Question

In this problem, you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let S be the length of a slot. Suppose all frames are of constant length L = kRS, where R is the transmission rate of the channel and k is a large integer. Suppose there are N nodes, each with an infinite number of frames to send. We also assume that dprop < s,="" so="" that="" all="" nodes="" can="" detect="" a="" collision="" before="" the="" end="" of="" a="" slot="" time.="" the="" protocol="" is="" as="">

⢠If, for a given slot, no node has possession of the channel, all nodes contend for the channel; in particular, each node transmits in the slot with probability p. If exactly one node transmits in the slot, that node takes possession of the channel for the subsequent k â 1 slots and transmits its entire frame.

If some node has possession of the channel, all other nodes refrain from transmitting until the node that possesses the channel has finished transmitting its frame. Once this node has transmitted its frame, all nodes contend for the channel.

Note that the channel alternates between two states: the productive state, which lasts exactly k slots, and the nonproductive state, which lasts for a random number of slots. Clearly, the channel efficiency is the ratio of k/(k + x), where x is the expected number of consecutive unproductive slots

a. For fixed N and p, determine the efficiency of this protocol.
b. For fixed N, determine the p that maximizes the efficiency.
c. Using the p (which is a function of N) found in (b), determine the efficiency as N approaches infinity
d. Show that this efficiency approaches 1 as the frame length becomes large

Answer 1

While the original problem is about the efficiency of a CSMA/CD-like protocol, the provided information relates to an unrelated highway statistics problem using the exponential distribution, which describes the average interarrival time for cars and the probabilities for certain time intervals between car arrivals.

Step-by-step explanation:

In relation to the problem of the CSMA/CD-like multiple access protocol which is not part of the provided car highway scenario, the efficiency of the protocol can be determined by calculating the expected number of unproductive slots, x, and using the formula for efficiency, which is k/(k + x). However, the student's question seems to be redirected or mistaken with a reference to a highway problem from a statistics exercise, which is unrelated to CSMA/CD protocols. Thus, we are not in a position to directly answer the original efficiency question without the correct context or information on the functioning of the particular CSMA/CD-like protocol in question.

For the highway problem, we can find the answers using the properties of the exponential distribution. The average time between two successive cars (interarrival time) with a rate of five cars per minute can be found by taking the inverse of the rate:

Average interarrival time = 1 / Rate = 1 / 5 minutes = 12 seconds.

To find the average time for another seven cars to pass, you simply multiply the average interarrival time by seven:

Average time for seven cars = 7 * Average interarrival time = 7 * 12 seconds = 84 seconds.

The probability of the next car passing within the next 20 seconds can be calculated using the exponential distribution function:

Probability = 1 - e^(-rate * time) = 1 - e^(-5/60 * 20)

For the probability that the next car will not pass for at least another 15 seconds, it's the complement of it passing within those 15 seconds:

Probability = e^(-rate * time) = e^(-5/60 * 15)Q

Answer 2

He is correct

Step-by-step explanation: