Two CSMA/CD stations are each trying to transmit long (multiframe) files. After each frame is sent, they contend for the channel, using the binary exponential backoff algorithm. What is the probability that the contention ends on round k, and what is the mean number of rounds per contention period

Answer and Step-by-step explanation:

a.

A round can can be defined as each calculation of the backoff period which is often followed by transmission attempt in which the number rounds starting with 1 – at round 1, in which each of the two stations will often attempt to transmit in the immediately following slot (no backoff), thus collision is inevitable.

While at round 2, there are often two slots in which each station might try after waiting 0 slots (no wait) and 1 slot.

Similarly the contention at round iis over 2i-1 slots.

b.

Therefore the “contention period” is the whole period between two frame transmissions that are successful which is why at around i, the first station might likely pick the slot 0 with probability 1/2i-1, so might the second station which is why the probability of a collision by both stations picking this slot is 1/22(i-1). Although it doesn't happen to only slot 0, but every slot from 0 to 2i-1, so the total probability of collision at the i-th round is 1/2i-1. Therefore the probability of having exactly k rounds is the multiplication of the probabilities of collision at round 1, round 2, round 3, round k-1, and the probability of not having a collision at round k.

Hence, to find the expectation of the same quantity. The answer is then simply ΣkPk. The probability Pk should be “frozen” after the k= 10, and the summation carried out only to k = 16, according to the details of the standard.

answer: asdfghjkl,mnbv

step-by-step explanation:

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8