Collisions, Probability, and Ethernet

Disclaimer: I am a student and don't expect a direct answer.

Goal: Understand what to do with the problem.
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JCW2Asked:
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cardrayCommented:
After the 5th collision, the adapter chooses from {0, 1, 2, ..., 31}. The probability that it chooses 4 is 1/32. It waits 4 x 512 x 0.1 = 204.8 microseconds.
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JCW2Author Commented:
How does the algorithm work?
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peterelvidgeCommented:
in the real world -- people use full duplex,  collisions/half duplex operation are pretty much thing of the past when coax lans and ethernet hubs were state of the art.

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JCW2Author Commented:
I'm still having trouble understanding where to put the numbers.
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Steve JenningsIT ManagerCommented:
Three "interesting" links (especially the last one):

http://www.techfest.com/networking/lan/ethernet3.htm (you'd probably better understand this)

http://www.comp.nus.edu.sg/~cs2105/tut/tut11.pdf (Singapore???)

http://www-users.itlabs.umn.edu/classes/Fall-2009/csci4211/Assignments/Ass4-key.pdf (BINGO!!!)

a. Read Chapter 5.5.2 on CSMA/CD. After the fifth collision, what is the probability
that a node chooses K= 6. (5 pts) The result of K = 6 corresponds to a delay of how
many seconds on a 10BAST-T Ethernet? (5 pts)

Answer>

After the n-th collision for this frame, the Ethernet adapter chooses a value for K at random from {0, 1, 2, … , 2^m -1} where m = min (n, 10). Thus, after the 5th collision, K is one of 32 values, such as {0, 1, 2, … , 2^5 - 1}. The probability of K=6 is 1/32 = 0.0313. Therefore, the adapter waits K*512 bit times = 6*512 bits / 10 Mbps = 3.072 * 10^-4 sec.

NOTE: In this answer the K value is different from your example.

You should probably use "google" a little more diligently. That's where I found this stuff, including your answer.

Good luck,
SteveJ
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JCW2Author Commented:
Thank you for your help.
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Networking Protocols

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