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Suppose I want to send an IP datagram from node D to node C . Each step I also want to list the MAC addresses of the source and the destination , as well as the source and the destination of the IP datagram .

As a newcomer in networking and in efforts to understand how this would look like , I have came up with a solution which I would like you to criticize and give me pointers to further my knowledge.

Step 1 :

Ethernet(MAC) source : FD-0B-1F-63-32-88 ( MAC of D ) IP source : 128.119.109.251 ( IP of D )

Step 2 :

Ethernet(MAC) destination :FA-C3-3B-23-12-56 ( MAC of inner router) IP destination : 128.119.109.13 ( IP of C )

Step 3 :

Ethernet(MAC) source : FA-C3-3B-23-12-56 ( MAC of inner router) IP source : 128.119.109.251 ( IP of D )

Step 4 :

Ethernet(MAC) destination : C4-E6-48-38-CE-AF ( MAC of C ) IP destination : 128.119.109.13 ( IP of C )

Is my way of thinking correct in this situation?

Thanks in advance.

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  • You don't say where your "steps" are, but generally, you're not right. IP addresses do not change as the packet moves from hop to hop. Only MAC addresses change. – Ron Trunk Sep 2 '20 at 16:29
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That is not correct. C and D share a common L2 segment (=broadcast domain) since they're connected by switches. Assuming their network prefixes are /24 (255.255.255.0 mask) they also share a common IP network (subnet).

So,

  1. D resolves C's IP to C's MAC address (via ARP for IPv4), if it doesn't already know.
  2. D encapsulates the IP packet in a frame addressed to that destination MAC.
  3. D sends the frame.
  4. The frame is forwarded upwards by the lower right-hand switch.
  5. The frame is forwarded towards C by the upper right-hand switch.
  6. The frame arrives at C.
  7. C decapsulates the original IP packet.

Neither IP addresses nor MAC addresses change in any way on that path. Nothing else in the diagram is required or used for D-to-C traffic.

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