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.

step-by-step explanation:

it will be a cylinder. with a diameter of 8 and 4 single pieces high

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation: