Consider a LAN with four nodes π1, π2, π3 and π4. Time is divided intoβ¦
2015
Consider a LAN with four nodes π1, π2, π3 and π4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by π1, π2, π3 and π4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _____________.
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Correct answer: 0.4404
Key idea: exactly one station must transmit in the slot (that station sends a frame) and all other stations must not transmit.
Probability S1 succeeds: 0.1 * (1-0.2) * (1-0.3) * (1-0.4) = 0.1 * 0.8 * 0.7 * 0.6 = 0.0336
Probability S2 succeeds: 0.2 * (1-0.1) * (1-0.3) * (1-0.4) = 0.2 * 0.9 * 0.7 * 0.6 = 0.0756
Probability S3 succeeds: 0.3 * (1-0.1) * (1-0.2) * (1-0.4) = 0.3 * 0.9 * 0.8 * 0.6 = 0.1296
Probability S4 succeeds: 0.4 * (1-0.1) * (1-0.2) * (1-0.3) = 0.4 * 0.9 * 0.8 * 0.7 = 0.2016
Total probability of a frame being sent in the first slot without collision = 0.0336 + 0.0756 + 0.1296 + 0.2016 = 0.4404 (β 0.44).
Note: the value 0.4 by itself would be just the probability that S4 generates a frame (but does not account for the requirement that all other stations stay silent), so it is not the correct answer for a successful, collision-free transmission by any station.
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