Suppose nodes A and B are on opposite ends of a 1000 meter cable, and they each
ID: 3691313 • Letter: S
Question
Suppose nodes A and B are on opposite ends of a 1000 meter cable, and they each have one frame of 1,500 bits (including all headers and preambles) to send to each other. Both nodes attempt to transmit at time t = 0. Suppose there are three repeaters between A and B, each inserting a 20 bit delay. Assume a transmission rate of 100 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. Ignore the Jam signal and the 96 bit time delay.
1 ) What is the one-way propagation delay (including repeater delays) between A and B in seconds? (Assume a propagation speed of 2*108 meters/second)
2) At what time (in seconds) is A’s packet completely delivered at B?
3) Now suppose that only A has a packet to send and the repeaters are replaced with switches. Suppose that each switch has a 20-bit processing delay in addition to a store and forward delay. At what time, in seconds, is A’s packet fully delivered at B?
Explanation / Answer
Answer for Question A:
Step 1: 1000 m / 2 * 10^8 m/sec + 4 * 20 bits/ 100 * 10^6bps
Step 2: 10 * 10 ^-6 + 0.8 * 10 ^-6
Step 3: 10.8 sec
Answer for Question B:
frame transmission time is (1000bits/10*106bps)=100 sec
At time t = 0 , both A and B transmit.
At time t = 10.8 sec , A detects a collision.
At time t = 21.6 sec last bit of B 's aborted transmission arrives at A .
At time t = 32.4 sec first bit of A 's retransmission arrives at B .
At time t= 32.4 sec + 100 sec =132.4 secA 's packet is completely delivered at B .
Answer for Question C:
10.8 sec + 5 * 100 sec = 510.8 sec