Hi, hoping I could check my solution to tutorial Q2c with someone here as I couldn't make it to the last two tutes :) (c) What is the probability that all 3 devices can successfully transmit their frames in a maximum of 3 time slots after the original collision? P(maximum of 3 time slots after the original collision) = P(any one node is successful in slot-1) * P(any one node is successful in slot-2) * P(final node successful in slot-3) Slot 1: three device with backoff times of 0, 1 at a probability of 1/2 each P(any one node is successful in slot-1) = 3C1 * P(0) * P(1) * P(1) = 3 * 1/2 * 1/2 * 1/2 = 3/8 Slot 2: two devices with backoff times of 0, 1, 2, 3 at a probability of 1/4 each If k1 is the backoff time chosen by one device, and k2 is backoff time chosen by the other device P(any one node is successful in slot-2) = 2C1 * P(k1 < k2) P(k1 < k2) consists of the 6 pairs [(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)] and each pair occurs with a probability of 1/4 * 1/4 = 1/16 P(k1 < k2) = 6 * 1/16 = 6/16 P(any one node is successful in slot-2) = 2 * 6/16 = 12/16 = 3/4 Slot 3: one device with backoff times of 0, 1, 2, 3, 4, 5, 6, 7 at a probability of 1/8 each P(final node successful in slot-3) = 1 Finally: P(maximum of 3 time slots after the original collision) = 3/8 * 3/4 * 1 = 9/32

ANONYMOUS wrote: For this to occur, each of the 3 slots must host exactly one transmission. If the 1st slot hosts a successful re-transmission by one node, the other two nodes must have chosen to not immediately retransmit, but to back-off for one slot. Then, in the 2nd slot, the two nodes that had backed-off will now *both* re-transmit => another collision! Then, in the 3rd slot, we still have two nodes wishing to re-transmit. So, it's not possible, and the probability of it happening is 0.