Revision Questions
1. If dynamic routing is used, the potential exists for a datagram to loop indefinitely through the internet. How could this problem be overcome in IP architecture?
•Inconsistent routing tables cause problems such as routing loops and counting to infinity.
•In a routing loop packets never reach their destination, but simply cycle repeatedly through a constant series of network nodes.
Diagram courtesy from CISCO.COM
Answer:
•One way to solve routing loops is by defining a maximum hop count
•With this approach, the routing protocol permits the routing loop to continue until the metric exceeds its maximum allowed value
•The distance vector default maximum in RIP (a standard distance-vector protocol) is 15 hops
2a. If datagrams can be fragmented in the course of their travels, the question arises as to where they should be reassembled. What are implications of allowing immediate router reassembly (at the next router after fragmentation)?
Answer:
1)Large buffers are required at routers and there is the risk that all of the buffer space will be used up storing partial datagrams.
2)All fragments of a datagram must pass through the same router. This inhibits the use of dynamic routing.
2b. What are the implications of performing reassembly of a fragmented datagram at the destination only?
Answer:
So that all the datagrams meant for the destination router will be all received and well. Once the datagrams have arrived and complete, then the reassembly will begin.
Reflective Questions
1. For connection-oriented transfer, it is sometimes desirable to use only a connection identifier during the data transfer phase. Reflect on the advantages of using a connection identifier.
Answer:
A channel number which is attached to data frames to tell the network how to route the data. A 13-bit field that defines the destination address of a packet. The address is local on a link-by-link basis
- Faster packets transfer through the routes.
- Less susceptible to packet loss.
- Guarantees that the packets arrive in proper sequence.
2. Typically, a connection-oriented transport service awaits an acknowledgment until a timeout expires, at which time the block of data will be retransmitted. In general, longer times are required for successful delivery across multiple networks. Reflect on the implications of this scenario.
Answer:
Time-out period often relatively long:
– Long delay before re-sending lost packet
Detect lost segments via duplicate ACKs.
– Sender often sends many segments back-to-back
– If segment is lost, there will likely be many duplicate ACKs.
When receiver receives out-of-order segments, it re-ACK the last in-order byte
Thus, the more networks spans more routes and possibility of more lost packets and duplicate ACKs will occur.
If sender receives 3 ACKs for the same data, it supposes that segment after ACKed data was lost:
– Fast retransmission: re-send segment before timer expires, restart timer
3. For the purpose of routing, each end system and router maintains a routing table that lists, for each possible destination network, the next router to which the datagram should be sent. Reflect on this
Answer:
A routing table is a set of rules. It contains the information necessary to forward a packet along the best path toward its destination. Each packet contains information about its origin and destination. When a packet is received, a network device examines the packet and matches it to the routing table entry providing the best match for its destination. The table then provides the device with instructions for sending the packet to the next hop on its route across the network.