieved by choosing the largest backoff value for ra (or for

rc). There are 14 scenarios:

Scenarios 1-7 rc =7 and ra = 6, 5, 4, 3, 2 ,1, 0

Scenarios 8-14 ra =7 and rc = 6, 5, 4, 3, 2 ,1, 0

t=21rc= 1

t=3ra = 6

rc = 7

DIFS + 6 DIFS +1

ACKACK

t=25rc = 0

SIFS

Scenario 1

rc=7, ra= 6

Maximum finish time = 37

Probability {Time = 37} = 14 (1/8) (1/8)

= 14/64

= 7/32

3 13 15 25 35 37 time

t=15

rc= 7

t=3

ra = 0

rc = 7

DIFS DIFS +7

ACKACK

t=25

rc = 0

SIFS

Scenario 7

rc=7, ra=0

Solution for Legacy 802.11

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