This is a sample Lab report from ECE 461 from previous years. L A B 6

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This is a sample Lab report from ECE 461 from previous years. L A B 6 Part 1 1. o the source and destination M A C/IP addresses change when a packet traverses a bridge? Provide an explanation and include an example for the captured data. Suppose that PC2 was configured as an IP router, which differences would you observe in the Ethernet and IP headers? The source and destination MAC and IP addresses did not change when packets traversed the bridge, because that bridge was a transparent bridge; transparent bridges operate in a way that is oblivious to the hosts of the network, so that is why it makes sense that the addresses did not change. If PC2 was an IP router, then it would be expected that the MAC addresses would be changed when they pass through PC2, but the IP addresses would still stay the same. The following shows the MAC addresses and IP addresses not changed when PC2 was a bridge: 3 0.000159 10.0.1.11 10.0.1.31 ICMP Echo (ping) request Frame 3 (98 bytes on wire, 98 bytes captured) Ethernet II, Src: LinksysG_7a:c8:da (00:04:5a:7a:c8:da), st: LinksysG_7b:44:94 (00:04:5a:7b:44:94) Internet Protocol, Src: 10.0.1.11 (10.0.1.11), st: 10.0.1.31 (10.0.1.31) Internet Control Message Protocol 4 0.000306 10.0.1.31 10.0.1.11 ICMP Echo (ping) reply Frame 4 (98 bytes on wire, 98 bytes captured) Ethernet II, Src: LinksysG_7b:44:94 (00:04:5a:7b:44:94), st: LinksysG_7a:c8:da (00:04:5a:7a:c8:da) Internet Protocol, Src: 10.0.1.31 (10.0.1.31), st: 10.0.1.11 (10.0.1.11) Internet Control Message Protocol 2. Include the output of the traceroute command from Step 5. Provide an explanation why PC2 does not appear in the output of the traceroute command in Step 5. Include the answers to the questions in Step 5. The following is the output of the tracroute command: traceroute to 10.0.1.31 (10.0.1.31), 30 hops max, 38 byte packets 1 10.0.1.31 (10.0.1.31) 0.286 ms 0.161 ms 0.148 ms - Why is PC2 not visible from PC1? PC2 is a transparent bridge, meaning that its operation and existence is supposed to be

completely oblivious to PC1 and that it operates at the data-link layer, so the packets sent by traceroute just passed straight through without ever having their TTL values decremented and causing a ICMP error message to be sent. - If PC2 was configured as an IP router, how would the output differ? PC2 would have showed up in this case, because routers work at the network layer and a PC2 acting as a router would have decremented the TTL fields of the sent packets and have caused ICMP error messages to be sent and thus revealing PC2. Part 2 1. Include the output of the traceroute command. traceroute to 10.0.1.31 (10.0.1.31), 30 hops max, 38 byte packets 1 10.0.1.31 (10.0.1.31) 0.286 ms 0.161 ms 0.148 ms Part 5 2. Provide the answers to the questions in Step 5. - Compare the results to the outcome of the traceroute command in Exercise 1(C). The results are identical: The bridge is transparent and operates at the data-link layer and thus just passes the packets of the traceroute command through without ever touching the TTL field and thus never causing ICMP error messages to be sent. - Why is it not possible to issue a ping command to outer1? The command works at the IP Layer and outer1 in this exercise is set to be a bridge, which works at the data-link layer, and thus the router's IP addresses for the interfaces were not used; this made the router in some sense invisible to IP and thus ping could not find the router. 1. Use the data saved in Step 7 to document for a single packet, that the packet is forwarded in a loop. We sent a single ping request and we saw on PC4 many ping requests and ping replies. The following shows this: 1 0.000000 10.0.1.11 10.0.1.41 ICMP Echo (ping) request 2 0.000089 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply 3 0.000093 10.0.1.11 10.0.1.41 ICMP Echo (ping) request

4 0.000095 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply 5 0.000244 10.0.1.11 10.0.1.41 ICMP Echo (ping) request 6 0.000286 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply 7 0.000254 10.0.1.11 10.0.1.41 ICMP Echo (ping) request 8 0.000292 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply 9 0.000266 10.0.1.11 10.0.1.41 ICMP Echo (ping) request 10 0.000302 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply 11 0.000278 10.0.1.11 10.0.1.41 ICMP Echo (ping) request 12 0.000306 10.0.1.41 10.0.1.11 ICMP Echo (ping) reply All the addresses in the packets are the same since the bridges do not modify them, so they are just duplicate packets. 2. Use the output from ethereal and the M A C forwarding tables to explain why some packets are looping indefinitely. If one were to analyze the MAC forwarding table entries and use the specific MAC destination addresses in the ethereal output, one can find that the forwarding table entries are such that a loop will occur. For example, for the case of a ping request packet of ours with MAC destination 00:04:5a:7b:21:cc, we saw the following: PC1 sent packets that reached FastEthernet0/0 of outer1. outer1 sent these packets out FastEthernet0/1 which reached FastEthernet0/1 of outer4. The following shows this: 0004.5a7b.21cc forward FastEthernet0/1 1 1 0 outer4 sent these packets out FastEthernet0/0 which reached FastEthernet0/1 of outer3. The following shows this: 0004.5a7b.21cc forward FastEthernet0/0 0 18 4

outer3 sent these packets out FastEthernet0/0 which reached of PC2. The following shows this: 0004.5a7b.21cc forward FastEthernet0/0 3 12 0 PC2 sent these packets out its and completing the loop. The following shows this (Note: the number 2 one the left is the port number and in our case port number 2 refers to ): 3. The network topology of Figure 6.10 has more than one cycle. Explain why the I C MP packets continue looping in the same cycle. As the ping from PC1 traverses the network, the bridges will learn on which port PC1 is. Likewise, as the echo reply from PC4 traverses the network, the bridges will learn on which ports PC4 lies. As a result, when the looping echo requests reach outer 3 & 4, they will not be forwarded, since it will now know which side PC4 lies on. The only bridge that will continue looping is outer 2. outer 2 will receive messages in such a way that it believes that PC4 lies on both sides of its ports. Hence, the loop will continue only through outer 2. 4. Use the saved data to draw the spanning tree for the network in Figure 6.10 as seen by the bridges. For each bridge, include information on the root bridge, the root port, the designated ports, and the blocked ports. The topology in Figure 6.10 is shown below with the hubs and PCs removed. designates a root port, designated a destination port and B designated a blocked port. Note that PC 2 is the root bridge: PC2 outer 3 outer 2 outer 1 B outer 4 B 5. Identify the bridges that transmit BPUs after the spanning tree protocol has

converged. For each Ethernet segment, determine the following fields for the BPU sent on that segment: root I, root path cost, and bridge I. Explain how these messages are interpreted. Show how the entire spanning tree can be constructed from these messages. The bridges that transmit BPUs are all the routers and PC 2. By capturing a packet sent from outer 3 at PC 4. The root I can be found under the spanning tree protocol field: oot Identifier: 0 / 128 / 00:04:5a:7a:c6:6f The oot I is the 128 followed by the MAC address of the root which is 00:04:5a:7a:c6:6f. The root path cost is given in this field: oot Path Cost: 19 and the bridge I is given in this field: Bridge Identifier: 0 / 128 / 00:1c:58:7d:ff:f0. These BPUs are analyzed and each bridge will determine whether the advertised BPU is better than its current BPU. As BPUs from every bridge reach every other bridge, the root bridge will be discovered, and the root and designated ports will be determined as well. We can use these messages in a similar fashion. By studying these messages, we can determine that the root is PC 2 and hence, router 3 will have as its root port and as its designated port. If we were to study the Ethernet segments of all the bridges we would be able to construct the spanning tree by determining which ports are root, designated and blocked. 6. Use Figure 6.10 to trace the path of the packets that are sent as a result of the ping command. Justify your answer with saved data from Step 10. The path taken is from PC1 to PC2 to outer 3 and finally to PC 4. The path back is just the reverse. This is the data saved on PC 4 which shows the request and reply: 30 113.726844 10.0.1.11 10.0.1.41 ICMP 98 Echo (ping) request 31 113.726885 10.0.1.41 10.0.1.11 ICMP 98 Echo (ping) reply There were no ICMP packets captured on PC 3, hence this can be the only route taken.

Part 6 1. Use the saved output to draw the spanning tree as seen by the bridges after the spanning tree has been rebuilt. For each bridge, indicate the root port, the designated ports, and the blocked ports. Briefly explain the changes of the spanning tree. We disconnected the cable on PC 2. This is the resulting spanning tree, PC 2 is still the root bridge: PC2 outer 3 outer 2 outer 1 outer 4 B The new tree has modified to make outer 1 have no blocked ports, because it now needs and to connect PC 1 to the root which is PC 2. outer 4 has also been modified to have a new blocked port at and a root port at. 2. raw the spanning tree as seen by the bridges after a new root bridge has been configured. outer 1 is the new root bridge; the following is the new tree. Note that at PC 2 is still disconnected as instructed: PC2 B outer 3 outer 2 outer 1 outer 4

Part 7 1. escribe which of the ping commands are successful and which fail. Use the data you captured to determine the route of the ping. For each route, provide an explanation why the path is taken. Ping from PC 1 to PC 3, 4, 2 failed and it succeeded from PC 4 to PC 1. The route from PC 4 to PC 1 is through outer 3 to outer 2 to PC1. The return route is from PC1 to its default gateway outer 2, and from outer 2 to outer 3 and finally to PC4. This path to PC1 is taken because when PC4 asks for the MAC address of PC1, outer 3 will respond since it does not know PC4 has another route to PC1. The return route goes through the IP routers because PC1 has a /24 prefix and it believes that it is on a different LAN than PC4, and thus it will send the packet to its default gateway.