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Q # 3 in CSMA/CD, after the fifth collision, what is the probability that a node

ID: 3588602 • Letter: Q

Question

Q # 3 in CSMA/CD, after the fifth collision, what is the probability that a node chooses K-4? The result K-4 corresponds to a delay of how many seconds on a 10 Mbps Ethernet? Q # 4 Consider a 100 Mbps 100BASE-T Ethernet with all nodes directly connected to a hub. To have an efficiency of 0.50, what should be the maximum distance between a node and the hub? Assume a frame length of 64 bytes and that there are no repeaters. Does this maximum distance also ensure that a transmitting node A will be able to detect whether any other node transmitted while A was transmitting? Why or why not? How does your maximum distance compare with the actual 100 Mbps standard?

Explanation / Answer

3)Answer:

After fifth collision, node will choose a value of K at random from {0, 1, 2,..., 25 - 1} i.e. 0 to 31.

So P(K=4) =1/32

Now, after choosing K=4, node will wait for K*512 bit times. Bit time for 10 Mbps ethernet is 0.1microsec. Therefore total waiting time for node = 4 * 512 * 0.1 microsec = 204.8 microsec