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Experimental Analysis of Performance of 

Wireless LAN

U. Ahmad, S.A. Raza, T. MustafaDepartment of Computer Science.

University of Agriculture, Faisalabad.

Dr. Mohsin NazirDept. of Information and Communications Technologies

Lahore College for Women University

Abstract:Wireless local-area networks based on IEEE 802.11 a/b/g standards are growing rapidly.WLANs can provide the benefits of network connectivity without the restrictions of 

being tied to a location or restricted by wires. Despite the convenience of mobility, the

performance of a WLAN must be addressed carefully before it can be adopted and

deployed in any environment. In our research, we addressed the impact of various keyparameters on the actual performance of IEEE 802.11g. We performed series of 

experiments to assess the performance of IEEE802.11g, in the presence of interferences,and finding the maximum through-put under realistic conditions. In addition the impactof co-channel, adjacent channel interferences and noises on the quality of WLAN speed

was also exposed. Overall, we conducted independent set of experiments to measure the

IEEE 802.11g’s effective application-level throughput. The analysis results andmeasurement provided insights into the required provisioning for 802.11g WLAN to

ensure that it will provide the needed coverage and capacity for the intended users.

I- Introduction:Wireless technologies enable freedom of mobility for users by releasing the constraint of 

physical connections  –  network connections become cable-free. Wireless technologies

use radio frequency (RF) as the medium of transmission, and allow organizations toeliminate cables for simpler network management at effective costs. The IEEE 802.11

standard establishes several requirements for the RF transmission characteristics of an

802.11 radio. Included in these are the channelization scheme as well as the spectrumradiation of the signal (that is, how the RF energy spreads across the channel

frequencies). In IEEE 802.11g, channels 1, 6 and 11 are considered to be non-overlapping

and hence the premise that these channels can be used such that multiple networks canoperate in close proximity without interfering with each other [6]. Interference has

always been considered as an unavoidable peril in wireless networks. Based on channel

of origin, interference can be categorized into co-channel (from transmissions on thesame channel as the receiver) and adjacent-channel (transmissions on adjacent and

overlapping channels). The IEEE 802.11 b/g standards operate in the unlicensed ISM 2.4Ghz spectrum which has 11 out of 14 channels available for use in the US [4, 5]. These

overlapping channels degrade network performance. Few researches have beenconducted in this area. In this paper we present a full scale performance study and

analysis on IEEE 802.11g, to measure its effective application-level throughput under

different scenarios. In order to improve its performance, a clear understanding of WLAN

behavior is needed, therefore measuring and analyzing the performance of system underrealistic conditions is of paramount importance. In my experiments, we studied the effect

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of interferences on TCP traffic.

The rest of paper is organized as follows. In section II, we discuss the related work.Experimental setup is described in Section III. Performance evaluation methodology is

explained in section IV. The results obtained are described in section V; we present our

conclusion and discuss future work in section VI.

II- Related Work:Stine et al. (2003) [7] stated that Interference has always been considered as anunavoidable peril in wireless networks. To reverse the effect they constructed simple

analytical and empirical models of such interference occurring in IEEE 802.11 networks,

and illustrated two scenarios where such interference can be exploited. Banerjee (2006) [1] stated that in 802.11 and other wireless networks, adjacent channel interference is

considered a peril. In order to avoid this peril, two simultaneously communicating nodes

that are in close proximity are assigned to different non-overlapping channels, i.e.,

channels 1, 6, and 11 in 802.11b are non-overlapping. Boulmalf  et al. (2006) [2]addressed the impact of various key parameters on the actual performance of IEEE

802.11g in the presence of interferences, and found the maximum through-put underrealistic conditions. The analysis results and measurement campaign provided insights

into the required provisioning for 802.11g WLAN to ensure that it provided the neededcoverage and capacity for the intended users. Liese et al. (2006) [3] studied the relative

performances of single and multiple channels in both single hop and multi hop wireless

mesh networks. In one of their experiments, they studied the effect of antenna placementon the access point and determine the impact on performance. Sharma et al. (2006) [6]

characterized the performance of multi-channel IEEE 802.11g wireless networks. They

conducted the experiments on a sample topology consisting of just two flows on non-overlapping channels and found that the expected increase in throughput was seen only

when the separation between the antennas of the radio devices was above a threshold

value.

III- Experimental Setup:One testing location in University of Agriculture Faisalabad was selected for research,

Testing environment consisted of two 802.11g APs, four 802.11g wireless cards, IBMThinkPad same configuration laptops running windows XP and two Dell Latitude dame

configurations laptops as shown in Figure 1.1. Laptops were placed at a distance of 5

meters from APs. Experiments were conducted by keeping channel of one AP constantwhile changing the channels of the other AP in order to check the interference caused by

channel overlapping. Iperf was used to generate TCP traffic and measure the throughput.

SPSS was used to perform analysis of the results that we got from my experiments.

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 Figure 1.1 Experimental Setup 

A. Hardware RequirementsSr. No. Brand Computer Operating System Processor RAM

1 IBM T22 ThinkPad Microsoft Windows XP 900 MHz 256 MB

2 IBM T22 ThinkPad Microsoft Windows XP 900 MHz 256 MB

3 Dell Latitude D 600 Microsoft Windows XP4 Dell Latitude D 600 Microsoft Windows XP

Sr. No. Access Points WLAN Standard

1 D Link DWL-700 IEEE 802.11 g

2 D Link DWL-900 IEEE 802.11 g

Sr. No. Wireless Network Cards WLAN Standard

1 D-Link DWL G650 IEEE 802.11 g

B. Software RequirementsSr. No. Tools

1 DU Meter

2 Iperf 

3 SPSS

IV- Performance Evaluation Methodology:In our experiments we characterized the results on the basis Throughput of TCP

generated through Iperf. We measured the effect on throughput by changing the

frequency channel schemes:

AP / Wireless Cards Frequency Channel

D-Link DWL 700AP 1D-Link DWL 2100AP 1

D-Link DWL G650 6

D-Link DWL G650 6

D-Link DWL G650 6

D-Link DWL G650 6Table 1.1 Frequency Channel Schemes

The frequency channel scheme showed in table 1.1 was changed as: Keeping 700AP

5 Meters 5 Meters

DWL 700 AP

CH 1 to 11

DWL 2100 AP

CH 1 to 11

DWL-G650

CH 6 fixDWL-G650

CH 6 fix

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TCP Throughut Graph,

Frequency Channel of AP-A=1

0

1

2

3

4

5

6

7

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H

 .   1   0

   F .   C

   H

 .   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=2

0

1

2

3

4

5

6

7

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=3

0

12

3

4

5

6

7

8

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

channel 1 and changing 900AP from 1 to 11, than changing 700AP channel to 2 and

again changing 900AP from 1 to 11, utilizing 700AP all channels one time and 900APchannels 11 times thus making 121 possible combinations.

V- Experimental Results:

The experiments were done 10 times in university environment, and the mean results of 10 sheets were calculated. The detail results in the form of graphs are given below:

Figure 2.1: TCP Throughput Graph 

Figure 2.2: TCP Throughput Graph 

Figure 2.3: TCP Throughput Graph

Figure 2.1 shows that when the channel of 

both 2100AP and 700AP is 1 TCP throughput

decreases while with all other channels itsvalue increase because same frequency causes

interference and decrease the throughput due

to overlapping.

Figure 2.2 shows when the channel of both 2100AP 700AP is 2 TCP throughput decreases while with all

other channels its value increased due to the fact that

same channels overlap each other and cause interfer

Highest possible throughput value is achieved when channel of AP2 is in the range of 6-11.

Figure 2.3 shows when the channel of both2100AP and 700AP is 3 TCP throughput

decreases while with all other channels its

value increased due to the fact that samechannels overlap each other and cause

interference. Highest possible throughputvalue is achieved when the channel of AP2 is

in the range of 6-11

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TCP Throughput Chart

Frequency Channel of AP-A=4

0

1

2

3

4

5

6

7

   F .   C   H

 .   1

   F .   C   H

 .   2

   F .   C   H

 .   3

   F .   C

   H .   4

   F .   C   H

 .   5

   F .   C   H

 .   6

   F .   C   H

 .   7

   F .   C   H

 .   8

   F .   C   H

 .   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Chart

Frequency Channel of AP-A=5

0

12

3

4

5

6

7

8

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=6

0

1

2

3

4

5

6

7

   F .   C   H .   1

   F .   C   H .   2

   F .   C   H .   3

   F .   C   H .   4

   F .   C   H .   5

   F .   C   H .   6

   F .   C   H .   7

   F .   C   H .   8

   F .   C   H .   9

   F .   C   H .   1   0

   F .   C   H .   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g

   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=7

0

1

2

3

4

5

6

7

8

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n

   M   B   i   t  s   /  s  e  c

Figure 2.4: TCP Throughput Graph 

Figure 2.5: TCP Throughput Graph 

Figure 2.6: TCP Throughput Graph

Figure 2.7: TCP Throughput Graph

Figure 2.4 shows that when the channel of 

both 2100AP and 700AP is 4 TCP

throughput decreases while with all otherchannels its value increase because same

frequency causes interference and decrease

the throughput due to overlapping. Thecombination of 4-3 and 5-3 gives smaller

throughput values as compared to othercombinations.

Figure 2.5 shows when the channel of both2100AP and 700AP is 2 TCP throughput

decreases while with all other channels its

value increased due to the fact that same

channels overlap each other and cause

interference. Highest possible throughputvalues are achieved with the combination of 2-

5, 10-5.

Figure 2.6 shows that when the channel of both

2100AP and 700AP is 6 TCP throughputdecreases while with all other channels its

value increase because same frequency causes

interference and decrease the throughput dueto overlapping

Figure 2.7 shows when the channel of both

2100AP and 700AP is 7 TCP throughput

decreases while with all other channels its

value increased due to the fact that samechannels overlap each other and cause

interference. Lowest possible throughput

value is achieved when the channel of AP2 is

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TCP Throughput Graph

Frequency Channel of AP-A=8

0

1

2

3

4

5

6

7

8

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n

   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=9

0

1

2

3

4

5

6

7

8

   F .   C   H .   1

   F .   C   H .   2

   F .   C   H .   3

   F .   C   H .   4

   F .   C   H .   5

   F .   C   H .   6

   F .   C   H .   7

   F .   C   H .   8

   F .   C   H .   9

   F .   C   H .   1   0

   F .   C   H .   1   1

Frequency Channel of AP-B

   T

   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=10

0

1

2

3

4

5

6

7

8

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h  r  o  u  g   h  p  u   t   i  n   M   B   i   t  s   /  s  e  c

TCP Throughput Graph

Frequency Channel of AP-A=11

0

1

2

3

4

5

6

7

8

9

   F .   C

   H .

   1

   F .   C

   H .

   2

   F .   C

   H .

   3

   F .   C

   H .   4

   F .   C

   H .

   5

   F .   C

   H .

   6

   F .   C

   H .

   7

   F .   C

   H .

   8

   F .   C

   H .

   9

   F .   C

   H .

   1   0

   F .   C

   H .

   1   1

Frequency Channel of AP-B

   T   C   P   T   h   r   o   u   g   h   p   u   t   i   n

   M   B   i   t   s   /   s   e   c

 Figure 2.8: TCP Throughput Graph 

Figure 2.9: TCP Throughput Graph

Figure 2.10: TCP Throughput Graph

Figure 2.11: TCP Throughput Graph

Figure 2.8 shows when the channel of both2100AP and 700AP is 8 TCP throughput

decreases while with all other channels its

value increased due to the fact that same

channels overlap each other and causeinterference. Lowest possible throughput

value is achieved when the channel of AP2 is

in the ran e of 5-11.

Figure 2.9 shows that when the channel of both 2100AP and 700AP is 9 TCP

throughput decreases while with all other

channels its value increase because same

frequency causes interference and decreasethe throughput due to overlapping. Highest

throughput values are obtained from 1-5

Figure 2.10 shows that when the channel of 

both 2100AP and 700AP is 10 TCP throughputdecreases while with all other channels its

value increase because same frequency causes

interference and decrease the throughput due

to overlapping. Throughput is decreased as thelevel of frequency channels moves from 1 to

11 accept at 10 which gives lowest value due

to interference. 

Figure 2.11 shows that when the channel of both

2100AP and 700AP is 11 TCP throughputdecreases while with all other channels its value

increase because same frequency causesinterference and decrease the throughput due to

overlapping. Throughput is decreased as the level

of frequency channels moves from 1 to 11.

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