This test has six questions. The test runs the entire class period, starting at 4:30 and 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. Answers should be no longer than 75 to 100 words.• An agreed-upon bit pattern serving as the divisor in calculations.• The frame-forwarding structure defined by bridge forwarding tables.• The amount of time a channel spends transmitting a signal divided by the total time.• A (fully-qualified domain name, IP address) pair.• Translates a bit stream into a bit stream more suitable for transmission.
Keeping in mind that giving an example of X is not defining X, briefly (in one or two sentences, 25 to 30 words) define:
• An agreed-upon bit pattern serving as the divisor in calculations:The generator polygon (not cyclic redundancy check, which contains a generator polygon, but also contains much else).• The frame-forwarding structure defined by bridge forwarding tables:A spanning tree.• The amount of time a channel spends transmitting a signal divided by the total time:The effective transmission rate, or the utilization.• A (fully-qualified domain name, IP address) pair:A DNS A record for IPv4 or an AAAA record for IPv6.• Translates a bit stream into a bit stream more suitable for transmission:A line encoding.
Following FTP's lead, to establish a connection with out-of-band control, HTTP would need two connections, one for data and the other for control. The control connection lasts the life of the connection, while the data connection can be created as needed or created once at connection set-up and kept for the life of the connection. HTTP can use a single continuously connected data channel because HTTP response messages are sized delimited and don't need an explicit end signal.
TCP congestion control is based on a non-trivial sliding window. Stop-and-wait has a trivial sliding window; using stop-and-wait in TCP breaks congestion control. Assuming congestion appears to the end-points as dropped packets, under normal operation stop-and-wait has at most one packet (data or ack) in flight at any time. If congestion is detected, stop-and-wait has two options: drop packets into the network less often, or drop smaller packets into the network (assuming acks are already as small as possible, the data packets would be smaller).
Assuming the router forwarding algorithm uses maximal prefix matching, what are the address ranges sent to each output port? Justify your answer.
output prefix port 01 3 100 3 101 2 1101 0 * 1
Addresses in the range 000000002 through 001111112 match no prefix and are sent to the default port 1.
Addresses in the range 010000002 through 011111112 match prefix 01 and are sent to port 3.
Addresses in the range 100000002 through 100111112 match prefix 100 and are sent to port 3.
Addresses in the range 101000002 through 101111112 match prefix 101 and are sent to port 2.
Addresses in the range 110000002 through 110011112 match no prefix and are sent to the default port 1.
Addresses in the range 110100002 through 110111112 match prefix 1101 and are sent to port 0.
Addresses in the range 111000002 through 111111112 match no prefix and are sent to the default port 1.
Assuming there are at least two constituent LANs in the network (otherwise why have bridges or hubs in the first place), LB has a higher aggregate bandwidth than does LH. A bridge preserves the constituent LANs’ collision domains, which means transmissions can occur simultaneously on each constituent LAN. Hubs, on the other hand, don't preserve collision domains, and only one host at a time on the combined LAN can transmit at a time.
This question is lightly tricky because “bandwidth” was given without units. Assuming units in Hz, the bandwidth is fv - fr = 0.32 PHz, where
and
fv = (c m/sec)/(400·10-9 m/cyc) = (300·106 m/sec)/(400·10-9 m/cyc) = (0.75·1015 cyc/sec)
If, on the other hand, the units are in bits/sec, a few more assumptions are needed. First, assume an errorless channel, which enables Nyquist's equation. Second, assume a two-symbol alphabet, which makes the log equal to 1. The bandwidth is 0.32 Pbits/sec.
fr = (c m/sec)/(400·10-9 m/cyc) = (300·106 m/sec)/(700·10-9 m/cyc) = (0.43·1015 cyc/sec)