A and B are the only two stations on an Ethernet. Each has a steady queue of…
2004
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
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Correct answer: B
Key insight: after the first collision A won the backoff and transmitted successfully, so A's backoff stage resets while B has already experienced one collision.
Consequently, at the second collision:
A will choose a backoff uniformly from {0,1} (2 choices).
B will choose a backoff uniformly from {0,1,2,3} (4 choices) because it has already experienced one collision earlier.
Compute the probability that A's backoff is strictly smaller than B's:
If A picks 0 (probability 1/2), probability B picks a value >0 is 3/4. Contribution: (1/2)*(3/4)=3/8.
If A picks 1 (probability 1/2), probability B picks a value >1 is 2/4 = 1/2. Contribution: (1/2)*(1/2)=1/4.
Total probability = 3/8 + 1/4 = 5/8 = 0.625.
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