wireless proj 2

Intro to Wireless NetworkingECE 477

Spring 2014

Investigation on Performance of 802.11 using OPNET Simulation

Project #2

Syed Mudassir Rehman (16160367)

Mohammed Ateeq Ahmed (12419549)

[email protected]@mail.umkc.edu

TASK 1

Understanding reduced throughput when more nodes are sending.

2


Question 1.1: What is the average throughput for each node when destinations are set to random? State your answers here and elsewhere in Mbps, rounded off to tenths of an Mbps.

Answer:

When destinations are random,

Node Average Throughput (Mbps)

1 25.598

2 36.986

3 24.964

4 24.700

3


Question 1.2: Now set the destination for all traffic to go to 2 (this means Node 2). The throughput for Nodes 1, 3, and 4 should now be zero. What is the throughput for Node 2?

Answer:

When the destination was set to Node 2,


1 0

2 437.178

3 0

4 0

4


Question 1.3: Now manually set the destinations so that we have 1 to 2, 3 to 4, and 4 to 1. Now what is the total throughput?

Answer: Destinations were set: 1-->2, 3-->4, 4-->1

Total throughput obtained is 435.354 Mbps

Question 1.4: Now what is the total throughput per node?


1 154.0021

2 132.874

3 0

4 148.478

Total--> 435.354

5


Question 1.5: The results should be very different in Q1.3 and Q1.4 than Q1.1. In reality, the total network throughput is nearly the same when one or three nodes are sending and node destinations are specified instead of using the random destination option. This means that the definition of “random” as a destination in OPNET is important to understand. Go under “Edit Attributes” and click the question mark next to the Destination Address option. It gives an explanation of what it means for a destination to be “random”. How does this definition help us to understand why the results are so different?

Answer: When we consider Random Destination setting, the Throughput of the system is obtained less unlike in the case of setting a destination to a particular node, where the Throughput increases and comparitively higher.

Question 1.6: Now compare the scenarios using random destinations as in Q1.1 and the fixed destinations set as in Q1.3. Compute the average results for the statistic called “WLAN HT/MPDUs per PPDU (packets)”.

Answer: The average results for nodes 1, 3 and 4 are obtained in the excel sheet below in both cases.

Random destinations

6


Fixed destinations

Table for comparison between random and fixed destinations

Node 1 Node 2 Node 3 Node 4

Destination setup Random (HT/MPD per PPDU)

1.162664887 - 1.172650385 1.549499778

Destination setup Fixed (HT/MPD per PPDU)

15.08222 - 15.10616 15.12048

Question 1.7: What do your answers in Q1.6 tell you about the reason why throughput is so much lower for the case of using random destinations?

Answer: The Throughput of the system is higher when the destination is set to a particular fixed node compared to the random destination setting. Hence,

7


TASK 2

Medium Access Delay for an Overloaded Node

8


Question 2.1: What interarrival time should you use to generate a constant rate of 500 Mbps?

Answer:

(1/X packets/sec)*(10000 bits/packet) = 500 Mbps.

X = (10000/ (500*10e6)) = 0.00002 seconds.

Therefore the interarrival time to generate a constant rate of 500 Mbps is 0.00002 sec.

Plot 2.1: Implement this new interarrival time for 500 Mbps and run the simulation. Create a new plot and this time select items under Object Statistics/Independent BSS/DCF_wkstn1. Plot a curve of the Data Dropped (Buffer Overflow) (bits/sec) and Queue Size. Queue size shows how many packets are in the queue. Show these as “Stacked Statistics” (on separate subplots) instead of “Overlaid Statistics.”

Data Dropped vs Queue Size

Average Data Dropped is 45275.8 packets/sec

Average Queue size is 201.5052245 packets

9


Plot 2.2: Then plot a different curve, the Medium Access Delay (seconds) as a separate plot.

Medium Access Delay Curve

Question 2.2: By looking into the “Edit Attributes” to find information about the queue and Plot 2.1, compute the average packet size that is transmitted.

Answer:

From the "Edit Attributes", the buffer size is 2.024 Mbits.

Also, the total packet size is 10000 bits.

Queue Size = 2024000/10000 = 202.4 =~ 202 packets

Also, from Plot 2.1 the average Queue size estimated was 201.5052245 packets.

Hence, the max queue size should be around 202 packets.

10


Question 2.3: The amount of time it takes to transmit each packet can be computed from the plots given above. Those packets which are transmitted must wait for every other packet in the queue to go before it, so their media access delay consists of their own transmission time plus the transmission times of every other packet in the queue ahead of them. Assume each packet takes the same amount of time. How much time do you compute as an estimate of what it takes to transmit each packet? This means the combination of (1) delays due to contentions and backoffs, and (2) transmission delays due to IFS’s, etc., and (3) actual transmit time over the wireless link. This should come from a formula you use to compute this. In this case there is no delay due to contentions and backoffs since only one node is sending.

Answer:

From plot 2.2, the average Media Access Delay in seconds is 0.043539931.

From question 2.2, we have the max queue size as 202 packets.

The time taken to transmit each packet = 0.043539931/202 = 0.00021554421 seconds

Question 2.4: We know how to find what the packet transfer time should be for a packet size in Q2.2 over a 600 Mbps link. We can compute that. So, compute how much extra time from Q2.3 is needed for all of the other delays (backoffs, IFS’s, etc.).

Answer:

Packet transfer time for 10000 bit packet over 600Mbps link = 10000/600*10e6 = 0.000016667 seconds

Extra time for other delays = Time taken to transmit each packet – packet transfer time over 600 Mpbs link

Extra time = 0.00021554421 - 0.000016667= 0.00019887754 seconds.

11


TASK 3

Effect of RTS/CTS and Contention on Packet Transfer Time

12


Plot 3.1: Show the plot for Media Access Delay when RTS/CTS is used.

Medium Access Delay Curve with RTS/CTS

Question 3.1: Given your analysis in previous questions, what do you compute as the extra time added to each packet transmission (more than the case without RTS/CTS) because of the RTS/CTS procedure?

Answer:

From the plot 3.1, the average Media Access Delay in seconds is 0.066002995 seconds.

We already have the max queue size of 202 packets.

The time taken to transmit each packet (for the case of Media Access Delay with RTS/CTS) = 0.066002995/202 = 0.0003267475 seconds.

Extra time for other delays with RTS/CTS = Time taken to transmit each packet with RTS/CTS – packet transfer time over 500 Mpbs link

Extra time = 0.0003267475 - 0.00002 = 0.0003067475 seconds.

Extra Time with RTS/CTS = Time taken to transmit with RTS/CTS – Time taken to transmit without RTS/CTS

13



Plot 3.2: Now turn on Node 3 and set all of its parameters the same as for Node 1, except set its destination to be Node 1. Do not use RTS/CTS for any of the nodes. Plot the media access delay.

Media Access Delay Curve

Question 3.2: Given your analysis in previous questions, what do you compute as the extra time added to each packet transmission (more than the case of one node sending without RTS/CTS) because of the contention procedure?

14


Answer:

From the plot 3.2, the average Media Access Delay is 0.232420808 seconds.


The time taken to transmit each packet (for the case of Media Access Delay without RTS/CTS) = 0.232420808/202 = 0.0011505980 seconds.

Extra time for each packet transmission = The time taken due to contention procedure - the time taken to transfer packet before


Plot 3.3: Now turn on Node 4 and set all of its parameters the same as for Node 1, except set its destination to be Node 3. Do not use RTS/CTS. Plot the media access delay.

Medium Access Delay Curve

Question 3.3: Given your analysis in previous questions, what do you compute as the extra time added to each packet transmission (more than the case of one node sending without RTS/CTS) because of the contention procedure?

15


Answer:

From the plot 3.3, the average Media Access Delay is 0.432800731 seconds.


The time taken to transmit each packet (for the case of Media Access Delay with RTS/CTS) = 0.432800731/202 = 0.0021425778 seconds.

Extra time for each packet transmission = The time taken due to contention procedure - the time taken to transfer packet before


TASK 4

Your Own Study Of Media Access Delay

16


Part 4.1: Produce 4 more plots and a table of the results.

Case 1.) Conditions: Recovering 2 nodes (1 and 2), Failing 2 nodes (3 and 4), destination of 1 is 2

CWmin = 31, CWmax = 511

17


Case 2.) Conditions: Recovering 2 nodes (1 and 2), Failing 2 nodes (3 and 4), destination of 1 is 2


Case 3.) Conditions: Recovering 3 nodes (1, 2 and 3), Failing 1 node 4), destinations are random


18


Case 4.) Conditions: Recovering 3 nodes (1, 2 and 3), Failing 1 node 4), destinations are random


Table of Results:

X Congestion Window values

Media Access Delay in seconds

Case CWmin CWmax Node 1 Node 2 Node 3 Node 4

1 31 511 0.000867147 N/A - -

2 15 2047 0.000696269 N/A - -

3 63 1023 0.015021932 N/A 0.014886329 -

4 7 255 0.012621655 N/A 0.012814111 -

Part 4.2: Discuss what the results demonstrate and what we learn from those results.

Answer:

19


As observed from the table above, we can say that, as the CWmin value decreases from 31 to 15, the Media Access Delay at Node 1 decreases. Similar change can be seen when CWmin decreases from 63 to 7, the Media Access Delay for Node 1 and Node 3 also decreases. Hence, we can come to conclusion that the Media Access Delay is reduced as CWmin value decreases. This is because the congestion window avoids collision of packets when the channel is busy.

20

wireless proj 2

Documents