1)We Have 3 Nodes, A, B C. the Weights of the Directed Edges Are: (A,B), (B, C), (C,B)

1)We have 3 nodes, a, b c. The weights of the directed edges are: (a,b), (b, c), (c,b), (c,a) are 1. (a,c) is 10. SPT is (a,b) and (b,c). RPF tree is (a,b), (a,c).

2) NAT distinguishes nodes behind it based on their Port number and when 2 of them use the same port number then the mapping is not possible. And the intermediate router expects to connect to the server on port 80, but with mappings both servers can’t use the same port number.

There is a problem when a client outside the NAT wants to establish a connection with one of these servers. In this case when the request packet reaches to the router it doesn’t know to which router it must forward the packet because both of them listen on port 80.

3)For node the RIP table after convergence is:

Destination / Next hop / Cost
/ / 1
/ / 2
…
…
/ / 1

After breaking the link between and , and detects after 180 s that the link is breaking and the cost of the path is now ∞. Then they advertise that there is no connection to anymore to their neighbors and they advertise this to their neighbors until it reaches to the nodes . This node will advertise to , but here this node will advertise back that it has connection to with cost and then will updates its table and put this route with next hop and cost to its table. And advertise back to and the advertisements going backward until reach to and it knows that there is a path between itself and with cost k and next hop . The same scenario is valid for in the reverse way. And the number of advertisements is 4*.

4)There are posibilities for address.

For interface A considering longest prefix matching we must consider the group of addresses in row 1 and subtract row 2 from it and add addresses in row 4:

Interface C, the addresses from row 3 minus addresses in row 4 because they have longest matching:

And the rest are going to link B: 1-0.34-0.37=0.28