Chapter 1 Homework

P3.

A.  The network can hold 4n simultaneous connections.
B.  It would support 2n simultaneous connections.

P5.

A.  d(prop) = m / s
B.  d(trans) = L / R
C.  d(end-end) = N(d[prop] + d[trans])    (where N is equal to the number of routers)
D.  Ignoring the other delays, the last bit of the packet should just be arriving at the router at time t = d(trans)
E.  The first bit of the packet should be nearing it's destination at time t = d(trans)
F.  At time t = d(trans), the first bit of the packet should just leaving to it's destination
G.  d(prop) and d(trans) will be approximately equal at a distance of 50,000,000 meters.


P18.

A. The average round trip time was 6ms.
B.  There were always six different routers.
C.  There appeared to be only two different ISPs.
D.  The inter-continental connection was much more complex. There were fourteen different routers and three different ISPs on each trial. The average round trip time was 73ms.



P30.

A.  It should take ~200 minutes.
B.  It should take 3 seconds for the first packet to reach its final destination. The second packet should arrive at 4 seconds.
C. It should take ~67 minutes for the final packet to reach the final destination. This is much faster than the non-message switched file.
D. There are a couple downsides to message segmentation. First of all, the packets have to be reassembled at the final destination, taking time. Also, since the header size of packets is usually the same, regardless of size, you'll end up with much more data to transfer with multiple packets.