If your lab does not have enough IP, if you expect to use the network number of the largest private address Class B to e
Posted: Mon Jun 06, 2022 5:27 pm
If your lab does not have enough IP, if you expect to use the network number of the largest private address Class B to establish a Class C network 140.116.18.0 to establish the number of these 3 LANs in the laboratory. The host requirements are 120, 60, and 60 respectively. (It must be greater than or equal to the required number when allocating IP) Please provide your plan for each department in the future. a) Network number, b) Netmask (expressed in CIDR), and c) The range of IPs that each department can actually assign.
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ii. Please write out the content of the routing table with the 140.116.96.1 interface Router (including network number, netmask, delivery handler) (10%)
iii. In this network environment, please explain the purpose of the Address Resolution Protocol? (5%) Please use this diagram to discuss if 140.116.99.11 is to be connected to mailserver 1, please draw the 4 possible ARP Layer2 during the transmission process. The header content of the packet, refer to the picture for the content of the header field. (10%) (Please draw 5 fields in sequence for the ARP packet, Layer 2: Dest Mac, Source Mac, Type, Layer 3: Sender IP Target IP)
46-1500 bytes of payload 6-byte 2-byte type header details source address Figure 15.1 Illustration of the Ethernet frame format and header details. 5A:3C:11:AB:CC:01 Net2 Netl 140.116.96.0 140.116.99.0/ header 6-byte destination address. 140.116.99.1/140.116.96. 00:04:AA:0C:89:12 / 00:0A:AA:0C:89:13 24 PROTOCOL ADDRESS TYPE OPERATION (first 4 octets) SENDER PADDR (first 2 octets) TARGET HADDR (first 2 octets) 140.116.99.11 00:0A:AC:12:23:1A 0 HARDWARE ADDRESS TYPE HADDR LEN PADOR LEN SENDER HADDR SENDER HADDR (last 2 octets) SENDER PADDR (last 2 octets) 16 TARGET HADDR (last 4 octets) TARGET PADDR (all 4 octets) 4-byte CRC 5.53.250 23:1B TY 140.116.96.96/192.168.[ID].253 00:0A:AA:0C:89:14 /00:0A:AA:0C:89:15 Mail Server1 140.116.96.5 00:0A:AC:12:23:1C 31 192.168.[ID].0) Mail Server2 192.168.[ID].250 00:0A:AC:12:23:ID New PC 00:0A:AC:12:23:1E
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- If Your Lab Does Not Have Enough Ip If You Expect To Use The Network Number Of The Largest Private Address Class B To E 1 (70.97 KiB) Viewed 30 times
iii. In this network environment, please explain the purpose of the Address Resolution Protocol? (5%) Please use this diagram to discuss if 140.116.99.11 is to be connected to mailserver 1, please draw the 4 possible ARP Layer2 during the transmission process. The header content of the packet, refer to the picture for the content of the header field. (10%) (Please draw 5 fields in sequence for the ARP packet, Layer 2: Dest Mac, Source Mac, Type, Layer 3: Sender IP Target IP)
46-1500 bytes of payload 6-byte 2-byte type header details source address Figure 15.1 Illustration of the Ethernet frame format and header details. 5A:3C:11:AB:CC:01 Net2 Netl 140.116.96.0 140.116.99.0/ header 6-byte destination address. 140.116.99.1/140.116.96. 00:04:AA:0C:89:12 / 00:0A:AA:0C:89:13 24 PROTOCOL ADDRESS TYPE OPERATION (first 4 octets) SENDER PADDR (first 2 octets) TARGET HADDR (first 2 octets) 140.116.99.11 00:0A:AC:12:23:1A 0 HARDWARE ADDRESS TYPE HADDR LEN PADOR LEN SENDER HADDR SENDER HADDR (last 2 octets) SENDER PADDR (last 2 octets) 16 TARGET HADDR (last 4 octets) TARGET PADDR (all 4 octets) 4-byte CRC 5.53.250 23:1B TY 140.116.96.96/192.168.[ID].253 00:0A:AA:0C:89:14 /00:0A:AA:0C:89:15 Mail Server1 140.116.96.5 00:0A:AC:12:23:1C 31 192.168.[ID].0) Mail Server2 192.168.[ID].250 00:0A:AC:12:23:ID New PC 00:0A:AC:12:23:1E