This test has six questions. The test runs the entire period, ending at 6:20. When you finish the test, you may leave.
Always justify answers unless the question explicitly states it’s not necessary. Some questions require reasonable assumptions in order to be answered. In such cases the answer should clearly state the assumptions made.• Multicast routing• Autonomous system• Dynamic routing algorithm• Tunneling• Head-of-line blocking
• Multicast routing:Routing in which PDUs are copied and distributed efficiently to a subset of hosts.• Autonomous system:A collection of networks under the authority of a single entity.• Dynamic routing algorithm:A routing algorithm that continuously recomputes the network model to adjust for changes in network conditions.• Tunneling:Encapsulating one PDU as the payload of another PDU.• Head-of-line blocking:Blocking that occurs when a PDU at the head of a queue is blocked and prevents unblocked PDUs behind it from progressing.
A routing algorithm’s operation is independent of its use in a datagram or VC network, and will not change based on the network type. A routing algorithm’s operation has three functions: build a model for the network, find the best paths through the network and generate the forwarding table. There is no need to change any of these functions based on network type.
VC and datagram forwarding tables are different, but the difference has no effect on routing-algorithm operation. Datagram and VC networks may have different views of best paths through the network (hop-count vs bandwidth, for example), but these differences have no significant effect on routing-algorithm operation.
Each input port gets the same shadow forwarding table. There are at least two ways to justify this claim. First, any PDU entering a router has to be forwarded in the same way regardless of the input port on which the PDU enters the router. This means each input port needs the same shadow routing table. Second, shadow forwarding tables were developed to relieve the performance-degrading memory contention resulting from all input ports accessing the same forwarding table.
There is a one-to-one correspondence between the number of interfaces a router has and the number of IP addresses the router has: a router with N interfaces has N IP addresses, and vice versa.
The maximally deep network topology under the assumptions given in the problem is a complete binary tree of depth 32 (or 31, or 30, depending on how you want to count). At the root (depth 0), the router forwards on a.b.c.d/1 and has two children (neighbors), one handling 0.b.c.d/1 and the other handling 128.b.c.d/1. A router at depth i forwards to its two children on a.b.c.d/(i + 1).
It is true that there are at most 32 (or 31, or 30 depending on your assumptions) possible prefix lengths. However, prefix of length i bits can encompass up to 2i different IP addresses. For example, there are 26 = 64 IP addresses covered by a six-bit prefix.