Assignment 2
Solutions for
forAssignment
18. Suppose we want to transmit the message 11001001 and protect it from errors using the
CRC polynomial x3+1.
(a) Use polynomial long division to determine the message that should be transmitted.
(b) Suppose the leftmost bit of the message is inverted due to noise on the transmission link.
What is the result of the receiver
receiver’’s CRC calculation? How does the receiver know that an
error has occurred?
Answer:
(a) In this case, x3+1 correspond to 1001. We multiply 11001001 since the degree of CRC is 3 and
we get 11001001000.
11010011
1001 / 11001001000
1001
1011
1001
1000
1001
1100
1001
1010
1001
011
The remainder is 11, so we should send 11001001000 + 011 = 1100 1001 011
(b) The inverting of the first bit gives us 0100 1001 011. The receiver divides the message by 1001,
which is x3+1, and finally gets 01000001 with a remainder of 10. With the remainder occurring,
the receiver will know that an error has occurred.
43. Suppose the round-trip propagation delay for Ethernet is 46.4 μs. This yields a minimum
packet size of 512 bits (464 bits corresponding to propagation delay + 48 bits of jam signal).
(a) What happens to the minimum packet size if the delay time is help constant, and the
signaling rate rises to 100 Mbps?
(b) What are the drawbacks to so large a minimum packet size?
(c) If compatibility were not an issue, how might the specifications is written so as to permit
a smaller minimum packet size?
Answer:
(a) If the signaling rate rises to 100 Mbps, there would be a 46.4μs × 100 Mbps = 4640 bits
corresponding to propagation delay. And if the jam signal still exits, the minimum packet size
would be 4640 + 48 = 4688 bits.
(b) The packet size is larger than many high level packet sizes. It also results in wasting a lot of
bandwidth.
(c) A smaller minimum packet size can be achieved if maximum collision diameter diameter could
be reduced (other tolerances also need to be tightened up).
61. How can a wireless node interfere with the communications of another node when the
two nodes are separated by a distance greater than the transmission range of either node?
Answer:
For example, we have three nodes named A, B, C as the figure below.
Node A and B are both in the range of C but greater than the transmission range of each other.
Suppose A is transmitting with C while B is going to communicate with C. A and B can not aware
of each other since their signals do not carry far. If B starts to send signal, it will cause collision
with the signal from A. This is called hidden nodes problem.