Homework #10 Solutions
cs349 -- Networks


14)

a) All the bridges know where X is now, and yes, Y's interface
   saw the packet because B2 has to broadcast.

b) All the bridges learn, but Y doesn't see the packet, because
   B2 knows where X is and does not have to broadcast.

c) Only B1 and B2 learn where Y is, Z does not see this packet.

d) All bridges already know where Z is.  W does see this packet
   because B3 does not know where Y is and has to broadcast.


15) Give a spanning tree for the LAN in Figure 3.44 and explain
    how any ties are resolved.

    The oddity here is that B2 abd B3 are going to get the
    same offer <B1, 0, B1> on two ports.  As the algorithm
    appears in the book, they will choose whichever comes
    first, which means that B2 and B3 will forward packets
    to whichever network B1 sends an offer to first.

21) What percentage of ATM bandwidth is consumed by cell headers?

    5 bytes / 53 bytes = 9.4% of the total bandwidth consumed
    by headers

In AAL3/4, what percentage of the total bw is consumed by
all non-payload bits when the user data is 512 bytes?

    The Adaption Layer packet is 512 bytes + 3 bytes of padding
    + 8 bytes of header = 523 bytes

    Chopped into 44-byte pieces, that's 12 cells.

    Total bytes sent = 12 * 53 = 636, of which 512 are user data.

    Wasted bw = 124 / 636 = 19.5%

    Actually, a better metric of waste is overhead, which
    is non-payload / payload = 124 / 512 = 24.2%

In AAL5, what percentage of the total bw is consumed by
all non-payload bits when the user data is 512 bytes?

    512 bytes packed into full 48-byte cells takes 
    10 cells plus 32 bytes of the last cell.  That
    leaves 16 bytes in the last cell, which is 8 bytes
    of padding and 8 bytes of trailer.

    Total bytes sent = 12 * 53 = 583, of which 512 are user data.

    Wasted bw = 71 / 583 = 12.2%

    Actually, a better metric of waste is overhead, which
    is non-payload / payload = 71 / 512 = 13.8%