Using the saved output, describe the time interval between each ARP Request issued by PC1. Describe the method used by ARP to determine the time between retransmissions of an unsuccessful ARP Request. Include relevant data to support your answer.


What will be an ideal response?

The clocks of the ARP requests are 17:46:33 (Frame1), 17:46:34 (Frame 2), 17:46:35 (Frame 3) and 17:46:42 (Frame 4). Therefore, the time difference of the ARP requests are 1 second, 1 second, and 7 seconds. In this case, we are trying to connect an unreachable host using telnet protocol. These ARP requests correspond one-to-one with the initial TCP SYN(synchronize) segment that TCP is trying to send. Therefore, it follows the TCP's timeout and retransmission algorithm. Basic idea is it doubles its timeout value for each retransmission, with an upper limit of 64 seconds. This doubling is called an exponential backoff.
```
Frame 1 (42 on wire, 42 captured)
Arrival Time: Aug 28, 2002 17:46:33.849018000
Time delta from previous packet: 0.000000000 seconds
Time relative to first packet: 0.000000000 seconds
Frame Number: 1
Packet Length: 42 bytes
Capture Length: 42 bytes
Ethernet II
Destination: ff:ff:ff:ff:ff:ff (ff:ff:ff:ff:ff:ff)
Source: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Type: ARP (0x0806)
Address Resolution Protocol (request)
Hardware type: Ethernet (0x0001)
Protocol type: IP (0x0800)
Hardware size: 6
Protocol size: 4
Opcode: request (0x0001)
Sender MAC address: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Sender IP address: 10.0.1.11 (10.0.1.11)
Target MAC address: 00:00:00:00:00:00 (00:00:00:00:00:00)
Target IP address: 10.0.1.10 (10.0.1.10)
Frame 2 (42 on wire, 42 captured)
Arrival Time: Aug 28, 2002 17:46:34.847718000
Time delta from previous packet: 0.998700000 seconds
Time relative to first packet: 0.998700000 seconds
Frame Number: 2
Packet Length: 42 bytes
Capture Length: 42 bytes
Ethernet II
Destination: ff:ff:ff:ff:ff:ff (ff:ff:ff:ff:ff:ff)
Source: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Type: ARP (0x0806)
Address Resolution Protocol (request)
Hardware type: Ethernet (0x0001)
Protocol type: IP (0x0800)
Hardware size: 6
Protocol size: 4
Opcode: request (0x0001)
Sender MAC address: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Sender IP address: 10.0.1.11 (10.0.1.11)
Target MAC address: 00:00:00:00:00:00 (00:00:00:00:00:00)

11
Target IP address: 10.0.1.10 (10.0.1.10)
Frame 3 (42 on wire, 42 captured)
Arrival Time: Aug 28, 2002 17:46:35.847717000
Time delta from previous packet: 0.999999000 seconds
Time relative to first packet: 1.998699000 seconds
Frame Number: 3
Packet Length: 42 bytes
Capture Length: 42 bytes
Ethernet II
Destination: ff:ff:ff:ff:ff:ff (ff:ff:ff:ff:ff:ff)
Source: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Type: ARP (0x0806)
Address Resolution Protocol (request)
Hardware type: Ethernet (0x0001)
Protocol type: IP (0x0800)
Hardware size: 6
Protocol size: 4
Opcode: request (0x0001)
Sender MAC address: 00:00:c0:52:d2:f3 (00:00:c0:52:d2:f3)
Sender IP address: 10.0.1.11 (10.0.1.11)
Target MAC address: 00:00:00:00:00:00 (00:00:00:00:00:00)
Target IP address: 10.0.1.10 (10.0.1.10)
```

Computer Science & Information Technology

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A company hires a third-party firm to conduct an assessment of vulnerabilities exposed to the Internet. The firm informs the company that an exploit exists for anFTP server that had a version installed from eight years ago. The company has decided to keep the system online anyway, as no upgrade exists form the vendor.Which of the following BEST describes the reason why the vulnerability exists?

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