GATE | GATE-CS-2004 | Question 90

A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is:
(A) 0.5
(B) 0.625
(C) 0.75
(D) 1.0

Answer: (B)

Explanation: This is basically the question related to unfairness of exponential back-off algorithm called ‘capture effect’. You can find more info about it here:

The solution to the above problem goes like this:

At every attempt to transit a frame, both A and B chooses value of ‘k’ randomly. Based on the value of ‘k’, back-off time is calculated as a multiple of ‘k’. The station or node having the smaller back-off time gets to send the frames eariler.

1st attempt: Value of ‘k’ would be k=0 or k=1 (0 <= k <= 2^n-1; where n=nth attempt). Since A won the first race, A must have chosen k=0 and B must have chosen k=1 (A wins here with probability 0.25). As A won, A will again choose k=0 or k=1 for its 2nd frame, but B will choose k=0,1,2 or 3 as B failed to send its first frame in the first attempt.

2nd attempt: Let kA= value of k chosen by A and kB = value of k chosen by B. We will use notation (kA,kB) to show the possible values. Now the sample space for the 2nd attempt is (kA,kB) = (0,0),(0,1),(0,2),(0,3),(1,0),(1,1),(1,2) or (1,3) i.e. 8 possible outcomes. For A to win, kA should be less than kB (kA < kB). Thus, our event space is (kA, kB) = (0,1),(0,2),(0,3),(1,2),(1,3) i.e. 5 possible outcomes.

Thus the probability that A wins the 2nd back-off race = 5/8 = 0.625

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