Project Report Qualnet Project Phase 1 Karan & Abhinav

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Concordia University *

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6851

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Electrical Engineering

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Dec 6, 2023

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Project Report Qualnet Project Phase- 1 (ELEC 6851) Submitted to: Prof. M. Mehmet Ali Department of Electrical and Computer Engineering Concordia University Submitted by: ID Name 40230432 Karandeep Singh 40221579 Abhinavpreet Singh

Abstract We learned how to utilize Qualnet software to construct nodes and connect them in this project. The parameters of each connection/packet were supplied, allowing the results to be examined using an analyzer. The tables include information about each layer of each node as well as packets of data going between the designated connections. Project Introduction The Qualnet Developer IDE is a graphical user interface (GUI) software included with Qualnet 7.3 for designing network scenarios. It may be used to create network situations visually and then conduct simulations of these networks. We'll start with a small three-node wireless network that uses the CSMA/CA technique described by the 802.11 standard to access the channel. Ad-hoc networking scenarios presume that nodes be manually planted on the landscape.

PART 1 STEP:1 With three nodes, we established a single wireless subnetwork. These nodes are distributed at random, as indicated in Figure 1, so that all nodes in the subnetwork may hear each other's broadcasts. As per given figure 1 Node 1 is connected to Node 2 Node 2 is connected to Node 3 Node 3 is connected to Node 1 Screen Shot of Network Figure 1 STEP: 2 Following the successful creation of traffic between the nodes, the application layer attributes of each connection were specified. At the source, we create a poisson process of data packets and make the packet length exponentially dispersed. The parameters must be adjusted so that we may create more than 1000 packets per node. These are shown in the table below:

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Node Number Average packet length in (bytes) Average time interval for packet generation in (seconds) Node 1 512 1 Node 2 512 1 Node 3 512 1 Figure 2 STEP: 3 After modifying the application parameters to obtain the needed minimum packet count of 1000, we must select a simulation time longer than CT in the Scenario properties. In addition to unicast traffic, the system creates routing packets known as broadcast traffic during the simulation.

Figure 3 Simulation time is 1600 seconds iii) Number of packets generated during simulation. Packets generated from the node 1: 1374 Packets generated from the node 2: 1307 Packets generated from the node 3: 1261 (2) The number of user packets received from the higher layer and number of user packets sent to the lower layer at each layer is shown in table 1, table 2 Table 1 Layer Function Traffic type Nodes 1 2 3 Application sent unicast 1374 1308 1261 broadcast 161 160 160 Transport received unicast 1374 1308 1261 broadcast 161 160 160 sent unicast 1374 1308 1261 broadcast 161 160 160 Network received unicast 1374 1308 1261 broadcast 161 160 160 sent unicast 1406 1334 1283 broadcast 161 160 160 unicast 1406 1334 1283

MAC received broadcast 161 160 160 sent unicast 1406 1334 1283 broadcast 164 163 163 ACK 1283 1406 1334 RTS 1406 1334 1283 CTS 1283 1406 1334 Physical received unicast 1406 1334 1283 broadcast 164 163 163 ACK 1283 1406 1334 RTS 1406 1334 1283 CTS 1283 1406 1334 sent unicast 1406 1334 1283 broadcast 164 163 163 ACK 1283 1406 1334 RTS 1406 1334 1283 CTS 1283 1406 1334 Number of Unicast and Broadcast packets received by each layer from the layer above and sent to the layer below by each transmitting node. Table 2 Layer Function Traffic type Nodes 1 2 3 Application received unicast 1261 1374 1308 broadcast 320 321 321 Transport sent unicast 1261 1374 1308 broadcast 320 321 321 received unicast 1261 1374 1308 broadcast 320 321 321 Network sent unicast 1261 1374 1308 broadcast 320 321 321 received unicast 1283 1406 1334 broadcast 320 321 321 MAC sent unicast 1283 1406 1334 broadcast 320 321 321 received unicast 1283 1406 1334 broadcast 326 327 327 ACK 1406 1334 1283 RTS 1283 1406 1334 CTS 1406 1334 1283 Physical sent unicast 1283 1406 1334 broadcast 326 327 327 ACK 1406 1334 1283 RTS 1283 1406 1334 CTS 1406 1334 1283 received unicast 1283 1406 1334 broadcast 326 327 327 ACK 1406 1334 1283 RTS 1283 1406 1334 CTS 1406 1334 1283

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Table 2. Number of Unicast and Broadcast packets received by each layer from the layer below and sent to the layer above by each receiving node. 3) If the number of packets received and sent by a layer is not equal to each other it means that either packet fragmentation or re-assembly has occurred in the layer. Determine the number of fragmented or re-assembled packets.  Node 1 Total unicast fragments sent (application layer) = 1374 Total unicast fragments received (application layer) =1261 Packets fragmented (Network IP layer) = 30 Fragments received (Network IP layer) = 42 Fragments reassembled (Network IP layer) =42 Fragments created (Network IP layer) =62 Packets created after reassembling (Network IP layer) = 20 Newly Generated fragments = Fragments created – Packets fragments =62-30= 32  Node 2 Total unicast fragments sent (application layer) = 1308 Total unicast fragments received (application layer) = 1374 Packets fragmented (Network IP layer) = 26 Fragments received (Network IP layer) = 62 Fragments reassembled (Network IP layer) = 62 Fragments created (Network IP layer) = 52 Packets created after reassembling (Network IP layer) = 30 Newly Generated fragments = Fragments created – Packets fragments =52-26= 26  Node 3 Total unicast fragments sent (application layer) = 1261 Total unicast fragments received (application layer ) = 1308 Packets fragmented (Network IP layer) = 20 Fragments received (Network IP layer) = 52 Fragments reassembled (Network IP layer) = 52 Fragments created (Network IP layer) = 42 Packets created after reassembling (Network IP layer) = 26 Newly Generated fragments = Fragments created – Packets fragments =42-20= 22

4) Relationship between the numbers of ACK, RTS, CTS packets and number of unicast packets by retransmissions- When one node wishes to communicate data to another, it sends an RTS (Request to Send) packet. The receiving node responds with a CTS (Cleared to Send) message. After receiving the CTS packet, the transmitter node sends the data packets. When the receiver node receives a data packet, it sends an ACK (Acknowledgement Packet) packet back to the sending node. In the MAC layer the,  Number of received RTS = Number of sent CTS  Number of Received CTS = Number of unicast packets + Number of Retransmission packets due to ACK timeout If the number of retransmissions is equal to 0, then  Number of CTS = Number of unicast packets  Number of received ACK = Number of sent unicasts packets (5 ) Determine and explain packet losses at each layer o Packet drops due to retransmission limit at node 1 there is 0 packet loss o Packet drops due to retransmission limit at node 2 there is 0 packet loss o Packet drops due to retransmission limit at node 3 there is 0 packet loss (6) Number of signals transmitted and received at the physical layer at each node-  Node 1 Number of signals transmitted = Unicast sent + Broadcast sent + CTS sent + RTS sent + ACK sent + Number of retransmission packets Number of signals transmitted = 1406+164+1283+1406+1283 = 5542 Number of signals received = Sum of Signal transmitted by (Node 2+Node 3) Number of signals received = Signals detected = 11040  Node 2 Number of signals transmitted = 1334+163+1406+1334+1406 = 5643 Number of signals received = Sum of Signal transmitted by (Node 3+Node 1) Number of signals received = Signals detected = 10939  Node 3 Number of signals transmitted = 1283+163+1334+1283+1334 = 5397 Number of signals received = Sum of Signal transmitted by (Node 2+Node 3) Number of signals received = Signals detected = 11185

PART 2 STEP:1 In Part 2, Node 3 is separated from Nodes 1 and 2 in order to notice packet errors and retransmit the packets. As the distance between Node 3 and the other nodes, i.e. Node 1, Node 2, grows, so do the packet errors. The Figure depicts the identical network as Figure 1, except that the distance among Node 1 and Node 3 has been increased. The distance between Nodes 1 and 2 stays unchanged. Figure 4 in which position of node 3 is changed from existing location STEP: 2 Following the successful creation of traffic between the nodes, the application layer attributes of each connection were specified. At the source, we create a poisson process of data packets and make the packet length exponentially dispersed. The parameters must be adjusted so that we may create more than 1000 packets per node. These are shown in the table below: Node Number Average packet length in (bytes) Average time interval for packet generation in (seconds) Node 1 512 1 Node 2 512 1 Node 3 512 1

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Figure 5 STEP: 3 After modifying the application parameters to obtain the needed minimum packet count of 1000, we must select a simulation time longer than CT in the Scenario properties. In addition to unicast traffic, the system creates routing packets known as broadcast traffic during the simulation.

iii) Number of packets generated during simulation. Packets generated from the node 1: 1374 Packets generated from the node 2: 1308 Packets generated from the node 3: 1261 The number of user packets received from the higher layer and number of user packets sent to the lower layer at each layer is shown in table 2.1, table 2.2 Table 2.1 Layer Function Traffic type Nodes 1 2 3 Application sent unicast 1374 1308 1261 broadcast 161 160 160 Transport received unicast 1374 1308 1261 broadcast 161 160 160 sent unicast 1374 1308 1261 broadcast 161 160 160 Network received unicast 1374 1308 1261 broadcast 161 160 160 sent unicast 1406 1334 1283 broadcast 161 160 160 MAC received unicast 1406 1334 1283 broadcast 161 160 160 sent unicast 1413 1278 1220 broadcast 164 163 163 ACK 1264 1416 1351 RTS 1416 2045 2119

CTS 2107 1418 2055 Physical received unicast 1413 1278 1220 broadcast 164 163 163 ACK 1264 1416 1351 RTS 1416 2045 2119 CTS 2107 1418 2055 sent unicast 1413 1278 1220 broadcast 164 163 163 ACK 1264 1416 1351 RTS 1416 2045 2119 CTS 2107 1418 2055 Number of Unicast and Broadcast packets received by each layer from the layer above and sent to the layer below by each transmitting node. Table 2.2

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Number of Unicast and Number Broadcast packets received by each layer from the layer below and sent to the layer above by each receiving node. (4) If the number of packets received and sent by a layer is not equal to each other it means that either packet fragmentation or re-assembly has occurred in the layer. Determine the number of fragmented or re-assembled packets.  Node 1 Total unicast fragments sent (application layer) = 1374 Total unicast fragments received (application layer) = 1197 Packets fragmented (Network IP layer) = 30 Fragments received (Network IP layer) = 33 Fragments reassembled (Network IP layer) =24 Fragments created (Network IP layer) = 62 Packets created after reassembling (Network IP layer) = 12 Fragments dropped (Network IP layer) = 8 Newly Generated fragments = Fragments created – Packets fragments =62-30= 32 Layer Function Traffic type Nodes 1 2 3 Application received unicast 1197 1374 1261 broadcast 304 310 288 Transport sent unicast 1197 1374 1261 broadcast 304 310 288 received unicast 1197 1374 1261 broadcast 304 310 288 Network sent unicast 1197 1374 1261 broadcast 304 310 288 received unicast 1218 1406 1287 broadcast 304 310 288 MAC sent unicast 1218 1406 1287 broadcast 304 310 288 received unicast 1218 1416 1294 broadcast 310 316 294 ACK 1413 1278 1220 RTS 2107 1418 2055 CTS 1413 2045 2119 Physical sent unicast 1218 1416 1294 broadcast 310 316 294 ACK 1413 1278 1220 RTS 2107 1418 2055 CTS 1413 2045 2119 received unicast 1218 1416 1294 broadcast 310 316 294 ACK 1413 1278 1220 RTS 2107 1418 2055 CTS 1413 2045 2119

 Node 2 Total unicast fragments sent (application layer) = 1308 Total unicast fragments received (application layer) = 1374 Packets fragmented (Network IP layer) = 26 Fragments received (Network IP layer) = 62 Fragments reassembled (Network IP layer) = 62 Fragments created (Network IP layer) = 52 Packets created after reassembling (Network IP layer) = 30 Newly Generated fragments = Fragments created – Packets fragments =52-26= 26  Node 3 Total unicast fragments sent (at the application layer) = 1261 Total unicast fragments received (at application layer = 1261 Packets fragmented (Network IP layer) = 20 Fragments received (Network IP layer) = 42 Fragments reassembled (Network IP layer) 32 Fragments created (Network IP layer) = 42 Packets created after reassembling (Network IP layer)=16 Fragments dropped (Network IP layer) = 10 Fragments in Buffer (Network IP layer) = 0 Newly Generated fragments = Fragments created – Packets fragments =42-20= 22 (5 ) Determine and explain packet losses at each layer o Packet drops due to retransmission limit at node 1 there is 0 packet loss o Packet drops due to retransmission limit at node 2 there is 56 packets loss o Packet drops due to retransmission limit at node 3 there is 73 packet loss (6) Number of signals transmitted and received at the physical layer at each node-  Node 1 - Number of signals transmitted = Unicast sent + Broadcast sent + CTS sent + RTS sent + ACK sent + Number of retransmission packets = 1413+164+1264+1416+2107+3 = 6367 Number of signals received = Signals detected = 14894  Node 2 – Number of signals transmitted = 1278+163+1416+2045+1418+767 =7087 Number of signals received = Signals detected = 14174  Node 3 – Number of signals transmitted = 1220+163+1351+2119+2055+899 = 7807 Number of signals received = Signals detected = 1345 Relationship between the numbers of ACK, RTS, CTS packets and number of unicast packets by retransmissions-  Number of received RTS = Number of sent CTS

 Number of Received CTS = Number of unicast packets + Number of Retransmission \packets due to ACK timeout If the number of retransmissions is equal to 0, then Number of CTS = Number of unicast packets  Number of received ACK = Number of sent unicasts packets  The number of retransmissions in – Node 1 - 3 Node 2 - 767 Node 3 -899 The number of retransmissions increases as the distance between Node 3 increases, as do the received unicast packets and the Application layer owing to packet fragmentation and packet loss. Broadcast Messages are increased because of Control Frames

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