Homework #13 Solutions
cs349 -- Networks

Chapter 4

6) The ident field contains 16 bits, so there can be 65536
   fragments in flight before it wraps around.  So the question
   is, how fast do we have to send 576 bytes in order to wrap
   around in 60 seconds:

   65536 * 576 bytes * 8 bits/byte
   -------------------------------  = 60 seconds
           x bits/second

           x = 5.033 Mbps

   If this bandwidth is exceeded, then it is possible for
   two fragments with the same ident field to arrive out
   of order and be reassembled incorrectly.

8) Two problems with reassembling at an intermediate router:

   a) it might be inefficient if the packets have to get
      fragmented again later in the path

   b) it requires routers to keep fragments in buffers until
      the packet can be reassembled.  Buffer space on routers
      is a precious commodity -- it is generally unacceptable
      to do anything that ties up buffer space for a long time.

      IPv6 abandoned fragmentation in order to force the sending
      host to do MTU discovery, because it is advantageous to
      discover the MTU once and then use it many times rather
      than obstinately send oversize packets into the network.

9) If the timeout value for ARP entries is too small, then
   hosts will have to broadcast ARP requests more often, which
   increases packet latency and network traffic.

   If it is too large, the system takes longer to respond to
   changes.