SY0-301 · Question #241
Review the following diagram depicting communication between PC1 and PC2 on each side of a router. Analyze the network traffic logs which show communication between the two computers as captured by…
The correct answer is D. The router implements NAT. The key clue is in the traffic logs: the source IP shown is 10.2.2.1 (the router's OUTSIDE interface), NOT 192.168.1.30 (PC1's actual private IP). PC1 initiated the connection from the private 192.168.1.x subnet, but by the time the packets reach PC2's network, the source IP…
Question
Review the following diagram depicting communication between PC1 and PC2 on each side of a router. Analyze the network traffic logs which show communication between the two computers as captured by the computer with IP 10.2.2.10. DIAGRAM PC1 PC2 [192.168.1.30]--------[INSIDE 192.168.1.1 router OUTSIDE 10.2.2.1]----- ----[10.2.2.10] LOGS 10:30:22, SRC 10.2.2.1:3030, DST 10.2.2.10:80, SYN 10:30:23, SRC 10.2.2.10:80, DST 10.2.2.1:3030, SYN/ACK 10:30:24, SRC 10.2.2.1:3030, DST 10.2.2.10:80, ACK Given the above information, which of the following can be inferred about the above environment?
Options
- A192.168.1.30 is a web server.
- BThe web server listens on a non-standard port.
- CThe router filters port 80 traffic.
- DThe router implements NAT.
How the community answered
(46 responses)- A7% (3)
- B22% (10)
- C13% (6)
- D59% (27)
Explanation
The key clue is in the traffic logs: the source IP shown is 10.2.2.1 (the router's OUTSIDE interface), NOT 192.168.1.30 (PC1's actual private IP). PC1 initiated the connection from the private 192.168.1.x subnet, but by the time the packets reach PC2's network, the source IP has been replaced with the router's public-facing IP (10.2.2.1). This is the defining behavior of Network Address Translation (NAT) - it rewrites the source IP of outbound packets from a private address to a public one. Option A is wrong because port 80 (HTTP) is the destination on PC2, meaning PC2 is the web server, not PC1. Option B is wrong because port 80 is the standard HTTP port. Option C is wrong because there is no evidence of filtering - the connection completes successfully (SYN → SYN/ACK → ACK shows a full TCP handshake).
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