Networking-Kurose ross

Question

in CSMA/CD after Fifth collision ,what is the probability that a Node choose K=4? The result k=4 correspond to how many seconds delay on a 10 Mbps Ethernet?

k is standard notation here

Answer 1

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) = $\frac{1}{32}$

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.

Comments

nicely explained Brother

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Why node will wait for 512*k bits time