Suppose two nodes, A and B, are attached to opposite ends of an 800 m cable, and
ID: 3767095 • Letter: S
Question
Suppose two nodes, A and B, are attached to opposite ends of an 800 m cable, and that they each have one frame of 1 024 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are five repeaters between A and B, each inserting a 20-bit delay. Assume the transmission rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is used. After the first collision, A draws K = 0 and B draws K = 1 in the exponential backoff protocol after sending the 48-bit jam signal.
a. What is the one-way propagation delay (including repeater delays) between A and B in seconds? Assume that the signal propagation speed is 2 * 10^8 m/sec
b. At what time (in seconds) is A’s packet completely delivered at B?
c. Now suppose that only A has a packet to send, and that the repeaters are replaced with switches. Suppose that each switch has a 16-bit processing delay in addition to a store-and-forward delay. At what time, in seconds, is A’s packet delivered at B?
Q1.3 What is CSMA/CD? How does it work? Explain why RTT on a LAN is an important parameter for CSMA/CD to work.
Q1.4 What techniques can be used for error-detection in the data link layer?
Explanation / Answer
b)
First note, the transmission time of a single frame is give by 1500/(100Mbps)=15 micro sec, longer than the propagation delay of a bit.
>At time t = 0 , both A and B transmit.
>At time t = 4.8µ sec , both A and B detect a collision, and then abort.
>At time t = 9.6µ sec last bit of B 's aborted transmission arrives at A.
>At time t =14.4µ sec first bit of A 's retransmission frame arrives at B .
>At time t =14.4µ + 1500bits/100x10 pow 6 bps = 29.4µsec A 's packet is completely delivered at B.
a)
800m/2.10 pow 8 m/sec + 4.20bits/100x10 pow 6 bps
=(4x10 pow -6 + 0.8x10 pow -6)sec
=4.8 µsec
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