Question: Part 1. 1. On a PC, Open Command Prompt Window (CMD). 2. Start Wireshark and Start a packet capture by pressing Start 3. In the

 Part 1. 1. On a PC, Open Command Prompt Window (CMD).

2. Start Wireshark and Start a packet capture by pressing Start 3.

Part 1. 1. On a PC, Open Command Prompt Window (CMD). 2. Start Wireshark and Start a packet capture by pressing Start 3. In the Command Prompt Window type ipconfig /flushdns followed by ping humber.ca -n 2 4. Stop the capture by pressing the red square. 5. In Wireshark, use a filter to display only ICMP and DNS frames. Type: icmp or dns 6. Answer the following questions: 1. What protocol is used to send the ping request? 2. What protocol is used to lookup IP address of Humber.ca? 3. In Ping command, can the target-name be IP address or Host Name? 4. Why did the Ping command use DNS? 5. What does the "-n 2 option signify in the Ping command? PART 2 1. On a PC, Open Command Prompt Window (CMD). 2. Start a Wireshark capture by pressing Start 3. From the Command Prompt Window clear the cache and obtain a new IP address: arp - * ipconfig /release ipconfig /renew arp -a 4. Stop the Wireshark capture. 5. Use a filter to display only arp or dhcp You should see at least 2 ARP and 2 DHCP frames similar to the following screen: 54012557000 18:66:da:Od:46:04 JuniperN_91:d9.c1 ARP 42 Who has 142.214 223.129? Tell 142.214.223.137 Frame 59 42 bytes on wire (356 bits), 42 bytes captured (336 bits) on interface o Ethernet II, Src: 18:66:da: Od:46:04 (18:66: da: 00:46:04). Dst: JuniperN_91:09:01 (54:e0:32:91:09:01) Destination: Juniper N_91:09:01 (54:e0:32:91:09:01) Address: Juniper N_91:09:01 (54:e0:32:91:09:01) ..0. - LG bit: Globally unique address (factory default) ....................... - IG bit: Individual address (unicast) Source: 18:66:da:od:46:04 (18:66: da:od:46:04) Address: 18:66:da:00:46:04 (18:66:da:od:46:04) ..0. - LG bit: Globally unique address (factory default) - IG bit: Individual address (unicast) Type: ARP (0X0806) Address Resolution Protocol (request) Hardware type: Ethernet (1) Protocol type: IP (OX0800) Hardware size: 6 Protocol 5ize: 4 opcode: request (1) Sender MAC address: 18:66da:00:46:04 (18:66:da:00:46:04) Sender IP address: 142.214.223.137 (142.214.223.137) Target MAC address: Juniper N_91:09:01 (54:e0:32:91:09:01) Target IP address: 142.214.223.129 (142.214.223.129) 2 6. Answer the following questions: 6. Does ARP use TCP or IP? 7. Does DHCP use TCP or UDP transport? 8. What is the value in the protocol type field? 9. What is the value in the protocol size field? 10. What does the arp-d* command do? PART 3 Based on the DHCP frame in LAB part B answer the following: 11. What does DHCP stand for? 12. What is the value in the hardware type field? 13. What is the value in the hardware address length field? 14. What is the IP address of your DHCP server? 15. What device is hosting your DHCP server? PART 3 16. In the following diagram, label the frame flow between a PC and the IP gateway PC 17. In the following diagram, label the frame flow between a PC and the DHCP server that enables the PC to obtain an IP address : PC 3

Step by Step Solution

There are 3 Steps involved in it

1 Expert Approved Answer
Step: 1 Unlock blur-text-image
Question Has Been Solved by an Expert!

Get step-by-step solutions from verified subject matter experts

Step: 2 Unlock
Step: 3 Unlock

Students Have Also Explored These Related Databases Questions!