2. The topology of a computer network is shown in Figure Q2 as below. Network 152.121.0.0 Network 155.65.0.0 Interface 3

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answerhappygod
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2. The topology of a computer network is shown in Figure Q2 as below. Network 152.121.0.0 Network 155.65.0.0 Interface 3

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2 The Topology Of A Computer Network Is Shown In Figure Q2 As Below Network 152 121 0 0 Network 155 65 0 0 Interface 3 1
2 The Topology Of A Computer Network Is Shown In Figure Q2 As Below Network 152 121 0 0 Network 155 65 0 0 Interface 3 1 (117.52 KiB) Viewed 56 times
2. The topology of a computer network is shown in Figure Q2 as below. Network 152.121.0.0 Network 155.65.0.0 Interface 3 Interface Interface 2 Router D Interface 2 Router Interface 1 Interface 1 Router Router A Interface 2 Interface 3 Interface Interface 3 Router Interface 2 Network Client Server 159.170.0.0 (in private network) (in public network) Figure Q2 In the above network, there are five routers, i.e., Router A to Router E. One client is in a private network connected to Router A, and a server is in a public network connected to Router C. The IP addresses and MAC addresses of the interfaces belonging to Router A to Router E, the client, and the server are listed in Table Q2-1 to Table Q2-6, respectively. Table 02-1: Router Interface IP Address MAC Address Interface IP Address MAC Address Table Q2-2. Router B 1 134.123.0.1 AA-BB-CC-11-22-33 1 172.124.1.2 BB-AA-CC-11-22-33 2 AA-BB-CC-44-55-66 3 131.92.0.1 BB-AA-CC-44-55-66 172.124.0.1 192.168.0.1 3 AA-BB-CC-77-88-99 Table Q2-3: Router Interface IP Address MAC Address Interface Table Q2-4: Router D IP Address MAC Address 147.62.0.1 DD-AA-BB-11-22-33 134.123.1.4 DD-AA-BB-44-55-66 1 136.71.0.1 2 CC-AA-BB-11-22-33 CC-AA-BB-44-55-66 2 131.92.1.4 3 Table Q2-5: Router E Table Q2-6: Client and Server Interface IP Address MAC Address Device IP Address MAC Address 1 147.62.1.4 192.168.0.5 EE-AA-CC-11-22-33 EE-AA-CC-44-35-66 Client Server FF-AA-BB-11-22-33 FF-AA-CC-44-55-66 2 136.71.1.4 139.170.6.5 a) (a) Answer the following questions related to the network shown in Figure Q2. Based on the shortest path approach, construct the routing table of Router A using the following format. (6 marks)
Gateway Interface Network Destination 152.121.0.0 155.65.0.0 159.170.0.0 (6) Consider the network address translation (NAT) in Figure Q2. Via Routers A, B and C, the client with the private IP address 192.168.0.5, which is connected to Router A, wants to communicate to the server with the public IP address 159.170.6.5, which is in the network 159.170.0.0 connected to Router C. In this process, Router A will perform NAT to protect the private IP address of the client. The public IP address used by Router A for NAT is assumed to be the public IP address of its Interface 2, which is shown in Table Q2-1. (0) First, the client will send a message to the server via the shortest path routing strategy "client - Router A → Router B Router C → server”. In this process, just consider the frame sent from the client to Router A and the frame sent from Router A to Router B. In the following table, write down the source IP address, destination IP address, source MAC address, and destination MAC address in each of the above frames. (8 marks) Frame Source IP Destination IP Source MAC Destination MAC Address Address Address Address From Client to Router A From Router A to Router B (ii) After receiving the message sent from the client, the server will reply to the client by sending a message via the shortest path routing strategy "server Router C → Router B → Router A → client". In this process, just consider the frame sent from Router C to Router B. In the following table, write down the source IP address, destination IP address, source MAC address, and destination MAC address of the above frame. (4 marks) Frame Source IP Destination IP Source MAC Destination MAC Address Address Address Address From Router C to Router B c) Consider the error checking strategy in the data link layer. Assume that the server in Figure Q2 receives a sequence 011011010 from the client. (1) If an even parity bit is used by the client, is the transmission in error? (1 mark) (ii) If an odd parity bit is used by the client, is the transmission in error? (1 mark)
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